ELG4139: Rectifiers and Controlled Rectifiers
AC to DC Converters
Linear Rectifier
Consist of:
•
Transformer: steps ac voltage up or down.
•
Rectifier Diodes: change ac to “bumpy” dc.
•
Filter Network: includes capacitors and inductors,
smooths out the bumps.
•
Voltage Regulator: keeps the voltage constant.
•
Protection: usually a zener diode circuit.
Example: Computer Power Supply
Example: Adjustable Motor Speed Drive
Power Supply Specifics: Half Wave Rectifier
Source: ARRL
Half-Wave Rectifier
High ripple factor.
Low rectification
efficiency.
Low transformer
utilization factor.
Power Supply Specifics
Full Wave Center-Tapped Rectifier
Source: ARRL
Power Supply: Full Wave Bridge Rectifier
Source: ARRL
Filtering
Capacitors are used in power supply filter networks. The
capacitors smooth out the rippled AC to DC.
Source: ARRL
Rectifier Performance Parameters
  Pdc / Pac Rectification Efficiency
Vac 
2
Vrms
2
 Vdc
FF  Vrms /Vdc
Ripple factor
𝑃𝑎𝑐 = 𝑉𝑟𝑚𝑠 𝐼𝑟𝑚𝑠
Form Factor
2
2
Vrms
 Vdc2
Vac
Vrms
2
RF 



1

FF
1
2
Vdc
Vdc
Vdc
Example 1: A half-wave rectifier has a pure resistive load of R
Determine (a) The efficiency, (b) Form factor (c) Ripple factor.

Vdc
V
V
1

Vm sin(t ) dt  m ( cos  cos(0))  m
2
2


0
Vm
I rms 
2R

Vrms
V
1

(Vm sin t ) 2  m
2
2

0
Vm

Pdc
V *I
 dc dc
Pac Vrms * I rms
Vrms
FF 
Vdc
RF 
Vm
 R

 40 .53 %
Vm Vm
*
2 2R
*
Vm

2

  1.57
Vm 2

I dc
Vdc Vm


R R
Vac
 FF 2  1  1.57 2  1  1.211
Vdc
.
Three-Phase Diode Bridge Rectifier
Waveforms and Conduction Times of Three-Phase Bridge Rectifier
Three-Phase Full-Wave Rectifier
Example 2: A 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) and, (d) Input power factor.
Vdc 
1

Vm sin t dt 


2Vm

0

 190 .956 V
I dc
2Vm

 12 .7324 A
 R
1/ 2
1

2
Vrms    Vm sin t  dt 
  0


Vm
 212 .132 V
2
Pdc
Vdc I dc


 81.06 %
Pac Vrms I rms
Vrms
FF 
 1.11
Vdc
2
2
Vrms
 Vdc2
Vac
Vrms
2
RF 



1

FF
 1  0.482
2
Vdc
Vdc
Vdc
Input power factor =
VS I S cos 
Real Power

1
Apperant Power
VS I S
Alternative! Controlled Switching Mode
• By using linear regulator, the AC to DC converter is not efficient and
of large size and weight!
• Using Switching-Mode
• High efficiency
• Small size and light weight
• For high power (density) applications.
• Use Power Electronics!
Thyristors and Controlled Rectifiers
Controlled Rectifier Circuit
𝑉𝑑𝑐
1
=
2𝜋
𝑉𝑟𝑚𝑠
𝜋
𝛼
1
=
2𝜋
𝑉𝑝 1
=2 𝜋
𝑉𝑝
𝑉𝑝 𝑠𝑖𝑛𝜔𝑡𝑑𝜔𝑡 =
1 + 𝑐𝑜𝑠𝛼
2𝜋
𝜋
𝛼
𝜋−𝛼+
𝑉𝑝2 𝑠𝑖𝑛2 𝜔𝑡𝑑𝜔𝑡
𝑠𝑖𝑛2𝛼
2
1/2
1/2
Example: Consider the following SCR-based variable voltage supply. For RL=240 Ohm, derive the
RMS value of the load voltage as a function of the firing angle, and then calculate the load power
when the firing angle  is 0, /2, and .
Full-Wave Rectifiers Using SCR
𝑉𝑑𝑐
2
=
2𝜋
𝑉𝑟𝑚𝑠
=
𝑉𝑝
2
𝜋+𝛼
𝛼
2
=
2𝜋
2𝑉𝑝
𝑉𝑝 𝑠𝑖𝑛𝜔𝑡𝑑𝜔𝑡 =
𝑐𝑜𝑠𝛼
𝜋
𝜋+𝛼
𝛼
𝑉𝑝2 𝑠𝑖𝑛2 𝜔𝑡𝑑𝜔𝑡
1/2
= 𝑉𝑠
With a purely resistive load, SCRs S1 and S2 can conduct from  to , and SCRs S3
and S4 can conduct from + to 2.