CH2

18

Chapter 2

Diode Circuits or Uncontrolled Rectifier 2.1 Introduction The only way to turn on the diode is when its anode voltage becomes higher than cathode voltage as explained in the previous chapter. So, there is no control on the conduction time of the diode which is the main disadvantage of the diode circuits. Despite of this disadvantage, the diode circuits still in use due to it’s the simplicity, low price, ruggedness, ….etc.

Because of their ability to conduct current in one direction, diodes are used in rectifier circuits. The definition of rectification process is “ the process of converting the alternating voltages and currents to direct currents and the device is known as rectifier” It is extensively used in charging batteries; supply DC motors, electrochemical processes and power supply sections of industrial components.

The most famous diode rectifiers have been analyzed in the following sections. Circuits and waveforms drawn with the help of PSIM simulation program [1].

There are two different types of uncontrolled rectifiers or diode rectifiers, half wave and full wave rectifiers. Full-wave rectifiers has better performance than half wave rectifiers. But the main advantage of half wave rectifier is its need to less number of diodes than full wave rectifiers. The main disadvantages of half wave rectifier are:

1- High ripple factor, 2- Low rectification efficiency, 3- Low transformer utilization factor, and, 4- DC saturation of transformer secondary winding.

2.2 Performance Parameters In most rectifier applications, the power input is sine-wave voltage provided by the

electric utility that is converted to a DC voltage and AC components. The AC components are undesirable and must be kept away from the load. Filter circuits or any other harmonic reduction technique should be installed between the electric utility and the rectifier and between the rectifier output and the load that filters out the undesired component and allows useful components to go through. So, careful analysis has to be done before building the rectifier. The analysis requires define the following terms: The average value of the output voltage, dcV , The average value of the output current, dcI , The rms value of the output voltage, rmsV , The rms value of the output current, rmsI

The output DC power, dcdcdc IVP *= (2.1)

The output AC power, rmsrmsac IVP *= (2.2)

The effeciency or rectification ratio is defiend as ac

dcPP

=η (2.3)

The output voltage can be considered as being composed of two components (1) the DC component and (2) the AC component or ripple. The effective (rms) value of the AC component of output voltage is defined as:-

Diode Circuits or Uncontrolled Rectifier

19

22dcrmsac VVV −= (2.4)

The form factor, which is the measure of the shape of output voltage, is defiend as shown in equation (2.5). Form factor should be greater than or equal to one. The shape of output voltage waveform is neare to be DC as the form factor tends to unity.

dc

rmsVVFF = (2.5)

The ripple factor which is a measure of the ripple content, is defiend as shown in (2.6). Ripple factor should be greater than or equal to zero. The shape of output voltage waveform is neare to be DC as the ripple factor tends to zero.

11 22

222−=−=

−== FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac (2.6)

The Transformer Utilization Factor (TUF) is defiend as:-

SS

dcIV

PTUF = (2.7)

Where SV and SI are the rms voltage and rms current of the transformer secondery

respectively. Total Harmonic Distortion (THD) measures the shape of supply current or voltage. THD

should be grearter than or equal to zero. The shape of supply current or voltage waveform is near to be sinewave as THD tends to be zero. THD of input current and voltage are defiend as shown in (2.8.a) and (2.8.b) respectively.

121

2

21

21

2−=

−=

S

S

S

SSi

II

IIITHD (2.8.a)

121

2

21

21

2−=

−=

S

S

S

SSv

VV

VVVTHD (2.8.b)

where 1SI and 1SV are the fundamental component of the input current and voltage, SI

and SV respectively. Creast Factor CF, which is a measure of the peak input current IS(peak) as compared to its

rms value IS, is defiend as:-

S

peakS

II

CF )(= (2.9)

In general, power factor in non-sinusoidal circuits can be obtained as following:

φcossVoltampereApparent

PowerR===

SS IVPealPF (2.10)

Where, φ is the angle between the current and voltage. Definition is true irrespective for

any sinusoidal waveform. But, in case of sinusoidal voltage (at supply) but non-sinusoidal current, the power factor can be calculated as the following:

Chapter Two Dr. Ali M. Eltamaly, King Saud University

20

Average power is obtained by combining in-phase voltage and current components of the same frequency.

FaactorntDisplacemeFactorDistortionII

IVIV

IVPPF

S

S

SSSS*coscos

1111 ==== φφ (2.11)

Where 1φ is the angle between the fundamental component of current and supply voltage. Distortion Factor = 1 for sinusoidal operation and displacement factor is a measure of displacement between ( )tv ω and ( )ti ω .

2.3 Single-Phase Half-Wave Diode Rectifier

Most of the power electronic applications operate at a relative high voltage and in such cases; the voltage drop across the power diode tends to be small with respect to this high voltage. It is quite often justifiable to use the ideal diode model. An ideal diode has zero conduction drops when it is forward-biased ("ON") and has zero current when it is reverse-biased ("OFF"). The explanation and the analysis presented below are based on the ideal diode model.

2.3.1 Single-Phase Half Wave Diode Rectifier With Resistive Load

Fig.2.1 shows a single-phase half-wave diode rectifier with pure resistive load. Assuming sinusoidal voltage source, VS the diode beings to conduct when its anode voltage is greater than its cathode voltage as a result, the load current flows. So, the diode will be in “ON” state in positive voltage half cycle and in “OFF” state in negative voltage half cycle. Fig.2.2 shows various current and voltage waveforms of half wave diode rectifier with resistive load. These waveforms show that both the load voltage and current have high ripples. For this reason, single-phase half-wave diode rectifier has little practical significance.

The average or DC output voltage can be obtained by considering the waveforms shown in Fig.2.2 as following:

∫ ==π

πωω

π0

sin21 m

mdcVtdtVV (2.12)

Where, mV is the maximum value of supply voltage.

Because the load is resistor, the average or DC component of load current is:

RV

RVI mdc

dc π== (2.13)

The root mean square (rms) value of a load voltage is defined as:

2sin

21

0

22 mmrms

VtdtVV == ∫π

ωωπ

(2.14)

Similarly, the root mean square (rms) value of a load current is defined as:

RV

RVI mrms

rms 2== (2.15)


21

It is clear that the rms value of the transformer secondary current, SI is the same as that of the load and diode currents

Then R

VII mDS 2

== (2.15)

Where, DI is the rms value of diode current.

Fig.2.1 Single-phase half-wave diode rectifier with resistive load.

Fig.2.2 Various waveforms for half wave diode rectifier with resistive load.


22

Example 1: The rectifier shown in Fig.2.1 has a pure resistive load of R Determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) TUF (e) Peak inverse voltage (PIV) of diode D1 and (f) Crest factor. Solution: From Fig.2.2, the average output voltage dcV is defiend as:

ππ

πωω

π

πmm

mdcVVtdtVV =−−== ∫ ))0cos(cos(

2)sin(

21

0

Then, R

VR

VI mdcdc π

==

2)sin(

21

0

2 mmrms

VtVV == ∫π

ωπ

, R

VI mrms 2

= and, 2m

SVV =

The rms value of the transformer secondery current is the same as that of the load:

RVI m

S 2= Then, the efficiency or rectification ratio is:

rmsrms

dcdc

ac

dcIVIV

PP

**

==η %53.40

2*

2

*==

RVV

RVV

mm

mmππ

(b) 57.12

2 ====π

πm

m

dc

rmsV

V

VVFF

(c) 211.1157.11 22 =−=−== FFVVRF

dc

ac

(d) %6.28286.0

22

====

RVV

RVV

IVPTUF

mm

mm

SS

dc ππ

(e) It is clear from Fig2.2 that the PIV is mV .

(f) Creast Factor CF, 22//)( ===

RVRV

II

CFm

m

S

peakS

2.3.2 Half Wave Diode Rectifier With R-L Load

In case of RL load as shown in Fig.2.3, The voltage source, SV is an alternating sinusoidal voltage source. If ( )tVv ms ωsin= , sv is positive when 0 < ω t < π, and sv is negative when π < ω t <2π. When sv starts becoming positive, the diode starts conducting and the source keeps the diode in conduction till ω t reaches π radians. At that instant defined by ω t =π radians, the current through the circuit is not zero and there is some energy stored in the inductor. The voltage across an inductor is positive when the current


23

through it is increasing and it becomes negative when the current through it tends to fall. When the voltage across the inductor is negative, it is in such a direction as to forward-bias the diode. The polarity of voltage across the inductor is as shown in the waveforms shown in Fig.2.4.

When sv changes from a positive to a negative value, the voltage across the diode changes its direction and there is current through the load at the instant ω t = π radians and the diode continues to conduct till the energy stored in the inductor becomes zero. After that, the current tends to flow in the reverse direction and the diode blocks conduction. The entire applied voltage now appears across the diode as reverse bias voltage.

An expression for the current through the diode can be obtained by solving the deferential equation representing the circuit. It is assumed that the current flows for 0 < ω t < β, where β > π ( β is called the conduction angle). When the diode conducts, the driving function for the differential equation is the sinusoidal function defining the source voltage. During the period defined by β < ω t < 2π, the diode blocks current and acts as an open switch. For this period, there is no equation defining the behavior of the circuit. For 0 < ω t < β, the following differential equation defines the circuit:

βωω ≤≤=+ ttViRdtdiL m 0),(sin* (2.17)

Divide the above equation by L we get:

βωω ≤≤=+ ttL

ViLR

dtdi m 0),(sin* (2.18)

The instantaneous value of the current through the load can be obtained from the

solution of the above equation as following:

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡+= ∫

∫∫−Adtt

LVeeti mdt

LRdt

LR

ωsin*)( (2.19)

Where A is a constant.

Then; ⎥⎥

⎦

⎤

⎢⎢

⎣

⎡+= ∫

−Adtt

LVeeti mt

LRt

LR

ωsin*)( (2.20)

By integrating (2.20) we get:

( )t

LR

m AetLtRLwR

Vti−

+−+

= ωωω cossin)( 222 (2.21)


24

Fig.2.3 Half Wave Diode Rectifier With R-L Load

Fig.2.4 Various waveforms for Half wave diode rectifier with R-L load.

Assume wLjRZ +=∠φ

Then 2222 LwRZ += ,

φcosZR = , φω sinZL = and RLωφ =tan

Substitute these values into (2.21) we get the following equation:

( )t

LR

m AettZ

Vti−

+−= ωφωφ cossinsincos)(

Then, ( )t

LR

m AetZ

Vti−

+−= φωsin)( (2.22)

R

wLZ

Φ


25

The above equation can be written in the following form:

( ) ( ) φω

ωω φωφω tansinsin)(

tm

tL

Rm Aet

ZVAet

ZVti

−−+−=+−= (2.23)

The value of A can be obtained using the initial condition. Since the diode starts

conducting at ω t = 0 and the current starts building up from zero, ( ) 00 =i (discontinuous conduction). The value of A is expressed by the following equation:

( )φsinZ

VA m=

Once the value of A is known, the expression for current is known. After evaluating A, current can be evaluated at different values of tω .

( ) ( )⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+−=

−φ

ω

φφωω tansinsin)(t

m etZ

Vti (2.24)

Starting from ω t = π, as tω increases, the current would keep decreasing. For some

value of tω , say β, the current would be zero. If ω t > β, the current would evaluate to a negative value. Since the diode blocks current in the reverse direction, the diode stops conducting when tω reaches β. The value of β can be obtained by substituting that

βω == wtti 0)( into (2.24) we get:

( ) ( ) 0sinsin)( tan =⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+−=

−φ

β

φφββ eZ

Vi m (2.25)

The value of β can be obtained from the above equation by using the methods of

numerical analysis. Then, an expression for the average output voltage can be obtained. Since the average voltage across the inductor has to be zero, the average voltage across the resistor and the average voltage at the cathode of the diode to ground are the same. This average value can be obtained as shown in (2.26). The rms output voltage in this case is shown in equation (2.27).

)cos1(*2

sin*2

0

βπ

ωωπ

β

−== ∫ mmdc

VtdtVV (2.26)

)2sin(1(5.0*2

)sin(*21

0

2 ββπ

ωπ

β

−+== ∫VmdwttVV mrms (2.27)


26

2.3.3 Single-Phase Half-Wave Diode Rectifier With Free Wheeling Diode

Single-phase half-wave diode rectifier with free wheeling diode is shown in Fig.2.5. This circuit differs from the circuit described above, which had only diode D1. This circuit shown in Fig.2.5 has another diode, marked D2. This diode is called the free-wheeling diode.

Fig.2.5 Half wave diode rectifier with free wheeling diode.

Fig.2.6 Various voltage waveforms of diode rectifier with free-wheeling diode.

Fig.2.7 Various current waveforms of diode rectifier with free-wheeling diode.


27

Let the source voltage sv be defined as ( )tVm ωsin which is positive when πω << t0 radians and it is negative when π < ω t < 2π radians. When sv is positive, diode D1 conducts and the output voltage, ov become positive. This in turn leads to diode D2 being reverse-biased during this period. During π < wt < 2π, the voltage ov would be negative if diode D1 tends to conduct. This means that D2 would be forward-biased and would conduct. When diode D2 conducts, the voltage ov would be zero volts, assuming that the diode drop is negligible. Additionally when diode D2 conducts, diode D1 remains reverse-biased, because the voltage across it is sv which is negative.

When the current through the inductor tends to fall (when the supply voltage become

negative), the voltage across the inductor become negative and its voltage tends to forward bias diode D2 even when the source voltage sv is positive, the inductor current would tend to fall if the source voltage is less than the voltage drop across the load resistor.

During the negative half-cycle of source voltage, diode D1 blocks conduction and diode D2 is forced to conduct. Since diode D2 allows the inductor current circulate through L, R and D2, diode D2 is called the free-wheeling diode because the current free-wheels through D2.

Fig.2.6 shows various voltage waveforms of diode rectifier with free-wheeling diode. Fig.2.7 shows various current waveforms of diode rectifier with free-wheeling diode.

It can be assumed that the load current flows all the time. In other words, the load current is continuous. When diode D1 conducts, the driving function for the differential equation is the sinusoidal function defining the source voltage. During the period defined by π < ω t < 2π, diode D1 blocks current and acts as an open switch. On the other hand, diode D2 conducts during this period, the driving function can be set to be zero volts. For 0 < ω t < π, the differential equation (2.18) applies. The solution of this equation will be as obtained before in (2.20) or (2.23).

( ) ( )⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+−=

−φ

ω


m etZ

Vti πω << t0 (2.28)

For the negative half-cycle ( πωπ 2<< t ) of the source voltage D1 is OFF and D2 is

ON. Then the driving voltage is set to zero and the following differential equation represents the circuit in this case.

πωπ 20* <<=+ tforiRtdidL (2.29)

The solution of (2.29) is given by the following equation:

φπω

ω tan)(−

−=

t

eBti (2.30)

The constant B can be obtained from the boundary condition where Bi =)(π is the starting value of the current in πωπ 2<< t and can be obtained from equation (2.23) by substituting πω =t


28

Then, ( ) ( ) BeZ

Vi m =+−=

−)sin(sin)( tanφ

π

φφππ

The above value of )(πi can be used as initial condition of equation (2.30). Then the load current during πωπ 2<< t is shown in the following equation.

( ) ( ) φπω

φπ

φφπω tantansinsin)(−

−−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+−=

tm eeZ

Vti for πωπ 2<< t (2.31)

For the period πωπ 32 << t the value of )2( πi from (2.31) can be used as initial

condition for that period. The differential equation representing this period is the same as equation (2.28) by replacing ω t by πω 2−t and the solution is given by equation (2.32). This period ( πωπ 32 << t ) differ than the period π<< wt0 in the way to get the constant A where in the πω << t0 the initial value was 0)0( =i but in the case of πωπ 32 << t the initial condition will be )2( πi that given from (2.31) and is shown in (2.33).

( ) φπω

φπωω tan2

2sin)(−

−+−−=

tm AetZ

Vti for πωπ 32 << t (2.32)

The value of ( )π2i can be obtained from (2.31) and (2.32) as shown in (2.33) and (2.34) respectively.

( ) ( ) φπ

φπ

φφππ tantansinsin)2(−−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+−= ee

ZVi m (2.33)

( ) AZ

Vi m +−= φπ sin)2( (2.34)

By equating (2.33) and (2.34) the constant A in πωπ 32 << t can be obtained from the following equation:

( ) ( )φπ sin2Z

ViA m+= (2.35)

Then, the general solution for the period πωπ 32 << t is given by equation (2.36):

( ) ( ) ( ) φπω

φπφπωω tan2

sin22sin)(−

−⎟⎠⎞

⎜⎝⎛ ++−−=

tmm eZ

VitZ

Vti πωπ 32 << t (2.36)

Where ( )π2i can be obtained from equation (2.33).

Example 2 A diode circuit shown in Fig.2.3 with R=10 Ω, L=20mH, and VS=220 2 sin314t.

(a) Determine the expression for the current though the load in the period πω 20 << t and determine the conduction angle β .

(b) If we connect free wheeling diode through the load as shown in Fig.2.5 Determine the expression for the current though the load in the period of πω 30 << t .


29

Solution: (a) For the period of πω << t0 , the expression of the load current can be obtained from (2.24) as following:

.561.010

10*20*314tantan3

11 radRL

===−

−− ωφ and 628343.0tan =φ

Ω=+=+= − 8084.11)10*20*314(10)( 23222 LRZ ω

( ) ( )

( )[ ]t

tm

et

etZ

Vti

ω

φω

ω

φφωω

5915.1

tan

*532.0561.0sin8084.11

2220

sinsin)(

−

−

+−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+−=

( ) tetti ωωω 5915.1*0171.14561.0sin3479.26)( −+−= The value of β can be obtained from the above equation by substituting for 0)( =βi . Then, ( ) ββ 5915.1*0171.14561.0sin3479.260 −+−= e

By using the numerical analysis we can get the value of β. The simplest method is by using the simple iteration technique by assuming

( ) ββ 5915.1*0171.14561.0sin3479.26 −+−=Δ e and substitute different values for β in the region πβπ 2<< till we get the minimum value of Δ then the corresponding value of β is the required value. The narrow intervals mean an accurate values of β . The following table shows the relation between β and Δ:

β Δ

1.1 π 6.49518 1.12 π 4.87278 1.14 π 3.23186 1.16 π 1.57885 1.18 π -0.079808 1.2 π -1.73761

It is clear from the above table that πβ 18.1≅ rad. The current in πβ 2<< wt will be

zero due to the diode will block the negative current to flow.

(b) In case of free-wheeling diode as shown in Fig.2.5, we have to divide the operation of this circuit into three parts. The first one when πω << t0 (D1 “ON”, D2 “OFF”), the second case when πωπ 2<< t (D1 “OFF” and D2 “ON”) and the last one when

πωπ 32 << t (D1 “ON”, D2 “OFF”).

In the first part ( πω << t0 ) the expression for the load current can be obtained as In case (a). Then:

( ) wtetwti 5915.1*0171.14561.0sin3479.26)( −+−= ω for πω << t0 the current at πω =t is starting value for the current in the next part. Then


30

( ) Aei 1124.14*0171.14561.0sin3479.26)( 5915.1 =+−= − πππ

In the second part πωπ 2<< t , the expression for the load current can be obtained from (2.30) as following:

φπω

ω tan)(−

−=

t

eBti where AiB 1124.14)( == π

Then ( )πωω −−= teti 5915.11124.14)( for ( πωπ 2<< t ) The current at πω 2=t is starting value for the current in the next part. Then

Ai 095103.0)2( =π

In the last part ( πωπ 32 << t ) the expression for the load current can be obtained from (2.36):

( ) ( ) ( ) φπω

φπφπωω tan2

sin22sin)(−

−⎟⎠⎞

⎜⎝⎛ ++−−=

tmm eZ

VitZ

Vti

( ) ( ) ( )πωωω 25915.1532.0*3479.26095103.08442.6sin3479.26)( −−++−=∴ tetti

( ) ( )πωωω 25915.11131.148442.6sin3479.26)( −−+−=∴ tetti for ( πωπ 32 << t ) 2.4 Single-Phase Full-Wave Diode Rectifier The full wave diode rectifier can be designed with a center-taped transformer as shown in Fig.2.8, where each half of the transformer with its associated diode acts as half wave rectifier or as a bridge diode rectifier as shown in Fig. 2.12. The advantage and disadvantage of center-tap diode rectifier is shown below: Advantages

• The need for center-tapped transformer is eliminated, • The output is twice that of the center tapped circuit for the same secondary voltage,

and, • The peak inverse voltage is one half of the center-tap circuit.

Disadvantages

• It requires four diodes instead of two, in full wave circuit, and, • There are always two diodes in series are conducting. Therefore, total voltage drop

in the internal resistance of the diodes and losses are increased. The following sections explain and analyze these rectifiers.


31

2.4.1 Center-Tap Diode Rectifier With Resistive Load In the center tap full wave rectifier, current flows through the load in the same direction

for both half cycles of input AC voltage. The circuit shown in Fig.2.8 has two diodes D1 and D2 and a center tapped transformer. The diode D1 is forward bias “ON” and diode D2 is reverse bias “OFF” in the positive half cycle of input voltage and current flows from point a to point b. Whereas in the negative half cycle the diode D1 is reverse bias “OFF” and diode D2 is forward bias “ON” and again current flows from point a to point b. Hence DC output is obtained across the load.

Fig.2.8 Center-tap diode rectifier with resistive load.

Fig.2.9 Various current and voltage waveforms for center-tap diode rectifier with resistive load.

In case of pure resistive load, Fig.2.9 shows various current and voltage waveform for converter in Fig.2.8. The average and rms output voltage and current can be obtained from the waveforms shown in Fig.2.9 as shown in the following:


32

πωω

π

πm

mdcVtdtVV 2sin1

0

== ∫ (2.36)

RVI m

dc π2

= (2.37)

( )2

sin1

0

2 mmrms

VtdtVV == ∫π

ωωπ

(2.38)

RVI m

rms 2= (2.39)

PIV of each diode = mV2 (2.40)

2m

SVV = (2.41)

The rms value of the transformer secondery current is the same as that of the diode:

RVII m

DS 2== (2.41)

Example 3. The rectifier in Fig.2.8 has a purely resistive load of R Determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) TUF (e) Peak inverse voltage (PIV) of diode D1 and(f) Crest factor of transformer secondary current. Solution:- The efficiency or rectification ratio is

%05.81

2*

2

2*2

**

====

RVV

RVV

IVIV

PP

mm

mm

rmsrms

dcdc

ac

dc ππη

(b) 11.1222

2 ====π

πm

m

dc

rmsV

V

VVFF

(c) 483.0111.11 22 =−=−== FFVVRF

dc

ac

(d) 5732.0

222

22

2===

RVV

RVV

IVPTUF

mm

mm

SS

dc ππ

(e) The PIV is mV2

(f) Creast Factor of secondary current, 2

2

)( ===

RVR

V

II

CFm

m

S

peakS


33

2.4.2 Center-Tap Diode Rectifier With R-L Load Center-tap full wave rectifier circuit with RL load is shown in Fig.2.10. Various voltage

and current waveforms for Fig.2.10 is shown in Fig.2.11. An expression for load current can be obtained as shown below:

Fig.2.10 Center-tap diode rectifier with R-L load

Fig.2.11 Various current and voltage waveform for Center-tap diode rectifier with R-L load


34

It is assumed that D1 conducts in positive half cycle of VS and D2 conducts in negative half cycle. So, the deferential equation defines the circuit is shown in (2.43).

)sin(* tViRtdidL m ω=+ (2.43)

The solution of the above equation can be obtained as obtained before in (2.24)

( ) ( )⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+−=

−φ

ω


m etZ

Vti for πω << t0 (2.44)

In the second half cycle the same differential equation (2.43) and the solution of this

equation will be as obtained before in (2.22)

( ) φπω

φπωω tansin)(−

−+−−=

tm AetZ

Vti (2.45)

The value of constant A can be obtained from initial condition. If we assume that i(π)=i(2π)=i(3π)=……..=Io (2.46) Then the value of oI can be obtained from (2.44) by letting πω =t

( ) ( )⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+−==

−φ

π

φφππ tansinsin)( eZ

ViI mo (2.47)

Then use the value of oI as initial condition for equation (2.45). So we can obtain the

value of constant A as following:

( ) φππ

φπππ tansin)(−

−+−−== Ae

ZVIi m

o

Then; ( )φsinZ

VIA mo += (2.48)

Substitute (2.48) into (2.45) we get:

( ) ( ) φπω

φφπωω tansinsin)(−

−⎟⎠⎞

⎜⎝⎛ ++−−=

tm

om e

ZVIt

ZVti , then,

( ) ( ) φπω

φπω

φφπωω tantansinsin)(−

−−

−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+−−=

t

o

tm eIetZ

Vti (for πωπ 2<< t ) (2.49)

In the next half cycle πωπ 32 << t the current will be same as obtained in (2.49) but we

have to take the time shift into account where the new equation will be as shown in the following:

( ) ( ) φπω

φπω

φφπω tan2

tan2

sin2sin)(−

−−

−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+−−=

t

o

tm eIewtZ

Vti (for πωπ 32 << t ) (2.50)


35

2.4.3 Single-Phase Full Bridge Diode Rectifier With Resistive Load Another alternative in single-phase full wave rectifier is by using four diodes as shown

in Fig.2.12 which known as a single-phase full bridge diode rectifier. It is easy to see the operation of these four diodes. The current flows through diodes D1 and D2 during the positive half cycle of input voltage (D3 and D4 are “OFF”). During the negative one, diodes D3 and D4 conduct (D1 and D2 are “OFF”).

Fig.2.12 Single-phase full bridge diode rectifier.

Fig.2.13 Various current and voltage waveforms of Full bridge single-phase diode rectifier.


36

In positive half cycle the supply voltage forces diodes D1 and D2 to be "ON". In same time it forces diodes D3 and D4 to be "OFF". So, the current moves from positive point of the supply voltage across D1 to the point a of the load then from point b to the negative marked point of the supply voltage through diode D2. In the negative voltage half cycle, the supply voltage forces the diodes D1 and D2 to be "OFF". In same time it forces diodes D3 and D4 to be "ON". So, the current moves from negative marked point of the supply voltage across D3 to the point a of the load then from point b to the positive marked point of the supply voltage through diode D4. So, it is clear that the load currents moves from point a to point b in both positive and negative half cycles of supply voltage. So, a DC output current can be obtained at the load in both positive and negative halves cycles of the supply voltage. The complete waveforms for this rectifier is shown in Fig.2.13

Example 4 The rectifier shown in Fig.2.12 has a purely resistive load of R=15 Ω and, VS=300 sin 314 t and unity transformer ratio. Determine (a) The efficiency, (b) Form factor, (c) Ripple factor, (d) TUF, (e) The peak inverse voltage, (PIV) of each diode, (f) Crest factor of input current, and, (g) Input power factor. Solution: 300=mV V

VVtdtVV mmdc 956.1902sin1

0

=== ∫ πωω

π

π

, AR

VI mdc 7324.122

==π

( ) VVtdtVV mmrms 132.212

2sin1

2/1

0

2 ==⎥⎥⎦

⎤

⎢⎢⎣

⎡= ∫

π

ωωπ

, AR

VI m

rms 142.142

==

(a) %06.81===rmsrms

dcdc

ac

dcIVIV

PPη

(b) 11.1==dc

rmsVVFF

(c) 482.011 22

222=−=−=

−== FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac

(d) %81142.14*132.212

7324.12*986.190===

SS

dcIV

PTUF

(e) The PIV= mV =300V

(f) 414.1142.14

15/300)( ===S

peakS

II

CF

(g) Input power factor = 1*Re 2==

SS

rmsIV

RIPowerApperant

Poweral


37

2.4.4 Full Bridge Single-phase Diode Rectifier with DC Load Current The full bridge single-phase diode rectifier with DC load current is shown in Fig.2.14.

In this circuit the load current is pure DC and it is assumed here that the source inductances is negligible. In this case, the circuit works as explained before in resistive load but the current waveform in the supply will be as shown in Fig.2.15. The rms value of the input current is oS II =

Fig.2.14 Full bridge single-phase diode rectifier with DC load current.

Fig.2.15 Various current and voltage waveforms for full bridge single-phase diode rectifier with

DC load current.


38

The supply current in case of pure DC load current is shown in Fig.2.15, as we see it is odd function, then na coefficients of Fourier series equal zero, 0=na , and

[ ]

[ ] .............,5,3,14

cos0cos2

cos2

sin*20

0

==−=

−== ∫

nforn

In

nI

tnn

ItdtnIb

oo

oon

ππ

π

ωπ

ωωπ

ππ

(2.51)

Then from Fourier series concepts we can say: )..........9sin

917sin

715sin

513sin

31(sin*4)( +++++= tttttIti o ωωωωω

π (2.52)

%46151

131

111

91

71

51

31))((

2222222=⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=∴ tITHD s or we can obtain

))(( tITHD s as the following:

From (2.52) we can obtain the value of is π2

41

oS

II =

%34.4814

21

24

1))((2

2

2

1=−⎟⎟

⎠

⎞⎜⎜⎝

⎛=−

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

=−⎟⎟⎠

⎞⎜⎜⎝

⎛=∴

π

πo

o

S

Ss I

IIItITHD

Example 5 solve Example 4 if the load is 30 A pure DC Solution: From example 4 Vdc= 190.986 V, Vrms=212.132 V

AIdc 30= and rmsI = 30 A

(a) %90===rmsrms

dcdc

ac

dcIVIV

PPη

(b) 11.1==dc

rmsVVFF

(c) 482.011 22

222=−=−=

−== FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac

(d) %9030*132.21230*986.190

===SS

dcIV

PTUF

(e) The PIV=Vm=300V

(f) 13030)( ===

S

peakS

II

CF

(g) AII oS 01.27

230*4

24

1 ===ππ

Input Power factor= =PowerApperant

PoweralRe

LagI

IIV

IV

S

S

SS

SS 9.01*30

01.27cos*cos* 11 ====φφ


39

2.4.5 Effect Of LS On Current Commutation Of Single-Phase Diode Bridge Rectifier. Fig.2.15 Shows the single-phase diode bridge rectifier with source inductance. Due to

the value of LS the transitions of the AC side current Si from a value of oI to oI− (or vice versa) will not be instantaneous. The finite time interval required for such a transition is called commutation time. And this process is called current commutation process. Various voltage and current waveforms of single-phase diode bridge rectifier with source inductance are shown in Fig.2.16.

Fig.2.15 Single-phase diode bridge rectifier with source inductance.

Fig.2.16 Various current and voltage waveforms for single-phase diode bridge rectifier with source

inductance.

Let us study the commutation time starts at t=10 ms as indicated in Fig.2.16. At this time the supply voltage starts to be negative, so diodes D1 and D2 have to switch OFF and


40

diodes D3 and D4 have to switch ON as explained in the previous case without source inductance. But due to the source inductance it will prevent that to happen instantaneously. So, it will take time tΔ to completely turn OFF D1 and D2 and to make D3 and D4 carry the entire load current ( oI ). Also in the time tΔ the supply current will change from oI to

oI− which is very clear in Fig.2.16. Fig.2.17 shows the equivalent circuit of the diode bridge at time tΔ .

Fig.2.17 The equivalent circuit of the diode bridge at commutation time tΔ .

From Fig.2.17 we can get the following equations

0=−dtdiLV S

sS (2.53)

Multiply the above equation by tdω then, SsS diLtdV ωω = (2.54)

Integrate both sides of the above equation during the commutation period ( tΔ sec or u rad.) we get the following:

SsS diLtdV ωω =

∫∫−+

=o

o

I

ISs

u

m diLtdtV ωωωπ

π

sin (2.55)

Then; ( )[ ] osm ILuV ωππ 2coscos −=+− Then; ( )[ ] osm ILuV ω2cos1 −=+−

Then; ( )m

osV

ILu ω21cos −=

Then; ⎟⎟⎠

⎞⎜⎜⎝

⎛−= −

m

osV

ILu ω21cos 1 (2.56)

And ⎟⎟⎠

⎞⎜⎜⎝

⎛−==Δ −

m

osV

ILut ωωω

21cos1 1 (2.57)

It is clear that the DC voltage reduction due to the source inductance is the drop across the source inductance.

dtdiLv S

srd = (2.58)


41

Then oS

I

ISS

u

rd ILdiLtdvo

o

ωωωπ

π

2−== ∫∫−+

(2.59)

∫+u

rd tdvπ

π

ω is the reduction area in one commutation period tΔ . But we have two

commutation periods tΔ in one period of supply voltage. So the total reduction per period

is: oS

u

rd ILtdv ωωπ

π

42 −=∫+

(2.60)

To obtain the average reduction in DC output voltage rdV due to source inductance we have to divide the above equation by the period time π2 . Then;

oSoS

rd ILfILV 42

4−=

−=

πω (2.61)

The DC voltage with source inductance tacking into account can be calculated as following:

osm

rdceinducsourcewithoutdcactualdc IfLVVVV 42tan −=−=

π (2.62)

To obtain the rms value and Fourier transform of the supply current it is better to move the vertical axis to make the waveform odd or even this will greatly simplfy the analysis. So, it is better to move the vertical axis of supply current by 2/u as shown in Fig.2.18. Moveing the vertical axis will not change the last results. If you did not bleave me keep going in the analysis without moveing the axis.

Fig. 2.18 The old axis and new axis for supply currents.

Fig.2.19 shows a symple drawing for the supply current. This drawing help us in getting the rms valuof the supply current. It is clear from the waveform of supply current shown in Fig.2.19 that we obtain the rms value for only a quarter of the waveform because all for quarter will be the same when we squaret the waveform as shown in the following equation:


42

]2

[2 2/

0

2

2/

22

∫ ∫+⎟⎠⎞

⎜⎝⎛=

u

uo

os tdItdt

uI

I

π

ωωωπ

(2.63)

Then; ⎥⎦⎤

⎢⎣⎡ −=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−+=

322

228342 23

2

2 uIuuu

II oos

ππ

ππ

(2.64)

2u

−

oI

oI− 2u

−π2u

π

2u

+π

22 u

−π

π2

usI

2π

Fig.2.19 Supply current waveform

To obtain the Fourier transform for the supply current waveform you can go with the classic fourier technique. But there is a nice and easy method to obtain Fourier transform of such complcated waveform known as jump technique [ ]. In this technique we have to draw the wave form and its drevatives till the last drivative values all zeros. Then record the jump value and its place for each drivative in a table like the table shown below. Then; substitute the table values in (2.65) as following:


43

2u

−

oI

oI−2u

−π

uIo2

uIo2

−

sI ′

2u

π

2u

+π

22 u

−π

π2

2u

−

2u

−π

2u

π

2u

+π

22 u

−π

usI

Fig.2.20 Supply current and its first derivative.

Table(2.1) Jumb value of supply current and its first derivative.

sJ 2u

− 2u

2u

−π 2u

+π

sI 0 0 0 0

sI ′ uIo2

uIo2

− uIo2

− uIo2

It is an odd function, then 0== no aa

⎥⎥⎦

⎤

⎢⎢⎣

⎡′−= ∑∑

==

m

sss

m

sssn tnJ

ntnJ

nb

11sin1cos1 ωω

π (2.65)

⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−

−=

2sin

2sin

2sin

2sin2*11 unununun

uI

nnb o

n πππ

2sin*8

2nu

unIb o

nπ

= (2.66)

2sin*8

1u

uIb o

π= (2.67)


44

Then; 2

sin*28

1u

uII o

S π= (2.68)

( )

⎥⎦⎤

⎢⎣⎡ −

=

⎥⎦⎤

⎢⎣⎡ −

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

=

⎟⎠⎞

⎜⎝⎛

⎥⎦⎤

⎢⎣⎡ −

=⎟⎠⎞

⎜⎝⎛=

32

sin2

32

2cos

2sin4

2cos

322

2sin*

28

2cos*

21

uu

uuu

uu

u

uI

uu

Iu

IIpf

o

o

S

S

ππππ

ππ

π

(2.69)

Example 6 Single phase diode bridge rectifier connected to 11 kV, 50 Hz, source inductance mHX s 5= supply to feed 200 A pure DC load, find:

i. Average DC output voltage. ii. Power factor.

iii. Determine the THD of the utility line current.

Solution: (i) From (2.62), VVm 155562*11000 ==

osm

rdceinducsourcewithoutdcactualdc IfLVVVV 42tan −=−=

π

VV actualdc 9703200*005.0*50*415556*2=−=

π

(ii) From (2.56) the commutation angle u can be obtained as following:

.285.015556

200*005.0*50**2*21cos21cos 11 radV

ILum

os =⎟⎠⎞

⎜⎝⎛ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛−= −− πω

The input power factor can be obtained from (2.69) as following

( ) ( ) 917.0

3285.

2285.0

285.0sin*2

32

sin*22

cos*1 =

⎥⎦⎤

⎢⎣⎡ −

=

⎥⎦⎤

⎢⎣⎡ −

=⎟⎠⎞

⎜⎝⎛=

ππππ uu

uuIIpf

S

S

AuII oS 85.193

3285.0

2200*2

322 22

=⎥⎦⎤

⎢⎣⎡ −=⎥⎦

⎤⎢⎣⎡ −=

ππ

ππ

Auu

II oS 46.179

2285.0sin*

285.0*2200*8

2sin*

28

1 =⎟⎠⎞

⎜⎝⎛==

ππ

%84.40146.17985.1931

22

1=−⎟

⎠⎞

⎜⎝⎛=−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

S

Si I

ITHD


45

2.5 Three Phase Diode Rectifiers 2.5.1 Three-Phase Half Wave Rectifier

Fig.2.21 shows a half wave three-phase diode rectifier circuit with delta star three-phase transformer. In this circuit, the diode with highest potential with respect to the neutral of the transformer conducts. As the potential of another diode becomes the highest, load current is transferred to that diode, and the previously conduct diode is reverse biased “OFF case”.

Fig.2.21 Half wave three-phase diode rectifier circuit with delta star three-phase transformer. For the rectifier shown in Fig.2.21 the load voltage, primary diode currents and its FFT

components are shown in Fig.2.22, Fig.2.23 and Fig.2.24 respectively.

Fig.2.22 Secondary and load voltages of half wave three-phase diode rectifier.

6π

65π


46

Fig.2.23 Primary and diode currents.

Fig.2.24 FFT components of primary and diode currents.

Primary current

Diode current


47

By considering the interval from6π to

65π in the output voltage we can calculate the

average and rms output voltage and current as following:

mm

mdc VVtdtVV 827.0233sin

23 6/5

6/

=== ∫ πωω

π

π

π

(2.70)

RV

RVI mm

dc*827.0

**233

==π

(2.71)

( ) mmmrms VVtdtVV 8407.08

3*321sin

23 6/5

6/

2 =+== ∫ πωω

π

π

π

(2.72)

RVI m

rms8407.0

= (2.73)

Then the diode rms current is equal to secondery current and can be obtaiend as following:

RV

RVII mm

Sr 4854.03

08407=== (2.74)

Note that the rms value of diode current has been obtained from the rms value of load current divided by 3 because the diode current has one third pulse of similar three pulses in load current. ThePIV of the diodes is mLL VV 32 = (2.75) Example 7 The rectifier in Fig.2.21 is operated from 460 V 50 Hz supply at secondary side and the load resistance is R=20 Ω. If the source inductance is negligible, determine (a) Rectification efficiency, (b) Form factor (c) Ripple factor (d) Transformer utilization factor, (e) Peak inverse voltage (PIV) of each diode and (f) Crest factor of input current. Solution:

(a) VVVV mS 59.3752*58.265,58.2653

460====

mm

dc VVV 827.0233

==π

,

RV

RVI mm

dc0827

233

==π

mrms VV 8407.0=

RVI m

rms8407.0

=


48

%767.96===rmsrms

dcdc

ac

dcIVIV

PPη

(b) %657.101==dc

rmsVVFF

(c) %28.1811 22

222=−=−=

−== FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac

(d) RVII m

rmsS 38407.0

31

==

%424.66

38407.0*2/*3

/)827.0(*3

2===

RVV

RVIV

PTUFm

m

m

SS

dc

(e) The PIV= 3 Vm=650.54V

(f) 06.2

38407.0

/)( ===

RV

RVI

ICF

mm

S

peakS


49

2.5.2 Three-Phase Half Wave Rectifier With DC Load Current and zero source inductance In case of pure DC load current as shown in Fig.2.25, the diode current and primary current are shown in Fig.2.26.

Fig.2.25 Three-phase half wave rectifier with dc load current

Fig.2.26 Primary and secondary current waveforms and FFT components of three-phase half wave

rectifier with dc load current

New

axi

s


50

To calculate Fourier transform of the diode current of Fig.2.26, it is better to move y axis to make the function as odd or even to cancel one coefficient an or bn respectively. If we put Y-axis at point ot 30=ω then we can deal with the secondary current as even functions. Then, 0=nb of secondary current. Values of na can be calculated as following:

321 3/

3/0

oo

ItdIa ∫−

==π

π

ωπ

(2.76)

[ ]

harmonicstrepleanallfor

nfornI

nfornI

tnnI

dwttnIa

o

oo

on

0

17,16,11,10,5,43*

,....14,13,8,7,2,13*sin

cos*1

3/3/

3/

3/

=

=−=

===

=

−

−∫

π

πω

π

ωπ

ππ

π

π

(2.77)

⎟⎠⎞

⎜⎝⎛ −−++−−++= ...8cos

817cos

715cos

514cos

412cos

21cos3

3)( ttttttIItI OO

s ωωωωωωπ

(2.78)

%24.1090924.119

*21

23

3/1))((2

2

2

1==−=−

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=−⎟⎟⎠

⎞⎜⎜⎝

⎛=

π

πO

o

S

Ss I

IIItITHD

It is clear that the primary current shown in Fig.2.26 is odd, then, an=0,

[ ]

harmonicstrepleanallfor

nfornI

tnnItdtnIb

o

oon

0

,....14,13,11,10,8,7,5,4,2,13

cos2sin*2 3/20

3/2

0

=

==

−== ∫

π

ωπ

ωωπ

ππ

(2.79)

⎟⎠⎞

⎜⎝⎛ −−+++++= ...8sin

817sin

715sin

514sin

412sin

21sin3)( ttttttIti O

P ωωωωωωπ

(2.80)

The rms value of oP II32

= (2.81)

%983.67133

21

2332

1))((2

2

2

1=−⎟

⎠⎞

⎜⎝⎛=−

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

=−⎟⎟⎠

⎞⎜⎜⎝

⎛=

π

πO

o

P

PP I

I

IItITHD (2.82)


51

Example 8 Solve example 7 if the load current is 100 A pure DC

Solution: (a) VVVV mS 59.3752*58.265,58.2653

460====

VVVV mm

dc 613.310827.0233

===π

, AIdc 100=

VVV mrms 759.3158407.0 == , AIrms 100=

%37.98100*759.315100*613.310

====rmsrms

dcdc

ac

dcIVIV

PPη

(b) %657.101==dc

rmsVVFF

(c) %28.1811 22

222=−=−=

−== FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac

(d) AII rmsS 735.57100*31

31

===

%52.67735.57*2/*3

100*613.310*3

===mSS

dcVIV

PTUF


(f) 732.1735.57

100)( ===S

peakS

II

CF

2.5.3 Three-Phase Half Wave Rectifier With Source Inductance The source inductance in three-phase half wave diode rectifier Fig.2.27 will change the

shape of the output voltage than the ideal case (without source inductance) as shown in Fig.2.28. The DC component of the output voltage is reduced due to the voltage drop on the source inductance. To calculate this reduction we have to discuss Fig.2.27 with reference to Fig.2.28. As we see in Fig.2.28 when the voltage bv is going to be greater than the voltage av at time t (at the arrow in Fig.2.28) the diode D1 will try to turn off, in the same time the diode D2 will try to turn on but the source inductance will slow down this process and makes it done in time tΔ (overlap time or commutation time). The overlap time will take time tΔ to completely turn OFF D1 and to make D2 carry the entire load current ( oI ). Also in the time tΔ the current in bL will change from zero to oI and the current in aL will change from oI to zero. This is very clear from Fig.2.28. Fig.2.29 shows the equivalent circuit of three phase half wave diode bridge in commutation period tΔ .


52

Fig.2.27 Three-phase half wave rectifier with load and source inductance.

Fig.2.28 Supply current and output voltage for three-phase half wave rectifier with pure

DC load and source inductance.

Fig.2.29 The equivalent circuit for three-phase half wave diode rectifier in commutation period.


53

From Fig.2.29 we can get the following equations

01 =−− dcD

aa Vdt

diLv (2.83)

02 =−− dcD

bb Vdt

diLv (2.84)

subtract (2.84) from(2.83) we get:

012 =⎟⎠⎞

⎜⎝⎛ −+−

dtdi

dtdiLvv DD

ba

Multiply the above equation by tdω the following equation can be obtained: ( ) ( ) 012 =−+− DDba didiLtdvv ωω

substitute the voltage waveforms of av and bv into the above equation we get:

( ) ( )2132sinsin DDmm didiLtdtVtV −=⎟

⎠

⎞⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ −− ωωπωω

Then; ( )216sin3 DDm didiLtdtV −=⎟

⎠

⎞⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ + ωωπω

Integrating both parts of the above equation we get the following:

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=⎟

⎠⎞

⎜⎝⎛ + ∫∫∫

+o

o

I

DI

D

u

m didiLtdtV0

2

0

1

65

65 6

sin3 ωωπω

π

π

Then; om LIuV ωππππ 266

5cos66

5cos3 −=⎟⎠

⎞⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ ++−⎟

⎠⎞

⎜⎝⎛ +

Then; ( ) ( )( ) om LIuV ωππ 2coscos3 −=+− Then; ( )( ) om LIuV ω2cos13 −=+−

Then; ( )m

oVLIu

32cos1 ω

=−

Then; ( )m

oVLIu

321cos ω

−=

Then ⎟⎟⎠

⎞⎜⎜⎝

⎛−= −

m

oVLIu

321cos 1 ω (2.85)

⎟⎟⎠

⎞⎜⎜⎝

⎛−==Δ −

m

oVLIut

321cos1 1 ω

ωω (2.86)

It is clear that the DC voltage reduction due to the source inductance is equal to the drop across the source inductance. Then;

dtdiLv D

rd =


54

Then, o

I

D

u

rd LIdiLtdvo

ωωω

π

π

== ∫∫+

0

65

65

(2.87)

∫+u

rd tdv6

5

65

π

πω is the reduction area in one commutation period tΔ . But, we have three

commutation periods, tΔ in one period. So, the total reduction per period is:

o

u

rd LItdv ωω

π

π3*3

65

65

=∫+

To obtain the average reduction in DC output voltage rdV due to source inductance we have to divide the total reduction per period by π2 as following:

oo

rd ILfLIV 32

3==

πω (2.88)

Then, the DC component of output voltage due to source inductance is: o

ceinducsourcewithout

dcActualdc ILfVV 3tan

−= (2.89)

om

Actualdc ILfVV 3233

−=π

(2.90)

Example 9 Three-phase half-wave diode rectifier connected to 66 kV, 50 Hz , 5mH supply to feed a DC load with 500 A DC, fined the average DC output voltage.

Solution: Vvm 538892*3

66000=⎟

⎠

⎞⎜⎝

⎛=

(i) oceinduc

sourcewithout

dcActualdc ILfVV 3tan

−=

VILfVV om

Actualdc 44190500*005.0*50*32

53889*3*33233

=−=−=ππ

2.5 Three-Phase Full Wave Diode Rectifier

The three phase bridge rectifier is very common in high power applications and is shown in Fig.2.30. It can work with or without transformer and gives six-pulse ripples on the output voltage. The diodes are numbered in order of conduction sequences and each one conduct for 120 degrees. These conduction sequence for diodes is 12, 23, 34, 45, 56, and, 61. The pair of diodes which are connected between that pair of supply lines having the highest amount of instantaneous line to line voltage will conduct. Also, we can say that, the highest positive voltage of any phase the upper diode connected to that phase conduct


55

and the highest negative voltage of any phase the lower diode connected to that phase conduct. 2.5.1 Three-Phase Full Wave Rectifier With Resistive Load

In the circuit of Fig.2.30, the AC side inductance LS is neglected and the load current is pure resistance. Fig.2.31 shows complete waveforms for phase and line to line input voltages and output DC load voltages. Fig.2.32 shows diode currents and Fig.2.33 shows the secondary and primary currents and PIV of D1. Fig.2.34 shows Fourier Transform components of output DC voltage, diode current secondary current and Primary current respectively.

For the rectifier shown in Fig.2.30 the waveforms is as shown in Fig.2.31. The average output voltage is :-

LLmLLm

mdc VVVVtdtVV 3505.1654.12333sin33 3/2

3/

===== ∫ ππωω

π

π

π

(2.91)

RV

RV

RV

RVI LLLLmm

dc3505.123654.133

====ππ

(2.92)

( ) LLmmmrms VVVtdtVV 3516.16554.14

3*923sin33 3/2

3/

2==+== ∫ π

ωωπ

π

π

(2.93)

RVI m

rms6554.1

= (2.94)

Then the diode rms current is

RV

RVI mm

r 9667.03

6554.1== (2.95)

RVI m

S 29667.0= (2.96)

1 3 5

4 6 2

b

c

IL

VLIs

Ip

a

Fig.2.30 Three-phase full wave diode bridge rectifier.


56

Fig.2.31 shows complete waveforms for phase and line to line input voltages and output DC load

voltages.

Fig.2.32 Diode currents.


57

Fig.2.33 Secondary and primary currents and PIV of D1.

Fig.2.34 Fourier Transform components of output DC voltage, diode current secondary current and

Primary current respectively of three-phase full wave diode bridge rectifier.

Example 10 The rectifier shown in Fig.2.30 is operated from 460 V 50 Hz supply and the load resistance is R=20 Ω. If the source inductance is negligible, determine (a) The efficiency, (b) Form factor (c) Ripple factor (d) Transformer utilization factor, (e) Peak inverse voltage (PIV) of each diode and (f) Crest factor of input current.


58

Solution: (a) VVVV mS 59.3752*58.265,58.2653

460====

VVVV mm

dc 226.621654.133===

π, A

RV

RVI mm

dc 0613.31654.133===

π

VVVV mmrms 752.6216554.14

3*923

==+=π

, AR

VI mrms 0876.316554.1

==

%83.99===rmsrms

dcdc

ac

dcIVIV

PPη

(b) %08.100==dc

rmsVVFF

(c) %411 22

222=−=−=

−== FF

VV

VVV

VVRF

dc

rms

dc

dcrms

dc

ac

(d) RV

RVII mm

rmsS 352.16554.1*8165.032

===

%42.95352.1*2/*3

/)654.1(*3

2===

RVV

RVIV

PTUFm

m

m

SS

dc


(f) 281.1352.1

/3)( ===

RV

RVI

ICF

m

m

S

peakS

2.5.2 Three-Phase Full Wave Rectifier With DC Load Current

The supply current in case of pure DC load current is shown in Fig.2.35. Fast Fourier Transform of Secondary and primary currents respectively is shown in Fig2.36. As we see it is odd function, then an=0, and

[ ]

....,.........15,14,12,10,9,8,6,4,3,2,0

.....),........3(132),3(

112

)3(7

2),3(5

2,32

cos2

sin*2

1311

751

6/56/

6/5

6/

==

==

−=−==

−=

= ∫

nforb

IbIb

IbIbIb

tnn

I

tdtnIb

n

oo

ooo

o

on

ππ

πππ

ωπ

ωωπ

ππ

π

π

(2.97)

⎟⎠⎞

⎜⎝⎛ ++−−= tttt vtItI o

s ωωωωωπ

13sin13111sin

1117sin

715sin

51sin32)( (2.98)


59

%31251

231

191

171

131

111

71

51))((

22222222

=

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=tITHD s

Also ))(( tITHD s can be obtained as following:

oS II32

= , oS IIπ

3*21 =

%01.311/3*23/21))(( 2

2

1=−=−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

πS

Ss I

ItITHD

Fig.2.35 The D1 and D2 currents, secondary and primary currents.


60

Fig2.36 Fast Fourier Transform of Secondary and primary currents respectively.

For the primary current if we move the t=0 to be as shown in Fig.2.28, then the function

will be odd then, 0=na , and

....,.........15,14,12,10,9,8,6,4,3,2,0

......,.........13,11,7,5,13*23

2cos3

coscos12

sin*sin*2sin*2

1

1

3/21

3/2

3/1

3/

01

==

==

⎟⎠⎞

⎜⎝⎛ −+−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛++= ∫∫∫

nforb

nforn

Ib

nnnnI

tdtnItdtnItdtnIb

n

n

n

π

ππππ

ωωωωωωπ

π

π

π

π

π

(2.99)

⎟⎠⎞

⎜⎝⎛ ++++= tttttItIP ωωωωω

π13sin

13111sin

1117sin

715sin

51sin3*2)( 1 (2.100)

%30251

231

191

171

131

111

71

51))((

22222222

=

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=tITHD P

Power Factor =S

S

S

SII

II 11 )0cos(* =

2.5.4 Three-Phase Full Wave Diode Rectifier With Source Inductance

The source inductance in three-phase diode bridge rectifier Fig.2.37 will change the shape of the output voltage than the ideal case (without source inductance) as shown in Fig.2.31. The DC component of the output voltage is reduced. Fig.2.38 shows The output DC voltage of three-phase full wave rectifier with source inductance.


61

Fig.2.37 Three-phase full wave rectifier with source inductance

Fig.2.38 The output DC voltage of three-phase full wave rectifier with source inductance

Let us study the commutation time starts at t=5ms as shown in Fig.2.39. At this time cV starts to be more negative than bV so diode D6 has to switch OFF and D2 has to switch ON. But due to the source inductance will prevent that to happen instantaneously. So it will take time tΔ to completely turn OFF D6 and to make D2 carry all the load current ( oI ). Also in the time tΔ the current in bL will change from oI to zero and the current in cL will change from zero to oI . This is very clear from Fig.2.39. The equivalent circuit of the three phase diode bridge at commutation time tΔ at mst 5= is shown in Fig.2.40 and Fig.2.41.


62

Fig.2.39 Waveforms represent the commutation period at time t=5ms.

Fig.2.40 The equivalent circuit of the three phase diode bridge at commutation time tΔ at

mst 5=


63

Fig.2.41 Simple circuit of the equivalent circuit of the three phase diode bridge at commutation

time tΔ at mst 5=

From Fig.2.41 we can get the following defferntial equations:

061 =−−−− bD

bdcD

aa Vdt

diLVdt

diLV (2.101)

021 =−−−− cD

cdcD

aa Vdt

diLVdt

diLV (2.102)

Note that, during the time tΔ , 1Di is constant so 01 =dt

diD , substitute this value in (2.101)

and (2.102) we get the following differential equations:

dcD

bba Vdt

diLVV =−− 6 (2.103)

dcD

cca Vdt

diLVV =−− 2 (2.104)

By equating the left hand side of equation (2.103) and (2.104) we get the following differential equation:

dtdiLVV

dtdiLVV D

ccaD

bba26 −−=−− (2.105)

026 =−+−dt

diLdt

diLVV Dc

Dbcb (2.106)

The above equation can be written in the following manner: ( ) 026 =−+− DcDbcb diLdiLdtVV (2.107) ( ) 026 =−+− DcDbcb diLdiLtdVV ωωω (2.108)

Integrate the above equation during the time tΔ with the help of Fig.2.39 we can get the limits of integration as shown in the following:

( ) 00

2

0

6

2/

2/

=−+− ∫∫∫+ o

o

I

DcI

Db

u

cb diLdiLtdVV ωωωπ

π

( ) 03

2sin3

2sin2/

2/

=−−+⎟⎠

⎞⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ −∫

+

ocob

u

mm ILILtdtVtV ωωωπωπωπ

π

assume Scb LLL ==


64

oS

u

m ILttV ωπωπωπ

π2

32cos

32cos

2/

2/=⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛ −−

+

oS

m

IL

uuV

ω

ππππππππ

23

22

cos3

22

cos3

22

cos3

22

cos

=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ +++⎟

⎠⎞

⎜⎝⎛ −+−

oSm ILuuV ωππππ 26

7cos6

cos6

7cos6

cos =⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛ −

+⎟⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛ −−

( ) ( ) ( ) ( )

m

oSV

IL

uuuu

ω

ππππ

2

23

23

67sinsin

67coscos

6sinsin

6coscos

=

⎥⎦

⎤⎢⎣

⎡++⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−

( ) ( ) ( ) ( )m

oSV

ILuuuu ω23sin5.0cos23sin5.0cos

23

=⎥⎦

⎤⎢⎣

⎡++−−−

( )[ ]m

oSV

ILu ω2cos13 =−

( )LL

oS

LL

o

m

oV

ILVLI

VLIu ωωω 21

221

321cos −=−=−=

⎥⎦

⎤⎢⎣

⎡−= −

LL

oSV

ILu ω21cos 1 (2.109)

⎥⎦

⎤⎢⎣

⎡−==Δ −

LL

oSV

ILut ωωω

21cos1 1 (2.110)

It is clear that the DC voltage reduction due to the source inductance is the drop across the source inductance.

dtdiLv D

Srd = (2.111)

Multiply (2.111) by tdω and integrate both sides of the resultant equation we get:

oS

I

D

u

rd ILLditdvo

ωωω

π

π== ∫∫

+

0

2

2

(2.112)

∫+u

rd tdv2

2

π

πω is the reduction area in one commutation period tΔ . But we have six

commutation periods tΔ in one period so the total reduction per period is:


65

oS

u

rd ILtdv ωω

π

π66

2

2

=∫+

(2.113)

To obtain the average reduction in DC output voltage rdV due to source inductance we have to divide by the period time π2 . Then,

oo

rd fLILIV 62

6==

πω (2.114)

The DC voltage without source inductance tacking into account can be calculated as following:

dLLrdceinducsourcewithoutdcactualdc fLIVVVV 635.1tan −=−= (2.115)

Fig.2.42 shows the utility line current with some detailes to help us to calculate its rms value easly.

u

oI

oI− 32π

sI

u+3

2π

262 u

+π

Fig.2.42 The utility line current

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+⎟⎠⎞

⎜⎝⎛= ∫ ∫

+u

u

ud

os tdItdt

uII

0

232

22π

ωωωπ ⎥⎦

⎤⎢⎣⎡ −++= uuu

uIo

23312 3

2

2 ππ

Then ⎥⎦⎤

⎢⎣⎡ −=

632 2 uII o

Sπ

π (2.116)

Fig.2.43 shows the utility line currents and its first derivative that help us to obtain the Fourier transform of supply current easily. From Fig.2.43 we can fill Table(2.2) as explained before when we study Table (2.1).


66

26u

−π

u

oI

oI−26

5 u−

π

267 u

−π 26

11 u−

π

sI

u

uIo

uIo−

sI ′

26u

−π

265 u

−π

267 u

−π

2611 u

−π

Fig.2.43 The utility line currents and its first derivative.

Table(2.2) Jumb value of supply current and its first derivative.

sJ

26u

−π

26u

+π

265 u

−π

265 u

+π

267 u

−π

267 u

+π

2611 u

−π

2611 u

+π

sI 0 0 0 0 0 0 0 0

sI ′ uIo

uIo−

uIo−

uIo

uIo−

uIo

uIo

uIo−

It is an odd function, then 0== no aa

⎥⎥⎦

⎤

⎢⎢⎣

⎡′−= ∑∑

==

m

sss

m

sssn tnJ

ntnJ

nb

11sin1cos1 ωω

π (2.117)

⎥⎦

⎤⎟⎠

⎞⎟⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛ −−

⎢⎣

⎡⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ ++⎟

⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ −

−=

2611sin

2611sin

267sin

267sin

265sin

265sin

26sin

26sin*11

unununun

ununununuI

nnb o

n

ππππ

πππππ

⎥⎦⎤

⎢⎣⎡ +−−=

611cos

67cos

65cos

6cos

2sin*2

2ππππ

πnnnnnu

unIb o

n (2.118)

Then, the utility line current can be obtained as in (2.119).

( ) ( ) ( ) ( )

( ) ( ) ⎥⎦

⎤++−−⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+

⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=

tutu

tututuu

ti

ωω

ωωωπ

ω

13sin2

13sin13

111sin2

11sin11

1

7sin2

7sin715sin

25sin

51sin

2sin34

22

22 (2.119)

Then; ⎟⎠⎞

⎜⎝⎛=

2sin62

1u

uII o

S π (2.120)

The power factor can be calculated from the following equation:


67

⎟⎠⎞

⎜⎝⎛

⎥⎦⎤

⎢⎣⎡ −

⎟⎠⎞

⎜⎝⎛

=⎟⎠⎞

⎜⎝⎛=

2cos

632

2sin62

2cos

21 u

uI

uuI

uIIpf

o

o

S

S

ππ

π

Then; ( )

⎥⎦⎤

⎢⎣⎡ −

=

63

sin*3uu

upfππ

(2.121)

Example 11 Three phase diode bridge rectifier connected to tree phase 33kV, 50 Hz supply has 8 mH source inductance to feed 300A pure DC load current Find;

(i) commutation time and commutation angle. (ii) DC output voltage. (iii) Power factor. (iv) Total harmonic distortion of line current.

Solution: (i) By substituting for 50**2 πω = , AId 300= , HL 008.0= , VVLL 33000=

in (2.109), then oradu 61.14.2549.0 ==

Then, msut 811.0==Δω

.

(ii) The the actual DC voltage can be obtained from (2.115) as following: dLLrdceinducsourcewithoutdcactualdc fLIVVVV 635.1tan −=−=

VVdcactual 43830300*008.*50*633000*35.1 =−= (iii) the power factor can be obtained from (2.121) then

( ) ( ) 9644.0

62549.0

3*2549.0

2549.0sin3

63

sin*3=

⎥⎦⎤

⎢⎣⎡ −

=

⎥⎦⎤

⎢⎣⎡ −

=ππππ uu

upf Lagging

(iv) The rms value of supply current can be obtained from (2.116)as following

AuII ds 929.239

62549.0

3*300*2

632 22

=⎟⎠⎞

⎜⎝⎛ −=⎥⎦

⎤⎢⎣⎡ −=

ππ

ππ

The rms value of fundamental component of supply current can be obtained from (2.120) as following:

Auu

II oS 28.233

22549.0sin*

2*2549.0*300*3432*

2sin

234

1 =⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

ππ

9644.02

2549.0cos*929.23928.233

2cos*1 =⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

uIIpf

s

S Lagging.

%05.24128.233

929.239122

1=−⎟

⎠⎞

⎜⎝⎛=−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

S

Si I

ITHD


68

2.7 Multi-pulse Diode Rectifier Twelve-pulse bridge connection is the most widely used in high number of pulses

operation. Twelve-pulse technique is using in most HVDC schemes and in very large variable speed drives for DC and AC motors as well as in renewable energy system. An example of twelve-pulse bridge is shown in Fig.2.33. In fact any combination such as this which gives a 30o-phase shift will form a twelve-pulse converter. In this kind of converters, each converter will generate all kind of harmonics described above but some will cancel, being equal in amplitude but 180o out of phase. This happened to 5th and 7th harmonics along with some of higher order components. An analysis of the waveform shows that the AC line current can be described by (2.83).

( ) ( ) ( ) ( ) ( ) ⎟⎠⎞

⎜⎝⎛ +−+−+−= ...25sin

25123sin

23113sin

1315sin

111sin32)( tttttIti dP ωωωωω

π(2.83)

%5.13351

351

251

231

131

111))((

222222

=

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=tITHD P

As shown in (10) the THDi is about 13.5%. The waveform of utility line current is shown in Fig.2.34. Higher pulse number like 18-pulse or 24-pulse reduce the THD more and more but its applications very rare. In all kind of higher pulse number the converter needs special transformer. Sometimes the transformers required are complex, expensive and it will not be ready available from manufacturer. It is more economic to connect the small WTG to utility grid without isolation transformer. The main idea here is to use a six-pulse bridge directly to electric utility without transformer. But the THD must be lower than the IEEE-519 1992 limits.

2N :1

32 N :1

Vda

c

b

a1

b1

c1

a2

b2

c2

Fig.2.33 Twelve-pulse converter arrangement


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(a) Utility input current. (b) FFT components of utility current.

Fig.2.34 Simulation results of 12.pulse system.


70

Problems 1- Single phase half-wave diode rectifier is connected to 220 V, 50 Hz supply to feed

Ω5 pure resistor. Draw load voltage and current and diode voltage drop waveforms along with supply voltage. Then, calculate (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current.

2- The load of the rectifier shown in problem 1 is become Ω5 pure resistor and 10 mH inductor. Draw the resistor, inductor voltage drops, and, load current along with supply voltage. Then, find an expression for the load current and calculate the conduction angle, β . Then, calculate the DC and rms value of load voltage.

3- In the rectifier shown in the following figure assume VVS 220= , 50Hz, mHL 10= and VEd 170= . Calculate and plot the current an the diode voltage drop along with supply voltage, sv .

sv

diodev+ -

Lv i

dE

+ -

+

-

4- Assume there is a freewheeling diode is connected in shunt with the load of the

rectifier shown in problem 2. Calculate the load current during two periods of supply voltage. Then, draw the inductor, resistor, load voltages and diode currents along with supply voltage.

5- The voltage v across a load and the current i into the positive polarity terminal are as follows:

( ) ( ) ( ) ( )tVtVtVVtv d ωωωω 3cos2sin2cos2 311 +++= ( ) ( ) ( )φωωω −++= tItIIti d 3cos2cos2 31

Calculate the following: (a) The average power supplied to the load. (b) The rms value of ( )tv and ( )ti . (c) The power factor at which the load is operating.

6- Center tap diode rectifier is connected to 220 V, 50 Hz supply via unity turns ratio center-tap transformer to feed Ω5 resistor load. Draw load voltage and currents and diode currents waveforms along with supply voltage. Then, calculate (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current.

7- Single phase diode bridge rectifier is connected to 220 V, 50 Hz supply to feed Ω5 resistor. Draw the load voltage, diodes currents and calculate (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current.


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8- If the load of rectifier shown in problem 7 is changed to be Ω5 resistor in series with 10mH inductor. Calculate and draw the load current during the first two periods of supply voltages waveform.

9- Solve problem 8 if there is a freewheeling diode is connected in shunt with the load. 10- If the load of problem 7 is changed to be 45 A pure DC. Draw diode diodes currents

and supply currents along with supply voltage. Then, calculate (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current. (f) input power factor.

11- Single phase diode bridge rectifier is connected to 220V ,50Hz supply. The supply has 4 mH source inductance. The load connected to the rectifier is 45 A pure DC current. Draw, output voltage, diode currents and supply current along with the supply voltage. Then, calculate the DC output voltage, THD of supply current and input power factor, and, input power factor and THD of the voltage at the point of common coupling.

12- Three-phase half-wave diode rectifier is connected to 380 V, 50Hz supply via 380/460 V delta/way transformer to feed the load with 45 A DC current. Assuming ideal transformer and zero source inductance. Then, draw the output voltage, secondary and primary currents along with supply voltage. Then, calculate (a) Rectfication effeciency. (b) Crest factor of secondary current. (c) Transformer Utilization Factor (TUF). (d) THD of primary current. (e) Input power factor.

13- Solve problem 12 if the supply has source inductance of 4 mH. 14- Three-phase full bridge diode rectifier is connected to 380V, 50Hz supply to feed

Ω10 resistor. Draw the output voltage, diode currents and supply current of phase a. Then, calculate: (a) The rectfication effeciency. (b) Ripple factor of load voltage. (c) Transformer Utilization Factor (TUF) (d) Peak Inverse Voltage (PIV) of the diode. (e) Crest factor of supply current.

15- Solve problem 14 if the load is 45A pure DC current. Then find THD of supply current and input power factor.

16- If the supply connected to the rectifier shown in problem 14 has a 5 mH source inductance and the load is 45 A DC. Find, average DC voltage, and THD of input current.

17- Single phase diode bridge rectifier is connected to square waveform with amplitude of 200V, 50 Hz. The supply has 4 mH source inductance. The load connected to the rectifier is 45 A pure DC current. Draw, output voltage, diode currents and supply current along with the supply voltage. Then, calculate the DC output voltage, THD of supply current and input power factor.

18- In the single-phase rectifier circuit of the following figure, 1=SL mH and VVd 160= . The input voltage sv has the pulse waveform shown in the following

figure. Plot si and di waveforms and find the average value of dI .


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+

-SV

dVSi

di

Hzf 50=

o60 o60 o60o120

o120o120

V200tω

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