PE 304 Lect 4 & 5 Uncontrolled Rectifier Circuits.pdf

Copyright © 2003 

by John Wiley & Sons, Inc. 

Lectures 4 & 5 

Diode Rectifiers 

Chapter 5 Line-Frequency Diode 

Rectifiers 

5-1



Diode Rectifier Block Diagram 

• Uncontrolled utility interface (ac to dc) 



5-2

• Resistive load 



A Simple Circuit 



5-3



A Simple Circuit (R-L Load) 

• Current continues to flows for a while even after the input 

voltage has gone negative 



5-4

• From t = 0, the diode is forward biased 

Vs = + 

Ri 

L 

di 

dt 

• For 0t 1 , V L → negative, current decreases 

•t>t 2 ,V L → negative, but the current is still positive 

•at t 3 , the current goes to zero 





5-5

But i(i 3 ) = i(0) = 0 



t 

1 

∫ 

0 

V 

L 

dt + 

1 

L 

t 

3 

∫ 

0 

∫ 

∫ 

V 

V 

V 

L 

L 

L 

dt 

= 

dt = di 

= 

Area A − Area B = 

t 

t 

t 

3 

0 

3 

1 

V 

V 

L 

L 

dv 

L 

dt 

i(t 

3 

∫ 

i(0) 

dt = 0 

dt = 0 

i(t 

) 

∫ 

i(0) 

di 

3 

) 

di 

For t>t 3 , V R and V L are zero. So a reverse polarity voltage 

appears across the diode. The waveforms repeat with the time 

period T=1/f 



5-6

A Simple Circuit (Load has a dc back-emf) 

• Current begins to flow when the input voltage exceeds the dc back-emf 

• Current continues to flows for a while even after the input voltage has 

gone below the dc back-emf 





5-7

Single-Phase Diode Rectifier Bridge 

• Large capacitor at the dc output for filtering and energy 

storage 





5-8



Diode-Rectifier Bridge Analysis 

• Two simple (idealized) cases to begin with 



5-9

Redrawing Diode-Rectifier Bridge 

• Two groups, each with two diodes 





5-10

• When v s positive, D1 and D2 conduct and v d = v s and i s = i d . 

When v s goes negative, D3 and D4 conduct, and therefore, 

v d = -v s and i s = -i d . At any time, dc-side output voltage of 

diode rectifier expressed as 

• Similarly, the ac-side current can be expressed as 



i 

s 

vd (t) = | vs 

⎧id 

if v 

= ⎨ 

⎩− 

id 

if 

s 



| 

> 0 

v < 0 

• and the transition between the two values is instantaneous 

due to the assumption of L s . The average value V do is 

2 

V do = 

2 Vs 

= 0.9 Vs 

π 

s 

5-11

•Also 

2 

Is1 

= 

2 Id 

= 0.9I 

d 

π 

⎧0 

for even values 

of h 

Ish 

= ⎨ 

⎩Is1/h 

for odd values of h 

THD = 48.43% 

DPF = 1.0 

Is 

PF = DPF 

I 



1 

s 

= 

0.9 



and 

5-12

Waveforms with a 

purely resistive load 

and a purely dc current 

at the output 

• In both cases, the dc-side 

voltage waveform is the same 





5-13

Diode-Rectifier Bridge Input Current 

• Idealized case with a purely dc output current 





5-14

Diode-Rectifier Bridge Analysis with AC- 

Side Inductance 

• Output current is assumed to be purely dc 





5-15

Understanding Current Commutation 

• Assuming inductance in this circuit to be zero 





5-16

Understanding Current Commutation (cont.) 

• Inductance in this circuit is included 





5-17





5-18 

s 

s 

s 

s 

s 

s 

L 

di 

L 

ω 

t) 

td(ω 

sinω 

V 

2 

t 

ω 

0 

dt 

di 

L 

t 

sinω 

V 

2 

v 

= 

< 

< 

= 

= u 

s 

s 

s 

2 

0 

do 

s 

d 

s 

d 

s 

s 

u 

s 

s 

u 

0 

u 

I 

0 

d 

s 

s 

s 

s 

u 

0 

V 

0.45 

V 

π 

2 

2 

2 

t) 

td(ω 

sinω 

V 

2 

π 

2 

1 

V 

V 

2 

I 

L 

ω 

1 

cosu 

I 

L 

ω 

cosu) 

(1 

V 

2 

A 

cosu) 

(1 

V 

2 

t) 

td(ω 

sinω 

V 

2 

A 

I 

L 

ω 

di 

L 

ω 

t) 

td(ω 

sinω 

V 

2 

d 

= 

= 

= 

− 

= 

= 

− 

= 

− 

= 

= 

= 

= 

∫ 

∫ 

∫ 

∫





5-19 

d 

s 

u 

d 

d 

s 

s 

u 

s 

d 

s 

u 

0 

s 

π 

0 

d 

I 

π 

2 

L 

ω 

π 

2 

A 

area 

ΔV 

I 

π 

2 

L 

ω 

V 

0.45 

π 

2 

A 

area 

V 

0.45 

V 

t) 

td(ω 

sinω 

V 

2 

π 

2 

1 

t) 

td(ω 

sinω 

V 

2 

π 

2 

1 

V 

= 

= 

− 

= 

− 

= 

− 

= ∫ 

∫ 

π 

I 

L 

ω 

2 

V 

0.9 

π 

A 

area 

V 

V 

V 

2 

I 

L 

ω 

2 

1 

cosu 

I 

L 

ω 

2 

di 

L 

ω 

t) 

ω 

td( 

sinω 

V 

2 

A 

area 

radian 

Volt 

i 

2 

I 

i 

i 

- 

I 

i 

i 

i 

i 

i 

d 

s 

s 

u 

do 

d 

s 

d 

s 

d 

s 

I 

I 

- 

s 

s 

u 

0 

s 

u 

u 

d 

s 

u 

d 

4 

D 

3 

D 

u 

2 

D 

1 

D 

d 

d 

− 

= 

− 

= 

− 

= 

= 

= 

= 

− 

+ 

− 

= 

= 

= 

= 

= 

∫ 

∫

Current Commutation Waveforms 

• Shows the volt-seconds needed to commutate current 





5-20

Current Commutation in Full-Bridge Rectifier 

• Shows the necessary volt-seconds 





5-21

Understanding Current Commutation 

• Note the current loops for analysis 





5-22

Rectifier with a dcside 

voltage 





5-23





5-24 

s 

s 

circuit 

short 

θ 

θ 

d 

d 

θ 

θ 

d 

s 

θ 

θ 

d 

s 

s 

d 

b 

θ 

θ 

d 

s 

θ 

θ 

d 

s 

d 

s 

d 

s 

L 

b 

p 

b 

s 

d 

L 

ω 

V 

I 

π 

dθ 

) 

(θ 

i 

I 

t) 

d(ω 

) 

V 

- 

t) 

sin(ω 

V 

2 

( 

0 

t) 

d(ω 

) 

V 

- 

t) 

sin(ω 

V 

2 

( 

L 

ω 

1 

θ) 

( 

i 

θ 

θ 

where 

t) 

d(ω 

) 

V 

- 

t) 

sin(ω 

V 

2 

( 

di 

L 

ω 

V 

- 

t) 

sin(ω 

V 

2 

dt 

di 

L 

v 

θ 

π 

θ 

sinθ 

V 

2 

V 

b 

b 

b 

b 

b 

= 

= 

= 

= 

> 

= 

= 

− 

= 

= 

∫ 

∫ 

∫ 

∫ 

∫

DC-Side Voltage and Current Relationship 

• Zero current corresponds to dc voltage equal to the peak of 

the input ac voltage 





5-25



Effect of DC-Side Current on 

THD, PF and DPF 

• Very high THD at low current values 



5-26

Crest Factor versus the Current Loading 

• The Crest Factor is very high at low values of current 





5-27

Diode-Rectifier with a Capacitor Filter 

• Power electronics load is represented by an equivalent load 

resistance 





5-28



Diode Rectifier Bridge 

• Equivalent circuit for analysis on one-half cycle basis 



5-29

Diode-Bridge Rectifier: Waveforms 





5-30

Diode-Bridge Rectifier: Waveforms 

• Analysis using PSpice- You may repeat with PSim 





5-31

• Analysis using PSpice- 



Input Line-Current Distortion 



5-32



Line-Voltage Distortion 

• PCC is the point of common coupling 



5-33

(v 



(v 

pcc 

v 

v 

pcc 

) 

pcc 

pcc 

) 

1 

dis 

= 

= 

= 

v 

v 

v 

s 

s 

s 

= − L 

− L 

− L 

− L 

s1 

s1 

s1 

s1 

∑ 

h≠1 

dis 

dt 

di 

dt 

s1 

dis1 

dt 

di 

dt 

sh 

− L 

s1 



∑ 

h≠1 

di 

dt 

sh 

5-34



Line-Voltage Distortion 

• Distortion in voltage supplied to other loads 



5-35



Voltage Doubler Rectifier 

• In 115-V position, one capacitor at-a-time is charged from the 

input. 



5-36

A Three-Phase, Four-Wire System 

• A common neutral wire is assumed 





5-37





5-38 

line 

n 

3 

s 

n 

2 

1) 

k 

3(2 

h 

2 

sh 

n 

1) 

k 

3(2 

h 

h 

h 

sh 

n 

c 

b 

a 

n 

1 

k 

2 

h 

h 

h 

sh 

1 

1 

1 

s 

c 

1 

k 

2 

h 

h 

h 

sh 

1 

1 

1 

s 

b 

1 

k 

2 

h 

h 

h 

sh 

1 

1 

1 

s 

1 

k 

2 

h 

ah 

1 

a 

a 

I 

3 

I 

I 

3 

I 

I 

3 

I 

) 

Φ 

- 

t 

sin(ω 

I 

2 

3 

i 

i 

i 

i 

i 

h) 

240 

- 

Φ 

- 

t 

sin(ω 

I 

240) 

- 

Φ 

- 

t 

sin(ω 

I 

2 

i 

and 

h) 

120 

- 

Φ 

- 

t 

sin(ω 

I 

2 

120) 

- 

Φ 

- 

t 

sin(ω 

I 

2 

i 

1,2,3..... 

k 

where 

) 

Φ 

- 

t 

sin(ω 

I 

) 

Φ 

- 

t 

sin(ω 

I 

2 

i 

i 

i 

= 

≅ 

⎟ 

⎟ 

⎠ 

⎞ 

⎜ 

⎜ 

⎝ 

⎛ 

= 

= 

+ 

+ 

= 

+ 

= 

+ 

= 

= 

+ 

= 

+ 

= 

∑ 

∑ 

∑ 

∑ 

∑ 

∑ 

∞ 

− 

= 

∞ 

− 

= 

∞ 

+ 

= 

∞ 

+ 

= 

∞ 

+ 

= 

∞ 

+ 

=

Current in A Three-Phase, Four-Wire 

System 

• The current in the neutral wire can be very high 





5-39

• Commonly used 

Three-Phase, Full-Bridge Rectifier 





5-40

Three-Phase, Full-Bridge Rectifier: Redrawn 

• Two groups with three diodes each 





5-41





5-42 

LL 

LL 

6 

π 

6 

π 

LL 

do 

LL 

6 

π 

6 

π 

LL 

LL 

ab 

d 

d 

d 

a 

v 

1.35 

v 

2 

π 

3 

t) 

ω 

td( 

cosω 

v 

2 

3 

π/ 

1 

V 

v 

2 

t) 

ω 

td( 

cosω 

v 

2 

A 

π 

6 

1 

t 

ω 

π 

6 

1 

- 

t 

cosω 

v 

2 

v 

v 

conducting 

is 

4 

or 

1 

diode 

neither 

when 

0 

conducting 

is 

4 

diode 

when 

I 

- 

conducting 

is 

1 

diode 

when 

I 

i 

= 

= 

= 

= 

= 

< 

< 

= 

= 

⎪ 

⎩ 

⎪ 

⎨ 

⎧ 

= 

∫ 

∫ 

− 

−



Is 

= 

2 

Id 

3 

= 0.816 I 

1 

Is1 

= 

π 

Is1 

Ish 

= 

h 

6 Id 

= 0.78I 

where h = 5.7.11.13, ..... 

DPF = 1.0 

PF 

= 

3 

π 

= 

0.955 



d 

d 

5-43

Three-Phase, Full-Bridge Rectifier Waveforms 

• Output current is 

assumed to be dc 





5-44

Three-Phase, Full-Bridge Rectifier: Input 

Line-Current 

• Assuming output current to be purely dc and zero ac-side 

inductance 





5-45


• Including the ac-side inductance 





5-46

3-Phase Rectifier: Current Commutation 

• output 

current is 

assumed to be 

purely dc 





5-47





5-48 

LL 

d 

s 

LL 

d 

s 

I 

0 

Ls 

LL 

cn 

an 

u 

0 

cn 

an 

I 

0 

s 

cn 

an 

s 

Ls 

Lc 

La 

cn 

an 

comm 

s 

c 

s 

Lc 

s 

a 

s 

La 

u 

d 

c 

u 

a 

v 

2 

I 

L 

ω 

2 

1 

cosu 

2 

cosu) 

(1 

v 

2 

I 

L 

ω 

diu 

ω 

t 

sinω 

v 

2 

v 

- 

v 

t) 

d(ω 

2 

v 

v 

diu 

L 

ω 

2 

v 

v 

dt 

diu 

L 

dt 

diu 

2 

v 

v 

v 

v 

v 

dt 

diu 

L 

dt 

di 

L 

v 

dt 

diu 

L 

dt 

di 

L 

v 

i 

I 

i 

i 

i 

d 

d 

− 

= 

− 

= 

= 

= 

− 

= 

− 

= 

= 

− 

= 

− 

= 

− 

= 

= 

= 

= 

− 

= 

= 

∫ 

∫ 

∫

v 

A 

Pn 

ΔV 

V 



u 

d 

d 

= 

v 

V 

an 

do 

− L 

= ω LsI 

ω LsI 

= 

π/ 3 

= 

d 

d 

s 

- ΔV 

diu 

dt 

= 

d 

3 

π 

= 

ω 

v 

L 

LL 



s 

an 

I 

d 

= 1.35V 

− v 

2 

- 

cn 

3 

π 

ω 

L 

s 

I 

d 

5-49

Rectifier with a Large Filter Capacitor 

• Output voltage is assumed to be purely dc 





5-50


• THD, PF and DPF as functions of load current 





5-51

Crest Factor versus the Current Loading 

• The Crest Factor is very high at low values of current 





5-52

Three-Phase Rectifier Waveforms 

• Can be simulated using PSpice or PSim 





5-53

PE 304 Lect 4 & 5 Uncontrolled Rectifier Circuits.pdf

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