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, R1 = R2 = R = 47 k  , C1 = C2 = C = 0.01 μF. Find the time period and frequency.

Fig.7p.1 Un-symmetric astable multivibrator

Solution: Given R1 = R2 = R = 47 k  ,C1 = C2 = 0.01 µF.

This is a symmetric astable multivibrator. TRC1.38 1.38 47 1036 0.01 10 0.648 ms 11 f  1.54 kHz. T 0.648 103

2. For the astable multivibrator in Fig.7p.1, R1 = 20 k  , R2 = 30 k  ,C1 = C2 = C = 0.01 μF. Find the time period, duty cycle and the frequency. Solution: This is an un-symmetric astable multivibrator. 36 TRC2110.69 0.69 20 10 0.01 10 0.2 ms 36 TRC1220.69 0.69 30 10 0.01 10 0.138 ms

TTT120.138  0.2  0.338 ms T 0.138 per cent D 1 100 per cent   100 per cent  59.17 per cent T 0338 11 f  2.95 kHz. T 0.338 103

© Dorling Kindersley India Pvt. Ltd 2010 1 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, VCC=10 V, RC = R3= 2 k  , R1 = R2 = 20 k , C = 0.1 F,

hFE(min) = 30. Show that the ON device is in saturation. Also find f. Assume suitable values for VCE(sat) and VBE(sat). Si transistors are used.

Fig.7p.3 Astable multivibrator with vertical edges Solution: Assume Q1 is OFF and Q2 is ON and in saturation. If Q2 is ON and in saturation, for silicon transistors, VC2 =VCE(sat) = 0.2 V, VB2 = Vσ = 0.7 V then D2 is ON. The collector load is R3//RC.

' 22 R = R3//RC =  1 kΩ C 4

VVCC CE(sat) 10 0.2 9.8 IC 2 '39.8 mA RC 1 kΩ 110 VV 10 0.7 9.3 I CC  0.465 mA B2 R 20 kΩ 20 103

IC 2 9.8 mA I B2min  0.326 mA hFE min 30 IB2 >> IB2min Hence Q2 is in saturation. To find f: 0.7 0.7 For a symmetric astable multivibrator: f = 350 Hz RC 20 1036 0.1 10

© Dorling Kindersley India Pvt. Ltd 2010 2 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 hFEmin = 30, VCE(sat) = 0.2 V, IC(sat) = 2 mA.

Solution:

Given VCE(sat) = 0.2 V, VBE(sat) = V  = 0.7 V, IC(sat) = 2 mA, f = 2 kHz, hFEmin = 30. As the output amplitude is specified as 12 V, choose VCC = 12 V. As f = 2 kHz, 11 T  0.1 ms f 10 103 The astable is symmetric, hence R1 = R2 = R and C1 = C2 =C T 0.1 TT 0.05 ms 1222 To calculate RC2:

VVCC CE(sat) 10 0.2 RC 2 3 4.9 kΩ IC(sat) 210

RC1 = RC2 = 4.9 kΩ. To calculate R2:

VVCC   R2  I B2(sat)

IC(sat) 2 mA I B2min 0.066 mA hFEmin 30 If Q2 is in saturation

IIBB2(sat)  1.5 2(min) 1.5 0.066  0.099 mA 10 0.7 R 93.9 kΩ 2 0.099 103

RR1293.9 kΩ As C1 = C2 T1 = 0.69R2C2 0.05 103 C 771.7 pF 2 0.69 93.9 103 C1 = C2 = 771.7 pF

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 hFE = 40 are used. The amplitude of the square wave is required to be 20 V. IC = 5 mA, VCE(sat) = 0.1 V and VBE(sat) = 0.3 V.

Solution:

For Ge transistors, VCE(sat) = 0.1 V, VBE(sat) = V  = 0.3 V

© Dorling Kindersley India Pvt. Ltd 2010 3 Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G.

Given IC(sat) = 5 mA, f = 5 kHz, hFEmin = 40, duty cycle = 40 per cent. As the output amplitude is specified as 20 V, choose VCC = 20 V. As f = 5 kHz, 11 T  0.2 ms f 510 3 The astable is unsymmetric, hence T1≠ T2 i.e. R1C1 ≠ R2C2 choose C1 = C2 =C then R1≠R2 T T Duty cycle = 1  1  TT 21 T T 0.4 = 1 102.0 3 3 T1 0.4 0.2 10  0.08 ms T2 = T – T1 = 0.2– 0.08 = 0.12 ms To calculate RC2:

VVCC CE(sat) 20 0.1 RC 2 3.98 kΩ IC(sat) 5 mA

Choose RC1 = RC2 =3.98 kΩ = RC To calculate R2:

VVCC   R2  I B2(sat)

IC(sat) 5 mA I B2min 0.125 mA hFEmin 40 If Q2 is in saturation

IIBB2(sat)  1.5 2(min) 1.5 0.125 mA = 0.187 mA 20 0.3  R 105.3 kΩ 2 0.187 103 As C1 = C2 = C T1 = 0.69R2C2 3 T1 0.08 10 C2 3 1.1 nF 0.69R2 0.69 105.3 10

CC121.1 nF T2 = 0.69R1C1 3 9 R1  101.169.01012.0 0.12 103 R 158.1 kΩ 1 0.69 1.1 109 hFERC = 40 3.98 kΩ = 159 kΩ

© Dorling Kindersley India Pvt. Ltd 2010 4 Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G.

The values R1 and R2 are less than hFERC. Hence the devices Q1 and Q2 are in saturation, when ON.

6. For an un-symmetric astable multivibratorR1 = 100 kΩ, R2 = 100 kΩ, C1 = 0.02 F, C2 = 0.01F. Find the frequency of oscillation and the duty cycle.

Solution: 36 TRC2110.69 0.69 100 10 0.02 10 1.38 ms 36 TRC1220.69 0.69 100 10 0.01 10 0.69 ms

TTT120.69  1.38  2.09 ms 11 f  483 Hz T 2.09 103 T 0.69 per cent D 1 100 per cent   100 per cent  33 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, IC(sat) = 5 mA, hFEmin = 50, f = 5 kHz, duty cycle = 0.6.

Fig.7p.1 Un-symmetric astable multivibrator Solution: Assume VCE(sat) = 0.2 V

VBE(sat) = V  = 0.7 V Given IC(sat) = 5 mA, f = 2 kHz, hFEmin = 50, duty cycle = 0.6. As the output amplitude is specified as 12 V, choose VCC = 12 V. As f = 2 kHz, 11 T  0.5 ms f 210 3 The astable is unsymmetric, hence T1≠ T2

© Dorling Kindersley India Pvt. Ltd 2010 5 Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G.

i.e. R1C1 ≠ R2C2 choose C1 = C2 =C then R1≠R2 T T Duty cycle = 1  1  TT 21 T T 0.6 = 1 T

TT1 0.6 0.6 0.5 0.3 ms T2 = T – T1 = 0.5 – 0.3 = 0.2 ms To calculate RC2:

VVCC CE(sat) 12 0.2 11.8 RC 2  IC(sat) 5 mA 5 mA

RC2 = 2.36 kΩ Choose RC1 = RC2 =2.2 kΩ = RC. To calculate R2:

VVCC   R2  I B2(sat)

IC(sat) 5 mA I B2min  hFE min 50 =0.1 mA If Q2 is in saturation

IIBB2(sat)  1.5 2(min) 1.5 0.1 0.15 mA 12 0.7 11.3 V  R  75.3 kΩ 2 0.15 0.15 mA As C1 = C2 = C T1 = 0.69R2C 0.3 ms 0.69 75.3 kΩ C 0.3 103 C 5.77 nF 0.69 75.3 103 T2 = 0.69R1C 3 9 R1  1077.569.0102.0 0.2 103 R 50 kΩ 1 0.69 5.77 109 hFERC = 50 2.2 K 110 K The values R1 and R2 are less than hFERC. Hence the devices Q1 and Q2 are in saturation, when ON.

8. A Voltage-to-frequency converter shown in Fig.7p.2 generates oscillations at a frequency f1 when VBB = VCC. Find the ratio of VCC /VBB at which the frequency f2 = 4f1.

© Dorling Kindersley India Pvt. Ltd 2010 6 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   CC )1ln(2 VBB Case1: The frequency of oscillations is f1 when VCC = VBB 1 1  f   1   )11ln(2  2ln2

Case 2: If a new frequency f2 = 4f1 is desired. 1 f  2 V   CC )1ln(2 VBB

 4 ff 12 1 2 4  2ln2  2ln 2 1   2ln V   CC )1ln(2 VBB V  CC )1ln(42ln VBB

VCC 4  2ln)1ln( VBB Taking antilog

© Dorling Kindersley India Pvt. Ltd 2010 7 Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G.

V CC 4  2)1( VBB

V CC 2  414.12)1( VBB

V 1 CC  189.1414.1 VBB

V CC  189.01189.1 VBB

© Dorling Kindersley India Pvt. Ltd 2010 8