Transcript Lecture6 - Faculty Of Engineering And Technology

Lecture 9
Controlled Rectifiers
The Controlled Half-wave Rectifier
• Normal rectifiers are considered as uncontrolled
rectifiers.
• Once the source and load parameters are
established, the dc level of the output and power
transferred to the load are fixed quantities.
• A way to control the output is to use SCR instead of
diode. Two condition must be met before SCR can
conduct:
– The SCR must be forward biased (VSCR>0)
– Current must be applied to the gate of SCR
Controlled, Half-wave R load
• A gate signal is
applied at t = ,
where  is the
delay/firing angle.
I o, rms 
Vo , rms
R
Vs

2R
Example
• Design a circuit to produce an average voltage
of 40V across 100 load resistor from a 120Vrms
60 Hz ac source. Determine the power absorbed
by the resistor and the power factor.
Briefly describe what happen if the circuit is
replaced by diode to produce the same average
output.
Example (Cont)
• Solution
In such that to achieved 40V
average voltage, the delay angle
must be
Vs
[1  cos ]
2
120 2
40 
[1  cos ]
2
  61.2o  1.07 rad
Vo 
Vo , rms 
Vm
 sin(2 )
1 
2

2
120 2
1.07 sin 2(1.07)
1

2

2
 75.6V

V 2 rms 75.62
P

 57.1W
R
100
pf 
57.1
 0.63
 75.6 
(120)

 100 
• If an uncontrolled diode is used,
the average voltage would be
Vo 
Vs


2 (120)

 54V
• That means, some reducing
average resistor to the design must
be made. A series resistor or
inductor could be added to an
uncontrolled rectifier, while
controlled rectifier has advantage
of not altering the load or
introducing the losses
Controlled, Half-wave R-L load
• The analysis of the circuit is very
much similar to that of uncontrolled
rectifier.
  t 
 V 


m
 [sin(wt   )  sin(   )e   for   t  
i ( wt )   Z 
0
otherwise

L
 L 
and Z  R 2  (L) 2 ,   tan1 
,  
R
 R 
rm s current,
I rms
1

2

2
1 2
i
(

t
)
d
(

t
)

i (t ) d (t )


2 
2
and, averagecurrent,

1
Io 
i (t ) d (t )

2 
Controlled, Half-wave R-L
load
The averageoutputvoltage,
Vm
[cos  cos  ]
1 Vm sin(t )dt 
Vo 
2

2 

The average power absorbed by load,
2
P  I rms
R ;
Controlled full-wave rectifiers
Resistive
load：
1 
Vo   Vm sin( wt )d ( wt )
 
Vm

( 1  cos  )

Io 
I rms
  delay angle
Vo Vm

(1  cos  )
R R
1

Vm
2

(
sin
wt
)
d ( wt )
  R

Vm 1  sin(2 )


R 2 2
4
The power delivered to the load
P  I 2 rm sR
The rms current in source is the same as the rms
the load.
current in
Full wave Controlled Rectifier
with RL load
Discontinuous and Continuous
Operations
Discontinuous Mode
discontinuous current :
io ( wt ) 

Vm
sin( t   )  sin(    )e ( t  ) /(  )
Z

for
  t  
Z  R 2  ( L )2
  tan1 (
L
)
R
, L
R
For discontinuous current
  
Analysis of the controlled full-wave rectifier operating in the
discontinuous current mode is identical to that of the controlled
half-wave rectifier, except that the period for the output current
is  .
Continuous Mode
continuous current
wt    
, i(   )  0
sin(     )  sin(   )e  (   ) /( )

sin(   ) 1  e  /( )
sin( -  )  0
( - )  0
    Tan (
-1
L

0
0

v0 ( wt )  Vo   Vn cos(nwt  n)
n 1
Vo 
)
R
for continuous current
1
 
 
Vm sin wt d ( wt ) 
Vn  an  bn
2
n  Tan -1 (
 bn
)
an
2Vm

cos
2
an 
2Vm  cos(n  1) cos(n  1) 


  n 1
n  1 
bn 
2Vm  sin(n  1) sin(n  1) 

  n  1
n  1 
n  2,4,6,....
In  Vn
Zn
 Vn
Irm s  Io 
2
| R  jnwL |


(
n  2 ,4...
Io  Vo
R
In
2
)2
R-L Source load ：
Fig.4-14
The SCRS may be turned on at any
forward biased, which is at an angle
  sin 1 (VdcVm )
time that they are
For continuous current case, the average bridge output voltage is
Vo 
2 Vm
cos 

average load current is
Io 
Vo  Vdc
R
The ac voltage terms are unchanged from the controlled rectifier
with an R-L load. The ac current terms are determined from
circuit.
Power absorbed by the dc voltage is
Pdc  Io Vdc
Power absorbed by resistor in the load is
P  I 2 rmsR  Io2 R if
L is l arge
Controlled Single-phase converter operating as
inverter
an
Average DC output voltage
Vd 
2 2

Vs cos 
• Assuming AC side inductance is zero
• Note that output voltage can go negative for
alpha > 90 degrees.
• This means negative power flow or inversion
Copyright © 2003
by John Wiley & Sons, Inc.
Chapter 6 Thyristor Converters