fullwaverectifier.pdf

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EE 435- Electric Drives Dr. Ali M. Eltamaly

Chapter 2

Diode ircuits or Uncon trolled Rectifier

2.1 Introduction

Because of their ability to conduct current in one direction, diodes are used in rectifiercircuits. 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.

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 ParametersIn 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, dc I ,

The rms value of the output voltage, rmsV ,

The rms value of the output current, rms I

The output DC power, dcdcdc I V P *= (2.1)

The output AC power, rmsrmsac I V P *= (2.2) The effeciency or rectification ratio is defiend as acdc P P /=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:-

22dcrmsac V V V −= (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.

dcrms V V FF /= (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.


2/21

Diode Circuits or Uncontrolled Rectifier 9

11 22

222

−=−=−

== FF V

V

V

V V

V

V RF

dc

rms

dc

dcrms

dc

ac (2.6)

Where S V and S I are the rms voltage and rms current of the transformer secondery respectively.

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

S S i

I

I

I

I I THD , and 12

1

2

21

21

2

−=−

=S

S

S

S S v

V

V

V

V V THD (2.8)

where 1S I and 1S V are the fundamental component of the input current and voltage, S I and S V

respectively.

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

φ cossVoltampereApparent

Power R ===

S S I V

P eal PF (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:

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

frequency.

Faactor nt Displaceme Factor Distortion I

I

I V

I V

I V

P PF

S

S

S S

S S

S S

*coscos

1111 ==== φ

φ (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 ( )t v ω and ( )t i ω .

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, V S 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. Thesewaveforms show that both the load voltage and current have high ripples. For this reason, single-

phase half-wave diode rectifier has little practical significance.

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


3/21

Chapter Two 10

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

The average or DC output voltage can be obtained by considering the waveforms shown in

Fig.2.2 as following: ∫ ==π

π ω ω

π 0

sin2

1 mmdc

V t d t V V (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: R

V

R

V I mdcdc

π ==

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

2sin

2

1

0

22 mmrms

V t d t V V == ∫

π

ω ω π

(2.14)

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

V

R

V I mrmsrms

2==

It is clear that the rms value of the transformer secondary current, S I is the same as that of the

load and diode currents

Then R

V I I m DS

2== (2.15)

Where, D I is the rms value of diode current.

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) Peak inverse voltage (PIV) of diode D1.

Solution: From Fig.2.2, the average output voltagedc

V is defiend as:

π π

π ω ω

π

π mm

mdc

V V t d t V V =−−== ∫ ))0cos(cos(2)sin(2

1

0

Then, R

V

R

V I mdcdc

π ==


4/21


2)sin(

2

1

0

2 mmrms

V t V V == ∫

π

ω π

, R

V I mrms

2= and,

2

mS

V V =

The rms value of the transformer secondery current is the same as that of the load: RV I mS 2.=

Then, the efficiency or rectification ratio is:

rmsrms

dcdc

ac

dc

I V

I V

P

P

*

*==η %53.40

2*

2

*

==

R

V V

R

V V

mm

mm

π π

(b) 57.12

2 ==== π

π

m

m

dc

rms

V

V

V

V FF

(c) 211.1157.1122 =−=−== FF

V

V RF

dc

ac

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

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, if ( )t V v m s ω sin= , sv is positivewhen 0 < ω t < π , and sv is negative when π < ω t


5/21

Chapter Two 12

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

Assume wL j R Z +=∠φ Then 2222 Lw R Z += ,

φ cos Z R = , φ ω sin Z L = and R

Lω φ =tan

( ) ( )

+−=

−φ

ω

φ φ ω ω tan

sinsin)(

t

m et Z

V t i (2.24)

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

reaches β . The value of β can be obtained by substituting that β =

=wt

t i 0)( into (2.24) we get:

( ) ( ) 0sinsin)( tan =

+−=

−φ

β

φ φ β β e Z

V i m (2.25)

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

analysis. This average value can be obtained as shown in (2.26). The rms output voltage in thiscase is shown in equation (2.27).

)cos1(*2

sin*2

0

β π

ω ω π

β

−== ∫ mmdcV

t d t V

V (2.26)

)2sin(1(5.0*2

)sin(*2

1

0

2 β β

π ω

π

β

−+== ∫ Vm

dwt t V V mrms (2.27)

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:

R

wL

Z

Φ


6/21


dvantages

• 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.

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.2.9 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:

π ω ω

π

π

mmdc V t d t V V 2sin1

0

== ∫ and RV I mdcπ 2= (2.36)

Fig.2.8 Center-tap diode rectifie


7/21

Chapter Two 14

( )2

sin1

0

2 mmrms

V t d t V V == ∫

π

ω ω π

and R

V I mrms

2= (2.38)

PIV of each diode = mV 2 (2.40)

2

mS

V V = (2.41)

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

R

V I I 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) Peak inverse voltage (PIV) of diode D1.

Solution:- The efficiency or rectification ratio is

%05.81

2*

2

2*

2

*

*====

R

V V

R

V V

I V

I V

P

P

mm

mm

rmsrms

dcdc

ac

dc π π η

(b) 11.1222

2 ==== π

π

m

m

dc

rms

V

V

V

V FF

(c) 483.0111.11 22 =−=−== FF V V RF

dc

ac

(d) The PIV is mV 2

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 theoperation 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”).

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 thediodes 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

Fig.2.12 Single-phase full bridge diode rectifier.


8/21


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.

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

Example 4 The rectifier shown in Fig.2.12 has a purely resistive load of R=15 Ω and, V S =300 sin314 t and unity transformer ratio. Determine (a) The efficiency, (b) Form factor, (c) Ripple factor,

(d) The peak inverse voltage, (PIV) of each diode, , and, (e) Input power factor.

Solution: 300=m

V V

V V

t d t V V mmdc 956.1902

sin1

0

=== ∫ π ω ω π π

, A R

V I mdc 7324.12

2==

π

( ) V V

t d t V V mmrms 132.2122

sin1

2/1

0

2 ==

= ∫

π

ω ω π

, A R

V I mrms 142.14

2==

(a) %06.81===rmsrms

dcdc

ac

dc

I V

I V

P

P η

(b) 11.1==

dc

rms

V

V

FF

(c) 482.011 22

222

=−=−=−

== FF V

V

V

V V

V

V RF

dc

rms

dc

dcrms

dc

ac

(d) The PIV=300V

(e) Input power factor = 1cosRe

==S S

S S

I V

I V

Power Apperant

Power al φ

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 thiscircuit 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. And the rms value of the input current is oS I I =


9/21

Chapter Two 16

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

Fig.2.15 Various current & voltage waveforms for single-phase diode bridge rectifier for pure DC load.

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*2

0

0

==−=

−== ∫

n for n

I n

n

I

t nn

I t d t n I b

oo

oon

π π

π

ω π

ω ω π

π π

(2.51)

Then from Fourier series concepts we can say:

)..........9sin9

17sin

7

15sin

5

13sin

3

1(sin*

4)( +++++= t t t t t I t i o ω ω ω ω ω

π (2.52)

%4615

1

13

1

11

1

9

1

7

1

5

1

3

1))((

2222222

=

+

+

+

+

+

+

=∴ t I THD s or we can obtain

))(( t I THD s as the following:

From (2.52) we can obtain the value of isπ 2

41

oS

I I =

%34.4814

21

2

41))((

2

2

2

1

=−

=−

=−

=∴ π

π

o

o

S

S s I

I

I

I t I THD


10/21


Example 5 solve Example 4 if the load is 30 A pure DC

Solution: From example 4 V dc= 190.986 V, V rms=212.132 V A I dc 30= and rms I = 30 A

(a) %90===rmsrms

dcdc

ac

dc

I V

I V

P

P η (b) 11.1==

dc

rms

V

V FF

(c) 482.011 22

222

=−=−=−

== FF V

V

V

V V

V

V RF

dc

rms

dc

dcrms

dc

ac

(d) The PIV=V m=300V

(e) A I

I oS 01.272

30*4

2

41 ===

π π

Input Power factor= = Power Apperant

Power al Re Lag

I

I

I V

I V

S

S

S S

S S 9.01*30

01.27cos*cos* 11 === φ φ

2.4.5 Effect Of L S 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 S i from a value of o I to o I − (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 & voltage waveforms for single-phase diode bridge rectifier with source inductance.


11/21

Chapter Two 18

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 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 t ∆ to

completely turn OFF D1 and D2 and to make D3 and D4 carry the entire load current ( o I ). Alsoin the time t ∆ the supply current will change from o I to o I − which is very clear in Fig.2.16.

Fig.2.17 shows the equivalent circuit of the diode bridge at time t ∆ .

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

−= −

m

o s

V

I Lu

ω 21cos

1 (2.56)

oS oS rd I L f I L

V 42

4

−=−

= π

ω (2.61)

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

o sm

rd ceinduc sourcewithout dcactual dc I fL

V V V V 4

2tan

−=−=π

(2.62)

−=

32

2 2 u I I o s

π

π (2.64)

2sin*

2

81

u

u

I I oS

π

= (2.68)

( )

−

=

32

sin2

uu

u pf

π π

(2.69)

Example 6 Single phase diode bridge rectifier connected to 11 kV, 50 Hz, source inductance

mH X s 5= supply to feed 200 A pure DC load, find: (i) Average DC output voltage, (ii) Power

factor. And (iii) Determine the THD of the utility line current.

Solution: (i) From (2.62), V V m

155562*11000 ==

o sm

rd ceinduc sourcewithout dcactual dc I fL

V V V V 4

2tan

−=−=π


12/21


V V actual dc

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*2

1cos

2

1cos11

rad V

I L

um

o s

=

−=

−=

−− π ω

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

( ) ( )917.0

3

285.

2285.0

285.0sin*2

32

sin*2

2cos*1 =

−

=

−

=

=

π π

π π

uu

uu

I

I pf

S

S

Au I

I oS 85.1933

285.0

2

200*2

32

2 22

=

−=

−=

π

π

π

π

Au

u

I I oS 46.1792

285.0sin*285.0*2

200*8

2sin*2

81 =

== π π

%84.40146.179

85.1931

22

1

=−

=−

=

S

S i

I

I THD

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”.

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.

6

π

6

5π


13/21

Chapter Two 20

Fig.2.23 Primary and diode currents.

Fig.2.24 FFT components of primary and diode currents.

By considering the interval from

6

π to

6

5π in the output voltage we can calculate the average

and rms output voltage and current as following:

mm

mdc V V

t d t V V 827.02

33sin

2

36/5

6/

=== ∫ π ω ω π π

π

and R

V

R

V I mmdc

*827.0

**2

33==

π (2.70)

( ) mmmrms V V t d t V V 8407.08

3*3

2

1sin

2

36/5

6/

2 =+== ∫ π ω ω π π

π

(2.72)

R

V I mrms

8407.0= (2.73)

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

R

V

R

V I I mmS r 4854.0

3

08407=== (2.74)

Primary current

Diode current


14/21


Note :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 m LL V V 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) Rectificationefficiency, (b) Form factor (c) Ripple factor (d) Transformer utilization factor, (e) Peak inverse

voltage (PIV) of each diode.

Solution:

(a) V V V V mS 59.3752*58.265,58.2653

460====

mm

dc V V

V 827.02

33==

π ,

R

V

R

V I mmdc

0827

2

33==

π

mrms V V 8407.0= , R

V I mrms

8407.0= Then, %767.96===

rmsrms

dcdc

ac

dc

I V

I V

P

P η

(b) %657.101==dc

rms

V

V FF

(c) %28.1811 22

222

=−=−=−

== FF V

V

V

V V

V

V RF

dc

rms

dc

dcrms

dc

ac

(e) The PIV= 3 V m=650.54V

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.

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:

32

13/

3/

0o

o

I t d I a ∫

−

==π

π

ω π

[ ]

harmonicstrepleanall for

n for n

I

n for

n

I

t nn

I dwt t n I a

o

o

oon

0

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

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

sincos*1 3/

3/

3/

3/

=

=−=

==

== −−∫

π

π

ω π

ω π

π

π

π

π

(2.77)


15/21

Chapter Two 22

−−++−−++= ...8sin

8

17sin

7

15sin

5

14sin

4

12sin

2

1sin

3

3)( t t t t t t

I I t I OO s ω ω ω ω ω ω

π (2.78)

%24.1090924.119

*21

2

3

3/1))((2

2

2

1

==−=−

=−

= π

π

O

o

S

S s

I

I I I t I THD

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

rectifier with dc load current

N e w a

x i s


16/21


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

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

460====

V V V

V mmdc 613.310827.02

33

=== π , A I dc 100=

V V V mrms 759.3158407.0 == , A I rms 100=

%37.98100*759.315

100*613.310====

rmsrms

dcdc

ac

dc

I V

I V

P

P η

(b) %657.101==dc

rms

V

V FF

(c) %28.1811 22

222

=−=−=−

== FF

V

V

V

V V

V

V RF

dc

rms

dc

dcrms

dc

ac

(d) The PIV= 3 V m=650.54V

2.5 Three-Phase Full Wave Diode Rectifier

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 pureresistance. Fig.2.31 shows complete waveforms for phase and line to line input voltages andoutput 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.

1 3 5

4 6 2

b

c

I L

V L

I s

I p

a

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

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

LLm LLm

mdc V V V V

t d t V V 3505.1654.12333

sin33

3/2

3/

===== ∫ π π ω ω π π

π

(2.91)

R

V

R

V

R

V

R

V I LL LLmmdc

3505.123654.133====

π π (2.92)

( ) LLmmmrms V V V t d t V V 3516.16554.143*9

2

3

sin3

33/2

3/

2

==+== ∫ π ω ω π π

π

(2.93)

R

V I mrms

6554.1= (2.94)


17/21

Chapter Two 24

Then the diode rms current is

R

V

R

V I mmr 9667.0

3

6554.1== (2.95)

R

V

I

m

S 29667.0= (2.96)

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

voltages.

Fig.2.32 Diode currents.


18/21


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) Peak inverse voltage (PIV) of each diode .

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

460====

V V V

V mm

dc 226.621654.133

===π

, A R

V

R

V I mmdc 0613.31

654.133===

π

V V V V mmrms 752.6216554.14

3*9

2

3==+=

π , A

R

V I mrms 0876.31

6554.1==


19/21

Chapter Two 26

%83.99===rmsrms

dcdc

ac

dc

I V

I V

P

P η and (b) %08.100==

dc

rms

V

V FF

(c) %411 2

2

222

=−=−=−

== FF V

V

V

V V

V

V RF

dc

rms

dc

dcrms

dc

ac and (d) The PIV= 3 V m=650.54V

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(

72),3(

52,32

cos2

sin*2

1311751

6/56/

6/5

6/

==

==−=−==

−== ∫

n for b

I b I b I b I b I b

t nn

I t d t n I b

n

ooooo

oon

π π π π π

ω π

ω ω π

π

π

π

π

(2.97)

++−−= t t t t vt

I t I o s ω ω ω ω ω

π 13sin

13

111sin

11

17sin

7

15sin

5

1sin

32)( (2.98)

%3125

1

23

1

19

1

17

1

13

1

11

1

7

1

5

1))((

22222222

=

+

+

+

+

+

+

+

=t I THD s

Also ))(( t I THD s can be obtained as following:

oS I I 32= , oS I I

π 3*21 = then, %01.311

/3*23/21))((

2

2

1

=−=−

=

π S

S s

I I t I THD

Power Factor =S

S

S

S

I

I

I

I 11 )0cos(* =

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


20/21


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

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 DCcomponent of the output voltage is reduced. Fig.2.38 shows The output DC voltage of three-

phase full wave rectifier with source inductance.

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


21/21

Chapter Two 28

Commutation angle is ,

−= −

LL

oS

V

I Lu

ω 21cos 1 (2.109)

The DC voltage reduction due to source inductance is : oo

rd fLI LI

V 6

2

6==

π

ω (2.114)

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

d LLrd ceinduc sourcewithout dcactual dc fLI V V V V 635.1

tan −=−= (2.115)

−=

63

22

u I I oS

π

π and

=

2sin

621

u

u

I I oS

π

The power factor can be calculated from the following equation:

( )

−

=

−

=

=

63

sin*3

2

cos

632

2sin

62

2

cos2

1

uu

uu

u I

u

u

I

u

I

I pf

o

o

S

S

π π π π

π

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 π ω = , A I d 300= , H L 008.0= , V V LL 33000= in

(2.109), then orad u 61.14.2549.0 ==

(ii) The the actual DC voltage can be obtained from (2.115) as following:

d LLrd ceinduc sourcewithout dcactual dc fLI V V V V 635.1

tan −=−=

V V dcactual 43830300*008.*50*633000*35.1 =−=

(iii) the power factor can be obtained from (2.121) then

( ) ( )9644.0

6

2549.0

3*2549.0

2549.0sin3

63

sin*3=

−

=

−

=π

π π

π u

u

u pf Lagging

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

Au I

I d s 929.2396

2549.0

3*

300*2

63

2 22=

−=

−=

π

π

π

π

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

following: Au

u

I I oS 28.233

2

2549.0sin*

2*2549.0*

300*3432*

2sin

2

341 =

=

=

π π

9644.02

2549.0cos*

929.239

28.233

2cos*1 =

=

=

u

I

I pf

s

S Lagging.

%05.24128.233

929.2391

22

1

=−

=−

=

S

S i

I

I THD

fullwaverectifier.pdf

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