Transcript 2_Rectifiers

Author: Ales Havel
E-mail: [email protected]
Phone number: 4287
Headquarters: E227
Web page: http://homen.vsb.cz/~hav278/
POWER SEMICONDUCTOR SYSTEMS I
Presentation contents

Main types of converters

AC-DC converters

Single phase rectifiers





Construction possibilities, wiring
Load voltage and current waveforms
Mathematics
Neat examples
Three phase rectifiers




Construction possibilities, wiring
Load voltage and current waveforms
Mathematics
Neat examples
DC-DC converters
 DC-AC converters
 AC-AC converters

Power converters

A power semiconductor converter is an electrical
device for converting electrical energy.
Division of AC/DC converters

Properties of rectifiers

Number of input phases




Number of pulses


Single phase rectifiers
Three phase rectifiers
n – phase rectifiers
P = 1, 2, 3, 6…
Possibilities of output voltage control

Without control


Half-controlled rectifiers


Diodes + Thyristors
Full-controlled rectifiers


Diode rectifiers
Thyristors
Wiring


Bridge rectifiers
Node rectifiers
Half-wave single phase diode rectifiers
Voltage
Resistive load
Current
Voltage
Inductive load
Current
Half-wave single phase diode rectifiers
Voltage
R+L load
Current
R+L+D0 load
Voltage
Current
Half-wave single phase diode rectifiers
Voltage
R+C load
Current
R+V load
Voltage
Current
Two pulse diode rectifiers
Voltage
Current
Two pulse thyristor controlled rectifiers
Important mathematical relations
The output AVG voltage of diode rectifier or thyristor rectifier with control angle α = 0°:
U L0  U m 
p

 sin

p
The output AVG voltage of thyristor rectifier with control angle α.
U L  U L0  cos 
Condition 1: The load current must be continuous
Condition 2: No parallel bypass diode on the load
Output to Input phase RMS voltage ratio
p[-]
UL / U1 [-]
1
0.45
2
0.90
3
1.17
6
2.34
Limit of load current continuity for resistive load
p[-]
α[°]
3
30
6
60
Example 1- submission
Calculate the output AVG voltage of the 2-pulse
bridge controlled rectifier, if you know the
following values:
α = 60°, U1 = 230V
Example 1- solution
a) General calculation approach:
2
UL 
T

 

2U m
U m sin(t )dt 
2
 

sin(t )dt 
Um

 
  cos(t )


U m 
2 U1 230 2
4
  U m  1  1   U m
cos

cos






 103.5V


 

 
3
3
  2  2  


b) Calculation utilizing the basic relation:
U L  Um 
p

 sin

p
 cos  
2 2 U1

 sin

2
c) Calculation utilizing the table values:
U L  0.9  U1  cos

3

0.9  230
 103.5V
2
 cos

3

2 U1


230 2

 103.5V
Example 1- question?
How will change the output AVG voltage value if we use the bypass diode parallel to the load?
Solution:

U L  BD



2U m
Um
2
  U m sin(t )dt 
sin(

t
)
d

t

  cos(t ) 
T
2 


U m 
  U m 
1  3U m 3 2 U1 3  230 2
cos


cos


1



 155V




 
3
 
2  2
2
2
Six pulse thyristor controlled rectifiers
Six pulse thyristor controlled rectifiers
Thank You for your attention