This document describes the design of a series voltage regulator circuit. It begins with an overview of series voltage regulators and their components. It then provides the specifications for a design example, including an output voltage of 25V and maximum load current of 1A. The document proceeds through each step of the design process, selecting components like transistors, a zener diode reference, and resistor values. It concludes by showing the completed circuit design that meets the given specifications.

1. 1
Design of
Series Voltage
Regulator
V. R. Gupta
Assistant Professor
Electronics Engineering, RCOEM, Nagpur.
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2. 2
Learning Objectives
• To understand the working principle of Series Voltage
Regulator.
• To understand the design process of SVR.
• To design Series Voltage regulator to meet the given
specification.
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3. 3
Series Voltage Regulator
Active Resistor
Error Amplifier
or Differential
Amplifier
Reference
Voltage
Sampling
Network
Vo
Unregulated
supply
Vin
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4. 4
Series Voltage Regulator
Fig: A semiconductor- regulated power supply. The series pass element is Q1, the
difference amplifier is Q2, and the reference voltage is obtained from Zener diode
D.
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5. 5
Analysis of Series Voltage
Regulator
Fig: Analysis of series voltage regulator
2
2
1 2
B o o
R
V V bV
R R
 

2 2 2BE B E o zV V V bV V   

6. 6
Analysis of Series Voltage
Regulator
Fig: Analysis of series voltage regulator
1 2o R RV V V 
1
2
1 2
o o B
R
V V V
R R
 

1
2
1 2
o o Z BE
R
V V V V
R R
  


7. 7
Analysis of Series Voltage
Regulator
Fig: Analysis of series voltage regulator
1
2
1 2
1o Z BE
R
V V V
R R
 
    
2
2
1 2
o Z BE
R
V V V
R R
 
   
 1
2
2
1o Z BE
R
V V V
R
 
   
 

8. 8
Analysis of Series Voltage
Regulator
Fig: Analysis of series voltage regulator
 1
2
2
1o Z BE
R
V V V
R
 
   
 
1CE in oV V V 
inV 
When
 oV   obV  2BV   2 2 2( )BE B EV V V  

2BI 2 2C BI I 1BI 1CEV 1( )o in CEV V V  

9. 9
Design Steps for Series
Voltage Regulator1) Selection of Transistor Q1.
2) Selection of reference element (Zener Diode).
3) Selection of transistor Q2.
4) Selection of Current limiting resistor RD.
5) Selection of Sampling elements (R1 & R2)
6) Selection of Current limiting resistor R3
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10. 10
Design of Series Voltage
Regulator
Design a series-regulated power supply to provide a
nominal output voltage of 25 V and supply load current
IL ≤ 1A. The unregulated power supply has the
following specifications:
Vi = 50 ± 5V and ro or Rs = 10 Ω.
Assume, hfe1 = 100, hFE1 = 125 and
hfe2 = 200, hFE2 = 220
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11. 11Design of Series Voltage Regulator
Given Specifications:
Output voltage Vo = 25 V and
Supply load current IL ≤ 1A.
The unregulated power supply has Vi = 50 ± 5V and
ro or Rs = 10 Ω.
Assume, hfe1 = 100, hFE1 = 125 and
hfe2 = 200, hFE2 = 220
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12. 12Design of Series Voltage Regulator
Solution:
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13. 13Design of Series Voltage Regulator
Solution:
1) Selection Series Pass Transistor Q1
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1(max) (max) 55 25 30CE in oV V V V    
1(max) (max) 1C LI I A 
1(max) 1(max) 1(max) 30D CE CP V I W  

14. 14Design of Series Voltage Regulator
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15. 15Design of Series Voltage Regulator
Solution:
1) Selection of Series Pass Transistor Q1
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1(max) (max) 55 25 30CE in oV V V V    
1(max) (max) 1C LI I A 
1(max) 1(max) 1(max) 30D CE CP V I W  
Therefore, if we select a Texas Instruments 2N1722 silicon power
transistor for Q1, we measure at IC1 = 1A the following
parameters:
hfe1 = 100 (ac current gain)
hFE1 = 125 (dc current gain)

16. 16Design of Series Voltage Regulator
Solution:
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17. 17Design of Series Voltage Regulator
Solution:
2) Selection of Zener Diode (reference element)
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We assume that approximately 50% of the voltage VB1 appears
across Q2 to keep it in the active region and the remaining 50%
appears across Zener diode.
Therefore,
10.5 0.5*25.6 12.8 15Z BV V V V   
1 1 0.6 25 25.6B BE oV V V V    

18. 18
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19. 19Design of Series Voltage Regulator
Solution:
2) Selection of Zener Diode (reference element)
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We assume that approximately 50% of the voltage VB1 appears
across Q2 to keep it in the active region and the remaining 50%
appears across Zener diode.
Therefore,
Therefore we select Two 1N755 Zener diodes connected in series
to provide the reference voltage of 15 V.
10.5 0.5*25.6 12.8 15Z BV V V V   
1 1 0.6 25 25.6B BE oV V V V    
(min)6 6 12 and 20z zr I mA      

20. 20Design of Series Voltage Regulator
Solution:
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21. 21Design of Series Voltage Regulator
Solution:
3) Selection of Transistor Q2.
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Since Q2 is employed as a difference amplifier, it is a low power
transistor with high current gain.
From the circuit diagram, choose IC2 ≈ IE2 = 10mA

22. 22
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23. 23Design of Series Voltage Regulator
3) Selection of Transistor Q2.
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Since Q2 is employed as a difference amplifier, it is a low power
transistor with high current gain.
From the circuit diagram, choose IC2 ≈ IE2 = 10mA
The Texas Instruments 2N930 silicon transistor can provide the
collector current of 10 mA.
The manufacturer specifies
(m ) (m )30 and 45C ax CE axI mA V V 
Let us consider IC2 = 10mA and do the further calculations based
on this value.
2 2220 and 200 @ 10FE fe Ch h I mA  

24. 24Design of Series Voltage Regulator
Solution:
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25. 25Design of Series Voltage Regulator
4) Selection Current limiting resistor RD
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Since, the minimum current flowing through Zener diode is
Iz(min) = 20 mA and the minimum current flowing through the
collector terminal of Q2 is 10mA, therefore
Let us choose ID = 10mA.
Therefore, 25 15
1
10
o z
D
D
V V
R K
I mA
 
   
Power dissipation in RD, PDR = (ID)2 RD = 0.1 W

26. 26
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27. 27Design of Series Voltage Regulator
4) Selection Current limiting resistor RD
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Since, the minimum current flowing through Zener diode is
Iz(min) = 20 mA and the minimum current flowing through the
collector terminal of Q2 is 10mA, therefore
Let us choose ID = 10mA.
Therefore, 25 15
1
10
o z
D
D
V V
R K
I mA
 
   
Power dissipation in RD, PDR = (ID)2 RD = 0.1 W
Therefore, select a resistor RD = 1K of 1/8 W

28. 28Design of Series Voltage Regulator
Solution:
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29. 29Design of Series Voltage Regulator
5) Selection of Sampling Elements R1 and R2
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Each resistor is determined as follows:
2
2
2
10
45
220
C
B
FE
I mA
I A
h
  
 1
2
2
Since, 1o Z BE
R
V V V
R
 
   
 
 
1
2 2
, 1o
Z BE
R V
We get
R V V
 

Since we require 2 1BI I=
We select I1 = 10 mA

30. 30Design of Series Voltage Regulator
5) Selection of Sampling Elements R1 and R2
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Since we require 2 1BI I=
Therefore,
Then, voltage at base of Q2 is
2 2 0.6 15 15.6B BE ZV V V V    
2
1 3
1
25 15.6
940
10 10
oV V
R
I 
 
   

We select I1 = 10 mA

31. 31Design of Series Voltage Regulator
5) Selection of Sampling Elements R1 and R2
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Since, 2 1BI I=
Then, the resistor is determined by
2
2 3
1
15.6
1560
10 10
V
R
I 
   

Therefore I1 ≈ I2
Thus,
1 2940 and 1.5R R K   

32. 32Design of Series Voltage Regulator
6) Selection of Current limiting resistor R3
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The current I3 flowing through resistor R3 is
3(min) 1(max) 2(min)B CI I I 
2(min)Sinc 0e 1, CI mA
1(max) 1
1(max)
1 1
1000 10 10
125
8
and, C L D
B
FE FE
I I I I
I
h h
mA
 
 
 



33. 33Design of Series Voltage Regulator
6) Selection of Current limiting resistor R3
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The current I3 flowing through resistor R3 is
3(min) 1(max) 2(min)B CI I I 
3(min) 8 10 18I mA mA mA  
The value for resistor R3 corresponding to
Vi(min) = 45V and IL = 1A is given by
(min) 1 (min) 1
3
3 3
( )in B in BE oV V V V V
R
I I
  
 

34. 34Design of Series Voltage Regulator
6) Selection of Current limiting resistor R3
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The value for resistor R3 corresponding to
Vi(min) = 45V and IL = 1A is given by
(min) 1
3
3
3
3
( )
45 25.6
18 10
1.077
in BE oV V V
R
I
R K

 




 
Thus, the value for resistor R3 = 1.1 kΩ (Standard Value.)

35. 35
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36. 36
Fig: Series Voltage Regulator designed to meet the given specification.
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37. 37
Reference
Chapter – 18, Page no. 699 -705
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38. 38
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39. 39
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