CHAPTER 2 CIRCUITS WITH SWITCHES AND DIODESEce410/Chapter-2.pdfFig. 2.9 Half-wave diode rectifier. 1 ( cos sin) n vo Vo an n t bn n t V (2.38) The angle at which diode starts to conduct

Bhat, A.K.S. ELEC410 CHAPTER 2 Page 2-1

CHAPTER 2 CIRCUITS WITH SWITCHES AND DIODES

We assume that all elements are ideal. Switches: INFINITE or ZERO resistance to current, and INSTANTANEOUS TRANSITION from one state to another. (However, we will also discuss non-ideal devices at a later stage when we discuss power devices in more detail). 2.1 SWITCHES AND DIODES:

vAK

iA

0

D “ON”

D

R

+

+

-

-

vAK

vR

iA

D “OFF”

V2

R

V2

tVtv sin2)(

v(t)

IDEAL DIODE

Fig. 2.1 A simple circuit with ideal diode and ac source. OPERATION of a switch may

(1) Apply an energy source. (2) Remove an energy source. (3) Change the configuration in the network in other ways (for example: commutation network in a force commutated thyristor, soft-switching converters).

2.2 SWITCHED DC SOURCE: 2.2.1 R-LOAD When switch is ON (conducting): Voltage across the switch vsw = 0,

R

Vi A (2.1)

When switch is OPEN (OFF or in non-conducting state): vsw = V

Fig. 2.2 Switch with DC source and resistive load.

2.2.2 RC LOAD CIRCUIT


Fig. 2.3(a) Switch with DC source and RC load circuit. When SW is closed at t = 0, Kirchoff’s Voltage Law (KVL) gives V = vC + vR

= RividtC C

t

)0(

1

0 V (2.2)

Here, assume initial condition (I.C.), vc(t = 0) = vc(0) = 0. Differentiate (2.2):

0RC

i

dt

di A/s (2.3)

Solution for (2.3) is:

)/(RCtKei A (2.4) K is a constant of integration. I.C., vC(t = 0) = 0. vC CAN NOT CHANGE INSTANTANEOUSLY, [since vC = q/C, if vC changes instantaneously q changes instantly, i.e., infinite current] Immediately after SW is closed, t = 0+, vC(0+) = vC(0-) = vC(0) = 0. V = 0 + vR (0) = Ri(0) V (2.5)

i.e., R

Vi )0( A (2.6)

Substitute t = 0 and (2.6) in (2.4) R

VK A (2.7)

RC

t

R

Vi exp A (2.8)

As vR falls, vC rises till capacitor is fully charged, so that i = 0 A, vC = V V (2.9)


If the switch were opened at t = t1, before capacitor is fully charged, then the voltage across SW would be vSW = V – vC V (2.10)

Fig. 2.3(b) Waveforms i and vc in Fig. 2.3(a).

If R is VERY SMALL, initial current would be EXTREMELY HIGH, and flow of current will form a pulse of very short duration. This may destroy the power semiconductor switch. 2.2.3 RL LOAD CIRCUIT

Fig. 2.4(a) Switch with DC source and RL load circuit. If SW is closed, KVL:

V = vL + vR = Ridt

diL V (2.11)

L

Vi

L

R

dt

di A/s (2.12)

Two components of i: (i) FORCED or steady-state component (after infinite time) iF : is given by particular integral solution. (ii) NATURAL or TRANSIENT component (vanishes after infinite time) iN :

- Determined by circuit parameters and I.C.s (t = 0) - Complementary function of the solution.

R

ViF A (2.13)


0 NN i

L

R

dt

di A/s (2.14)

t

L

RKiN .exp A (2.15)

K = constant of integration. Therefore, total response or complete solution:

i = iF + iN = tLRKeR

V )/( A (2.16)

At t = 0, i = 0 R

VK A (2.17)

tLReR

Vi )/(1 A (2.18)

tLR

L Vedt

diLv )/( V (2.19)

If SW is opened, since L tends to maintain the current constant; and i → 0 instantaneously (if SW is opened) [Note: CURRENT THROUGH AN INDUCTOR CAN NOT CHANGE INSTANTANEOUSLY]

dt

di , Lv (see Fig. 2.4(b))

Power semiconductor switch will be destroyed.

Fig. 2.4(b) Waveforms in Fig. 2.4(a). 2.2.4 INDUCTIVE LOAD CIRCUIT


Fig. 2.5 If R is very small; L is large, then

L

V

dt

di A/s (2.20)

tL

Vi . A (2.21)

How to open the switch? Still a problem. ONE SOLUTION:

Fig. 2.6 Connection of a FWD. Connect a diode D [called free-wheeling diode (FWD)] across load. At t = 0, close SW. At t = t1, open SW. SW is closed, 0 < t ≤ t1: Operation is the same as earlier.

vSW = 0, vL = vD = V, i = tL

V. , iL = i, iD = 0

SW is open, D is ON, t ≥ t1 : 0dt

diLvL

At t = t1, when SW is opened since when vL becomes negative, D turns ON and vSW = V.

vL = vD = 0, i = 0, iD = 1.tL

V , iL = iD


PRACTICAL CIRCUIT: Energy stored in L will be dissipated in resistance of L and D, i will decay exponentially. If L is very large (or R is very small), subsequent closures of SW will increase i. 2.2.4 DC SOURCE AND RLC CIRCUIT

Fig. 2.7 Switch with DC source and RLC load circuit. SW is closed, KVL: V = vL + vR + vC

= Ridt

diL +

)0(

1

0c

tvidt

C V (2.22)

01

2

2 i

LCdt

di

L

R

dt

id A/s2 (2.23)

Steady-state solution: iF = 0 A (2.24) Transient solution: tsts

N eKeKi 2121 A (2.25)

s1 and s2 are roots of auxiliary equation

012

LCs

L

Rs (2.26)

Define:

Damping Factor, L

R

2 (2.27)

Resonance Frequency, LC

10 rads/sec (2.28)

Then (2.23) becomes:

02 202

2 i

dt

di

dt

id A/s2 (2.29)

Solution of is of the form:

ttt rr eKeKei 21 A (2.30)

where 20

2 r (ringing frequency) rads/sec (2.31)

If 2

02 , then i is made up of two decaying (or damped) sinusoid:


tBtBei rrt sincos 21 A (30a)

If 2

02 , then i is made up of two decaying exponential components.

2.2.5 AC SOURCE AND RLC CIRCUIT

Fig. 2.8 Switch with AC source and RLC load circuit.

tVtv sin2)( V (2.32) SW is closed at t = 0, KVL:

tVRividtCdt

diL C

t

sin2)0(

1

0 V (2.33)

)cos(21

2

2t

L

Vi

LCdt

di

L

R

dt

id A/s2 (2.34)

Natural response is

tstsN eKeKi 21

21 A (2.35) Same as with DC sources. Forced response (from STEADY-STATE AC ANALYSIS USING PHASOR CIRCUIT):

Z

tViF

)sin(2 A (2.36)

where

22 1

CLRZ (2.37a)

RC

L1

tan 1 rads (2.37b)

Total response (or complete solution): i = iF + iN

Apply I.C.s to obtain constants K1 and K2.


2.3 HALF-WAVE RECTIFIER When a diode is introduced between AC source and load, the excitation applied to the load is no longer sine-wave, and forced response can not be found by sinusoidal steady-state analysis. But such excitation is periodic, and, therefore (1) can be represented by FOURIER SERIES (sum of sine waves) (2) then find forced response by sinusoidal steady-state analysis (3) since load is linear, superposition gives the forced response. Half-wave rectifier introduces DC on the AC side. Therefore, if line-side transformer is used then it gets saturated. Hence, not used widely in practice, but study of half-wave rectifier can be extended to other configurations. FOURIER SERIES

tVv sin2 If v – vo > 0, vAK = 0 (diode conducts since forward biased) and current flows.

Fig. 2.9 Half-wave diode rectifier.

1)sincos(

nnnoo tnbtnaVv V (2.38)

The angle at which diode starts to conduct (firing angle) is and that at which it ceases to conduct (the extinction angle) , then the conduction angle is = – rads. (2.39) For passive load: = 0, = . Extinction angle depends on the nature of the load circuit. Average output voltage (DC output):

2

0)(

2

1tdvV oo =

)(

2

1tdvo V (2.40)

Fourier coefficients an and bn are defined as

2

0)(.cos

1tdtnva on =

)(.cos

1tdtnvo V (2.41)

2

0)(.sin

1tdtnvb on =

)(.sin

1tdtnvo V (2.42)


RMS value of nth harmonic is given by:

2/122

2

1nnnR baV V (2.43)

RMS value of rectified voltage is:

2/1

1

222 )(.2

1

nnRooR VVtdvV V (2.44)

RMS value of the harmonic components or RIPPLE VOLTAGE is:

2/1222/1

1

2oR

nnRRI VVVV

V (2.45)

The VOLTAGE RIPPLE FACTOR is defined as:

o

RIv V

VK (2.46)

For the current, all the equations corresponding to (2.38) to (2.46) can be written (here RL load circuit is assumed for illustration purpose):

1)sin()cos(

nnnnnoo tndtncIi V (2.47)

where

R

VI o

o A (2.48)

n

nn Z

ac A;

n

nn Z

bd A;

R

Lnn

1tan rad (2.49)

Zn is impedance for nth harmonic (i.e., n). Here nnn ZZ

22 LnRZn . (2.50)

The RMS value of nth harmonic current is:

2/122

2

1nnnR dcI A (2.51)

RMS value of the harmonic components of ripple current is:

2/1222/1

1

2oR

nnRRI IIII

A (2.52)

Current Ripple Factor; Ki


o

RIi I

IK (2.53)

R-LOAD

Fig. 2.10 Half-wave diode rectifier with R-

load and waveforms.

tVv sin2

= 0, = .

vo = v V; R

vi A: t0 rad. (2.54a)

vo = 0 V; i = 0 A: 2t rad. (2.54b) Average output voltage:

0)(sin2

2

1ttdVVo =

V2

V (2.55)

RMS output voltage:

2/1

0

22 )(sin22

1

ttdVVR =

2

V V (2.56)

RMS value of the harmonic components or RIPPLE VOLTAGE is:

2/122oRRI VVV V

=

2/122/12

14

1

o

o

Ro V

V

VV = 1.211Vo V (2.57)

= (1.211)V2

V (2.58)


o

RIv V

VK = 1.211 (2.59)

R

VI o

o = R

V

2

A (2.60)

R

VI R

R = R

V

2 A (2.61)

Ki = Kv = 1.211 (2.62) RC LOAD CIRCUIT

tVv sin2 Fig. 2.11 Half-wave rectifier with RC load.

KVL: vC + vR = v V (2.63)

tVRividtC C

t

sin2)0(

1

0 V (2.64)

Z

tViF

)sin(2 A (2.65)

where

22 1

C

RZ (2.66a)

CR

1tan 1 rads (2.66b)

)/( RCt

N Kei A (2.67) Total response (or complete solution):

i = iF + iN A (2.68) Close SW during negative half-cycle and capacitor has no initial charge.


t = 0, D starts conducting; vC(0) = 0, i = 0

sin.)sin(2 )/(RCtetZ

Vi A (2.69)

)0(1

0c

t

C vidtC

v

=

1sin

cos)cos(2 )/(RCteRCt

CZ

V V (2.70)

Substitute

sin1

ZC

and cosZR

)cos(.cossin2 )/( teVv RCtC V (2.71)

At the end of pulse current, when t = > /2; vC is positive. Therefore, vC is positive for next pulse of current. During succeeding cycles, D can conduct only when v > vC and eventually vC = V2 V. (2.72) And conduction ceases completely if R = 0, then (2.72) is reached at the end of first pulse.


RL LOAD CIRCUIT

tVv sin2

Fig. 2.12 Half-wave rectifier with RL load and waveforms.

SW is closed and D is conducting. KVL:

tVRidt

diL sin2 V (2.73)

If SW is closed during negative half-cycle,

Z

tViF

)sin(2 A (2.74)

where

22 LRZ (2.75a)


R

L1tan rads (2.75b)

tLR

N Kei )/( A (2.76) Total response (or complete solution): i = iF + iN A (2.77) Using I.C., i = 0 at t = 0:

sin.)sin(2 )/( tLRetZ

Vi A: 0 < t < rad. (2.78)

i = 0 < t < 2 rad (2.79) i = 0, at t = in (2.78) we get :

0sin.)sin( )/( LRe (2.80) This is a transcendental equation, can be solved numerically for given values of , R and L. Average Current:

tVvRidt

diL sin2 V (2.81)

dt

di

R

L

R

vi . A (2.82)

)(

sin2

td

di

R

L

R

tV

A (2.83)

Average current is given by

0)(

2

1tidIo

)()(

.sin.2

2

1

0td

td

di

R

Lt

R

V

A (2.84)

Second term is zero.


)cos1(2

2

R

VIo A (2.85)

Since no average voltage can appear across L,

)cos1(2

2

VRIV oo V (2.86)

For satisfactory design; ripple voltage VRI, Ripple current IRI, RMS value of rectified current IR and Kv, Ki are required. These values may be obtained by Fourier series analysis. Normalization:

First select base values: Here VB = V2 , ZB = Z , then Z

V

Z

VI

B

BB

2

Then normalize all parameters. Normalized value of instantaneous current is defined as:

Z

V

iiN

2 p.u. (2.87)

Then from (2.78):

Z

V

etZ

V

i

tLR

N2

sin.)sin(2 )/(

= sin.)sin( )/( tLRet p.u. : 0 < t < rad. (2.88)

Normalized average rectified current

)(sin.)sin(2

1 )/(

0tdetI tLR

oN

p.u. (2.89)

For given , can be obtained [using (2.80)], curve obtained is shown in Fig. 2.14(a). Then IoN can be calculated from (2.89). IoN versus plotted in Fig. 2.14(b). Normalized RMS current is given by

2/1

2)/(

0)(sin.)sin(

2

1

tdetI tLR

RN p.u. (2.90)

IRN versus is given in Fig. 2.14(b). RMS output voltage is


2/12

0)()sin2(

2

1

tdtVVR V (2.91)

Particular case, If L >> R

Fig. 2.13 voltage across inductor and current for the case of negligible resistance.

From (2.75a), (2.75b) and (2.78):

)cos1(2

tL

Vi

A (2.92)

From (2.92), average current is:

L

VIo

2

A (2.93)

The only harmonic present is fundamental, its RMS values is:

21

oR

I

L

VI

A (2.94)

RMS values of rectified current is

2/121

2RoR III = 1.225Io A (2.95)

vL = v for the entire cycle. Vo = 0 (2.96)

o

R

o

RIv V

V

V

VK 1 (2.97)

o

Ri I

IK 1 = 0.707 (2.98)


Fig. 2.14(a) versus for circuit of Fig. 2.12.

Fig. 2.14(b) IRN and IN (= IoN) versus for circuit of Fig. 2.12.

BHAT ELEC410

BHAT ELEC410

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CHAPTER 2 CIRCUITS WITH SWITCHES AND DIODESEce410/Chapter-2.pdfFig. 2.9 Half-wave diode rectifier. 1 ( cos sin) n vo Vo an n t bn n t V (2.38) The angle at which diode starts to conduct