Power Electronics

Diode Rectifier (Uncontrolled)

Dr Taosif Iqbal
College of E&ME, NUST
taosifiqbal@ceme.nust.edu.pk
Power Electronics 3. Uncontrolled Rectifier
1. Half wave rectification
2. Performance Parameters
3. Disadvantages of HW rectifier
4. HW Rectifier with RL Load
5. Half wave Rectifier with a Battery
6. Full wave rectifier
7. Comparison between Single Phase Rectifiers
8. Total Harmonic Distortion (THD)
9. Full Wave Bridge Rectifier with RL Load
10. FW Rectifier with C Filter
11. Three phase HW Rectifier
12. Three phase FW Bridge Rectifier
13. LC Filter Design
Power Electronics 3.1. Half wave rectification

Rectification: The process of converting the alternating

voltages and currents to direct currents
Power Electronics 3.1. Half wave rectification

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

Power Electronics 3.1. Half wave rectification


1 Vm Vm Vdc Vm
2 0
Vdc  V sin(  t ) d  t  (  cos   cos( 0 ))  I dc  
m
2  R  R

1 Vm Vrms Vm
2 0
Vrms  V 2
m sin 2
 t d  t  I rms  
2 R 2R

the load and diode currents

Vm
IS  ID 
2R
Power Electronics 3.2. Performance Parameters
  Pdc / Pac Rectification efficiency
Vac  Vrms
2
 Vdc2 Ripple in rectified output
FF  Vrms /Vdc Form factor
Vac
2
Vrms  Vdc2 2
Vrms
RF    2
 1  FF 2  1 Ripple factor
Vdc Vdc Vdc
Total Harmonic Distortion (Voltage)
VS2  VS21 VS2
THDv   1
VS21 VS21 Vs  [Vsdc
2
 Vs21  Vs22  ...  Vsn2 ]1/ 2

I S2  I S21 I S2 Total Harmonic Distortion (Current)

THDi   1
I S21 I S21
Power Factor
P V I cos 1 I S 1
PF   S S1  cos 1
VS I S VS I S IS

Pdc Transformer Utilization Factor

TUF 
Vs I s
 3.2. Performance Parameters of HW
Rectifier
Power Electronics

Example 1: The rectifier shown in Figure 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.

1 Vm Vm
2 0
Vdc  V sin(  t ) d  t  (  cos   cos( 0 )) 
2 
m

 Vm Vdc Vm
1 V I rms  I dc  
Vrms 
2 
(Vm sin t ) 2  m
2 2R R R
0
Vm Vm
*
Pdc V *I  R
  dc dc   40 .53 %
Pac Vrms * I rms Vm Vm
*
Vm 2 2R
Vrms 
FF   2   1.57
Vdc Vm 2

 3.2. Performance Parameters of HW
Rectifier
Power Electronics

Vac
RF   FF 2  1  1.57 2  1  1.211
Vdc
.
It is clear from the Figure that the PIV is Vm
• PVA = VsIs = 0.707 Vm x 0.5 Vm /R
• TUF = Pdc/VsIs = (0.318)2 / (0.707 x 0.5)= 0.286
• CF = Is-peak/ Is = 1/0.5 =2
• PF = Pac/VA = 0.52 / (0.707X0.5) = 0.707
Power Electronics 3.3. Disadvantages of HW rectifier

• High ripple factor 21%,

• Low rectification efficiency 40%,
• Low transformer utilization factor 0.286
Power Electronics 3.4. HW Rectifier with RL Load

RL load (without free wheeling diode)

Vm   Vm
2 0
 
Vdc  sin td ( t )  { cos  t}
2
0

Vm
 [1  cos(   )]
2
Power Electronics 3.4. HW Rectifier with RL Load

Highly inductive load (with free wheeling diode)

Power Electronics 3.5. HW Rectifier with a Battery

Diode conducts when input

voltage is higher than E
Power Electronics 3.6. Full wave rectifier

FW Rectifier with center-tapped Transformer

Power Electronics 3.6. Full wave rectifier
Power Electronics 3.6. Full wave rectifier

FW Bridge Rectifier
Power Electronics 3.6. Full wave rectifier
Performance of Bridge Rectifier
Example 4 single-phase diode bridge rectifier has a purely resistive load
of R=15 ohms and, VS=300 sin 314 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.

1 2 Vm
Vdc 
  Vm sin t dt  
 190 .956 V
0
 1/ 2
1 
   Vm sin t  dt 
Vm
Vrms 2
  212 .132 V
  0  2

Pdc Vdc I dc
   81.06 % Vrms
Pac Vrms I rms FF   1.11
Vdc

Vac
2
Vrms  Vdc2 2
Vrms
RF    2
 1  FF 2
 1  0.482
Vdc Vdc Vdc
Power Electronics 3.6. Full wave rectifier
Performance of Bridge Rectifier
Example 4: single-phase diode bridge rectifier has a purely resistive load of
R=15 ohms and, VS=300 sin 314t 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.

2Vm
I dc   12.7324 A
 R The PIV=300V

2 Vm2
Pdc 0.636 R  81%
TUF   Pdc is 81% of Pva  Pva needed
PVA 0.707 2 Vm2 1.23 times larger to supply Pdc to
R load

2 Vm2
Pac 0.707 R 1
PF   2
PVA 0.707 2 Vm
R
 3.7. Comparison between Single Phase
Rectifiers
Power Electronics

Half wave Full wave Fullwave

center-tap bridge
• Rectification efficiency 0.405 0.81 0.81
• Form factor 1.57 1.11 1.11
• Ripple factor 1.21 0.482 0.482
• Transformer rating secondary VA 3.49Pdc 1.75Pdc 1.23Pdc
• Output ripple frequency fr 1 fs 2 fs 2 fs
Power Electronics 3.8. Total Harmonics Distortion
Fourier Series Analysis of HW Rectifier

f ( x )  a0    an cos nx  bn sin nx 
1  1 
 
n 1
an  u (t ) cos ntdt  Vm sin t cos ntdt
1  Vm    0
a0 
2  
f ( x)dx 
 1  sin(1  n)t  sin(1  n)t

 Vm [ ]dt
1  2

0
an  f ( x) cos nxdx 
 
V  cos(1  n)t cos(1  n)t 
 m   (1  n)  (1  n) 
1  2  0
bn 
   f ( x) sin nxdx

Vm 1  cos(1  n) 1  cos(1  n) 
an  
1  1  2  (1  n) (1  n) 

  
bn  u (t )sin ntdt  Vm sin t sin ntdt
 0
  0 n  1,3,5,
V  sin(1  n)t sin(1  n)t  
 m   an   Vm  2 2 
2  (1  n) (1  n)  0  2  (1  n)  (1  n)  n  2, 4, 6,
  
Vm  sin(1  n) sin(1  n) 
 
2  (1  n) (1  n)  Vm Vm 2V 1 1 
vo (t )   sin t  m  cos 2t  cos 4t 
Vm  2  3 15 
 n 1
bn   2
0 n  2,3, 4, lim
sin(1  n)
 lim
 cos(1  n)

n 1 (1  n) n 1 1
Power Electronics 3.8. Total Harmonics Distortion
Fourier Series Analysis of FW Rectifier

f ( x )  a0    an cos nx  bn sin nx 
1  2 
 
n 1
an  u (t ) cos ntdt  Vm sin t cos ntdt
1  2Vm    0
a0 
2  
f ( x)dx 
 2  sin(1  n)t  sin(1  n)t

 Vm [ ]dt
1  2

0
an  f ( x) cos nxdx 
  V  cos(1  n)t cos(1  n)t 
 m  
1    (1  n) (1  n)  0
bn 
   f ( x) sin nxdx

Vm 1  cos(1  n) 1  cos(1  n) 
an  
1  2    (1  n) (1  n) 

  
bn  u (t )sin ntdt  Vm sin t sin ntdt
 0

2  cos(1  n)t  cos(1  n)t  0 n  1,3,5,


  0
Vm [
2
]dt 
an   4Vm  1 
Vm  sin(1  n)t sin(1  n)t 

   (n  1)(n  1)  n  2, 4, 6,
    
  (1  n) (1  n)  0
2Vm 4Vm  1 1 
V  sin(1  n) sin(1  n)  vo (t )     
  3 
cos 2 t cos 4 t
 m   
(1  n) 
15
  (1  n)
bn  0 n  1, 2,3,...
 3.9. Full Wave Bridge Rectifier with
RL Load
Power Electronics
 3.9. Full Wave Bridge Rectifier with
RL Load
Power Electronics
Power Electronics 3.10. FW Rectifier with C Filter
Power Electronics 3.10. FW Rectifier with C Filter
Discharging of capacitor
Vm  t / RCe
Io  e
R
vo  RI o  Vm e  t / RCe

Vr ( pp )  vo  t1   vo  t2   Vm  Vm e  t2 / Rce  Vm 1  e  t2 / Rc 
e  t2 / RCe  1 t2 / RCe , t2  T / 2
Vr ( pp ) VmT
VmT Vdc  Vm   Vm 
Vr ( pp )  Vm 1  1  T / 2 RCe   2 4 RCe
2 RCe
Vr ( pp )
Vdc  Vm  Vac  Vm 
2
VmT  1 
 Vm   Vm 1  
4 RCe  4 fRC e 

 4 fRCe  1 
 Vm  
 4 fRCe 
Power Electronics 3.10. FW Rectifier with C Filter
Find the value of filter capacitor for RF =0.05

RF0.05
Vr ( pp ) VmT
Vdc  Vm   Vm 
2 4 RCe R = 500Ω
Vm  120 2  169.7V
t Vm
Vr (pp)  Vm  Vo (min)  Vm d 
RL Ce 2 fRLCe

Vr ( pp )
V 1
RF  ac  2  Cf   166uF
Vdc Vdc (4 fR  1) RF

Vm 169.7
Vdc  169.7   169.7   169.7  11.21  158.49V
4 fRCe 4*60*500*126.2*10 6
Power Electronics 3.11. Three phase HW Rectifier
Power Electronics 3.11. Three phase HW Rectifier

Dark black – rectified output

Power Electronics 3.11. Three phase HW Rectifier

PIV
Power Electronics 3.11. Three phase HW Rectifier

  cos t5/6/6 
3 5 /6 3Vm
Vdc 
2  /6
Vm sin td t 
2
3Vm
 ( cos 5 / 6  cos  / 6)
2
3Vm 3 3Vm
 (2*0.866)   0.827Vm
2 2
3 3Vm 0.827 *Vm
I dc  
2*  * R R
 3.11. Three phase HW Rectifier
3 5 /6 3 5 /6
Vrms   Vm sin t  dt   Vm2 sin 2 td t
2
Power Electronics
2 /6 2 /6

3 5 /6 (1  cos 2 t ) 3V 2
5 /6

2  /6
Vm2
2
d t 
4
m
 1  cos 2t dt
/6


3Vm2
4   /6
5 /6
dt  
5 /6

 /6
 cos 2t dt 
3Vm2

4
 t5/6/6  sin 2t5/6/6 

3Vm2
 (5 / 6   / 6  sin 2*5 / 6  sin 2*  / 6)
4
3Vm2
 (4 / 6  sin 5 / 3  sin  / 3)
4 0.8407Vm
I rms 
3Vm2 R
 (4 / 6  2*0.866)  0.8407Vm
4
Power Electronics 3.11. Three phase HW Rectifier
1 5 /6 I m2 5 /6 (1  cos 2t )
I   I sin td t   d t
2 2 2

2 2
s m
/6 /6 2
5 /6
I m2 5 /6  I m2 sin 2t 

4  /6 (1  cos 2t )d t 


5
4
/ 6   / 6 
2  /6


I m2  4 sin 2*5 / 6 sin 2*  / 6  I m2  4 0.866 0.866 
        
4  6 2 2  4  6 2 2 
I m2  4  2.96 I m
2
   0.866    0.2355 I m2  I s  0.4854 I m
4  6  4


The PIV of the diodes is VLL,ms  3Vms * 2  Vm,3 ph  3Vm 
 3.11. Three phase HW Rectifier
Example 3.5 A 3-phase HW (Star) rectifier is operated with
Power Electronics
R load. Determine (a) Rectification efficiency, (b) Form factor
(c) Ripple factor (d) Peak inverse voltage (PIV) of each diode
(e) the peak current through diode if Idc = 30A at Vdc = 140V
(f) TUF (g) PF

Pdc Vdc I dc 0.827 2 Vm2 / R

   2 2
 96.767%
Pac Vrms I rms 0.8407 Vm / R

Vrms 0.8407Vm
FF    101.657%
Vdc 0.827Vm

Vac
RF   FF 2  1  18.28%
Vdc

The PIV  3Vm

 3.11. Three phase HW Rectifier
Example 3.5 A 3-phase HW (Star) rectifier is operated with
Power Electronics
R load. Determine (a) Rectification efficiency, (b) Form factor
(c) Ripple factor (d) Peak inverse voltage (PIV) of each diode
(e) the peak current through diode if Idc = 30A at Vdc = 140V
(f) TUF (g) PF

I d ,avg 
1 5 /6
2  /6
I m sin  td  t 
Im
2
  cos  t /6 
5 /6

I I
 m ( cos 5 / 6  cos  / 6)  m 2*0.866  0.276 I m
2 2
I d  30 / 3  10  I m  10 / 0.276  36.27 A
VA= 3 Vs  I s = 3  0.707Vm  0.4854Vm / R
Pdc 0.8242Vm2 / R
TUF = = = 0.66
VA 3  0.707  0.4854Vm / R
2

Pac 0.8412 Vm2 / R

PF   * * 2
 0.687
VA 3 0.707 0.4854Vm / R
Power Electronics 3.12. Three phase FW Rectifier

Conduction sequence of Diodes:

0  t  30  Vc  Va  Vb  D5 D6 conducts
30  t  90  Va  Vc  Vb  D1 D6 conducts
90  t  150  Va  Vb  Vc  D1 D2 conducts
150  t  210  V0  Va  Vc  D3 D2 conducts
210  t  270  Vb  Vc  Vc  D3 D4 conducts
270  t  330  Vc  Vb  Va  D5 D4 conducts
330  t  360  Vc  Va  Vb  D5 D6 conducts
 3.12. Three phase FW Rectifier
0  t  30 :
Power Electronics
 Vc  Va  Vb  D5 D6 conducts
30  t  90 :
 Va  Vc  Vb  D1 D6 conducts
90  t  150 :
 Va  Vb  Vc  D1 D2 conducts
150  t  210 :
Conduction sequence of Diodes:  V0  Va  Vc  D3 D2 conducts
210  t  270 :
 Vb  Vc  Vc  D3 D4 conducts
270  t  330 :
 Vc  Vb  Va  D5 D4 conducts
330  t  360 :
 Vc  Va  Vb  D5 D6 conducts
Power Electronics 3.12. Three phase FW Rectifier

Van (t)  Vm sin( t) Vab (t)  3Vm sin( t   / 6)

Vbn (t)  Vm sin( t  2 / 3) Vbc (t)  3Vm sin( t   / 2)
vcn (t)  Vm sin( t  2 / 3) Vca (t)  3Vm sin( t  5 / 6)
Power Electronics 3.12. Three phase FW Rectifier
Power Electronics 3.12. Three phase FW Rectifier

 cos(t   6)  
3  /2 3 3Vm

 /2
Vdc  3Vm sin(t   / 6)dt   /6
 /6 
3 3Vm 3 3Vm
 (cos 2 / 3  cos  / 3)  (1)  1.654Vm
 
V 1.654Vm π /2
I dc  dc 
 
3 2
V2 = 3Vm sin(ωt + π /6 ) dωt
R R rms π π /6

3  3V 2 π/2
 1  cos2(ωt + π /6 ) 
=
π
m
 
π /6 
2
 dωt

9V 2  sin2(ωt + π /6 )ππ/6
/2

m
 ωt π/2
π/6  
2π  2 
9V 2  sin2(π /2+π /6 )+ sin2(π /3 ) 
= m
 π / 3  
2π  2 
9V 2
 m
1.05 + 0.866   Vrms = 1.6554Vm
2π
Power Electronics 3.12. Three phase FW Rectifier
Peak current is ratio of max line voltage and R load
I m  VLL / R  3Vm / R
Vm Im
I rms  1.6554  1.6554  0.956 I m
R 3
Source current flows only for 4/6* 2pi in one cycle
4 2 2
I s  I rms  I S 
2
I rms  0.816*0.956 I m  0.784 I m
6 3
2 diode current flows only for
I d ,rms  I rms  0.552 I m 2/6 * 2pi in one cycle
6
2 I DC 1.654Vm 1.654 I m 0.955I m
I d , AVG      0.318 I m
6 3R 3 3 3
 3.12. Three phase FW Rectifier
Example 10 The 3-phase bridge rectifier is operated from line
Power Electronics
voltage VLL,rms=460 V 50 Hz supply and the load resistance is
R=20ohms. If the source inductance is negligible, determine (a) The
efficiency, (b) Form factor (c) Ripple factor (d) Peak inverse voltage
(PIV) of each diode (e) TUF (f) PF
2
Vm = 460 = 375.6V
3

3 3 Vm
Vdc = = 1.654Vm = 621.226 V
π
3 9* 3
Vms   Vm  1.6554Vm  621.752V
2 4
3 3Vm 1.654Vm
I dc    31.0613A
R R
Power Electronics 3.12. Three phase FW Rectifier
1.6554 Vm
I rms = = 31.0876 A
R

Pdc Vdc I dc
η= = = 99.83%
Pac Vrms I rms

Vrms 1.6554Vm
FF = = = 100.08%
Vdc 1.654Vm

Vac
2
Vrms  Vdc2 2
Vrms
RF = = = 2
 1 = FF 2
 1 = 4%
Vdc Vdc Vdc
Power Electronics 3.12. Three phase FW Rectifier
PIV = 3Vm = 650.54V

Vs , ph = 0.707Vm
2
I S, = I m = 0.82I m
3

VA = 3  VS  I S = 3  0.707Vm  0.82I m
3  0.707Vm  0.82  3  Vm / R = 3  Vm2 / R = 21.16KVA
Pdc 1.6542Vm2 / R
TUF = = 2
= 0.91
VA 3Vm / R
PAc 1.65542Vm2 / R
PF = = 2
= 0.913
VA 3Vm / R
Power Electronics 3.12. Three phase FW Rectifier
Power Electronics 3.13. LC Filter Design
Example 3.17: Finding the values of LC filter
An LC filter shown in figure is used to reduce the ripple content of output
voltage for a single phase full wave rectifier. The load resistance is R=40
ohms, load inductance is L=10mH, and source frequency is 60Hz.
Determine the values of Le and Ce so that the RF of the output voltage is
10%.

Solution:
To make it easier for the nth harmonic ripple current to pass through the
filter capacitor, the load impedance must be greater than capacitor
impedance;
1 10
R 2 +  nωL  >> R 2 +  nωL  =
2 2

nωCe nωCe
Power Electronics 3.13. LC Filter Design
The RMS value of the nth harmonic component appearing on the output
can be found by voltage divider rule:

 1/  nωCe    1 
Von =   Vnh =   Vnh
  nωLe   1/  nωC 
e     e e 
nω
2
L C  1
1/2
  2 
vac,ripple =   von 
 n= 2,4,6,... 
Fourier series of output voltage for FW rectifier:
2V 4V 4V 4V
vo (t)= m  m cos2ωt  m cos4ωt  m cos6ωt  ...
π 3π 15π 35π
Calculation can be simplified if dominant harmonic (which is the 2nd
harmonic) is considered. Its RMS and dc values are:
4Vm 2V
V2h = vdc = m
3 2π π
nd
Ripple voltage of 2 harmonic:
 1 
Vac = Vo2    V2h
  nω  LeCe  1 
2
 3.13. LC Filter Design
10 10
To calculate Ce: R 2 +  2ωL  =  Ce =
2
Power Electronics
= 326uF
2ωCe 4πf R +  2πfL 
2 2

To calculate Le:

Vac Vo2  1 V 4Vm π  1 

RF = = =  2h
=  
Vdc Vdc   2ω 2 LeCe  1  Vdc 3 2π 2Vm   2ω 2 LeCe  1 

2 1   1  3
        LeCe  1 = 4.71
2
= 0.1 = 0.1 2ω
3   2ω 2 LeCe  1    2ω 2 L C  1 
 e e  2

5.71
Le = = 30.83mH
 2ω 
2
Ce

For pure Resistive load:

10 10
R=  Ce = = 331uF
2ωCe 4πfR
5.71 5.71 10 106
Le = = = = 30.23mH
 2ω  Ce
6
2
16π f
2 2
Ce 331  10