Download Electronics Circuit Design

6. Unregulated Power Supply Design
6.1
6.2
6.3
6.4
Rectifier Circuit Classifications
Half-wave Rectifier Circuits Design
Full-wave Rectifier Circuits Design
Bridge Rectifier Circuits Design
6.5
6.6
6.7
Power Supply Classifications
Design equation of Power Supply with filter capacitor
Power Supply Circuits Design
1
6.1
Rectifier Circuit Classifications
rectifier = diode is used to convert input ac into output dc
220V
50Hz
Im Idc Irms
RS
n:1
Im
D
VSm
RL

• half-wave rectifier circuit
n:1+1
220V
50Hz
RS
Vsm
D1
Vsm
D2
Im Iav Irms
220V
50Hz
Vsm
D4
D2



Im
RL

D1
• full-wave rectifier circuit
RS

D
RS
n :1

D1
Im IdcIrms
D3
RL
• bridge rectifier circuit
D2

Im

D1
D2

D3
D4


2
EE3601-06
Electronics Circuit Design
rms and average (dc) values of rectified sine wave
RL
V  0.7
Im  sm
 Vm  Im  RL  Vsm  0.7  
Rs  RL
Rs  RL
220V
50Hz
• average (dc) value of half-wave rectified sine wave
Im Idc Irms
RS
n:1
D
VSm
RL

V
V
V
1
Vdc  Vav 
Vm sin d  m  cos 0  m 1  1 m
2
2
2


0
Vm=ImxRL
• rms value of half-wave rectified sine wave

1
2


V
1
Vrms 
Vm2 sin 2 d  m  sin 2 d
2

2 0
0




1
2

Vm  1  cos 2


d 
2


2 0




V 
1

 m 0   sin 2  
2  
2
 0 
1
2
Vm 
1


0

sin 2  0
2
2  





half-wave rectified sine wave


Vm 
 1d  cos 2d


2 
0
0

1
 2

1
2

1
 2
Vm 
1
1




0

sin
2


sin 0 


2
2
2 


V  Vm
 m

2
2 
3
EE3601-06
Electronics Circuit Design
n:1+1
• average (dc) value of full-wave rectified sine wave
Vdc  Vav 
Vsm
220V
50Hz

V
1
  Vm 1  1 2Vm
Vm sin d  P  cos 0





RS
Im Iav Irms
RL
Vsm
0
RS
RL
V  0.7
Im  sm
 Vm  Im  RL  Vsm  0.7  
Rs  RL
Rs  RL
n :1
• rms value of full-wave rectified sine wave

220V
50Hz

Vm 
1
2
2
 sin 2 d
Vm sin d 



0
0


Vrms 


1
2
V
1  cos 2d
 m 
2


0



Vm
V   1

 m 0
  sin 2  
2 
2
 0 
1
2
Vm 
1


0

sin 2  0 
2
2 





1
2
 1d  cos 2d

2 0
0


1
 2

1
2
RS
Im Idc Irms
Vsm
RL
RL
V  2  0.7 
Im  sm
 Vm  Im  RL  Vsm  1.4  
Rs  RL
Rs  RL
1
Vm=ImxRL
V 
1
1
 2
 m   0    sin 2  sin 0 
2
2 
2

V  Vm
 m

2
2




full-wave and bridge rectified sine wave
4
EE3601-06
Electronics Circuit Design
rms and average (dc) values of rectified sine wave
wave
Vm
t
Vm
t
Vm
t
Sine wave
Half-wave
Full-wave
rms
dc
Vm
0
2
Vm
2
Vm
Vm
2Vm
2


rms value of the
dc (average)
wave is used to value of the wave
find the heat
is used to find
power such as
the dc power
heater etc.
such as running
the dc motor
etc.
5
EE3601-06
Electronics Circuit Design
6.2
Half-wave Rectifier Circuits Design
• half-wave rectifier circuits
n:1
220V
50Hz
RS
Vsm sin  =
Vsm sin t
Vm=ImxRL
Im Idc Irms
RL




RL
V  0.7
Im  sm
 Vm  Im  RL  Vsm  0.7  
Rs  RL
Rs  RL
• Peak Inverse Voltage (PIV) of the diode
+
220V
50Hz
• PIV of the diode is found across the diode
when the diode is not conducting
• here in half-wave rectifier, PIV = Vsm
Vsm
RL
-
RS
n:1
220V
50Hz
Im
RS
n:1
-
Vsm
Im=0
-
PIV
+
RL
+
6
EE3601-06
Electronics Circuit Design
6.3
Full-wave Rectifier Circuits Design
• full-wave rectifier circuits
n:1+1
220V
50Hz
RS
Vsm
Vsm
Vm=ImxRL
Im Iav Irms
RL

RS
• Peak Inverse Voltage (PIV) of the diode
• PIV of the diode is found across the diode
when the diode is not conducting
• here in full-wave rectifier, if the voltage drop
due to Rs and diode are neglected, PIV = 2Vsm
220V
50Hz



n:1+1 + RS +
V
- sm
+
V
- sm RS
neglected
Im Iav Irms
+
+
PIV
RL
7
EE3601-06
Electronics Circuit Design
6.4
Bridge Rectifier Circuits Design
• bridge rectifier circuits
n :1
220V
50Hz
RS
Vsm
Im Idc Irms
RL
VP

• Peak Inverse Voltage (PIV) of the diode
• PIV of the diode is found across the diode
220V
when the diode is not conducting
50Hz
• here in bridge rectifier, if the voltage drop due
to Rs and diode are neglected, PIV = Vsm

n :1


neglected
RS
+
Vsm
-
-
+
PIV
+
Im Idc Irms
RL
- PIV
8
EE3601-06
Electronics Circuit Design
Summary of Design Equations
Rectifier Circuits
n:1
220V
50Hz
VSm
n:1+1
Im Idc Irms
RS
220V
50Hz
D
RL
RS
Vsm
D1
Vsm
D2
Im Iav Irms
RL
RS
• half-wave rectifier circuit
PIV = Vsm
RL
V  0.7
Im  sm
 Vm  Im  RL  Vsm  0.7  
Rs  RL
Rs  RL
Vm
Vm
• full-wave rectifier circuit
RL
V  0.7
Im  sm
 Vm  Im  RL  Vsm  0.7  
Rs  RL
Rs  RL
n :1
Sine
wave
t
Half-wave
t
Vm
Full-wave
t
Vm
2
0
Vm
2
Vm
Vm
2Vm
2


PIV = 2Vsm
220V
50Hz
RS
Vsm
D4
D2
D1
D3
• bridge rectifier circuit
Im IdcIrms
RL
PIV = Vsm
RL
V  2  0.7 
Im  sm
 Vm  Im  RL  Vsm  1.4  
Rs  RL
Rs  RL
9
EE3601-06
Electronics Circuit Design
Design of half-wave rectifier-1
A half-wave rectifier is to deliver an average voltage of 40.5V to a dc load of
RL=90 from an ac supply of 220V, 50Hz. Design the transformer turn ratio
“n:1” , PIV of the diode and average current rating of the diode. Assume that
the secondary winding resistance of the transformer is 10.
n:1
220V
50Hz
RS 10
V
sm
Vav  Iav  90  40.5V  Iav 
Im Idc Irms
Vdc=
40.5V
RL
90
40.5V
 0.45A
90
I
Iav  0.45A  m  Im  0.45    1.41A

V  0.7V
Im  1.41A  sm
 Vsm  1.41(10  90)  0.7  141.7V
10  90
n 220V(rms )
2 220 2 220 2




 n  1.99  2
1
Vs (rms )
Vsm
141.7
2
PIV  Vsm  141.4V
10
EE3601-06
Electronics Circuit Design
Design of half-wave rectifier-2
A half-wave rectifier is to heat a resistive load of RL=90 with a power of 44.1W
from an ac supply of 220V, 50Hz. Design the transformer turn ratio “n:1” , PIV of
the diode and average current rating of the diode. Assume that the secondary
winding resistance of the transformer is 10.
n:1
220V
50Hz
RS 10
V
sm
2
PL  Irms
 90  44.1W  Irms 
Im Idc Irms
PRL=
44.1W
RL
90
44.1
 0. 7 A
90
I
Irms  m  0.7A  Im  2  0.7  1.4A
2
I
1 .4
Iav  m 
 0.446A


V  0. 7 V
Im  1.4A  sm
 Vsm  1.4(10  90)  0.7  140.7V
10  90
n 220V(rms ) 220 2 220 2



 n  2.21
1
Vs (rms )
Vsm
140.7
PIV  Vsm  140.7V
11
EE3601-06
Electronics Circuit Design
Design of full-wave rectifier-1
A full-wave rectifier is to heat a resistive load of RL=90 with a power of 44.1W
from an ac supply of 220V, 50Hz. Design the transformer turn ratio “n:1+1” , PIV
of the diode and average current rating of the diode. Assume that the secondary
winding resistance of the transformer is 10.
n:1+1
220V
50Hz
RS 10
Vsm
Vsm
RS 10
Im Iav Irms
PRL=
44.1W
2
PL  Irms
 90  44.1W  Irms 
RL
90
44.1
 0.7A
90
I
Irms  m  0.7A  Im  2  0.7  1A
2
2I
2
Iav  m   0.64A


V  0.7V
Im  1A  sm
 Vsm  1  (10  90)  0.7  100.7V
10  90
n 220V(rms ) 220 2 220 2



 n  3.09
1
Vs (rms )
Vsm
100.7
PIV  2Vsm  2  100.7  201.4V
EE3601-06
Electronics Circuit Design
12
Design of full-wave rectifier-2
A full-wave rectifier using a full average current rating of the diode of 5A is to
deliver a heating power of 1KW to a resistive load of RL from an ac supply of 220V,
50Hz. Design the value of RL , transformer turn ratio “n:1+1” , PIV of the diode.
Assume that the secondary winding resistance of the transformer is 1.
n:1+1
220V
50Hz
Idc=5A
Im Irms
RS 1
Vsm
Vsm
RS 1
PRL=
1kW
RL
2Im
I
5 
7.85
 Im 
 7.85A  Irms  m 
 5.55A

2
2
2
1000
2
1KW  PL  Irms
RL  RL 
 32.46
2
5.55
V  0.7V
Im  7.85A  sm
 Vsm  7.85  33.46  0.7  263.36V
1  32.46
5  Iav 
n 220V(rms) 220 2 220 2



 n  1.18
1
Vs (rms)
Vsm
263.46
PIV  2Vsm  2  263.36  526.72V
13
EE3601-06
Electronics Circuit Design
Design of bridge rectifier
A bridge rectifier using a full average current rating of the diode of 5A is to
deliver a heating power of 1KW to a resistive load of RL from an ac supply of 220V,
50Hz. Design the value of RL , transformer turn ratio “n:1” , PIV of the diode.
Assume that the secondary winding resistance of the transformer is 1.
n :1
220V
50Hz
RS
1
Vsm
Idc=5A
Im Irms
1kW
RL
2Im
I
5 
7.85
 Im 
 7.85A  Irms  m 
 5.55A

2
2
2
1000
2
1KW  PL  Irms
RL  RL 
 32.46
2
5.55
V  1.4V
Im  7.85A  sm
 Vsm  7.85  33.46  1.4  264V
1  32.46
5  Iav 
n 220V(rms) 220 2 220 2



 n  1.178
1
Vs (rms)
Vsm
264
PIV  Vsm  264V
14
EE3601-06
Electronics Circuit Design
6.5
Power Supply Classifications
DC Power Supply = filter C is used to smooth the output dc from the rectifier
Vm
Vm Vdc V
n:1
+
220V
50Hz
Vm
C
RL
-

• half-wave power supply circuit
n:1+1
Vm
220V
50Hz
Vm
Vm Vdc V
+
C
RL






Vm

-
• full-wave power supply circuit
n :1
220V
50Hz
Vm Vdc V
Vm
C
• bridge power supply circuit
RL
EE3601-06
Electronics Circuit Design
Vm

15
6.6
Design equation of Power Supply with filter capacitor
Vsm=Vmax
n:1
220V
50Hz
VO V= - m1t1
V= m2t2
V
Vmin
VSm
C
RL
VSm
VO
t1
Disch arg e equation is vc  Vmax e
Disch arg e slope is (m1 ) 
t
RLC
 Vmax  t
vc   Vmax



R
C
t
L 

RL C
m1 
T
(if t  RL C)
m2 
 Vmax
RL C
V t
Vmax
V T
and V  m2 t2  max 2  t2 

2
T
T
Vmax
2
2
 V
But at t  t1   V  m1 t1   max
 RL C
 VRL C
t1  T  t2
or
 VRL C
Vmax

T
V
2 
2 
Vmax
C 
t2
 1
V
  1 
 f 
2Vmax
p 

Vmax
T

 VRL C
  t1  t1 

Vmax

V  T T 
V
  2 
2Vmax
2
Vmax
 1
   V  2Vmax 
 f
p

Vmax
 Capacitor design equation
 VfpRL EE3601-06
Electronics Circuit Design
16




Ripple factor of the dc output voltage
VO
dc
Vmax
Ripple
Vdc V
Vmin
t1
T/2
t
t2
T/2
Ripple factor of the dc output voltage is the amount of rms. voltage of the
ripple with respect to the dc output voltage
V
( ripple )
r  rms
Vdc
ripple is a triangular wave , whose rms . value is
Vp
V
Vmin
 max
3 2 3
2 3
V
Vmin Vmax  Vmax  V  2Vmax  V
Vdc  max


2
2
2
V
V
V
2
 r  rms 


Vdc
2 3 2Vmax  V
3 2Vmax  V
Vrms 

V

17
EE3601-06
Electronics Circuit Design
Summary of Design Equations
Power Supply Circuits
Vm Vdc V
n:1
+
220V
50Hz
Vm
C
RL
-
• half-wave power supply circuit
n:1+1
Vm
220V
50Hz
Vm
Vm Vdc V
+
C
RL
• full-wave power supply circuit
n :1
Vm Vdc V
Vm
Vmax
 Capacitor design equation
 VfpRL
fp  2fs (full  wave )  fs (half  wave )
ripple is a triangular wave, whose rms. value is
-
220V
50Hz
C 
Vrms 
Vp
3

V
 Vmin
 max
2 3
2 3
V
V
 Vmin Vmax  Vmax   V  2Vmax   V
Vdc  max


2
2
2
V (ripple)
r  rms
Vdc
RL
C
• bridge power supply circuit
18
EE3601-06
Electronics Circuit Design
6.7
Power Supply Circuits Design
• Design Example 1
• Draw the full-wave power supply with filter capacitor.
• Design the size of the capacitor for the following conditions.
• Find Ripple factor of the dc output voltage.
• Transformer turn ratio is 1:2 (1:1+1) with 100V, 60Hz. ac at the primary
winding. Load resistor RL=2k. Required minimum output voltage is 70V.
1:2
100V
60Hz
VS 100V
C
VS 100V
RL
VO
n 1 100V 100 2
 

1 2
2Vs
2Vsm
 Vsm  100 2  141.4V, Vmin  70V,
 V  141.4  70  71.4V
fp  2fs  2  60  120 Hz ( full  wave )
C 
Vmax
141.4

 8.23 F
Vfp RL 71.4  120  2  10 3
Vdc 
141  70
 105.5V
2
V  71.4V  ripple rms 
ripple 
71.4
 20.88V
2 3
20.88
 0.198  19.8%
105.5
EE3601-06
Electronics Circuit Design
19
• Design Example 2
• Draw the half-wave power supply with filter capacitor.
• Design the size of the capacitor for the following conditions.
• Transformer turn ratio is 1:1 with 100V, 60Hz. ac at the primary winding.
Load resistor RL=2k. The required ripple factor is 10%
• what is the dc voltage output?
1:1
100V
60Hz
100V
C
RL
Vmax  100 2  141.4V
V
V
V
2
r  rms 


 0.1
Vdc
2 3 2Vmax  V
3 2Vmax  V 
V
3 2  141.4  V
C 

 0.1  V  41.76
Vmax
141.4

 28.2 F
Vfp RL 41.76  60  2  10 3
V
Vmin 141.4  141.4  41.76 
Vdc  max

 120.52V
2
2
20
EE3601-06
Electronics Circuit Design