Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
Chapter-7
Astable Multivibrators
1. For the multivibrator in Fig.7p.1, R 1 = R 2 = R = 47 k  , C 1 = C 2 = C = 0.01 μF.
Find the time period and frequency.
Fig.7p . 1 Un-symmetric astable multivibrator
Solution:
Given R 1 = R 2 = R = 47 k  ,C 1 = C 2 = 0.01 µF.
This is a symmetric astable multivibrator.
T  1.38RC  1.38  47  103  0.01  106  0.648 ms
1
1
 1.54 kHz.
f  
T 0.648  103
2. For the astable multivibrator in Fig.7p.1, R 1 = 20 k  , R 2 = 30 k  ,C 1 = C 2 = C =
0.01 μF. Find the time period, duty cycle and the frequency.
Solution:
This is an un-symmetric astable multivibrator.
T2  0.69 R1C1  0.69  20  103  0.01  106  0.2 ms
T1  0.69 R2 C2  0.69  30  103  0.01  106  0.138 ms
T  T1  T2  0.138  0.2  0.338 ms
T
0.138
 100 per cent  59.17 per cent
per cent D  1  100 per cent 
T
0338
1
1
f  
 2.95 kHz.
T 0.338  103
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Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
3. For the symmetric astable multivibrator that generates square waves with vertical
edges shown in Fig.7p.3, V CC =10 V, R C = R 3 = 2 k  , R 1 = R 2 = 20 k  , C = 0.1  F ,
h FE(min) = 30. Show that the ON device is in saturation. Also find f. Assume suitable
values for V CE(sat) and V BE(sat). Si transistors are used.
Fig.7p.3 Astable multivibrator with vertical edges
Solution:
Assume Q 1 is OFF and Q 2 is ON and in saturation. If Q 2 is ON and in saturation, for
silicon transistors, V C2 =V CE(sat) = 0.2 V, V B2 = Vσ = 0.7 V then D 2 is ON. The collector
load is R 3 //R C .
2 2
 1 kΩ
RC' = R 3 //R C =
4
VCC  VCE (sat) 10  0.2
9.8
IC 2 


 9.8 mA
'
1 kΩ
RC
1  103
V  V 10  0.7
9.3
I B 2  CC


 0.465 mA
R
20 kΩ
20  103
I
9.8 mA
I B 2min  C 2 
 0.326 mA
hFE min
30
I B2 >> I B2min
Hence Q 2 is in saturation.
To find f:
0.7
0.7

 350 Hz
For a symmetric astable multivibrator: f =
3
RC 20  10  0.1  106
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Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
4. Design a symmetric collector-coupled astable multivibrator to generate a square wave
of 10 kHz having peak-to-peak amplitude of 10 V where h FEmin = 30,
V CE(sat) = 0.2 V, I C(sat) = 2 mA.
Solution:
Given V CE(sat) = 0.2 V, V BE(sat) = V  = 0.7 V, I C(sat) = 2 mA, f = 2 kHz, h FEmin = 30.
As the output amplitude is specified as 12 V, choose V CC = 12 V.
As f = 2 kHz,
1
1
T  
 0.1 ms
f 10  103
The astable is symmetric, hence
R 1 = R 2 = R and C 1 = C 2 =C
T 0.1
T1  T2  
 0.05 ms
2
2
To calculate R C2 :
VCC  VCE (sat) 10  0.2
RC 2 

 4.9 kΩ
I C (sat)
2  103
R C1 = R C2 = 4.9 kΩ.
To calculate R 2 :
V  V
R2  CC
I B 2(sat)
I C (sat)
2 mA
 0.066 mA
hFE min
30
If Q 2 is in saturation
I B 2(sat)  1.5I B 2(min)
 1.5  0.066  0.099 mA
10  0.7
R2 
 93.9 kΩ
0.099  103
R1  R2  93.9 kΩ
As C 1 = C 2
T 1 = 0.69R 2 C 2
0.05  103
C2 
 771.7 pF
0.69  93.9  103
C 1 = C 2 = 771.7 pF
I B 2 min 

5. Design an un-symmetric astable multivibrator having duty cycle of 40 per cent. It is
required to oscillate at 5 kHz. Ge transistors with h FE = 40 are used. The amplitude of the
square wave is required to be 20 V. I C = 5 mA, V CE(sat) = 0.1 V and V BE(sat) = 0.3 V.
Solution:
For Ge transistors, V CE(sat) = 0.1 V, V BE(sat) = V  = 0.3 V
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Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
Given I C(sat) = 5 mA, f = 5 kHz, h FEmin = 40, duty cycle = 40 per cent.
As the output amplitude is specified as 20 V, choose V CC = 20 V.
As f = 5 kHz,
1
1
T  
 0.2 ms
f 5  103
The astable is unsymmetric, hence
T1≠ T2
i.e. R 1 C 1 ≠ R 2 C 2
choose C 1 = C 2 =C
then R 1 ≠R 2
T1
T
 1
Duty cycle =
T1  T2 T
T1
0.4 =
0.2  10 3
T1  0.4  0.2  103  0.08 ms
T 2 = T – T 1 = 0.2– 0.08 = 0.12 ms
To calculate R C2 :
VCC  VCE (sat) 20  0.1
RC 2 

 3.98 kΩ
I C (sat)
5 mA
Choose R C1 = R C2 =3.98 kΩ = R C
To calculate R 2 :
V  V
R2  CC
I B 2(sat)
I C (sat)
5 mA
 0.125 mA
hFE min
40
If Q 2 is in saturation
I B 2(sat)  1.5I B 2(min)
 1.5  0.125 mA = 0.187 mA
20  0.3
 R2 
 105.3 kΩ
0.187  103
As C 1 = C 2 = C
T 1 = 0.69R 2 C 2
T1
0.08  103
C2 

 1.1 nF
0.69 R2 0.69  105.3  103
C1  C2  1.1 nF
T 2 = 0.69R 1 C 1
0.12  10 3  0.69 R1  1.1  10 9
I B 2 min 

0.12  103
 158.1 kΩ
0.69  1.1  109
h FE R C = 40  3.98 kΩ = 159 kΩ
R1 
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Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
The values R 1 and R 2 are less than h FE R C . Hence the devices Q 1 and Q 2 are in
saturation, when ON.
6. For an un-symmetric astable multivibratorR 1 = 100 kΩ, R 2 = 100 kΩ, C 1 = 0.02 F,
C 2 = 0.01F. Find the frequency of oscillation and the duty cycle.
Solution:
T2  0.69 R1C1  0.69  100  103  0.02  106  1.38 ms
T1  0.69 R2 C2  0.69  100  103  0.01  106  0.69 ms
T  T1  T2  0.69  1.38  2.09 ms
1
1
 483 Hz
f  
T 2.09  103
T
0.69
 100 per cent  33 per cent .
per cent D  1  100 per cent 
T
2.09  103
7. Design an unsymmetrical astable multivibrator shown in Fig.7p.1 using silicon n–
p–n transistors having an output amplitude of 12 V. Given data,
I C(sat) = 5 mA, h FEmin = 50, f = 5 kHz, duty cycle = 0.6.
Fig.7p.1 Un-symmetric astable multivibrator
Solution:
Assume V CE(sat) = 0.2 V
V BE(sat) = V  = 0.7 V
Given I C(sat) = 5 mA, f = 2 kHz, h FEmin = 50, duty cycle = 0.6.
As the output amplitude is specified as 12 V, choose V CC = 12 V.
As f = 2 kHz,
1
1
T  
 0.5 ms
f 2  103
The astable is unsymmetric, hence
T1≠ T2
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Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
i.e. R 1 C 1 ≠ R 2 C 2
choose C 1 = C 2 =C
then R 1 ≠R 2
T1
T
 1
Duty cycle =
T1  T2 T
T
0.6 = 1
T
T1  0.6T  0.6  0.5  0.3 ms
T 2 = T – T 1 = 0.5 – 0.3 = 0.2 ms
To calculate R C2 :
VCC  VCE (sat ) 12  0.2 11.8
RC 2 


5 mA
5 mA
I C (sat)
R C2 = 2.36 kΩ
Choose R C1 = R C2 =2.2 kΩ = R C .
To calculate R 2 :
V  V
R2  CC
I B 2(sat)
I B 2 min 
I C (sat )

5 mA
50
hFE min
=0.1 mA
If Q 2 is in saturation
I B 2(sat )  1.5I B 2(min)
 1.5  0.1  0.15 mA
12  0.7
11.3 V
 R2 

 75.3 kΩ
0.15
0.15 mA
As C 1 = C 2 = C
T 1 = 0.69R 2 C
0.3 ms  0.69  75.3 kΩ  C
0.3  103
C
 5.77 nF
0.69  75.3  103
T 2 = 0.69R 1 C
0.2  10 3  0.69 R1  5.77  10 9
0.2  103
 50 kΩ
0.69  5.77  109
h FE R C = 50  2.2 K  110 K
The values R 1 and R 2 are less than h FE R C . Hence the devices Q 1 and Q 2 are in saturation,
when ON.
R1 
8. A Voltage-to-frequency converter shown in Fig.7p.2 generates oscillations at a
frequency f 1 when V BB = V CC . Find the ratio of V CC /V BB at which the frequency f 2 = 4f 1 .
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Pulse and Digital Circuits
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
Fig.7p.2 Voltage-to-frequency converter
Solution:
From Eq.(7.30), the expression for the frequency of oscillations is
1
f=
V
2 ln(1  CC )
V BB
Case1: The frequency of oscillations is f 1 when V CC = V BB
1
1
 f1 

2 ln(1  1) 2 ln 2
Case 2: If a new frequency f 2 = 4f 1 is desired.
1
f2 
V
2 ln(1  CC )
VBB
f 2  4 f1
 4
2

 ln 2
1
2 ln(1 
ln 2  4 ln(1 
ln(1 
1
2

2 ln 2  ln 2
VCC
)
VBB
VCC
)
VBB
VCC 4
)  ln 2
VBB
Taking antilog
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Pulse and Digital Circuits
(1 
VCC 4
) 2
V BB
(1 
VCC 2
)  2  1.414
VBB
1
Venkata Rao K., Rama Sudha K. and Manmadha Rao G.
VCC
 1.414  1.189
V BB
VCC
 1.189  1  0.189
V BB
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