Download Lecture 7

7. Design of BJT, Op. Amp., IC Regulated
Power Supplies
7.1
Zener diode and Zener Regulator Design
7.2
Zener regulated Power Supply design
7.3
BJT regulated Power Supply design
7.4
Operational Amplifier Regulated Power Supplies
7.5
IC Regulator Design
1
7.1
Zener diode and Zener regulator design
Regulated power supply = output dc is constant (stable) at different loads or
at varying ac supply conditions = battery source characteristics
Zener diode
IZ
IZF
Zener used as
regular diode
VZ
VR
VF
IZmin=0.1I Zmax
Zener used as
reverse
breakdown
voltage diode
0.7V
VZ
IZR
EE3601-07
Electronics Circuit Design
IZmax=PZ/VZ
2
Zener diode design conditions
Ri
VSmin
to
VSmax
IZ
VZ
• Whether input voltage is VSmin or VSmax , the output voltage will be constant
at Zener breakdown voltage VZ
• Constant output voltage is the regulated output voltage and the circuit is
Zener regulator circuit.
• Zener current will becomes less IZmin at VSmin and it will increase to IZmax
at VSmax
• Minimum Zener current IZmin should not less than 10% IZmax to maintain
constant VZ
• Maximum Zener current IZmax should not more than PZ/VZ not to burn the
Zener diode
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Electronics Circuit Design
Zener regulator design equation
Ri
VSmin to V Smax
ILmin to ILmax
VZ
RL
IZmin to IZmax
1. When the load draws more current (ILmax), Zener current will becomes less (IZmin)
and the supply voltage will becomes smaller (VSmin) then:
VS min VZ
VS min VZ  I L max  I Z min Ri or Ri 
I L max  I Z min
2. When the load draws less current (ILmin), Zener current will becomes more (IZmax)
and the supply voltage will increase (VSmax) then:
VS max VZ
VS max VZ  I L min  I Z max Ri or Ri 
I L min  I Z max
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Electronics Circuit Design
3. Then equating the two equations, we have:
VS min VZ
VS max VZ

taking I Z min  0.1 I Z max
I L max  I Z min I L min  I Z max
VS min VZ
VS max VZ
We have

or if simplified ,
I L max  0.1 I Z max
I L min  I Z max
 Ri 
V VS min   IL max VS max VZ
I
I Z max  L min Z
VS min  0.9VZ  0.1VS min

Zener rating design equation
Now I Z maxVZ  PZ
Value of Ri 
VS min VZ
VS max VZ
or Ri 
I L min  I Z max
I L max  I Z min
Power rating of Ri  PRi 
VZ 2
Ri
Because VS max
VZ   VS min VZ

Zener series resistor Ri rating design equation
Ri
VSmin to V Smax
VS max
ILmin to ILmax
VZ
RL
IZmin to IZmax
EE3601-07
Electronics Circuit Design
5
Summary of Design Equations
Zener Regulator
Ri
VSmin to V Smax
ILmin to ILmax
VZ
RL
IZmin to IZmax
Design Ri
Ri 
OR
Design IZmax
VS min  VZ
V
 VZ
 S max
IL max  0.1IZ max IL min  IZ max
V  VS min   IL max VS max  VZ 
I
IZ max  L min Z
VS min  0.9VZ  0.1VS min
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Electronics Circuit Design
• Design Example
• Draw the Zener regulator circuit
• The load current ranges from 100mA to 200mA and the input source voltage
ranges from 14V to 20V. Regulated output voltage is 10V
• Find the range of RL
• Design the required power rating of the Zener PZ
• Design the series resistor Ri and it’s power rating
Ri
14V to 20V
100mA to 200mA
VZ
RL
0.1IZmax to IZmax
Range of RL 
to
10V
 50 
200 mA
10V
 100 
100 mA
VO  VZ  10V
VS min VZ
VS max VZ
14  10
20  10



I L max  0.1 I Z max
I L min  I Z max
200  0.1 I Z max
100  I Z max
1600
 2000  I Z max  400  4 I Z max  I Z max 
 533 mA
3
Power rating of Zener  PZ  VZ I Z max  10  533 mA  5.33W
20  10
Value of Ri 
 15.8 
100  533
Power rating of Ri  PRi
EE3601-07
Electronics Circuit Design

20  10 2

15.8
 6.33W
7
7.2
Zener regulated Power Supply design
Zener regulated Power Supply design (full-wave)
V
n:1+1
Vac
fS
Ri
VS
VS
C
VSR
Full-wave Rectifier with capacitor filter
IL
VZ
RL
Zener regulator
Zener regulated Power Supply (full-wave)
Note that Ri is connected between VSmax and VZ,
Because the Zener voltage is constant V will appears across Ri
V
VZ
 C  S max
 Capacitor new design equation
Vfp Ri
fP = 2fS
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EE3601-07
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Zener regulated Power Supply design (bridge)
n :1
Vac
fS
Ri
VS
IL
RL
VZ
C
V
VZ
 C  S max
Vfp Ri
fP = 2fS
Zener regulator
Bridge Rectifier with capacitor filter
Zener regulated Power Supply (bridge)
Zener regulated Power Supply design (half-wave)
Ri
n:1
Vac
fS
VSmax
IL
VZ
C
V
VZ
 C  S max
Vfp Ri
RL
fP = fS
Half-wave Rectifier with capacitor filter
Zener regulator
Zener regulated Power Supply (half-wave)
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Electronics Circuit Design
7.3
BJT regulated Power Supply design
BJT regulated Power Supply design (full-wave)
n:1+1
Vac
fS

C
Ri
P
V
Z
IB
0.7
R
L
IL
In BJT regulated power supply,
current IL is divided by  to get
Zener output current now shown
as IB
Z
VS min  VZ
VS max  VZ
We have Ri 

taking IZ min  0.1IZ max
I
I
 L max   I



 Z min  L min    IZ max





VS min  VZ
VS max  VZ
Then

or if simplified,
 IL max   0.1I
 IL min   I


Z max
 
  Z max


V  VS min   IL max VS max  VZ 
1 I
IZ max   L min Z

VS min  0.9VZ  0.1VS max
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EE3601-07
Electronics Circuit Design
n:1+1
IL
Vac
fS

C
Ri
IB
0.7
RL
PZ
VZ
IL
REQ=RWorst case
Now the capacitor discharge resistance is not only Ri //( VS / IL ) at the BJT
collector where (VS/IL)<< Ri due to high collector current. Both VS and IL have
(max) and (min).
Designer should consider smallest REQ to satisfy the required VS range. Therefore
VSmin / ILmax is taken as worst case REQ
Note that REQ is across the capacitor and VS. Therefore VSmax is used to find C,
whereas Ri is across (VSmax -VZ) which is used to find C due to Ri
V
VZ
 C  S max
Vfp Ri
C 
V
VS max
where REQ  S min
Vfp REQ
I L max
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Summary of Design Equations
Regulated Power Supplies
Zener Regulated Power Supplies
n:1+1
Ri
Vsmax
Vac
fs
Vsmax
C
V
IL
PZ
VZ
RL
BJT Regulated Power Supplies
n:1+1
IL
Vac
fS

C
Ri
IB
PZ
VZ
0.7
RL
IL
REQ=RWorst case
V
 VZ
C  S max
 VfpRi
Ri 
fp  2fs (full  wave )  fs (half  wave )
VS min  VZ
V
 VZ
 S max
IL max  0.1IZ max IL min  IZ max
Design Ri
Design IZmax
C 
Ri 
V
VS max
where REQ  S min
 VfpREQ
IL max
VS min  VZ
VS max  VZ

 IL max   0.1I
 IL min   I



Z
max

  Z max


Design Ri
Design IZmax
12
EE3601-07
Electronics Circuit Design
• Design Example
Draw and Design a full-wave Zener regulated power supply using a 8:1+1 center-tapped
transformer and an 8V, 1W Zener diode that will provide a constant 8V to a load varying from 200
to 500. The input to the transformer is 120V,60Hz.,and ignore transformer losses and the diode
drop. Determine:(a)IZmax and IZmin (b) R i and VSmin (c)Size of capacitor needed (d) Percent regulation
if RZ=2
8:1+1
IL
Ri
15V
120V
60Hz
C
8V
20015V
VSR
R
L
1W
500
( a ) 1W  8V  I Z max  I Z max  1 / 8  125 mA  I Z min  0.1  125  12.5 mA
( b ) I L max  8V / 200   40 mA & I L min  8V / 500   16 mA
Ri 
VS max VZ
15 2  8

 93.6 
I L min  I Z max
16  125
Ri  93.6  
VS min VZ
 VS min  ( 93.6  40  0.1  125 )  8  12.9V
IL max  0.1 IZ max
( c ) V  VS max VS min  21.2  12.9  8.3V
V
VZ
21.2  8
 C  S max

 142 F
Vfp Ri
8.3  120  93.6
( d ) % Re g 
I Z max
 I Z min RZ
0.125  0.0125 2   2.81%

VZ
8
13
EE3601-07
Electronics Circuit Design
• Design Example
Draw and Design a full-wave BJT regulated power supply using a 3:1 center-tapped
transformer and an 12V Zener diode .The load current varies from 400 to 500mA. The input to the
transformer is 120V,60Hz., Take V = 30% and  = 100. Determine: (a) IZmax (b) R i (c) Size of
capacitor needed (d) Percent regulation if RZ=2
3:1+1
120V
60Hz.

Ri
C
IB
VZ=12V
VS max 
RL
IL
120 1
  2  28.28  VS min  28.28  0.3  28.28  19.8V
3
2
IZ max 

0.7
1 IL min VZ VS min   IL max VS max VZ


VS min  0.9VZ  0.1VS max

1
400 12  19.8   500 28.28  12 

 8.15 mA
100
19.8  0.9  12  0.1  28.28
V
V
19.8
28.28
REQ  S min 
 39.6  C  S max 
 701 F
IL max 0.5 A
Vfp REQ ( 28.28  19.8 )  120  39.6
Ri 
VS max VZ
28.28  12

 1.34 k
400
 I L min 

8
.
15
100

   I Z max

(d) % Re g 

IZ max
 IZ min RZ
VZ


8.15  0.815mA  2  0.0012%
12EE3601-07
Electronics Circuit Design
14
7.4
Operational Amplifier Regulated Power Supplies
(detail of Op. Amp. in Chapter 9)
+Vcc
Inverting
Input (-)
Non Inverting
Input (+)
ii = 0
v+ = v-
ii = 0
Properties of Op. Amp.
(1) ii = 0 (2) vi = 0
+
output
R0= 0
-Vcc
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Electronics Circuit Design
(a) Small IL less than 25mA
+Vcc
+Vcc
Ri
IZ
VZ
Ii=0 v+
+Vcc
Io=25mA(max)
Ri
v-Vcc
Ii=0
IZ
IL
+
RL
v+ = v-
VL
VZ
Rv
R2
R1
-
Op. Amp. regulator (direct fixed output)
v+
+Vcc
Io=25mA(max)
v-Vcc
Ii=0
IL
+
RL
v+ = v-
VL
-
Op. Amp. regulator (variable output with Rv)
IZ  50% IZ max
IZ  50% IZ max
IL  Io  25mA(max)
VL  v  v  VZ
Ii=0
IL  Io  25mA(max)
R
VL  v   v   VZ 1
Rv
Rv 
EE3601-07
Electronics Circuit Design
VZ
10%IZ
Variable R1
16
Example: Draw a 0 to 12V regulated output voltage Op. Amp. power supply.
The load current is IL< 25mA PZ =0.5W and VZ=12V. Design the values of Rv,
Ri, RL(min) if supply VCC=16V.
+Vcc
Ri
Ii=0
IZ
VZ
Rv
v+
+Vcc
Io=25mA(max)
v-
R2
-Vcc
R1
IL
Ii=0
+
RL
v+ = v-
VL
-
Op. Amp. regulator (variable output with Rv)
P
0.5
IZ  0.5 Z  0.5
 0.021A
VZ
12
V  VZ 16  12
Ri  cc

 190.5
IZ
0.021
Rv 
VZ
12

 5.714  5k
10%IZ 0.1  0.021
VZ  v   v   VL
 RL (min) 
VL
IL (max)

12
 480
25mA
17
EE3601-07
Electronics Circuit Design
(b) Large IL using BJT output
+Vcc
IL
Ri
IZ  50% IZ max
VZ
Rv 
VZ
10%IZ
Ii=0
IZ
Rv
R2
R1
25mA(max)
v+
Io=IL /
v-Vcc
Ii=0
IL  Io
IL
RL
v+ = v-
Op. Amp. regulator (BJT output)
+
VL
-
Variable R1
R
VL  v   v   VZ 1
Rv
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EE3601-07
Electronics Circuit Design
Example: Draw a 0 to 12V regulated output voltage Op. Amp. power supply.
The load current is IL= 100mA PZ =0.5W and VZ=12V. Design the values of
Rv, Ri, RL (min) and (min) if supply VCC=16V. Find actual Io if =50 for the
present BJT . +Vcc
IL
Ri
Ii=0
IZ
VZ
Rv
R2
R1
25mA(max)
v+
Io=IL /
v-Vcc
Ii=0
IL
RL
v+ = v-
+
VL
-
Op. Amp. regulator (BJT output)
VZ  v   v   VL
P
0.5
IZ  0.5 Z  0.5
 0.021A
VZ
12
V  VZ 16  12
Ri  cc

 190.5
IZ
0.021
Rv 
 RL (min) 
VL
IL (max)

12
 120
100mA
I
100mA
min  L 
4
Io
25mA
VZ
12

 5.714  5k
10%IZ 0.1  0.021
I
100mA
Actual Io  L 
 2mA

50
EE3601-07
Electronics Circuit Design
19
(c) Zener voltage multiplier BJT output
+Vcc
IL
Ri
Ii=0
VZ
10%IZ
Io=IL /
v-
IZ  50% IZ max IZ
Rv 
v+
25mA(max)
VZ
IL
-Vcc
Rv
R1
Ii=0
Variable R1
R
VZ 1  v  v  VR3
Rv
IL /   Io
R4
+
 R3  R4 

  VR3
V

VL L  R

3


R3
v+ = v-
-
IR4  IR3  10%IL
Op. Amp. regulator
(Zener voltage multiplier BJT output)
20
EE3601-07
Electronics Circuit Design
Example: Draw a 0 to 12V regulated output voltage Op. Amp. power supply.
The load current is IL= 100mA PZ =0.5W and VZ=9V. Design the values of Rv,
Ri, RL (min) and (min) if supply VCC=16V. Find actual Io if =50 for the
present BJT .
+Vcc
IL
Ri
Ii=0
v+
Io=IL /
v-
IZ
25mA(max)
VZ
P
0.5
IZ  0.5 Z  0.5
 0.028A
VZ
9
V  VZ 16  9
Ri  cc

 250
IZ
0.028
IR 4  IR3  10%IL  0.1  100  10mA
9
VR3  IR3  R3  R3 
 9k
10mA
IL
-Vcc
Rv
R1
Ii=0
+
R4
R4 
12  9
 3k
10mA
VL
R3
v+ = v-
-
VL
12
Op. Amp. regulator
RL (min) 

 120
I
(max)
100
mA
L
(Zener voltage multiplier BJT output)
IL 100mA
VZ
9



4
min
Rv 

 3.214  5k(s tan dard)
I
25
mA
o
10%IZ 0.1  0.028
I
100mA
Actual Io  L 
 2mA

50
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Electronics Circuit Design
(d) Output current limited Op. Amp voltage regulator
+Vcc
IL
Ri
IZ  50% IZ max
VZ
v+
v-
IZ
Rv
Io=(IL /1IC2
IC1
IC2
R1
V
RE1  BE2
IC1
Io=(IL /1
+
VBE2 RE1
-
IL Limited current
IL=IC1
+
RL
VZ
Rv 
10%IZ
I

VBE2  26mV  ln  C2 
 10fA 
Op. Amp. regulator (current limit)
VL
-
Variable R1
R
VL  v   v   VZ 1
Rv
IC2  Io (max)  IB1 (max)
 25mA  IB1 (max)
I (max)
 25mA  L
1
EE3601-07
Electronics Circuit Design
22
Example: Draw a 0 to 12V regulated output voltage Op. Amp. power supply.
The load current is limited to IL= 500mA PZ =0.5W and VZ=12V. Design the
values of Rv, Ri, RE1 if supply VCC=16V. Find IC2(max) if 1=100.
+Vcc
IL
Ri
v+
v-
IZ
P
0.5
IZ  0.5 Z  0.5
 0.021A
VZ
12
V  VZ 16  12
Ri  cc

 190.5
IZ
0.021
Rv 
VZ
Io=(IL /1IC2
Rv
R1
IC1
IC2
+
VBE2 RE1
-
IL
+
RL
Op. Amp. regulator (current limit)
VL
-
VZ
I (max)
12
500mA

 5.714  5k IC2  Io (max)  IB1 (max)  25mA  L
 25mA 
 20mA
1
100
10%IZ 0.1  0.021
 20  103 
0.736V
 20mA 
VBE2  26mV  ln 
 26mV  ln 
 1.47 
  0.736V  RE1 


15
500mA
 10fA 
 10  10

I
500
Io  L  IC2  IC2 (max)  Io (max) 
 25  5  20mA
1
100
(It can draw 25mA but actually it draws 20mA OK)
EE3601-07
Electronics Circuit Design
23
IC Regulator Design (Fixed Voltage )
7.5.1
Fixed Positive Voltage Regulator IC
n:1+1
Vac=220V
f=50Hz
Idc
Vmax
+Vi
C1
Vmax
Voltage regulator IC
1
7812
C2 is a high frequency
filter (C2 < 0.1F)
Regulated power supply +12V
C2
RL
3
Unregulated power supply
C1 is a ripple filter
capacitor (C1 > 100F)
2
Vo=+12V
Vdc
Vm
Vdc








IC Part Number
Output Voltage
Min input Voltage
Max input Voltage
7805
+5V
+7V
+25V
7806
+6V
+8V
+25V
7808
+8V
+10.5V
+25V
7810
+10V
+12.5V
+28V
7812
+12V
+14.5V
+30V
7815
+15V
+17.5V
+30V
7818
+18V
+21V
+33V
7824
+24V
+27V
+38V
24
EE3601-07
Electronics Circuit Design
Fixed Negative Voltage Regulator IC
n:1+1
Vac=220V
f=50Hz
Idc
Vmax
C1
Vmax
-Vi
Voltage regulator IC
1
7912
C2 is a high frequency
filter (C2 < 0.1F)


Regulated power supply -12V
C2
RL
3
Unregulated power supply
C1 is a ripple filter
capacitor (C1 > 100F)
2
Vo=-12V



-Vdc
-Vm



-Vdc
IC Part Number
Output Voltage
Min input Voltage
Max input Voltage
7905
-5V
-7V
-25
7906
-6V
-8V
-25
7908
-8V
-10.5V
-25
7910
-10V
-12.5V
-28
7912
-12V
-14.5V
-30
7915
-15V
-17.5V
-30
7918
-18V
-21V
-33
7924
-24V
-27V
-38
25
EE3601-07
Electronics Circuit Design
7.5.2
IC Regulator Design (Adjustable voltage)
The IC regulator LM317 (LM318 for negative voltage) can be used to produce any
regulated output voltage between 1.2V to 37V. Typical IC values are: Vref = 1.25V
and Iadj = 100A.
Adjustable Voltage regulator IC
n:1+1
Vac=220V
f=50Hz
Vmax
-Vi
Vmax
C1
Idc
VIN
1.2V < Vo< 37V
LM317
VOUT
ADJ
Iadj = 100A
Vref =
1.25V

R1

R 
Vo  1.25V1  2   100A  R2
R1 

R2

V


R 
Vo VR1  VR2  Vref  IR1  Iadj R2  Vref   ref  R2  Vref  1  2   Iadj R2
R1 
 R1 

Current through R1 and R2 can be taken for the design as 100Iadj=10mA
Therefore R1 design will be R1 = 1.25V/10mA = 0.125k=125
26
EE3601-07
Electronics Circuit Design
Summary of Design Equations
IC Regulators
-Vi
ILmax
Fixed Voltage IC
1
C1
7805
VSmin
Variable Voltage IC
Vo=-5V
ILmax
2
C1 LM317
VSmin
3
VOUT
ADJ
Iadj = 100A
Min input Voltage
Max input Voltage
-7V
-25
Vs max 
max V  min V
2
Vref =
1.25V
R1
R2

R 
Vo  1.25V1  2   100A  R2
R1 

V  Vs max  Vs min
V
REQ  S min
IL max
IR1=10mA
C1 
VS max
 VfpREQ
27
EE3601-07
Electronics Circuit Design
Example: Draw and design a -12V IC regulated bridge power supply using a MC7912
voltage regulator. Minimum and maximum rated voltages of the IC are -14.5V and -30V
respectively. Select the transformer turns ratio “n” , and the capacitor value using an
input of 115V at 60 Hz and an output current of 100mA to 500mA.
a :1
Vac
fS
-Vi
VS
Voltage regulator IC
1
7912
Vo=-12V
2
C
3
Vs max 
max V  min V
2

 30  ( 14.5)
2
 22.3V 
115 2
 7.29  n
22.3
V  Vs max  Vs min  22.3  (14.5)  7.8V
REQ 
VS min
VS max
14.5
 22.3

 29  C 

 821F
IL max
0.5A
 VfpREQ  7.8  120  29
28
EE3601-07
Electronics Circuit Design
Example:
Draw and deign the LM317 IC regulator circuit values of transformer turn ratio
“n”, C1, R1, and R2 if the maximum load current IL is 100mA, and if a ripple of peakto-peak voltage of V=2V is present at the input of the regulator. The output
voltage of the variable IC regulator is 14V dc. Take 220V ac supply at 50Hz. and
Vsmin = 15V
n:1+1
Vac=220V
f=50Hz
Idc
Vmax
Vmax
C1
VIN
LM317
VOUT
ADJ
Vs max  Vs min  V  15  2  17V 
REQ 
220 2
 18.3  n
17
VS min
VS max
15
17

 150  C1 

 566F
IL max
0.1A
 VfpREQ 2  100  150
Iadj = 100A
Vref =
1.25V
R1
R2
Current through R1 and R2 can be taken for the design as 100Iadj=10mA
Therefore R1 design will be R1 = 1.25V/10mA = 0.125k=125
Vo  14V  1.25V  10mA  100A  R2
R2 
14  1.25
 1.26k
10.1mA
29
EE3601-07
Electronics Circuit Design