Document3

31

3

HALF-WAVE RECTIFIERS The basics of analysis

31 INTRODUCTION

A rectifier converts ae to de The purpose ofa rectifier may be to produce an output that is purely de or the purpose may

be to produce a voltage or current waveform that has a specified de component

In practice the half-wave rectifier is used most often in low-power applications because the average current in the supply will not be zero and nonzero average current may cause problems in transformer performance While practical applications of this circuit are limited it is very worthwhile to analyze the half-wave rectifier in detail A thorough understanding of the half-wave rectifier circuit will enable the student to advance to the analysis of more complicated eircuits with a minimum of effort

The objectives of this chapter are to introduce general analysis techniques for power electronics circuits to apply the power computation concepts of the previous chapter and to illustrate PSpice solutions

32 RESlSTIVE LOAD

Creating It de Component Using an Electronic Switch A basic half-wave rectifier with a resistive load is shown in Fig 3middot a The source is ac and the objective is to

create a load voltage which has a nonzero de component The diode is a basic electronic switch that allows current in one direction only For the positive halfcycle ofthe source in this circuit the diode is on (forward biased) Considering the diode to be ideal the voltage across a forward-biased diode is zero and the current is positive

For the negative naif cycle of the source the diode is reverse biased making the current zero The voltage across the reverse-biased diode is the source voltage which has a negative value

The voltage waveforms across the source load and diode are shown in Fig 3-1 b Note that the units on the horizontal axis are in terms of angle (ret) This representation is useful because the values are independent offrequency

The dc component Vo oflhe output voltage is the average value of a half-wave rectified sinusoid

Vl-fbull V OOm) almt) (3-1)2 bull bulle

The de component ofthe current for the purely resistive load is

The lIoU-waYlt Reclifi Resistive Lood32

(3-2)

Average power absorbed by the resistor in Fig 3-1a can be computed from P= J2J = y 21R When the

voltage and current are half-wave rectified sine waves

bull v_IpVin(wt))d(ml) =2 2

(3-3)

v_ v =

R 2R

In the preceding discussion the diode was assumed to be ideal For a real diode the diode voltage drop will

cause the load voltage and current to be reduced but not appreciably ifYm is large For circuits that have voltages much

larger than the typical diode drop the improved diode model may have only seeond-order effects on the load voltage

and current computations

bull

-~

bull R middot0

J

)pn 31 (II Hllf-w ~uflCl wilh reilive kd_ (bl Ul~ wavcfor

EXAMPLE 3-1 Half-wave Rectifier with Resistive Load

For the half-wave rectifier of Fig 3-1a the source is a sinusoid of 120 volts rms at a frequency of60 Hz The

load resistor is 5 Q Determine (a) the average load current (b) the average power absorbed by the load (c) the power

factor of the circuit

Solution (a) The voltage across the resistor is a half-wave rectified sine wave with peak value Ym = 120 -12 = 1697

volts From Eq 3-2 the average voltage is Ymix and average current is

V v2 (120) = 108 A

R bull S

(b) From Eq 3-3 the nTIS voltage across the resistor for a half-wave rectified sinusoid is

e

Resistive Load 33

y = y 12(120) 849 V 2- 2

The power absorbed by the resistor is

v-R p-shy 849

4 1440 II

--- The rrnscurrent in the resistor is Vj(2R) = 170 amps and the power couldbe also be calculated fromfrmlR =(17ols

= 1440 watts

(e) The power factor is

1440 (120)([7)

pppi _

S

- 33 RESISTIVE-INDUCTIVE LOAD

Industrial loads typically contain inductanee as well as resistance As the source voltage goes through zero

becoming positive in the circuit of Fig 3-2a the diode becomes forward biased The Kirchhoffs voltage law equation

The solution can be obtained by expressing the current as the swn of the forced resporne and the natural

that describes the current in the circuit for the forward-biased ideal diode is

- v in(mt)bull

R i(t) + L di(t) di

(3-4)

-response-- i(t) ift) + i(t) (3-5) --

--------

bull -shy

- --

o

34

i(t) 0 iJt)

The forced response for thiscircuit is the current that exists after the natural response has decayed to zero In

this case the forced response is the steady-state sinusoidal current that would exist in the circuit if the diode were nor

present This steady-state current can be found from phasor analysis resulting in

ift) = ( )sin(mt - amp) (3-6)

-_ The natural response is the transient that occurs when the load is energized It is the solu~n to the

~

homogeneous differential equation for the circuit without the source or diode

Ri(l) bull L diet) = 0 (3-7)dt

For this first-order circuit the natural response has the Conn ~ -shy

gt_J

(3-8)

where r is the time constant UR and A is a constant which is determined from the initial condition Adding the forced

and natural responses to get the complete solution gt

(3-9)

The constant A is evaluated by using tile initial condition for current TIle initial condition of current in the

inductor is zero because it was zero before tile diode started conducting and it cannot change instantaneously

Using tile initial condition and Eq 3-9 to evaluate A

Vi(O) = ~sin(O - 9) -t- Aeo = 0

Z (3- O)

V V A --- sin(-9) = -- sin(9)

Z Z

Substituting for A in Eq 3-9

V V i(t) = - sin(mt - 9) + sin(amp) e- Ih

Z Z (3-11 )

v = -- [sin(lllt - amp) bull sin(9) e -w I

z

It is often convenient to write the function in terms of the angle cot rather than time This merely requires ult

to be tile variable instead oft Writing the above equation in terms of angle t in the exponential must be written as cot which requires r to be multiplied by co aISD The result is

~

35

V v i(mt) = sin(CDt 9) + --- sin(9) e-fDlkK Z Z (3-12) v = -[lUll - 0) + IO)e-~] Z

A typical graph of circuit current is shown in Fig 3~2b Equation 3-12 is valid for positive currents only

because of the diode in the circuit so current is zero when the function in Eq 3-12 is negative Wben the source voltage

again becomes positive the diode turns on and the positive part of the waveform in Fig 3-2b is repeated This occurs

at every positive half-cycle of the source The voltage waveforms for eaeh element are shown in Fig 3-2b

Note that the diode remains forward biased longer than 1t radians and that the source is negative for rbe last part

of the conduction interval This may seem unusual but an examination ofthe voltages reveals that Kirchhoffs voltage -law is satisfied and (here is no contradiction Also note that the inductor voltage is negative when the current is

decreasing (vL==ldidt)

The point when the current reaches zero in Eq 3-12 is when the diode I1lJTls off The first positive value of Olt

in Eq 3-12 that results in zero current is called the extinction angle p Substituting tor = pin Bq 3-12 the equation that must be solved is

V i(ll) = -[in(~ - 0)] (3-13)

Z -which reduces to-

~ lin(JJ - 9) bull sin(9) e -~ltlrr == 0 (3-14)1 _

There is no closed-form solution for p and some numerical method is required To summarize the current in the halfshy-wave rectifier circuit with R-L load (Fig 3-2) is expressed as

vVosmrltlgtl_O) bull ~(sin8)e- for O~mt~p

(wt) Z Z o for p~mt21t - (3-15)

- L = -

R

-The average power absorbed by the load is 12R since the average power absorbed by the inductor is zero

The rms value of the current is determined from the current function of Eq 3-15

--

I -

= J_Ii(UlI)ltIIltIgtI) 2_ e

1 bull-ltlt1gt1)lt11)2_ (3-16)

Average current is -(3-17)-

-I =

o

EXAMPLE 3-2 Half-wave Rectifier with R-L Load

For the half-wave rectifier of Fig 3-2a R = 1000 L =Ol H W = 377 rads and Vm= 100 V Determine (a)

an expression for the current in this circuit (b) the average current (c) the nns current (d) the power absorbed by the

R-L load and (e) the po wer factor

Solution For the parameters given

Z = (R2 + (mLll l = 10690 e= tanwUR) = 207 0 = 0361 radian

and WT = wUR = 0377 radian

(a) Eq 3-15 for current becomes

i(mt) = 0936sin(mt - 0361) + 0331e-oM(Imiddot37 A

forOsmtsPmiddot

Beta is found from Eq 3-14

sin(J - 0361) + sin(0361)e -jW37 = 0

Using a numerical root-finding program pis found to be 350 radians or 201

(b) Average current is oetermined from Eq 3-17

HO

I L f[0936sin(mt - 0361) + 0331 e--ltDWmiddotmjJ(mt) o

2 o

= 0308 A

(A numerical integration program is recommended)

(c) The rms currenr is found from Bq 3-16 as

sse ~ f [0936sin(mt - 0361) + 0331 e --ltDW377f~mt)I~ 2

(d) The power absorbed by the resistor is

P IR (0474) 100 224 W

The average po wer absorbed by the inductor is zero P can also be computed from the definition of average power

2 2

P _IP(rot)Jrot) 1-fgtltrot) i()laquoOOI)2n 211

c c

= 1- f [lOOsin(mt)][0936sin(mt - 0361) + 033le--mllOmiddot377]~cot) 2 e

224 II

Tho1laIfmiddotbullbull Roaifiltoshy 37

(e) The power factor is computed from the definition pf= PIS P is power supplied by the source which must be the same as that absorbed by the load

224pi = -p

= --- shyS (IOOIy2)(OA74)

Note that the power factor is no coste)

34 PSPICE SIMULATION

Using Simulation Software for Numerjeal Computations A computer simulation of the half-wave rectifier can be performed using PSpice PSpicc offers the advantage

of having the post-processor program Probe which can display the voltage and current waveforms in the circuit and perform numerical computations Quantities such as the rms and average currents average power absorbed by the load

and thc power factor can be determined directly withPSpice Harmonic content can be determined from the PSpice output

A transient analysis produces the desired voltages and eurrents One complete period is a sufficient time interval for the transient analysis

EXAMPLE 3-3 PSpice Analysis

Use PSpice to analyze the circuit of Example 3-2

Solution The circuit of Fig 3-2a is created using VSIN for the source and Dbreak fo r the diode In the simulation settings choose Time Domain (transient) for the analysis type and set the Run Time to 1667 ms for one period of the

source Set the Maximum Step Size to 1Dus to get adeqnate sampling ofrhe waveforms A transient analysis with a run time of 1667 IDS (one period for 60 Hz) and a maximum step size of 10 us is used for the simulation settings

Ifa diode model which approximates an ideal diode is desired for the purpose ofcomparing the simulation with analytical results editing the PSpice model and using n = 0001 will make the voltage drop across the forward-biased diode close (0 zero Alternatively a model for a power diode may be used to obtain a better representation of a real rectifier ctreuir For many circuits voltages and currents will not be affected significantly when different diode models

are nscd Therefore it may be convenient to use the Dbreak diode model for a preliminary analysis When the transient analysis is performed and the Probe screen appears display the current waveform by entering

the expression 1(RI) A method to display angle instead of time on the x axis is to use the x-variable option within the x-axis menu entering TIME60360 The factor of 60 converts the axis to periods (f = 60 Hz) and the factor 360 converts the axis to degrees Entering TIME6023 14 for the x variable converts the x axis to radians Figure 3-3a shows the result The extinction angle P is found to be 200 0 using the cursor option Note that nsing the default diode model in PSpice resulted in a value ofP very close to the 201 0 in Example 3-2

Probe ean be nsed to determine numerically the rms value ofa waveform While in Probe enter the expression RMS(I(RIraquo to obtain the rms value of the resistor cnrrent Probe displays a running valne ofthc integration in Eq

3-16 so the appropriate value is at the end ofone or more complete periods of the waveform Figure 3-3b shows how to obtain the rms eurrent The nns current is read as approximately 468 rnA Thiscompares very well with the 474 rnA

calculated in Example 3-2 Remember that the default diode model is used in PSpice and an ideal diode was used in

Example 3-2 The average current is found by entering AVG(I(RI)) resulting in 10 = 304 rnA

~ 1 I~

I ~

The Halfmiddot ve Recilielt PSpice38 --

~ 1000mA-------------- --

~

I

o

e BETA - 200 DEG (2pP4SS~61090u))

OAf--------o-------------=-----c~ 0 100

D 1 (Rl)

10DOmA---- ----

DBTERHININ~ RHS CURRENT I(Rl)

J I

(16 570457 962m)

SOOmA

READ RMS VALUE HERE

Irm - 46fl lIlAi bull-

10m 25m

Tim

Figure 3-3 a) Determining the extinction angle ~ in Probe The time axis is changed to angle using the x-variable

option and entering Time60360 b) Determining the rrns value ofcurrent in Probe

PSpiee is also useful in the design process For example the objective may be to design a half-wave reetifier

circuit to produce a specified value of average current by selecting the proper value ofL in an R-L load Since there is no closed-form solution a trial-and-error iterative methodmust be used A PSpice simulation that includes a parametric sweep is used to try several values ofL The following example illustrates this method

EXAMPLE 3-4 Half-wavc Rectifier Design Using PSpice Design a circuit to produce an average current of20 amps in a IO-il resistanee The source is 120 volts rms at 60 Hz

Solution A half-wave rcetifier is one circuit which can be used for this application Ifa simple half-wave rectifier with

39

the 10-0 resistance were used the average current would be (1201211)8 = 675 amps Some means must be found to

reduce the average current to the specified 2 amps A series resistance could be added to the load but resistances absorb

power An added series inductance will reduced the current without adding losses so an inductor is chosen Equations ]-15 and 3-17 describe the current function and its average for R-L loads There is no closed-form solution for L A trial-and-error technique in PSpice uses the parameter (PARAM) part and a parametric sweep to try a series of values

for L The PSpice circuit and the Simulation Settings box are shown in Fig 3-4

HALF-WAVE RECTIFIER WITH PARAMETRIC

01 R1

- --1Nl-- - ANv- -- [ Dbreak 10

V1 VOFF ~ 0 L1 VAIIlPL = (120sqrl(2l FREQ 60 T------=--_------- ILl

---

--

SWEEP

PARAMETERS L = 01

----

-

60A------------c=-------------- PARAHETPtC SWEEP

40A

01l~)

Till

Figure3-4 a) PSpice circuit for Example 3-4 b)A parametric sweep lsestablished inthe Simulation Senlngs box

c) L = 015 H for an average current of approximately 2 A-

TN Holfmiddot vlleciliampr 310

Average current in the resistor is found by entering AVG(I(Rl)) in Probe yielding a family of curves for different

inductance values (Fig 3-4c) Thethird inductance in the sweep (015 H) results in an average current of20 1118 amps

in the resistor which is very close to the design objective If further precision is necessary subsequent simulations can

be performed narrowing the range of L

35 R-L-SOURCE LOAD

Supplying Power 10 a de Source from an ac Source

Another variation ofthe half-wave rectifier is shown in Fig ]-5a The load consists ofa resistance inductance

and a de voltage Starting the analysis at (ill= 0 and assuming the initial current is zero recognize that the diode will

remain offas long as the voltage of the ac source is less than the de voltage Letting c be the value of (ill that causes the

source voltage to be equal to V ok

Vsino Vtk bull

usm - (3-18)-[ ~l ~

The diode starts to conduct at ltOt = c With the diode conducting Kirchhoffs voltage law for the circuit yields the

equation

L di(t)~sinlaquoQ) = Ri(t) bull + Va (3-19)

dt

- -

shy

shy

shyI

I

I

I

I

shy

shy

shy

shy

shy

shy

- shy

~

shy

shy

shy

shy-v

-v

--shy

-shy

shy

-shy

shy

shy

- shy

shy

shy

- shy

shy

shy

R-L-SoUJCC Load311

R L -VSltln(wI) + -=-

+ lk

IJ

R L R

+ V sin(CllI)

b (CJ

~ 2~

JoigUh 35 (al HIIlI-WIIie recntier wilh Rmiddotl source luad (bl tinuillur rorcec rcpon~~ (Uni lie SllIlfCe (l) Circuli for

forced response hom de soerce (d) wavefunns1middot

Total current is determined by summing the forced and natural responses

iC) = ijt) + iC)

The eurrent iit) is determined using superposition for the two sources The forced response from the ac source (Fig 3middot

Sb) is (VjZ)sin(rot 8) The forced response due to the de souree (Fig 35c) is YdjR The entire forced response is

v Vijt) - in( - 9) - (3-20)

Z R

The natural response is

i(f) = A emiddoth (3-21)

Adding the forced and naturaJ responses to get the complete response

c

__ 1

R-L-Source Lolld312

(J-22)

V V 2sin(mI-O) - ~ Ae-lhrrl for a Utlt S P

KmI) = Z R o otherwiJe

The extinction angle p is defined as the angle at which the current reaches zero as was done earlier in Eq 3-15 Using

the initial condition of i(u) = 0 and solving for A

V (J-2J)A = - -Zin(a-6)(

Fig 3-5d shows voltage and current waveforms for a half-wave rectifier with R-L-source load

The average power absorbed by the resistor is llR where

I = fi (mI)atmI) (3-24)

2bull bull

The average power absorbed by the de source is

(J-25)

where I is the average current that is 1 = fi(mI)d(mI) (J-2)bull 2bull

bull

Assuming the diode and the inductor to be ideal there is no average power absorbed by either The power supplied by the ac source is equal to the sum of the power absorbed by the resistor and the de source

(J-27)

or can be computed from

p ~

= _1 fvltmI)KmI)atmI) = _1 f(V inmt) I(mI)d(mI) (J-2)2 2 bull yen

~

EXAMPLE 3-5 Half-wave Rectifier with R-L-Source Load ~

~For the circuit of Fig 3middot5a R = 2 n L = 20 mH and Vde = 100 volts The ac source is 120 volts rms at60 Hz

(a) Determine an expression for the current in the circuit (b) Determine the power absorbed by the resistor (c) -~Determine the power absorbed by the dc source (d) Determine the power supplied by the ac source and the power factor

of the circuit

Solution From tile parameters given ~

- V

In = 1202 = 1697 V

Z = (R1 + (wL)p l = 780 n 9 = tanmiddot(wVR) = 131 rad

~

a = sinmiddot10011697) = 361 Q = 06]0 red

~

~

I

fIR Halfmiddot =lirr R-L-SOIJJce Load313

OJ = 377(00212) = 377 red

(a) Using Eq ]-22

T1i(Ull) = 218sin(mt-13l) - 50 + 753e-mlIJ A

The extinction angle p is found from the solution of

i) = 218sin(JJ -t31) - 50 + 753 e-jlI3middotT1 = 0

which results in p= 337 rad (193 0 ) using root-finding software1

(b) Using the preceding expression for i(wt) in Eq 3-24 and using a numerical integration program the rma current is

J_])1

f i(mt)laquomt) = 398 AI = 2

0- resulting in

(e) The power absorbed by the de source is lOYdlt Using Eq 3-26

~37

I = _I f i(mt)laquomt) = 225 A 2

06]

yielding

p = IY = (225)(100) = 225 w

(d) The power supplied by the ac source is the stun ofthe powers absorbed by the load

P =PR tPiE =312 t 225 = 256 W 8

The power factor is

Ppl= -P = -- shy

S V-I1fJIII

PSpicc Solution

The power quantities in this example can be determined from a PSpiee simulation of this circuit The cireulr

of Fig 3-5a is created using VSIN Dbreak R and L In the simulation settings choose Time Domain (transient) for the

analysis type and set theRun Time to 1667 ms for one period ofthe source Set the Maximum Step Size to 10 us to get

adequate sampling of the waveforms A transient analysis with a run time of 1667 ms (one period for 60 Hz) and a

maximum step size of 10 us is used for the simulation settings

Average power absorbed by the 2-Q resistor can be computed in Probe from thebasic definition ofthe average

ofp(t) by entering AVG(W(Rl) resulting in 297 watts or from 1R by entering RMS(I(R 1raquoRMS(I(RI))2 The

average power absorbed by the de source is computed from the Probe expression AVG(W(Vde)) yielding 217 watts

Ibe PSpice values differ slightly from thevalues obtained analytically because ofthe diode model However

the default diode is more realistic than the ideal diode in predicting actual circuit performance

Inductor_Source Load 314

36 INDUCTOR-SOURCE LOAD

Using Inductance to Limit Current

Another variation of the half-wave rectifier circuit has a load that consists ofan inductor and a de source as shown in Fig 3-6 Although a practical implementation ofthis circuit would contain some resistance the resistance may be negligible compared to other circuit parameters

V1l sin(Wl) +

F1gureJ6 Half-wave rectifier wnh inductor source load

Starting at rot=O and assuming zero initial current in the inductor the diode remains reverse biased until the ac source voltage reaches the de voltage The value of ex at which the diode starts to conduct is a calculated using Eq 3middot 18 With the diode conducting Kirchhoffs voltage law for the circuit is

diVsin(rot) L ) + v

0 ( I) _ L dt(rot) V sm ro - -- shy

bull ro dJ

Rearranging

Vsin(rot) - V

mL

Solving for iuct)

1 ~ - -fV 0(1)mLbull

Performing the integration

jVIII cosa - coKOt) + V(a_rot) for a~rot~p

irot) mL roL

o otherwise

(3-29)

(3-30)

(3-31)

(3-32)

(3-33) ~

~

A distinct feature of this circuit is that the power supplied by the source is the same as that absorbed by the de source less any losses associated with a nonideal diode and inductor If the objective is to transfer power from the ac - source to the desource losses are kept to a minimum by using this circuit -

~

~

-

Induc1ltlr-Sourre Load 315

EXAMPLE 3-6 Half-wave Rectifier with Inductor-Source Load

For the circuit of Fig 3-6 the ae source is 120 volts ems at 60 Hz L = 50 mH and Vde = 72 volts Determine

(a) an expression for the current (b) the power absorbed by the de source and (e) the power factor


a = sin [ 12~j21 = 251 = 0438 rad

(a) The equation for current is found from Eq 3-33

i(mt) = 983 - 900cos(mt) - 382w A for a s mt s p

where pis found to be 404 radians from the numerical solution of983 - 900cosp - 382p = O

(b) The power absorbed by the de source is Lvdo where

-flmt)at((JI)Jo 1 bull2bull

bull 0lt

_1 f [983 - 90000(01) - 382101) 21 Of31

resulting in

(e) The rrns current is found from

P (246)(72) 177 W

Therefore

1 bull-fj(OI~o1)2bull

bull 381 A

P 177

(120)(381)

37 THE FREEWHEELING DIODE

Creating a de Current A freewheeling diode D in Fig 3middot7a can be connected across an R-L load as shown The behavior of this

cireuit is somewhat different than the half-wave rectifier of Fig 3-2 The key to the analysis ofthis circuit is to determine

when each diode conducts First it is observed that both diodes cannot be forward biased at the same time Kirchhoffs

voltage law around the path containing the source and the two diodes shows that one diode must be reverse biased

Diode D will be on when the source is positive and diode D will be on when the souree is negative

101 t +

1- HoICmiddot Roadior Ind-lor-So~ wad316

For a positive source voltage

bull D1 is on

bull D1 is off bull the equivalent circuit is the same as that of Fig )middot2 shown again in Fig )-7b bull the voltage across the R-L load is the same as the source

For a negative source voltage

bull 0 1 is off bull O is on bull the equivalent circuit is the same at that of Fig )-7c

bull the voltage across the R-L load is zero

-int Dr

R

L

(j

+ shy R R

L L

I = 0

h)

Flgure J7 (a) Half-wave Rectifier with Freewheeling Diode (b) Equivalent Circuit for v gt 0 (b) Equivalent Circuit for v lt O

Since the voltage across the R-L load is the same as the source voltage when the source is positive and is zero when the source is negative the load voltage is a half-wave rectified sine wave

When the circuit is first energized the load current is zero and cannot change instantaneously The current reaches periodic steady-state after a few periods (depending on the LIR time constant) wbich means that the current at the end of a period is the same as the current at the beginning of the period as shown in Fig )-8 Tbe steady-state

current is usually of more interest than the transient that occurs when the circuit is first energized Steady-state load

source and diode currents are shown in Fig 3-9

-

The Half-WIIu 1l=nIishy317 Indwtor-Source Load

1111 TnlIsn

i I~

I t

i I

--

Figure 3~8 Load Current Reaching Steady State after the Circuit is Energized

-

o bull

Figure 3~9 Steady-state Load Voltage and Current Waveforms with Freewheeling Diode

The Fourier series for tbe half-wave rectified sine wave for the voltage across the load is

Y Y 2Y gt(t) = + ---sin(co i) E ---- cos(1KrV ) (3-34)2 0 246 (1 _ l)x

The current in tbe load can he expressed as a Fourier series by using superposition taking each frequency separately

The Fourier series method is illustrated in the following example

bull bull

InducOr-Sunc Load318

EXAMPLE ]middot7 Half-wave Rectifier with Freewheeling Diode

Determine the average load voltage and current and determine the power absorbed by the resistor in the cireuit

of Fig 3-73 where R 2 n and L = 25 mf-l V is 100 volts lind the frequency is 60 Hz

Solution The Fourier series for this half-wave rectified voltage the appears across tile load is obtained from Eq 3-34

The average load voltage is the de term in the Fourier series

[00v == V == = 318 V o

Average load current is

VoI =shy 159 A o R

Load power can be determined from IR and nns current is determined from the Fourier components of current The

amplitudes of the ae current components are determined from phasor analysis

I bull = V

Z

where Z = IR bull jI1 = 12 + jo377(025)1

The lie voltage amplitudes are determined from Eq 3-34 resulting in

v 100V=-=-=50 V

2 2

2VJi= = 212 V

(2 - 1raquolt

2V V 424 V

(4 - [raquolt

2V V 182 V

(6 - [raquolt

InduCllJl_SoWCe Load319

The nns current is obtained using Eq 2-64

[s~ [ 5r[ Ir(O~lr = 1634 A

Notice that the contribution to rms current from the harmonics decreases as n increases and higher-order terms are not

significant Power in the resistor is IR e= (1634)2 =-- 534 watts

Psptee Solution The circuit of Fig 3-7a is ereared using VSJN Dbreak R and L The PSpice model for Dbreak is changed to make n

= 0001 to approximate an ideal diode A transient analysis is ron with a run time of 150 ms with data saved after 100

ms to eliminate the slart-up transient from the data A maximum step size of lu us gives a smooth waveform

A portion of the output file is as follows

FOURIER ANALYSIS TEMPERATURE 27000 DEG C

FOURIER COMPONENTS OF TRA~SIENT RESPONSE V (OUTJ

DC COMPONENT = 3183002pound+01

HARMONIC FREQUENCY FOTJRIER NORMALIZED PHASE NORKALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)

1 6000E+01 5000E+01 1000E+00 -5604E-05 OOOOE+OO 2 1 200E+02 2122E+01 42UE~01 -9000E+01 -9000E+01 ) 1800E+02 5651E-05 1130E-06 -6631E+01 -6631E+01 2400E+02 4244E+00 6466E-02 -9 000E~01 -9000E+01 5 3000E+02 5699E-05 1 HOE-06 -9064pound+01 -9064E+01 6 3600E+02 1 819E+00 3638pound-02 -9000E+01 -9000E+01 4200E+02 5691E-05 1138pound~06 -9111pound+01 -9110pound+01 a 4800pound+02 1011E+00 2021pound-02 -9000pound+01 -9000pound+01 9 5400pound+02 5687E-05 1137pound-06 -90BOE+01 -9079E+01

FOURIER COMPONENTS OF TRANSIENT RESPONSE I (R_R1)

DC COMPONENT = 1591512E+01

HARMONIC FREQUpoundNCY fOURIER NORHJgtLIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (OEG) PHASE (DEG)

1 6000E+01 5189pound-1-00 1000E+00 -7802pound+01 OOOOE+OO 2 1200E02 1120pound00 2158pound-01 -1 739E02 -1 788E+01 3 1800E+0- 1963E-04 3782pound-05 -3719pound01 1969E+02

2400E0- 1123E-IH 2164E-02 -1770E+02 1 351E02 5 3000E02 7524E-05 1450E-05 6226E01 4524E+02 6 3600E+02 3217E-02 6200pound-03 -1 7pound11pound+02 2900E+02 7 4200E+0- B331E-05 1605pound-05 1693E+02 7154E+02

8 4800pound+02 1345E-02 2512E-03 -1783E+02 4458pound+02

9 5400pound+02 5435E-05 1047pound-05 -1074E+02 5948E+02

Note the close agreement between the analytically obtained Fourier terms and the PSpice output Average

current can be obtained in Probe by entering AVG(I(RI)) yielding l59 A Average power in the resistor can be

obtained by entering AVG(W(RI))) yielding P = 535 watts It is important that the simulation represents steady

state periodic current for the results to be valid

Induclor-SOlme Load 320

Redueing Load Current Harmonics The average current in the R-L load is a function of the applied voltage and the resistance but not the

inductance The Inductance affects only the ac terms in the Fourier series lfthe inductance is infinitely large the

impedance oflhe load to ac terms in the Fourier series is infmite and the load eurrent is purely de The load current

is then

v (3-35)

oR

A large inductor (LlRraquo T) with a freewheeling diode provides a means of establishing a nearly constant load

eurrent Zero-to-peak fluctuation in load current ean be estimated as being equal to the amplitude of the first ac term

in the Fourier series The peak-to-peak ripple is then

(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy

Figure 3-10 Waveforms for the Half-wave Rectifier with Freewheeling Diode of Example 3-9 with UR ~

~

EXAMPLE 3-8 Half-wave Rectifier with Freewheeling Diode LIR ~

For the half-wave rectifier with a freewheeling diode and RmiddotL load as shown in Fig 3-7a the souree is 240

volts rms at 60 Hz and R 8 n (a) Assume L is infinitely large Determine the power absorbed by the load and the

power factor as seen by the source Sketch v~ iDI and iD~ (b) Determine the average current in each diode (c) For

a fmite inductance determine L such that the peak-to-peak current is no more than 10 of the average current

Solution (a) The voltage across the R-L load is a half-wave rectified sine wave which has an average value of

Vj7t The load current is

~

~

~

~ ~

~

~

~

~

~

ThoHIIIfmiddotbull Ilocur Cpllelor FUl 321

v VJ i(mt) J = ~ 135 A J o R R

Power in the resistor is

p (I~) R (135) 8 1459 W

Source rms current is computed from

bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru

Voltage and current waveforms are shown in Fig 3-[0

(b) Each diode conducts for one-halfofthe time Average current for each diode is 12= 135t2 = 675 A

(c) The value of inductance required to limit the variation in load current to 10 can be approximated from the fundamental frequency of the Fourier series The voltage input to the load for n = I in Eq 3~34 has amplitude VJ2=

2(240)t2 = 170 V The peak-to-peak current must be limited to AI

Olt (Ol0X) lt (0IOXl35) 135 A

which corresponds to an amplitude of 13512 = 0675 A The load impedance at the fundamental frequency must then be

= VIZ ~ = 251 n I 0675

Thc load impedance is

Z = 251 = IR + jmLI 18 + 1177LI

Since the 8-Cl resistance is negligible eompared to the total impedance the inductance can be approximated as

L bull IZI = 251 = 067 H m 377

The inductance will have to be slightly larger than 067 H because Fourier terms higher than 11=1 were neglected in

this estimate

38 HALF~WAVE RECTIFIER WITH A CAPACITOR FILTER

Creating a de Voltage Irom an ae Source A common application of rectifier circuits is to convert an ac voltage input to a de voltage output The haJfshy

wave rectifier of Fig 3-11a has a parallel R-C load The purpose ofthe capacitor is to reduce the variation in the

output voltage making it more like de The resistance may represent an extemalload and the capacitor may be a

tilter which is part of the rectitier circuit

CpocilOr Filla322

Assuming the capacitor is initially uncharged and the circuit is energized at cot 0 the diode becomes

forward biased as the source becomes positive With the diode on the output voltage is the same as the source

voltage and the capacitor charges The capacitor is charged to Vm when the input voltage reaches its positive peak at rot = n2

As thesource decreases after rot = 1d2 the capacitor discharges into the load resistor At some point the

voltage ofthe source becomes less than the output voltage reverse biasing the diode and isolating the load from the

source The output voltage is a decaying exponential with time constant RC while the diode is off

The point when the diode turns off is determined by comparing the rates of change of the source and the

capacitor voltages The diode turns off when the downward rate of change of the source exceeds that permitted by

the time constant ofthe R-C load The angle rot = ais the point when the diode turns off in Fig 3-llb The output

voltage is described by

rs diode all

vo(mt) (3-37)ve--(ltIll - ffleBr diode off

where

V = Vsin8 (3-38)

Ii + ~ 10

II = V ~inU1r) C R v

[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)

Figure 3-11 (a) Half-wave Rectifier with R-C Load (b) Input and Output Voltages

The HolT_ Rali~ CapocilOf Filer323

The slopes of these functions are

_d_ (V sinmt) VCOsmt (3-39)d(mt) bull

aod

d ( sm 9 e -(~ - 4 - 9 [ ---1e (~- oRe-- - sin I (3-40)d(mt) bull bull raRC

At wt = a the slopes of the voltage functions are equal

V sina Vsin9V c0s9 -middot--e -(amp - ampYIIIilC = bull -mRC -raRC

=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)

When the source voltage comes back up to the value of the output voltage in the next period the diode

becomes forward biased and the output again is the same as the source voltage The angle at which the diode turns

on in the second period cot= 2n + a is the point when the sinusoidal source reaches the same value as the decaying

exponential output

Vsin(2s +0) (v sina)e-(hmiddot-UyR(

0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _

The preceding equation must he solved numerically for a

The current in the resistor is calculated from iR = vjR The current in the capacitor is calculated from

i-t) = C ltIvit) dt

which can also be expressed using err as the variable

ltIv (mt)-mt) = mC ---o~

d(mt)

cpacito Fil324

Using Vo from Eq 3-37

Vsin9 -mJ _ flYlDRC -~--e for 9s mt s 27t a

R (diade off) (3-44)

mCVcos(mt) for 2 a s rot s 2 9 (diode on)

The source current which is the same as the diode current is

(3-45)

The average capacitor current is zero so the average diode current is [he same as the average load current Since the

diode is on for a short time in each cycle the peak diode current is generally much larger than the average diode

current Peak capacitor current occurs when the diode turns 011 at cor = 271 + a From Eq 3~52

Ie _~ (Dey cos(2n a) == mCV cosc ~ (3-46)

Resistor current at cot = 271 + 11 is obtained from Eq )-)7

V sin(2mt a) Vsino = (3-47)

R R

Peak diode current is

R v ( coCcosa + S~) (H8)

The effectiveness of the capacitor filter is determined by the variation in output voltage This may be

expressed as the difference between the maximum and minimum output voltage which is the peak-to-peak ripple voltage For the half-wave rectifier of Fig 3middot11 a the maximum output voltage is Vm The minimum output voltage ~

occurs at lIt = 21[+0 which can be computed from Vmsin(a) The peak-to-peak ripple for the circuit of fig 3-11a is expressed as

Ii V V - Vsino V(1 - sina) (3-49)o

In circuits where the capacitor is selected to provide for a nearly constant de output voltage the R-C time constant is large compared to the period of the sine wave and Eq 3-42 applies Moreover the diode turns on dose to the peak of the sine wave when n JrIl The change in output voltage when the diode is off is described in Eq 3middot

37 In Eq 3-37 ifVe Vm and e JrIl then Bq 3-37 evaluated at n = n2 is

0(21 + a) Ve-C2a + 1Il - n 1aRC Ve-2aRC

The ripple voltage can then be approximated as

0-50)

Furthermore the exponential in the above equation can be approximated by the series expansion

eoFil

2_e -lrtJflJ1lC 1 shymRC

Substituting fOT the exponential in Eq 3-50 the peak-to-peak ripple is approximately

v ( 2_) ==~AV bull (3-51) III (JJRCo IRC

The output voltage ripple is reduced by increasing the filter capacitor C As C increases the conduction interval for the diode decreases Therefore increasing the capacitance to reduce the output voltage ripple results in a larger peak diode current

EXAMPLE 3-9 Half-wave Rectifier with R-C Load

The half-wave rectifier of Fig 3-11a has a l20-volt rms source at 60 Hz R == 500 fl and C = 100 JlF Determine (a) an expression for output voltage (b) the peak-to-peak voltage variation on the output and (c) an expression for capacitor current (d) Determine the peak diode current (e) Determine C such that AV is 1 of

v

Solution From the parameters given

V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0

The angle a is determined from Eq 3-41

9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V

The angle a is determined from the numerical solution ofEq 3-43

sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48

Output voltage is expressed from Eq 3-37

1697sin(mil vlaquo(JJ) =

1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a

Peak-to-peak output voltage is described by Eq 3-49

AVo V(I - sinal 0 1697(1 - sin(0843raquo = 43 V0

(c) The capacitor current is determined from Eq 3-44

1lIo Halfmiddot R~r eapu ~jltet326

-0339 e -( - reyrees A 9srots21t+a ic(cot) =

64 cos(Qlt) A 21t+aocoto2n+91 (d) Peak diode current is determined from Eq 3-48

SiD(843)] [D_ bull v2(l20~377(lormiddotCOS(O843) bull 500

426 + 034 = 450 A

(e) For 6Vo = 001 VIll Eq 3-51 can be used

Y Y Ibull - F bull 3333 iFCmiddot r-R-(-AY-) bull 7(6Oj-(5ccOOCj(OOlY-J 300 _

Note thai peak diode current can be determined from Eq 3-48 using an estimate of a from Eq 3-49

I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819

From Eq 3-48 peak diode current is 304 A

PSpke Solution

A PSpice circuit is created for Fig 3middotlla using VSIN Dbreak R and C The diode Dbreak used in this analysis

causes the results to differ slightly from the analytic solution based on the ideal diode The diode drop causes the

maximum output voltage to be slightly less than that of the source

The Probe output is shown in Fig 3middot]2 Angles 9 and a are determined directly by first modifying the xshy

variable to indicate degrees (x-variable = timc60360) and then using the eursor option The restrict data option is

used to compute quantities based on steady-state values (1661 IDS to 50 ms) Steady-state is characterized by

waveforms beginning and ending a period at the same values Note that the peak diode current is largest in the first

period because the capacitor is initially uncharged

shy 327 Capaoiar Filshy

shy- --shy-----_ _ ---shy-----~ shy~ -~~~-~~_ shy-~ -- --- _ --- --- --1

200V ~ -1 ~ ~)

-2001 --- -_bull - -- - --- -- - - -- --- -- _ -_ _ -- -- _- - _bullbullbull -- -- -- -- - bullbullbull --------~ bull 1(1) bull 11(2)

8 OA - -- - - - - - --- - - - - - - - - --- bull_- - - --- - _ - -- - - -_ -- ---- bullbullbull --- -- --- - _ - -- ~

00 ~ OI~ c~~m i

-u 00 bullbullbullbull-_~~__LL_~_~--J

0 10 20115 30 1(0)

Figure 3-12 Probe Output ForExample 3-9

Results rrom the Probe cursor

QUANTITY RESULT

40S (u~4S)

985deg ~ Vomax 1689 V

v min 126 V

6V 429 V

Iopeak 442 A steady state 636 A first period

IcPltak 412 A steady state 639 A first period

~

Results after restricting data to steady state

QUANTITY PROBE EXPRESSION RESULT

ID8 AVG(I(DI) 0295 A Ie RMS(I(Cl)) 0905 A

IRa8 AVG(W(Rl)) 438 W P AVG(W(V) -441 W PD AVG(W(Dl)) 345 mW

The IWfmiddotWYO IIltctifir Cpoltjor f~r328

In the preceding example the ripple or variation in output voltage is very large and the capacitor is not an

effective filter In many applications it is desirable to produce an output which is closer to dc This requires the

time constant RC to be large compared to the period ofrhe input voltage resulting in little decay of the output

voltage For an effective filter capacitor the output voltage is essentially the same as the peak voltage of the input

39 mE CONTROLLED HALF-WAVE RECTIFIER

The half-wave rectifiers analyzed previously in this chapter are classified as uncontrolled rectifiers Once the source

and load parameters are established the dc level of the output and the power transferred to the load are fixed

quantities

A way to control the output of a half-wave rectifier is to use an SCR I instead of a diode Figure 3-13a

shows a basic controlled half-wave rectifier with a resistive load Two conditions must be met before the SCR can

conduct

1) The SCR must be forward biased (VSCR gt 0) and

2) a current must be applied to tne gate of tne SCR

Unlike the diode the SCR will not begin to conduct as soon as the source becomes positive Conduction is delayed

until a gate current is applied whieh is the basis for using the SCR as a means of control Once the SCR is

conducting the gate current can be removed and the SCR remains on until the current goes to zero

Figure 3-13 (a) A Basic Controlled Rectifier (b) voltage Waveforms

Switehing with other controlled tum-on devices such as transistors can be used to control the output of a

converter

ThlUlfmiddot Rltaifter CopI Filer329

Resistive Load

Figure 3-13b shows the voltage waveforms for a controlled half-wave rectifier with a resistive load A gate signal is applied to the SCR at mt = a where (l is the delay angle The average (dc) voltage across the load resistor

in Fig 3-13a is

V = - V sin(fIlt)d(mr) = -[1 + cos a] f V (3-52) Q 2 211

The power absorbed by the resistor is vtRwhere the rms voltage across the resistor is computed from

v = ~ I f v(co)arrot) - 2 e

(3-53) f[v in(m)] arrot)

bull

V 1 _ a + sin(21l) 2 11 211

EXAMPLE 3-10 Controlled Half-wave Rectifier with Resistive Load

Design a circuit to produce an average voltage of 40 volts across a 1OO~o load resistor from a 120-Y rrns 60

Hz ac source Determine the power absorbed by the resistance and the power factor

Solution If an uncontrolled half-wave rectifier is used the average voltage would be Yjrc = 120J2rc= 54 volts

Some means of reducing the average resistor voltage to the design specification of 40 volts must be found A series

resistance or inductance could be added to an uncontrolled rectifier or a controlled rectifier could be used The

controlled rectifier of Fig 3-13a has the advantage of not altering the load or introdueing losses so it is selected for

this application

Eq 3-52 is rearranged to determine the required delay angle

a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~

The power factor of the circuit is

pplmiddot shys

p

(120)(756100) 571 middot063

R-L Load A controlled half-wave rectifier with an R~L load is shown in Fig 3-14a The analysis of this circuit is

similar to thai of tae uncontrolled rectifier The current is the sum ofthe forced and natural responses and Eq 3-9

applies

(WI) = iJ(JJJ) + i(cot)

= [ -J sin(cot - 8) + Ae -511 bull

The constant A is determined from the initial condition i(a) = 0

(3-54)

A = [- [ Jsin(a - 8)] emiddot~

Substituting for A and simplifying

sin(a - 8)e(1l - mlY-] 1[J[sin(Oll - 8)shy(Oll) bull (3-55)for asmtsp

o otherwise

The extinction angle Pis defined as the angle at which the current returns to zero as in the case of the

uncontrolled rectifier When rot = ~

I (Il) bull 0 bull [ -fJ[sin(P - 8) - sin(a - 8)e - py~J (3-56)

which must be solved numerically for Il The angle p- (J is called the conduction angle y Figure 3middot14b shows the

voltage waveforms The average (de) output voltage is

TheJ-blr_wvelcclirer 331

V y = fPy sin(cot)d(cot) ~[COSII - cosPl _ o 2 (1 2

The average current is computed from

I = ~J~)d(Olt)

where i(wt) is defined in Eq 3-55 Power absorbed by the load is 12

i

shybull

Capil Fill

(3-57)

(3-58)

rrnR where the rms current is computed from

(3-59)_1 fPi2(cot)~mt) 2 bull

+ +

+

(11)

- r----middotn

o

ibl

Figure 3-14 (a) Controlled Half-wave Rectifier with R-L load (b) Voltage Waveforms

cpunar Fill332

EXAMPLE 3-11 Controlled Half-wave Rectifier with R-L Load

For the circuit ofFig 3-14a the source is 120 volts nns at 60 Hz R =0 20 Q L = 004 H and the delay

angle is 45deg Detennine (a) an expression for i(wt) (b) the average current (c) the power absorbed by tbe load and (d) the power factor

Solution (a) From the parameters given

Vm = 1202 = 1697 volts

Z = (R2 + (roLlls = (202 + (377middotO04i)Ol = 250 Q

e laIJl(ll)lJR) = tan1(377O04)20) = 0646 radian (In = wLlR = 377004120 = 0754 a = 45deg = 0785 radian

Substituting the preceding quantities into Eq 3-55 current is expressed as

i(rot) 678 sin(rot - 0646) - 267e --tIDmiddot75 A

fOlasrotsPmiddot

The preceding equation is valid from a to ~ where pis found numerically by setting the equation to zero and solving

for rot with the result ~ = 379 radians (217deg) The conduction angle is y = p- u = 379 - 0785 = 30 I rad = 172

(b) Average current is determined from Eq 3-58

1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A

(c) The power absorbed by [he load is computed from l~moR where

= 326 A

yielding

P = IR = (326)(20) = 213 W

(d) The power factor is

PI= P S

213

(120)(326) = =~= 054

RmiddotL-Sour-ce Load A controlled rectifier with a series resistance inducranee and dc source is shown ill Fig 3-15 The analysis

of this circuit is very similar to that of the uncontrolled half-wave rectifier discussed earlier in this chapter The

major difference is that for the uncontrolled rectifier conduction begins as soon as the source voltage reaches the

level of the dc voltage For the controlled rectifier conduction begins when a gate signal is applied to the SCR

provided the SCR is forward biased Thus the gate signal may be applied at any time that the ac source is larger

than the de source

---

I

bull bullbullIt

333

- = m( ~J bullbullbullbull

cwren Il isex~~ ~ji~~ ~i~ eilhin~ ab~rdnge

o omewise where A is determined from Eq 3-6] ( A = [- ( 1 mea - 9) + ] eO

( (

R L (

( i (

(

c L L L Figure 3-15 Controlled Rectifier with RmiddotL-Source Load L L L

EXAMPLE 3-12 Controlled Rectifier with R-L-Source Load

The controlled half-wave rectifier of Fig 3-15 has an ac input of 120 volts ITI1S at 60 Hz R = mH and Vdo = 100 volts The delay angle a is 45~ (a) Determine an expression for the current (b)

power absorbed by the resistor (e) Determine the power absorbed by the de source in the load


Vm == 1202 = 1697 volts Z o~ (R2 + (Wl)2tS= (22 + (377middot002)l)Q = 780 n e = tanwUR) = tan- (377middot 002)12) = 1312 radian

W1 = wUR = 377middot00212 = 371

a = 45deg 0785 radian

(a) First use Eq 3-60 to determine ifa = 45 0 is allowable The minimum delay angle is

- = ml 1~~~iJ = 36

which indicates that 45 0 is allowable Equation 3-61 becomes

(3middot60)

(3middot61 )

2 n L = 20

Determine the

The Holfave Ream 334

i(CI)t) = 218 sin(mt - 1312) - 50 + 750e--ltll63middot71 A

for 0785s mt s 337 rad

where the extinction angle pis found numerically 10 be 337 radians from the equation i(P) = O

(b) Power absorbed by the resistor is IR where I is computed from Eq 3-59 using the preceding expression for

i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W

(c) Power absorbed by the dc souree is loYde where 10 is computed from Eq 3-58

1 = flli(mt)t(mt) = 219 A 2 II

p =1Y = (219)(100) = 219 W

310 PSPICE SOLUTIONS FOR CONTROLLED RECTIFIERS

Modeling the SCR in PSpiee To simulate the controlled half-wave rectifier m PSpice a model for the SCR must be selected An SCR

model available in a device library can be utilized in the simulation of a controlled half-wave rectifier A circuit for

Example 3-10 using the 2NI595 SCR in the PSpice demo version library of devices is shown in Fig 3-16a

An alternative model for the SCR is a voltage-controlled switch and a diode as described in Chapter 1 The

switch controls when the SCR begins to eonduct and the diode allows current in only one direction The switch

must be closed for at least the conduction angle of the current An advantage of using this SCR model is that the

device can be made ideal The major disadvantage ofthe model is that the switch control must keep the switch

closed for the entire conduction period and open the switch before the source becomes positive again A circuit for

the circuit in Example 3-11 is shown in Fig 3-l6b

(cliogt fill335

Controlled Half-wave Rectifier with SCR 2N1595 chllnge the BCR I1Ddr fo~ highee voltag tating

RL Source PARAMETERS 2 alpha e 45

2N1595 Teq 60 VOFF =0 VMlPL= 170 RG TO alphal360ilreq FREQ =60 Vl = II---Wvshy

Ik~ V2=5 TF = In

Vcontrol TR = 1n4Y I PIN=1mbull bull PER =llfreq

CONTROLLED HALnmVE aDTIPIER Switch and dJod4 fox SCR

____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm

VOFF 0 VA~PL 1TO FREQ =60

SOURCE

pound I

--------~-----------2

V1 = 0 ~o

2=5 TO = (ALPt-W360ffiOj PARAMETERS TA = 1n ALPHA-45 TF= tn Freq 60 PN (tlFreq-DLAY11 J PER 1Freq

Figure 3-16 a) A controlled half-wave rectifier using an SCR from the library of devices b) An SCR model

using a voltage-controlled switeh and a diode

EXAMPLE 3-13 Controlled Half-wave Rectifier Design Using PSpice

A load consists of a series-connected resistance inductance and de voltage source with R = 2 0 L 20 mH and Vde = 100 volts Design a circuit that will deliver 150 watts (0 the de voltage source from a 120-volt rrns 60

Hz ac source

336

Solution Power in the de souree of 150 watts requires an average load current of 15DW11 oov = 15 A An

uncontrolled rectifier with this souree and load will have an average current of225 A and an average power in the

de source of 225 W as was computed in Example 3-5 previously in this chapter A means of limiting the average

current to 15 A must be found Options include the addition of series resistance or inductance Another option that

is chosen for this application is the controlled half-wave rectifier of Fig 3-15 The power delivered to the load

eomponents is determined by the delay angle a Since there is no closed-form solution for a a trial-and-error

iterative method must be used A Pspice simulation which includes a parametric sweep is used to try several values

ofalpha The parametric sweep in established in the Simulation Setting menu (see Example 3-4) A PSpiee circuit is

CONTROLLED HAL~VE RECTIFIER parametric sweep for a1pha

r~om 100 -==shy

a

middot0

~v

V 0

VOFF 0 VAMPL 120sqrt(2)] FREO 60

e uDbre~k SOIJ~CE

TO NP1-W3601llDj PARMlETERS TR In AlPHA ~ 50 TF In Fq ~ 60 ~~ llFreq-DLAYl 1) PER [l1Fql

-

ow---------~TI--------------__

pARAMETRIC SWEKP FOR ALPHA

c

200W

~ (l66 70ru147531)

~io deg

p

J

O m 8m 12m 16ms 20m22ms 0 AVG(ioitVdc ) Time

Figure 3-17 a) PSpice circuit for Example 3-l3 b) Probe Output for showing a family of eurves for different delay

angles The parametric sweep in established in the Simulation Setting menu c = 70 degrees gives an average power

of approximately 148 W in the de source

337

When the expression AVG(W(Vderaquo is entered Probe produees a family of curves representing the results

for a number of values of c as shown in Fig 3-17b An II of 70 which results in about 148 watts delivered to the

load is the approximate solution

The following results are obtained from Probe for II = 70

QUANTITY EXPRESSION RESULT de Source Power AVG(W(Vdc)) 148 W (design objective of150 W)

IDl$ current RMS(l(RIraquo 287 A

ResistorPower AVGlaquoW(RI)) 165W

Source apparent power RMS(V(1 JmiddotRMS(I(VS) 344 V-A

Source 3VtfllgC JlOwer AVG1W(Vs)) 166 W

Power factor (PIS) 166344 048

311 COMMUTATION

The Errett of Source Iaductaece The preceding discussion on half-wave rectifiers assumed an ideal source In practical circuits the source

has an equivalent impedance which is predominantly induetive reactance For the single-diode half-wave rectifiers

of Figures 3-1 and 3-2 the nontdeal cireuit is analyzed by including the source induetance with the load elements

However the source inductance causes a fundamental change in circuit behavior for eircuns like the half-wave

rectifier with a freewheeling diode

A half-wave rectifier with a freewheeling diode and source inductance L is shown in Fig 3-18a Assume

that the load inductance is very large making the load current constant At t=O- the load current is Itgt OJ is off and

OJ is on As the source voltage becomes positive 0 1 turns on but the source current does not instantly equal the

load current because ofLbull Consequently O2 must remain on while the current in L and 01 increases to that of the

load The interval when both 0 1 and OJ are on is called the eommutation time or commutation angle Commutation

is the process ofturning offan electronic switch which usually involves transferring the load current from one switch 10 another J

When both 0 1 and D are on the voltage across L is

(3-62)

and current in L and the source is

1 ~ j~ = -fllLJl(Olt) + i(O) ~ f

~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is

V ----(I - cOBmt) coL

Commutation in tnis ease is an example of naturcl commutation or line commutation where the change in instantaneous line voltage results in a deviee turning off Other applications may useforced commutation where current in a device such as a thyristor is forced to zero by additional circuitry Load commutation makes use ofinherenl oscillating currents produced by the load to turn a device off

Commumlin338

The current in 0 1 starts at ILand decreases 10 zero Letting the angle at which the eurrent reaches zero be rot = u

V ~(I - cos 101) = 0 roL

Solving for u

bull = CO-[ 1 - l~1= C09[ 1 - l~1 (3-64)1----- _ where X = wL is the reactance of the source Figure 3-8b shows the effeet of the source reactance on the diode currents The commutation from D to D2 is analyzed similarly yielding an identical result for the commutation

angle u

The commutation angle affects the voltage across the load Since the voltage across the load is zero when

Dz is conducting the load voltage remains at zero through the commutation angle as shown in Fig 3-17b Recall

that the load voltage is a half-wave rectified sinusoid when the source is ideal

Average load voltage is

bull _IIvsin(mt)~) 2K

bull

v [-cos(mtlJI = ---(l +cosu)

2K

Using u from Eg 3-64

(3-65)

1 shy

Recall that the average of a half-wave rectified sine wave is VrrIt Source reactance thus reduces average load

voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__

Figure 3-18 (a) Half-wave Rectifier with Freewheeling Diode and Source Inductance (b) Diode Currents and Load Voltage Showing the Effects of Commutation

- 312 SUMMARY

bull A rectifier converts ac to de Power transfer is from the ae source to the de load

bull The half-wave rectifier with a resistive load has an average load voltage of VJx and an average load

current of VJnR

bull The current in a half-wave rectifier with an R-L load contains a natural and a forced response resulting in

VO DI(co _ a) i(cot) Z

o -

Land-r=shy R

The diode remains on as long as the current is positive Power in the R-L load is Inns1R

bull A half-wave rectifier with and R-L-Sourcc load does not begin to conduct until the ae source reaches the de

voltage in the load Power in the resistance is ImR and power absorbed by the de source is 10Vde where 10 is the average load current The load current is expressed as

340

v V-siD(coI-9) for ascotspZ R(cot) =

o otherwise

where

A = [ - 9in(a-O) + ealorlt bull1 bull A freewheeling diode forces the voltage across an RmiddotL load to be a half-wave rectified sine wave The load

current can be analyzed using Fourier analysis A large load inductance results in a nearly constant load

currenl

bull A large filter capacitor across a resistive load makes the load voltage nearly constant Average diode

current must be the same as average load current making the peak diode current large

bull An SCR in place of the diode in a half-wave rectifier provides a means ofcontrolling output current and

voltage

bull PSpice simulation is an effective way of analyzing circuit performance The parametric sweep in PSpice

allows several values of a circuit parameter to be tried and is an aid in eircult design

PROBLEMS (From the 1997 edition)

Hall-wave Rectifier with Reststtve Load

31) The half-wave rectifier circuit of Fig 3middotla has v(t) = 170sin(377t) V and R = 12 n Determine (a) the average

load current (b) the rms load current (c) the apparent power supplied by the source and (d) the power factor of the circuit

3-2) The half-wave rectifier circuit of Fig 3-1 a has a transformer inserted between the source and the remained of

the circuit The source is 240 volts rms at 60 Hz and the load resistor is 20 n (a) Determine the required turns

ratio of the transformer such that the average load current is 10 A (b) Determine the average current in the primary

winding of the transformer

3-3) For a half-wave rectifier with a resistive load (a) show that the power factor is 12 (b) determine the

displacement power factor and the distortion factor as defined in Chapter 2 The Fourier series for the half-wave

rectified voltage is given in Eq 3-34

Half-wave Reenner with R-L Load

3-4) A half-wave rectifier has a source of 120 volts rms at 60 Hzand an R-L load with R = 10 n and L = 10 mH

Determine (a) an expression for load current (b) the average current (c) the power absorbed by the resistor and (d)

the power factor Verify your answers with a PSplce simulation using an ideal diode model

3-5) A half-wave rectifier has a source of 120 volts rms at 60 Hz and an R-L load with R = 8 n and L = 15 ml-l

Determine (a) an expression for load current (b) the average current (c) the pltJwer absorbed by the resistor and (d)

the power factor Verify your answers with a PSpice simulation using an ideal diode model

3-6) A half-wave rectifier has a source of240 volts rms at 60 Hz and an R-L load with R = IS n and L = 100 mj-l

- I

341

Determine (a) an expression for load current (b) the average current (e) the power absorbed by the resistor and (d)

the power factor Use PSpiee 10simulate the circuit Use the default diode model and compare your PSpice results

with analytieal results

3-7) The inductor in Fig 3-2a represents an electromagnet modeled as a 01 henry inductance The source is 240

volts at 60 Hz Use PSpice to determine the value ofa series resistance such that the average current is 20 amperes

Half-wave Rectifier with R-L-Source Load

3-8) A half-wave rectifier of Fig 3-5a has a 240-volt rms 60 Hz ac source The load is a series inductance

resistance and de source with L = 100 rnlf R = 10 n and Yd = 100 volts Determine (a) the power absorbed by the de voltage source (b) the power absorbed by the resistance and (c) the power factor


resistance and de source with L = 100 mH R = 12 n and Vdlt = 48 volts Determine (a) the power absorbed by the

de voltage source (b) the power absorbed by the resistance and (c) the power factor

3-10) A half-wave rectifier of Fig 3-6 has a 120-volt rms 60 Hz ac source The load is a series inductance and de

voltage with L = 75 mH and Vde = 48 volts Determine the power absorbed by the de voltage source

3-11) A half-wave rectifier with a series inductor-source load has an ac source of 240middotvolts rms 60 Hz The dc

source is 125 volts Use PSpicc to determine the value of inductance which results in 150 watts absorbed by the dc

source Use the default diode model

3-12) A half-wave rectifter with a series inductor and de source has an ac source of 120-volls rms 60 Hz The dc

source is 24 volts Use PSpice to determine the value of inductance which results in 75 watts absorbed by the de

source Use the default diode

FreeWheeling Diode

3-13) The half-wave rectifier with a freewheeling diode (Fig 3-7a) has R = 12 n L = 30 ml-l The source is 120

volts rms at 60 Hz (a) From the Fourier series of the half-wave rectified sine wave that appears across the load

determine the dc component of the current (b) Determine the amplitudes of the first four nonzero ac terms in the

Fourier series Comment on the results

3-14) In Example 3-8 the inductance required to limit the peak-to-peak ripple in load current was estimated by

using the first ac term in the Fourier series USe PSpice to determine the peak-to-peak ripple with this inductance

and compare itto the estimate Use the ideal diode model (0 = 0001)

3-15) The half-wave rectifier with a freewheeling diode (Fig 3-7a) has R = 3 n and a source with V = 50 V at 60

Hz (a) Determine a value ofL such that the amplitude of the first ac current term in the Fourier series is Jess than

5 of the de current (b) Verify your results with PSpice and determine the peak-to-peak current

3-16) The circuit of Fig P3-16 is similar to the circuit of Fig 3-7a except that a de source has been added to the

load The circuit has v(t) = 170sin(377t) V R = 10 n and Vdoo = 24 V From the Fourier series (a) determine the

value of L such that the peek-to-peak variation in load current is no mare than I amp (b) Determine the power

absorbed by the de source (c) Determine the power absorbed by the resistor

342

R L

+

tilUlW Pll6

Hair-wave Rectifier with a Filter Capacitor

3middot17) A half-wave rectifier with a capacitor filter has V= 100 V R = 1 till C = 1000 IlF and (t) = 377 (a)

Determine the ratio ofthe R-C time constant 10 the period of the input sine wave What is the significance ofthis

ratio (b) Determine the peak-to-peak ripple voltage using the exact equations (c) Determine the ripple using the

approximate formula in Eq 3-51

3-18) Repeat Problem 3-17 with (a) R = 100 n and (b) R = Ion Comment on the results

3-19) A half-wave rectifier with a 1 k-Q load has a parallel capaeitor The source is 120-volls rms 60 Hz

Determine the peak-to-peak ripple of the output voltage when the capacitor is (a) 5000 IlF and (b) 10 J1F [5 the

approximation ofEq 3-5 reasonable in each case

3-20) Repeat Problem 3-19 with R = 500 n

3-21) A half-wave rectifier has a l20-volt rms 60 Hz ac source The load is 500 O Determine the value ofa filter

capacitor to keep the peak-to-peak ripple across the load to less than 2 volts Determine the average and peak values ofdiode current

3-22) A half-wave rectifier has a 120-volt rms 60 Hz ac source The load is 100 watts Determine the value ofa

filter capacitor to keep the peak-to-peak ripple across the load to less than 15 volts Determine the average and peak

values of diode current

Controlled Hair-wave Rectifier

3-23) Show that the controlled half-wave rectifier with a resistive load in Fig 3-13a has a power factor of

PI = ~ _-- + sffi(2a) 2 221 411

3-24) For the controlled half-wave rectifier with resistive load the source is 120 volts rms at 60 HL The resistance

is 100 n and the delay angle II is 60 Q (a) Determine the average voltage across the resistor (b) Determine the bull

power absorbed by the resistor (c) Detenninc the power factor as seen by the source

3-25) A controlled half-wave rectifier has an ac source of240 volts rms at 60 Hz The load is a 30-0 resistor (a)

Determine the delay angle such that the average load current is 30 amps (b) Determine the power absorbed by the

load (c) Detennine the power factor

3-26) A controlled half-wave rectifier has a l20-volt rms 60 Hz ae source The a series R-L load has R = 25 0 and

-

-----

--

-

TIlt Half- Rtotifitr 343

L = 50 mll The delay angle is 25deg Determine (a) an expression for load current (b) the average load current and (c) the power absorbed by the load

3-27) A controlled half-wave rectifier has a 120-volt rms 60 Hz ac source The a series R-L load has R = 40 n and

L = 75 mll The delay angle is 50deg Determine (a) an expression for load current (b) the average load current and (c) thc power absorbed by the load

3-28) A controlled naif-wave rectifier has an R-L load with R = 20 nand L = 40 ml-l The source is 120-volts rms

at 60 Hz Use PSpice to determine the delay angle required 1O produce an average current of20 amps in the load Use the default diode in the simulation

3-29) A controlled half-wave rectifier has an RmiddotL load with R = 16 n and L = 60 ml-l The source is 120-volts rms

at 60 Hz Use PSpice to determine the delay angle required to produce an average current of 18 amps in the load

Use the default diode in the simulation

3-30) A controlled half-wave rectifier has a 120-volt 60 Hz ac source The load is a series inductance resistance and dc source with L = 100 mH R = 12 n and Vde = 48 volts The delay angle is 45deg Determine (a) the power absorbed by the dc voltage source (b) the power absorbed by the resistance and (c) the power factor

3~31) A controlled half-wave rectifier has a 240-1011 rms 60 Hz ac source The load is a series resistance

inductance and de source with R = 100 n L = 150 mH and VdC = 125 volts The delay angle is 60deg Determine (a) the power absorbed by the dc voltage source (b) the power absorbed by the resistance and (c) the power factor

3~32) Use PSpice lo determine the delay angle required such that the dc source in Problem 3-3 absorbs 35 watts

3~33) A controlled half-wave rectifier has a series resistance inductance and de voltage source with R = 2 n L =

75 mH and Vde = 48 volts The source is 120 volts rms at 60 Hz The delay angle is 45deg Determine (a) an expression lor load current (b) the flower absorbed by the de voltage source and (c) the power absorbed by the resistor

3~34) Use PSpicc to determine the delay angle required such that the de source in Problem 3-33 absorbs 50 watts

3middot35) Develop an expression for current in a controlled half-wave rectifier circuit which has a load consisting of a series inductance L and de voltage Vdc The source is v = Vmsinrot and the delay angle is a Determine the average current if Vm = 100 L = 35 mH Vdc = 24 volts ro= 2n60rads and a = 80 0 Verify your result with PSpice bull

3-36) A controlled half-wave rectifier has an RmiddotL load A freewheeling diode is placed in parallel with the load The inductance is large enough to consider the load current to be constant Determine the load current as a function of the delay angle alpha Sketch the current in the SCR and the freewheeling diode Sketch the voltage across the

load

Commutation

3-37) The half-wave rectifier with freewheeling diode of Fig 3-188 has a 120-volt rms ac source which has an

inductance of 13 mll The load current is a constant 5 amps Determine the commutation angle and the average output voltage Use PSpicc to verify your results Use ideal diodes in the simulation Verify that the commutation

angle for D[ to D2 is the same as fOT D1 to D[

3-38) The half-wave rectifier with freewheeling diode of Fig 3-1amp has a (20middotvolt rms ac source which has an

344

inductance of 10 mHo The load is a series resistance-inductance with R = 20 n and L = 500 mli Use PSpice to

determine (a) the steady-state average load current (b) the average load voltage and (e) the commutation angle

Use the default diode in the simulation Comment on the results

3middot39) The half-wave rectifier with freewheeling diode of Fig 3- 18a has a 120-volt rms ae source which has an

inductance of5 ml-l The load is a series resistance-inductance with R = (5 n and L = 500 rnl-l Use PSpiee to

determine (a) the steady-state average load current ((b) the average load voltage and (e) the commutation angle

Use the default diode in the simulation ~

3-40) The commutation angle given in Eq 3-64 for the half-wave rectifier with a freewheeling diode was developed

for commutation of load current from 0 1 to D1bull Show that the commutation angle is the same for commutation from

0 1 to O2bull

3-41) Diode 0 1 in Fig 3-18a is replaced with an SCR to make a eontrolled half-wave rectifier Show that the angle

for commutation from the diode to the SCR is

u = cos- I( cosa _ I~) -a

where a is the delay angle of the SCR

Design Problems

3-42) A certain situation requires that either 160 watts or 75 watts be supplied to a 48-volt battery from an 120-volt

rrns 60 Hz ac source There is a two-position switch on a control panel set at either 160 or 75 Design a single

circuit to deliver both values of power and specify what the control switch will do Specify the values of all ofthe

components in your circuit The internal resistance of the battery is 01 n

3-43) Design a cireuit to produce an average current on amps in an inductance of 100 mH The ac source available

is 120 volts nns at 60 Hz Verify your design with PSpiee Give alternative circuits that could be used to satisfy the

design specifications and give reasons for your selection

3-44) Design a circuit whieh will deliver 100 watts to a 48-volt dc source from a 120-1011 rms 60 Hz ac source

Verify your design with PSpice Give alternative circuits that could be used to satisfy the design specifications and

give reasons for your selection

3-45) Design a circuit which will deliver 150 watts to a 100-volt de source from a 120-volt rrns 60 Hz ac source



The lIoU-waYlt Reclifi Resistive Lood32

(3-2)

Average power absorbed by the resistor in Fig 3-1a can be computed from P= J2J = y 21R When the

voltage and current are half-wave rectified sine waves

bull v_IpVin(wt))d(ml) =2 2

(3-3)

v_ v =

R 2R

In the preceding discussion the diode was assumed to be ideal For a real diode the diode voltage drop will

cause the load voltage and current to be reduced but not appreciably ifYm is large For circuits that have voltages much

larger than the typical diode drop the improved diode model may have only seeond-order effects on the load voltage

and current computations

bull

-~

bull R middot0

J

)pn 31 (II Hllf-w ~uflCl wilh reilive kd_ (bl Ul~ wavcfor

EXAMPLE 3-1 Half-wave Rectifier with Resistive Load

For the half-wave rectifier of Fig 3-1a the source is a sinusoid of 120 volts rms at a frequency of60 Hz The

load resistor is 5 Q Determine (a) the average load current (b) the average power absorbed by the load (c) the power

factor of the circuit

Solution (a) The voltage across the resistor is a half-wave rectified sine wave with peak value Ym = 120 -12 = 1697

volts From Eq 3-2 the average voltage is Ymix and average current is

V v2 (120) = 108 A

R bull S

(b) From Eq 3-3 the nTIS voltage across the resistor for a half-wave rectified sinusoid is

e

Resistive Load 33

y = y 12(120) 849 V 2- 2


v-R p-shy 849

4 1440 II


= 1440 watts


1440 (120)([7)

pppi _

S






- v in(mt)bull

R i(t) + L di(t) di

(3-4)


--------

bull -shy

- --

o

34

i(t) 0 iJt)




ift) = ( )sin(mt - amp) (3-6)


~




gt_J

(3-8)



(3-9)




Vi(O) = ~sin(O - 9) -t- Aeo = 0

Z (3- O)

V V A --- sin(-9) = -- sin(9)

Z Z



Z Z (3-11 )


z



~

35











V i(ll) = -[in(~ - 0)] (3-13)





(wt) Z Z o for p~mt21t - (3-15)

- L = -

R



--

I -




-I =

o










forOsmtsPmiddot


sin(J - 0361) + sin(0361)e -jW37 = 0



HO


2 o

= 0308 A





P IR (0474) 100 224 W


2 2


c c


224 II



224pi = -p



















~ 1 I~

I ~


~ 1000mA-------------- --

~

I

o

e BETA - 200 DEG (2pP4SS~61090u))

OAf--------o-------------=-----c~ 0 100

D 1 (Rl)

10DOmA---- ----


J I

(16 570457 962m)

SOOmA

READ RMS VALUE HERE


10m 25m

Tim







39






01 R1

- --1Nl-- - ANv- -- [ Dbreak 10


---

--

SWEEP

PARAMETERS L = 01

----

-


40A

01l~)

Till








35 R-L-SOURCE LOAD






Vsino Vtk bull

usm - (3-18)-[ ~l ~


equation


dt

- -

shy

shy

shyI

I

I

I

I

shy

shy

shy

shy

shy

shy

- shy

~

shy

shy

shy

shy-v

-v

--shy

-shy

shy

-shy

shy

shy

- shy

shy

shy

- shy

shy

shy

R-L-SoUJCC Load311


+ lk

IJ

R L R

+ V sin(CllI)

b (CJ

~ 2~




iC) = ijt) + iC)



v Vijt) - in( - 9) - (3-20)

Z R




c

__ 1

R-L-Source Lolld312

(J-22)





V (J-2J)A = - -Zin(a-6)(



I = fi (mI)atmI) (3-24)

2bull bull


(J-25)


bull


(J-27)


p ~


~




of the circuit


- V

In = 1202 = 1697 V


~

a = sinmiddot10011697) = 361 Q = 06]0 red

~

~

I


OJ = 377(00212) = 377 red

(a) Using Eq ]-22






J_])1


0- resulting in


~37


06]

yielding

p = IY = (225)(100) = 225 w


P =PR tPiE =312 t 225 = 256 W 8

The power factor is

Ppl= -P = -- shy

S V-I1fJIII

PSpicc Solution















V1l sin(Wl) +



diVsin(rot) L ) + v


bull ro dJ

Rearranging

Vsin(rot) - V

mL

Solving for iuct)




irot) mL roL

o otherwise

(3-29)

(3-30)

(3-31)

(3-32)

(3-33) ~

~


~

~

-






a = sin [ 12~j21 = 251 = 0438 rad






bull 0lt

_1 f [983 - 90000(01) - 382101) 21 Of31

resulting in


P (246)(72) 177 W

Therefore


bull 381 A

P 177

(120)(381)







101 t +



bull D1 is on





-int Dr

R

L

(j

+ shy R R

L L

I = 0

h)






-


1111 TnlIsn

i I~

I t

i I

--


-

o bull






bull bull







[00v == V == = 318 V o


VoI =shy 159 A o R



I bull = V

Z



v 100V=-=-=50 V

2 2

2VJi= = 212 V

(2 - 1raquolt

2V V 424 V

(4 - [raquolt

2V V 182 V

(6 - [raquolt



[s~ [ 5r[ Ir(O~lr = 1634 A

















8 4800pound+02 1345E-02 2512E-03 -1783E+02 4458pound+02

9 5400pound+02 5435E-05 1047pound-05 -1074E+02 5948E+02









is then

v (3-35)

oR




(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy


~








~

~

~

~ ~

~

~

~

~

~




p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














e

Resistive Load 33

y = y 12(120) 849 V 2- 2


v-R p-shy 849

4 1440 II


= 1440 watts


1440 (120)([7)

pppi _

S






- v in(mt)bull

R i(t) + L di(t) di

(3-4)


--------

bull -shy

- --

o

34

i(t) 0 iJt)




ift) = ( )sin(mt - amp) (3-6)


~




gt_J

(3-8)



(3-9)




Vi(O) = ~sin(O - 9) -t- Aeo = 0

Z (3- O)

V V A --- sin(-9) = -- sin(9)

Z Z



Z Z (3-11 )


z



~

35











V i(ll) = -[in(~ - 0)] (3-13)





(wt) Z Z o for p~mt21t - (3-15)

- L = -

R



--

I -




-I =

o










forOsmtsPmiddot


sin(J - 0361) + sin(0361)e -jW37 = 0



HO


2 o

= 0308 A





P IR (0474) 100 224 W


2 2


c c


224 II



224pi = -p



















~ 1 I~

I ~


~ 1000mA-------------- --

~

I

o

e BETA - 200 DEG (2pP4SS~61090u))

OAf--------o-------------=-----c~ 0 100

D 1 (Rl)

10DOmA---- ----


J I

(16 570457 962m)

SOOmA

READ RMS VALUE HERE


10m 25m

Tim







39






01 R1

- --1Nl-- - ANv- -- [ Dbreak 10


---

--

SWEEP

PARAMETERS L = 01

----

-


40A

01l~)

Till








35 R-L-SOURCE LOAD






Vsino Vtk bull

usm - (3-18)-[ ~l ~


equation


dt

- -

shy

shy

shyI

I

I

I

I

shy

shy

shy

shy

shy

shy

- shy

~

shy

shy

shy

shy-v

-v

--shy

-shy

shy

-shy

shy

shy

- shy

shy

shy

- shy

shy

shy

R-L-SoUJCC Load311


+ lk

IJ

R L R

+ V sin(CllI)

b (CJ

~ 2~




iC) = ijt) + iC)



v Vijt) - in( - 9) - (3-20)

Z R




c

__ 1

R-L-Source Lolld312

(J-22)





V (J-2J)A = - -Zin(a-6)(



I = fi (mI)atmI) (3-24)

2bull bull


(J-25)


bull


(J-27)


p ~


~




of the circuit


- V

In = 1202 = 1697 V


~

a = sinmiddot10011697) = 361 Q = 06]0 red

~

~

I


OJ = 377(00212) = 377 red

(a) Using Eq ]-22






J_])1


0- resulting in


~37


06]

yielding

p = IY = (225)(100) = 225 w


P =PR tPiE =312 t 225 = 256 W 8

The power factor is

Ppl= -P = -- shy

S V-I1fJIII

PSpicc Solution















V1l sin(Wl) +



diVsin(rot) L ) + v


bull ro dJ

Rearranging

Vsin(rot) - V

mL

Solving for iuct)




irot) mL roL

o otherwise

(3-29)

(3-30)

(3-31)

(3-32)

(3-33) ~

~


~

~

-






a = sin [ 12~j21 = 251 = 0438 rad






bull 0lt

_1 f [983 - 90000(01) - 382101) 21 Of31

resulting in


P (246)(72) 177 W

Therefore


bull 381 A

P 177

(120)(381)







101 t +



bull D1 is on





-int Dr

R

L

(j

+ shy R R

L L

I = 0

h)






-


1111 TnlIsn

i I~

I t

i I

--


-

o bull






bull bull







[00v == V == = 318 V o


VoI =shy 159 A o R



I bull = V

Z



v 100V=-=-=50 V

2 2

2VJi= = 212 V

(2 - 1raquolt

2V V 424 V

(4 - [raquolt

2V V 182 V

(6 - [raquolt



[s~ [ 5r[ Ir(O~lr = 1634 A

















8 4800pound+02 1345E-02 2512E-03 -1783E+02 4458pound+02

9 5400pound+02 5435E-05 1047pound-05 -1074E+02 5948E+02









is then

v (3-35)

oR




(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy


~








~

~

~

~ ~

~

~

~

~

~




p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














o

34

i(t) 0 iJt)




ift) = ( )sin(mt - amp) (3-6)


~




gt_J

(3-8)



(3-9)




Vi(O) = ~sin(O - 9) -t- Aeo = 0

Z (3- O)

V V A --- sin(-9) = -- sin(9)

Z Z



Z Z (3-11 )


z



~

35











V i(ll) = -[in(~ - 0)] (3-13)





(wt) Z Z o for p~mt21t - (3-15)

- L = -

R



--

I -




-I =

o










forOsmtsPmiddot


sin(J - 0361) + sin(0361)e -jW37 = 0



HO


2 o

= 0308 A





P IR (0474) 100 224 W


2 2


c c


224 II



224pi = -p



















~ 1 I~

I ~


~ 1000mA-------------- --

~

I

o

e BETA - 200 DEG (2pP4SS~61090u))

OAf--------o-------------=-----c~ 0 100

D 1 (Rl)

10DOmA---- ----


J I

(16 570457 962m)

SOOmA

READ RMS VALUE HERE


10m 25m

Tim







39






01 R1

- --1Nl-- - ANv- -- [ Dbreak 10


---

--

SWEEP

PARAMETERS L = 01

----

-


40A

01l~)

Till








35 R-L-SOURCE LOAD






Vsino Vtk bull

usm - (3-18)-[ ~l ~


equation


dt

- -

shy

shy

shyI

I

I

I

I

shy

shy

shy

shy

shy

shy

- shy

~

shy

shy

shy

shy-v

-v

--shy

-shy

shy

-shy

shy

shy

- shy

shy

shy

- shy

shy

shy

R-L-SoUJCC Load311


+ lk

IJ

R L R

+ V sin(CllI)

b (CJ

~ 2~




iC) = ijt) + iC)



v Vijt) - in( - 9) - (3-20)

Z R




c

__ 1

R-L-Source Lolld312

(J-22)





V (J-2J)A = - -Zin(a-6)(



I = fi (mI)atmI) (3-24)

2bull bull


(J-25)


bull


(J-27)


p ~


~




of the circuit


- V

In = 1202 = 1697 V


~

a = sinmiddot10011697) = 361 Q = 06]0 red

~

~

I


OJ = 377(00212) = 377 red

(a) Using Eq ]-22






J_])1


0- resulting in


~37


06]

yielding

p = IY = (225)(100) = 225 w


P =PR tPiE =312 t 225 = 256 W 8

The power factor is

Ppl= -P = -- shy

S V-I1fJIII

PSpicc Solution















V1l sin(Wl) +



diVsin(rot) L ) + v


bull ro dJ

Rearranging

Vsin(rot) - V

mL

Solving for iuct)




irot) mL roL

o otherwise

(3-29)

(3-30)

(3-31)

(3-32)

(3-33) ~

~


~

~

-






a = sin [ 12~j21 = 251 = 0438 rad






bull 0lt

_1 f [983 - 90000(01) - 382101) 21 Of31

resulting in


P (246)(72) 177 W

Therefore


bull 381 A

P 177

(120)(381)







101 t +



bull D1 is on





-int Dr

R

L

(j

+ shy R R

L L

I = 0

h)






-


1111 TnlIsn

i I~

I t

i I

--


-

o bull






bull bull







[00v == V == = 318 V o


VoI =shy 159 A o R



I bull = V

Z



v 100V=-=-=50 V

2 2

2VJi= = 212 V

(2 - 1raquolt

2V V 424 V

(4 - [raquolt

2V V 182 V

(6 - [raquolt



[s~ [ 5r[ Ir(O~lr = 1634 A

















8 4800pound+02 1345E-02 2512E-03 -1783E+02 4458pound+02

9 5400pound+02 5435E-05 1047pound-05 -1074E+02 5948E+02









is then

v (3-35)

oR




(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy


~








~

~

~

~ ~

~

~

~

~

~




p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














35











V i(ll) = -[in(~ - 0)] (3-13)





(wt) Z Z o for p~mt21t - (3-15)

- L = -

R



--

I -




-I =

o










forOsmtsPmiddot


sin(J - 0361) + sin(0361)e -jW37 = 0



HO


2 o

= 0308 A





P IR (0474) 100 224 W


2 2


c c


224 II



224pi = -p



















~ 1 I~

I ~


~ 1000mA-------------- --

~

I

o

e BETA - 200 DEG (2pP4SS~61090u))

OAf--------o-------------=-----c~ 0 100

D 1 (Rl)

10DOmA---- ----


J I

(16 570457 962m)

SOOmA

READ RMS VALUE HERE


10m 25m

Tim







39






01 R1

- --1Nl-- - ANv- -- [ Dbreak 10


---

--

SWEEP

PARAMETERS L = 01

----

-


40A

01l~)

Till








35 R-L-SOURCE LOAD






Vsino Vtk bull

usm - (3-18)-[ ~l ~


equation


dt

- -

shy

shy

shyI

I

I

I

I

shy

shy

shy

shy

shy

shy

- shy

~

shy

shy

shy

shy-v

-v

--shy

-shy

shy

-shy

shy

shy

- shy

shy

shy

- shy

shy

shy

R-L-SoUJCC Load311


+ lk

IJ

R L R

+ V sin(CllI)

b (CJ

~ 2~




iC) = ijt) + iC)



v Vijt) - in( - 9) - (3-20)

Z R




c

__ 1

R-L-Source Lolld312

(J-22)





V (J-2J)A = - -Zin(a-6)(



I = fi (mI)atmI) (3-24)

2bull bull


(J-25)


bull


(J-27)


p ~


~




of the circuit


- V

In = 1202 = 1697 V


~

a = sinmiddot10011697) = 361 Q = 06]0 red

~

~

I


OJ = 377(00212) = 377 red

(a) Using Eq ]-22






J_])1


0- resulting in


~37


06]

yielding

p = IY = (225)(100) = 225 w


P =PR tPiE =312 t 225 = 256 W 8

The power factor is

Ppl= -P = -- shy

S V-I1fJIII

PSpicc Solution















V1l sin(Wl) +



diVsin(rot) L ) + v


bull ro dJ

Rearranging

Vsin(rot) - V

mL

Solving for iuct)




irot) mL roL

o otherwise

(3-29)

(3-30)

(3-31)

(3-32)

(3-33) ~

~


~

~

-






a = sin [ 12~j21 = 251 = 0438 rad






bull 0lt

_1 f [983 - 90000(01) - 382101) 21 Of31

resulting in


P (246)(72) 177 W

Therefore


bull 381 A

P 177

(120)(381)







101 t +



bull D1 is on





-int Dr

R

L

(j

+ shy R R

L L

I = 0

h)






-


1111 TnlIsn

i I~

I t

i I

--


-

o bull






bull bull







[00v == V == = 318 V o


VoI =shy 159 A o R



I bull = V

Z



v 100V=-=-=50 V

2 2

2VJi= = 212 V

(2 - 1raquolt

2V V 424 V

(4 - [raquolt

2V V 182 V

(6 - [raquolt



[s~ [ 5r[ Ir(O~lr = 1634 A

















8 4800pound+02 1345E-02 2512E-03 -1783E+02 4458pound+02

9 5400pound+02 5435E-05 1047pound-05 -1074E+02 5948E+02









is then

v (3-35)

oR




(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy


~








~

~

~

~ ~

~

~

~

~

~




p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems























forOsmtsPmiddot


sin(J - 0361) + sin(0361)e -jW37 = 0



HO


2 o

= 0308 A





P IR (0474) 100 224 W


2 2


c c


224 II



224pi = -p



















~ 1 I~

I ~


~ 1000mA-------------- --

~

I

o

e BETA - 200 DEG (2pP4SS~61090u))

OAf--------o-------------=-----c~ 0 100

D 1 (Rl)

10DOmA---- ----


J I

(16 570457 962m)

SOOmA

READ RMS VALUE HERE


10m 25m

Tim







39






01 R1

- --1Nl-- - ANv- -- [ Dbreak 10


---

--

SWEEP

PARAMETERS L = 01

----

-


40A

01l~)

Till








35 R-L-SOURCE LOAD






Vsino Vtk bull

usm - (3-18)-[ ~l ~


equation


dt

- -

shy

shy

shyI

I

I

I

I

shy

shy

shy

shy

shy

shy

- shy

~

shy

shy

shy

shy-v

-v

--shy

-shy

shy

-shy

shy

shy

- shy

shy

shy

- shy

shy

shy

R-L-SoUJCC Load311


+ lk

IJ

R L R

+ V sin(CllI)

b (CJ

~ 2~




iC) = ijt) + iC)



v Vijt) - in( - 9) - (3-20)

Z R




c

__ 1

R-L-Source Lolld312

(J-22)





V (J-2J)A = - -Zin(a-6)(



I = fi (mI)atmI) (3-24)

2bull bull


(J-25)


bull


(J-27)


p ~


~




of the circuit


- V

In = 1202 = 1697 V


~

a = sinmiddot10011697) = 361 Q = 06]0 red

~

~

I


OJ = 377(00212) = 377 red

(a) Using Eq ]-22






J_])1


0- resulting in


~37


06]

yielding

p = IY = (225)(100) = 225 w


P =PR tPiE =312 t 225 = 256 W 8

The power factor is

Ppl= -P = -- shy

S V-I1fJIII

PSpicc Solution















V1l sin(Wl) +



diVsin(rot) L ) + v


bull ro dJ

Rearranging

Vsin(rot) - V

mL

Solving for iuct)




irot) mL roL

o otherwise

(3-29)

(3-30)

(3-31)

(3-32)

(3-33) ~

~


~

~

-






a = sin [ 12~j21 = 251 = 0438 rad






bull 0lt

_1 f [983 - 90000(01) - 382101) 21 Of31

resulting in


P (246)(72) 177 W

Therefore


bull 381 A

P 177

(120)(381)







101 t +



bull D1 is on





-int Dr

R

L

(j

+ shy R R

L L

I = 0

h)






-


1111 TnlIsn

i I~

I t

i I

--


-

o bull






bull bull







[00v == V == = 318 V o


VoI =shy 159 A o R



I bull = V

Z



v 100V=-=-=50 V

2 2

2VJi= = 212 V

(2 - 1raquolt

2V V 424 V

(4 - [raquolt

2V V 182 V

(6 - [raquolt



[s~ [ 5r[ Ir(O~lr = 1634 A

















8 4800pound+02 1345E-02 2512E-03 -1783E+02 4458pound+02

9 5400pound+02 5435E-05 1047pound-05 -1074E+02 5948E+02









is then

v (3-35)

oR




(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy


~








~

~

~

~ ~

~

~

~

~

~




p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems
















224pi = -p



















~ 1 I~

I ~


~ 1000mA-------------- --

~

I

o

e BETA - 200 DEG (2pP4SS~61090u))

OAf--------o-------------=-----c~ 0 100

D 1 (Rl)

10DOmA---- ----


J I

(16 570457 962m)

SOOmA

READ RMS VALUE HERE


10m 25m

Tim







39






01 R1

- --1Nl-- - ANv- -- [ Dbreak 10


---

--

SWEEP

PARAMETERS L = 01

----

-


40A

01l~)

Till








35 R-L-SOURCE LOAD






Vsino Vtk bull

usm - (3-18)-[ ~l ~


equation


dt

- -

shy

shy

shyI

I

I

I

I

shy

shy

shy

shy

shy

shy

- shy

~

shy

shy

shy

shy-v

-v

--shy

-shy

shy

-shy

shy

shy

- shy

shy

shy

- shy

shy

shy

R-L-SoUJCC Load311


+ lk

IJ

R L R

+ V sin(CllI)

b (CJ

~ 2~




iC) = ijt) + iC)



v Vijt) - in( - 9) - (3-20)

Z R




c

__ 1

R-L-Source Lolld312

(J-22)





V (J-2J)A = - -Zin(a-6)(



I = fi (mI)atmI) (3-24)

2bull bull


(J-25)


bull


(J-27)


p ~


~




of the circuit


- V

In = 1202 = 1697 V


~

a = sinmiddot10011697) = 361 Q = 06]0 red

~

~

I


OJ = 377(00212) = 377 red

(a) Using Eq ]-22






J_])1


0- resulting in


~37


06]

yielding

p = IY = (225)(100) = 225 w


P =PR tPiE =312 t 225 = 256 W 8

The power factor is

Ppl= -P = -- shy

S V-I1fJIII

PSpicc Solution















V1l sin(Wl) +



diVsin(rot) L ) + v


bull ro dJ

Rearranging

Vsin(rot) - V

mL

Solving for iuct)




irot) mL roL

o otherwise

(3-29)

(3-30)

(3-31)

(3-32)

(3-33) ~

~


~

~

-






a = sin [ 12~j21 = 251 = 0438 rad






bull 0lt

_1 f [983 - 90000(01) - 382101) 21 Of31

resulting in


P (246)(72) 177 W

Therefore


bull 381 A

P 177

(120)(381)







101 t +



bull D1 is on





-int Dr

R

L

(j

+ shy R R

L L

I = 0

h)






-


1111 TnlIsn

i I~

I t

i I

--


-

o bull






bull bull







[00v == V == = 318 V o


VoI =shy 159 A o R



I bull = V

Z



v 100V=-=-=50 V

2 2

2VJi= = 212 V

(2 - 1raquolt

2V V 424 V

(4 - [raquolt

2V V 182 V

(6 - [raquolt



[s~ [ 5r[ Ir(O~lr = 1634 A

















8 4800pound+02 1345E-02 2512E-03 -1783E+02 4458pound+02

9 5400pound+02 5435E-05 1047pound-05 -1074E+02 5948E+02









is then

v (3-35)

oR




(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy


~








~

~

~

~ ~

~

~

~

~

~




p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














~ 1 I~

I ~


~ 1000mA-------------- --

~

I

o

e BETA - 200 DEG (2pP4SS~61090u))

OAf--------o-------------=-----c~ 0 100

D 1 (Rl)

10DOmA---- ----


J I

(16 570457 962m)

SOOmA

READ RMS VALUE HERE


10m 25m

Tim







39






01 R1

- --1Nl-- - ANv- -- [ Dbreak 10


---

--

SWEEP

PARAMETERS L = 01

----

-


40A

01l~)

Till








35 R-L-SOURCE LOAD






Vsino Vtk bull

usm - (3-18)-[ ~l ~


equation


dt

- -

shy

shy

shyI

I

I

I

I

shy

shy

shy

shy

shy

shy

- shy

~

shy

shy

shy

shy-v

-v

--shy

-shy

shy

-shy

shy

shy

- shy

shy

shy

- shy

shy

shy

R-L-SoUJCC Load311


+ lk

IJ

R L R

+ V sin(CllI)

b (CJ

~ 2~




iC) = ijt) + iC)



v Vijt) - in( - 9) - (3-20)

Z R




c

__ 1

R-L-Source Lolld312

(J-22)





V (J-2J)A = - -Zin(a-6)(



I = fi (mI)atmI) (3-24)

2bull bull


(J-25)


bull


(J-27)


p ~


~




of the circuit


- V

In = 1202 = 1697 V


~

a = sinmiddot10011697) = 361 Q = 06]0 red

~

~

I


OJ = 377(00212) = 377 red

(a) Using Eq ]-22






J_])1


0- resulting in


~37


06]

yielding

p = IY = (225)(100) = 225 w


P =PR tPiE =312 t 225 = 256 W 8

The power factor is

Ppl= -P = -- shy

S V-I1fJIII

PSpicc Solution















V1l sin(Wl) +



diVsin(rot) L ) + v


bull ro dJ

Rearranging

Vsin(rot) - V

mL

Solving for iuct)




irot) mL roL

o otherwise

(3-29)

(3-30)

(3-31)

(3-32)

(3-33) ~

~


~

~

-






a = sin [ 12~j21 = 251 = 0438 rad






bull 0lt

_1 f [983 - 90000(01) - 382101) 21 Of31

resulting in


P (246)(72) 177 W

Therefore


bull 381 A

P 177

(120)(381)







101 t +



bull D1 is on





-int Dr

R

L

(j

+ shy R R

L L

I = 0

h)






-


1111 TnlIsn

i I~

I t

i I

--


-

o bull






bull bull







[00v == V == = 318 V o


VoI =shy 159 A o R



I bull = V

Z



v 100V=-=-=50 V

2 2

2VJi= = 212 V

(2 - 1raquolt

2V V 424 V

(4 - [raquolt

2V V 182 V

(6 - [raquolt



[s~ [ 5r[ Ir(O~lr = 1634 A

















8 4800pound+02 1345E-02 2512E-03 -1783E+02 4458pound+02

9 5400pound+02 5435E-05 1047pound-05 -1074E+02 5948E+02









is then

v (3-35)

oR




(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy


~








~

~

~

~ ~

~

~

~

~

~




p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














39






01 R1

- --1Nl-- - ANv- -- [ Dbreak 10


---

--

SWEEP

PARAMETERS L = 01

----

-


40A

01l~)

Till








35 R-L-SOURCE LOAD






Vsino Vtk bull

usm - (3-18)-[ ~l ~


equation


dt

- -

shy

shy

shyI

I

I

I

I

shy

shy

shy

shy

shy

shy

- shy

~

shy

shy

shy

shy-v

-v

--shy

-shy

shy

-shy

shy

shy

- shy

shy

shy

- shy

shy

shy

R-L-SoUJCC Load311


+ lk

IJ

R L R

+ V sin(CllI)

b (CJ

~ 2~




iC) = ijt) + iC)



v Vijt) - in( - 9) - (3-20)

Z R




c

__ 1

R-L-Source Lolld312

(J-22)





V (J-2J)A = - -Zin(a-6)(



I = fi (mI)atmI) (3-24)

2bull bull


(J-25)


bull


(J-27)


p ~


~




of the circuit


- V

In = 1202 = 1697 V


~

a = sinmiddot10011697) = 361 Q = 06]0 red

~

~

I


OJ = 377(00212) = 377 red

(a) Using Eq ]-22






J_])1


0- resulting in


~37


06]

yielding

p = IY = (225)(100) = 225 w


P =PR tPiE =312 t 225 = 256 W 8

The power factor is

Ppl= -P = -- shy

S V-I1fJIII

PSpicc Solution















V1l sin(Wl) +



diVsin(rot) L ) + v


bull ro dJ

Rearranging

Vsin(rot) - V

mL

Solving for iuct)




irot) mL roL

o otherwise

(3-29)

(3-30)

(3-31)

(3-32)

(3-33) ~

~


~

~

-






a = sin [ 12~j21 = 251 = 0438 rad






bull 0lt

_1 f [983 - 90000(01) - 382101) 21 Of31

resulting in


P (246)(72) 177 W

Therefore


bull 381 A

P 177

(120)(381)







101 t +



bull D1 is on





-int Dr

R

L

(j

+ shy R R

L L

I = 0

h)






-


1111 TnlIsn

i I~

I t

i I

--


-

o bull






bull bull







[00v == V == = 318 V o


VoI =shy 159 A o R



I bull = V

Z



v 100V=-=-=50 V

2 2

2VJi= = 212 V

(2 - 1raquolt

2V V 424 V

(4 - [raquolt

2V V 182 V

(6 - [raquolt



[s~ [ 5r[ Ir(O~lr = 1634 A

















8 4800pound+02 1345E-02 2512E-03 -1783E+02 4458pound+02

9 5400pound+02 5435E-05 1047pound-05 -1074E+02 5948E+02









is then

v (3-35)

oR




(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy


~








~

~

~

~ ~

~

~

~

~

~




p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems



















35 R-L-SOURCE LOAD






Vsino Vtk bull

usm - (3-18)-[ ~l ~


equation


dt

- -

shy

shy

shyI

I

I

I

I

shy

shy

shy

shy

shy

shy

- shy

~

shy

shy

shy

shy-v

-v

--shy

-shy

shy

-shy

shy

shy

- shy

shy

shy

- shy

shy

shy

R-L-SoUJCC Load311


+ lk

IJ

R L R

+ V sin(CllI)

b (CJ

~ 2~




iC) = ijt) + iC)



v Vijt) - in( - 9) - (3-20)

Z R




c

__ 1

R-L-Source Lolld312

(J-22)





V (J-2J)A = - -Zin(a-6)(



I = fi (mI)atmI) (3-24)

2bull bull


(J-25)


bull


(J-27)


p ~


~




of the circuit


- V

In = 1202 = 1697 V


~

a = sinmiddot10011697) = 361 Q = 06]0 red

~

~

I


OJ = 377(00212) = 377 red

(a) Using Eq ]-22






J_])1


0- resulting in


~37


06]

yielding

p = IY = (225)(100) = 225 w


P =PR tPiE =312 t 225 = 256 W 8

The power factor is

Ppl= -P = -- shy

S V-I1fJIII

PSpicc Solution















V1l sin(Wl) +



diVsin(rot) L ) + v


bull ro dJ

Rearranging

Vsin(rot) - V

mL

Solving for iuct)




irot) mL roL

o otherwise

(3-29)

(3-30)

(3-31)

(3-32)

(3-33) ~

~


~

~

-






a = sin [ 12~j21 = 251 = 0438 rad






bull 0lt

_1 f [983 - 90000(01) - 382101) 21 Of31

resulting in


P (246)(72) 177 W

Therefore


bull 381 A

P 177

(120)(381)







101 t +



bull D1 is on





-int Dr

R

L

(j

+ shy R R

L L

I = 0

h)






-


1111 TnlIsn

i I~

I t

i I

--


-

o bull






bull bull







[00v == V == = 318 V o


VoI =shy 159 A o R



I bull = V

Z



v 100V=-=-=50 V

2 2

2VJi= = 212 V

(2 - 1raquolt

2V V 424 V

(4 - [raquolt

2V V 182 V

(6 - [raquolt



[s~ [ 5r[ Ir(O~lr = 1634 A

















8 4800pound+02 1345E-02 2512E-03 -1783E+02 4458pound+02

9 5400pound+02 5435E-05 1047pound-05 -1074E+02 5948E+02









is then

v (3-35)

oR




(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy


~








~

~

~

~ ~

~

~

~

~

~




p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














- -

shy

shy

shyI

I

I

I

I

shy

shy

shy

shy

shy

shy

- shy

~

shy

shy

shy

shy-v

-v

--shy

-shy

shy

-shy

shy

shy

- shy

shy

shy

- shy

shy

shy

R-L-SoUJCC Load311


+ lk

IJ

R L R

+ V sin(CllI)

b (CJ

~ 2~




iC) = ijt) + iC)



v Vijt) - in( - 9) - (3-20)

Z R




c

__ 1

R-L-Source Lolld312

(J-22)





V (J-2J)A = - -Zin(a-6)(



I = fi (mI)atmI) (3-24)

2bull bull


(J-25)


bull


(J-27)


p ~


~




of the circuit


- V

In = 1202 = 1697 V


~

a = sinmiddot10011697) = 361 Q = 06]0 red

~

~

I


OJ = 377(00212) = 377 red

(a) Using Eq ]-22






J_])1


0- resulting in


~37


06]

yielding

p = IY = (225)(100) = 225 w


P =PR tPiE =312 t 225 = 256 W 8

The power factor is

Ppl= -P = -- shy

S V-I1fJIII

PSpicc Solution















V1l sin(Wl) +



diVsin(rot) L ) + v


bull ro dJ

Rearranging

Vsin(rot) - V

mL

Solving for iuct)




irot) mL roL

o otherwise

(3-29)

(3-30)

(3-31)

(3-32)

(3-33) ~

~


~

~

-






a = sin [ 12~j21 = 251 = 0438 rad






bull 0lt

_1 f [983 - 90000(01) - 382101) 21 Of31

resulting in


P (246)(72) 177 W

Therefore


bull 381 A

P 177

(120)(381)







101 t +



bull D1 is on





-int Dr

R

L

(j

+ shy R R

L L

I = 0

h)






-


1111 TnlIsn

i I~

I t

i I

--


-

o bull






bull bull







[00v == V == = 318 V o


VoI =shy 159 A o R



I bull = V

Z



v 100V=-=-=50 V

2 2

2VJi= = 212 V

(2 - 1raquolt

2V V 424 V

(4 - [raquolt

2V V 182 V

(6 - [raquolt



[s~ [ 5r[ Ir(O~lr = 1634 A

















8 4800pound+02 1345E-02 2512E-03 -1783E+02 4458pound+02

9 5400pound+02 5435E-05 1047pound-05 -1074E+02 5948E+02









is then

v (3-35)

oR




(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy


~








~

~

~

~ ~

~

~

~

~

~




p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














__ 1

R-L-Source Lolld312

(J-22)





V (J-2J)A = - -Zin(a-6)(



I = fi (mI)atmI) (3-24)

2bull bull


(J-25)


bull


(J-27)


p ~


~




of the circuit


- V

In = 1202 = 1697 V


~

a = sinmiddot10011697) = 361 Q = 06]0 red

~

~

I


OJ = 377(00212) = 377 red

(a) Using Eq ]-22






J_])1


0- resulting in


~37


06]

yielding

p = IY = (225)(100) = 225 w


P =PR tPiE =312 t 225 = 256 W 8

The power factor is

Ppl= -P = -- shy

S V-I1fJIII

PSpicc Solution















V1l sin(Wl) +



diVsin(rot) L ) + v


bull ro dJ

Rearranging

Vsin(rot) - V

mL

Solving for iuct)




irot) mL roL

o otherwise

(3-29)

(3-30)

(3-31)

(3-32)

(3-33) ~

~


~

~

-






a = sin [ 12~j21 = 251 = 0438 rad






bull 0lt

_1 f [983 - 90000(01) - 382101) 21 Of31

resulting in


P (246)(72) 177 W

Therefore


bull 381 A

P 177

(120)(381)







101 t +



bull D1 is on





-int Dr

R

L

(j

+ shy R R

L L

I = 0

h)






-


1111 TnlIsn

i I~

I t

i I

--


-

o bull






bull bull







[00v == V == = 318 V o


VoI =shy 159 A o R



I bull = V

Z



v 100V=-=-=50 V

2 2

2VJi= = 212 V

(2 - 1raquolt

2V V 424 V

(4 - [raquolt

2V V 182 V

(6 - [raquolt



[s~ [ 5r[ Ir(O~lr = 1634 A

















8 4800pound+02 1345E-02 2512E-03 -1783E+02 4458pound+02

9 5400pound+02 5435E-05 1047pound-05 -1074E+02 5948E+02









is then

v (3-35)

oR




(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy


~








~

~

~

~ ~

~

~

~

~

~




p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems















OJ = 377(00212) = 377 red

(a) Using Eq ]-22






J_])1


0- resulting in


~37


06]

yielding

p = IY = (225)(100) = 225 w


P =PR tPiE =312 t 225 = 256 W 8

The power factor is

Ppl= -P = -- shy

S V-I1fJIII

PSpicc Solution















V1l sin(Wl) +



diVsin(rot) L ) + v


bull ro dJ

Rearranging

Vsin(rot) - V

mL

Solving for iuct)




irot) mL roL

o otherwise

(3-29)

(3-30)

(3-31)

(3-32)

(3-33) ~

~


~

~

-






a = sin [ 12~j21 = 251 = 0438 rad






bull 0lt

_1 f [983 - 90000(01) - 382101) 21 Of31

resulting in


P (246)(72) 177 W

Therefore


bull 381 A

P 177

(120)(381)







101 t +



bull D1 is on





-int Dr

R

L

(j

+ shy R R

L L

I = 0

h)






-


1111 TnlIsn

i I~

I t

i I

--


-

o bull






bull bull







[00v == V == = 318 V o


VoI =shy 159 A o R



I bull = V

Z



v 100V=-=-=50 V

2 2

2VJi= = 212 V

(2 - 1raquolt

2V V 424 V

(4 - [raquolt

2V V 182 V

(6 - [raquolt



[s~ [ 5r[ Ir(O~lr = 1634 A

















8 4800pound+02 1345E-02 2512E-03 -1783E+02 4458pound+02

9 5400pound+02 5435E-05 1047pound-05 -1074E+02 5948E+02









is then

v (3-35)

oR




(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy


~








~

~

~

~ ~

~

~

~

~

~




p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems


















V1l sin(Wl) +



diVsin(rot) L ) + v


bull ro dJ

Rearranging

Vsin(rot) - V

mL

Solving for iuct)




irot) mL roL

o otherwise

(3-29)

(3-30)

(3-31)

(3-32)

(3-33) ~

~


~

~

-






a = sin [ 12~j21 = 251 = 0438 rad






bull 0lt

_1 f [983 - 90000(01) - 382101) 21 Of31

resulting in


P (246)(72) 177 W

Therefore


bull 381 A

P 177

(120)(381)







101 t +



bull D1 is on





-int Dr

R

L

(j

+ shy R R

L L

I = 0

h)






-


1111 TnlIsn

i I~

I t

i I

--


-

o bull






bull bull







[00v == V == = 318 V o


VoI =shy 159 A o R



I bull = V

Z



v 100V=-=-=50 V

2 2

2VJi= = 212 V

(2 - 1raquolt

2V V 424 V

(4 - [raquolt

2V V 182 V

(6 - [raquolt



[s~ [ 5r[ Ir(O~lr = 1634 A

















8 4800pound+02 1345E-02 2512E-03 -1783E+02 4458pound+02

9 5400pound+02 5435E-05 1047pound-05 -1074E+02 5948E+02









is then

v (3-35)

oR




(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy


~








~

~

~

~ ~

~

~

~

~

~




p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems



















a = sin [ 12~j21 = 251 = 0438 rad






bull 0lt

_1 f [983 - 90000(01) - 382101) 21 Of31

resulting in


P (246)(72) 177 W

Therefore


bull 381 A

P 177

(120)(381)







101 t +



bull D1 is on





-int Dr

R

L

(j

+ shy R R

L L

I = 0

h)






-


1111 TnlIsn

i I~

I t

i I

--


-

o bull






bull bull







[00v == V == = 318 V o


VoI =shy 159 A o R



I bull = V

Z



v 100V=-=-=50 V

2 2

2VJi= = 212 V

(2 - 1raquolt

2V V 424 V

(4 - [raquolt

2V V 182 V

(6 - [raquolt



[s~ [ 5r[ Ir(O~lr = 1634 A

















8 4800pound+02 1345E-02 2512E-03 -1783E+02 4458pound+02

9 5400pound+02 5435E-05 1047pound-05 -1074E+02 5948E+02









is then

v (3-35)

oR




(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy


~








~

~

~

~ ~

~

~

~

~

~




p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














101 t +



bull D1 is on





-int Dr

R

L

(j

+ shy R R

L L

I = 0

h)






-


1111 TnlIsn

i I~

I t

i I

--


-

o bull






bull bull







[00v == V == = 318 V o


VoI =shy 159 A o R



I bull = V

Z



v 100V=-=-=50 V

2 2

2VJi= = 212 V

(2 - 1raquolt

2V V 424 V

(4 - [raquolt

2V V 182 V

(6 - [raquolt



[s~ [ 5r[ Ir(O~lr = 1634 A

















8 4800pound+02 1345E-02 2512E-03 -1783E+02 4458pound+02

9 5400pound+02 5435E-05 1047pound-05 -1074E+02 5948E+02









is then

v (3-35)

oR




(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy


~








~

~

~

~ ~

~

~

~

~

~




p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems















1111 TnlIsn

i I~

I t

i I

--


-

o bull






bull bull







[00v == V == = 318 V o


VoI =shy 159 A o R



I bull = V

Z



v 100V=-=-=50 V

2 2

2VJi= = 212 V

(2 - 1raquolt

2V V 424 V

(4 - [raquolt

2V V 182 V

(6 - [raquolt



[s~ [ 5r[ Ir(O~lr = 1634 A

















8 4800pound+02 1345E-02 2512E-03 -1783E+02 4458pound+02

9 5400pound+02 5435E-05 1047pound-05 -1074E+02 5948E+02









is then

v (3-35)

oR




(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy


~








~

~

~

~ ~

~

~

~

~

~




p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














bull bull







[00v == V == = 318 V o


VoI =shy 159 A o R



I bull = V

Z



v 100V=-=-=50 V

2 2

2VJi= = 212 V

(2 - 1raquolt

2V V 424 V

(4 - [raquolt

2V V 182 V

(6 - [raquolt



[s~ [ 5r[ Ir(O~lr = 1634 A

















8 4800pound+02 1345E-02 2512E-03 -1783E+02 4458pound+02

9 5400pound+02 5435E-05 1047pound-05 -1074E+02 5948E+02









is then

v (3-35)

oR




(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy


~








~

~

~

~ ~

~

~

~

~

~




p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems
















[s~ [ 5r[ Ir(O~lr = 1634 A

















8 4800pound+02 1345E-02 2512E-03 -1783E+02 4458pound+02

9 5400pound+02 5435E-05 1047pound-05 -1074E+02 5948E+02









is then

v (3-35)

oR




(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy


~








~

~

~

~ ~

~

~

~

~

~




p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems


















is then

v (3-35)

oR




(3-36)

240V~ i

I3sA~S IC

o bull 2

us ~ 1----- ioI3J__-- ---shy


~








~

~

~

~ ~

~

~

~

~

~




p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems

















p (I~) R (135) 8 1459 W


bull _I(135)010gtI) 955 A 2

The power factor is

p 1459 0637 PI V I (240)(955) rru







= VIZ ~ = 251 n I 0675


Z = 251 = IR + jmLI 18 + 1177LI


L bull IZI = 251 = 067 H m 377


this estimate






CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














CpocilOr Filla322











rs diode all


where

V = Vsin8 (3-38)

Ii + ~ 10


[a

v

rvbull -

~

2~ ~~ 2+laquo 2

r ~

~

bull ~

tb)





aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems

















aod




=_1_ -raRC

1 1 tan9 -raRC

-(3-41 )

V (3-42)




exponential output


0 lin(a) - (sin9)e -(b - 8)ltIIR( 0 (3-43)1 _



i-t) = C ltIvit) dt



d(mt)

cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














cpacito Fil324






(3-45)







R R











0-50)


eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














eoFil







v


V 120v2 1697 V0 0

mRC 0 (2_60)(500)(10)~ 1885 ad0


9 == -tim-I (1885) + lit == 162 rad = 93

V llin9 == 1695 V


sin(a) - sin(162) e -z + - 1fi2)l1B83 0

yielding

a = 0843 rad == 48



1695 e -01gt1 - 161)11883 9sCIlIs2l1t+a








426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems


















426 + 034 = 450 A




I 0 ~u sin-I( 1 - A~l sm- ( I - ~c) = 819


PSpke Solution











200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems
















200V ~ -1 ~ ~)



00 ~ OI~ c~~m i


0 10 20115 30 1(0)



QUANTITY RESULT

40S (u~4S)


v min 126 V

6V 429 V



~













quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems






















quantities



conduct








converter


Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems















Resistive Load


in Fig 3-13a is





bull

V 1 _ a + sin(21l) 2 11 211








this application


a= co-H ~l 1]

= coH I2~~20)1 = 61ZO= 107 rad

Eq 3-53 gives

= 12(120) 11 _ 07 + in[2(107)] = 756 votu 2 ~ 11 211

-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














-

~

r

330 Cpllor FIlter

Load power is

p r

v ~

R bull (756)

100 571 w

~

~


pplmiddot shys

p

(120)(756100) 571 middot063



applies


= [ -J sin(cot - 8) + Ae -511 bull


(3-54)




o otherwise









I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems

















I = ~J~)d(Olt)


i

shybull

Capil Fill

(3-57)

(3-58)



+ +

+

(11)

- r----middotn

o

ibl


cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














cpunar Fill332





Vm = 1202 = 1697 volts





fOlasrotsPmiddot




1 = _I 1 379 [678 sin(mt - 0646) - 267e --to7S4]d(wt)

21t O7U

= 219 A


= 326 A

yielding

P = IR = (326)(20) = 213 W


PI= P S

213

(120)(326) = =~= 054






than the de source

---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














---

I

bull bullbullIt

333




( (

R L (

( i (

(







W1 = wUR = 377middot00212 = 371



- = ml 1~~~iJ = 36


(3middot60)

(3middot61 )

2 n L = 20

Determine the






i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems



















i(wt)

--f1(mt)t(cot) = 390 A P = (390)(2) = 304 W



p =1Y = (219)(100) = 219 W











(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














(cliogt fill335




Ik~ V2=5 TF = In



____ $~8J~9_dJlI_

Dllreak A1m

20

11

ltOm


SOURCE

pound I

--------~-----------2

V1 = 0 ~o






Hz ac source

336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














336










r~om 100 -==shy

a

middot0

~v

V 0


e uDbre~k SOIJ~CE


-

ow---------~TI--------------__


c

200W

~ (l66 70ru147531)

~io deg

p

J





337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














337











311 COMMUTATION













(3-62)



~

Vsin(mt)d((01) + 0 (JL 0

i V (1 _ ) (3-63)coL

Current in OJ is



Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














Commumlin338


V ~(I - cos 101) = 0 roL

Solving for u


angle u






bull


2K


(3-65)

1 shy


voltage

339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














339

Iv

LlL l-_--+-__l__

LI__L-__+-__L__


- 312 SUMMARY



current of VJnR



o -

Land-r=shy R




340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














340


o otherwise

where



currenl




voltage






















- I

341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














341




















FreeWheeling Diode















342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














342

R L

+

tilUlW Pll6


















PI = ~ _-- + sffi(2a) 2 221 411








-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














-

-----

--

-



















load

Commutation





344










0 1 to O2bull





Design Problems














344










0 1 to O2bull





Design Problems














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