Three phase supply waveform
Three phase Half wave rectification
Waveforms
Three phase bridge rectification
Comparism between half and full wave
rectification waveforms
Performance Parameters
rectification efficiency
  Pdc / Pac
Vac 
2
Vrms
2
 Vdc
form factor FF  Vrms /Vdc
ripple factor
Vac
RF 

Vdc
2
Vrms
2
 Vdc
Vdc

2
Vrms
2
Vdc
 1  FF 2  1
Vdc
3

2
I dc
Vrms
5 / 6

Vm sin t dt 
 /6
3 3 Vm 0.827 * Vm


2 * * R
R
3

2
I rms
3 3 Vm
 0.827Vm
2
5 / 6
 Vm sin t 
2
 /6
1 3* 3
dt 

Vm  0.8407 Vm
2 8
0.8407 Vm

R
08407 Vm
Vm
Ir  IS 
 0.4854
R
R 3
The PIV of the diodes is
2 VLL  3 Vm
Example A three phase rectifier is operated from 460 V 50
Hz supply at secondary side and the load resistance is
R=20. If the source inductance is negligible, determine :
(a) Rectification efficiency,
(b) Form factor
(c) Ripple factor
(d) Peak inverse voltage (PIV) of each diode.
460
VS 
 265.58 V , Vm  265.58 * 2  375.59 V
3
Vdc
3 3 Vm

 0.827 Vm
2
Vrms  0.8407 Vm
3 3 Vm 0827Vm
I dc 

2 R
R
0.8407 Vm
I rms 
R
Pdc
Vdc I dc


 96.767 %
Pac Vrms I rms
Vrms
FF 
 101.657 %
Vdc
2
2
2
Vrms
 Vdc
Vac
Vrms
2
RF 



1

FF
 1  18.28 %
2
Vdc
Vdc
Vdc
The PIV=
3 Vm= 650.54V
𝑉𝑑𝑐
𝐼𝑑𝑐
3
=
𝜋
2𝜋/3
𝜋/3
3 3𝑉𝑚 3 2𝑉𝐿𝐿
3𝑉𝑚 sin 𝜔 𝑡𝑑𝜔𝑡 =
=
= 1.654𝑉𝑚 = 1.3505𝑉𝐿𝐿
𝜋
𝜋
3 3𝑉𝑚 1.654𝑉𝑚 3 2𝑉𝐿𝐿 1.3505𝑉𝐿𝐿
=
=
=
=
𝜋𝑅
𝑅
𝜋𝑅
𝑅
𝑉𝑟𝑚𝑠 =
3
𝜋
2𝜋/3
2
3𝑉𝑚 sin 𝜔 𝑡 𝑑𝜔𝑡 =
𝜋/3
𝐼𝑟𝑚𝑠 =
𝐼𝑟 =
1.6554𝑉𝑚
𝑅
1.6554𝑉𝑚
𝑅 3
𝑉𝑚
= 0.9667
𝑅
𝑉𝑚
𝐼𝑆 = 0.9667 2
𝑅
3 9∗ 3
+
𝑉 = 1.6554𝑉𝑚 = 1.3516𝑉𝐿𝐿
2
4𝜋 𝑚
Example a three phase full wave rectifier is operated from 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 .
Vdc 
3 3 Vm

 1.654Vm  621.226 V
3 3 Vm 1.654Vm
I dc 

 31.0613 A
 R
R
3 9* 3
Vrms 

Vm  1.6554 Vm  621.752 V
2
4
I rms
1.6554 Vm

 31.0876 A
R
Pdc
Vdc I dc


 99.83 %
Pac Vrms I rms
Vrms
FF 
 100.08 %
Vdc
2
2
2
Vrms
 Vdc
Vac
Vrms
2
RF 



1

FF
1  4 %
2
Vdc
Vdc
Vdc
The PIV=
3 Vm=650.54V
Single-Phase Half-Wave Controlled Rectifier
u2
Resistive load
VT
T
u1
b)
u2
uVT
c)
id
ud
R
d)
a)
0
ug
 t1

t
t
0
ud
0
2

t
uVT
e)
0
t
• As shown in Fig above, the single-phase half-wave rectifier
uses a single thyristor to control the load voltage. The
thyristor will conduct, ON state, when the voltage vT is
positive and a firing current pulse iG is applied to the gate
terminal. Delaying the firing pulse by an angle α does the
control of the load voltage. The firing angle α is measured
from the position where a diode would naturally conduct. In
Fig, the angle a is measured from the zero crossing point of
the supply voltage vs . The load in is resistive and therefore
current id has the same waveform as the load voltage. The
thyristor goes to the non-conducting condition, OFF state,
when the load voltage and, consequently, the current try to
reach a negative value.
full-wave controlled rectifier
Full wave three-phase controlled
rectification