# Question 7.2: Consider the circuit in Fig. 7.16 (a) with VCC2 = 18 V, R1 =......

Consider the circuit in Fig. 7.16 (a) with$\ V_{CC2}$ = 18 V,$\ R_{1}$ = 250 kΩ,$\ R_{3}$ = 200 kΩ,$\ R_{2} = R_{4}$ = 1 MΩ,$\ R_{e1} = R_{e2}$ = 3 kΩ,$\ R_{C1}$ = 0.5 kΩ,$\ R_{C2}$ = 0.2 kΩ, C = 0.1 μF and$\ h_{FE}$ = 30. Si devices are used. Calculate the voltages in the circuit and plot the waveforms to the scale. Also obtain the frequency of oscillations.

Step-by-Step
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Given$\ V_{CC2}$ = 18 V,$\ R_{3}$ = 250 kΩ and$\ R_{4}$ = 1MΩ

$\ V_{CC1} = V_{CC2} × \frac{R_{4}}{R_{3} + R_{4}} = 18 × \frac{1}{1 + 0.200}$  = 15 V

$\ V_{BB} = V_{CC1} × \frac{R_{2}}{R_{1} + R_{2}} = 15 × \frac{1}{1 + 0.250}$  = 12 V.

When$\ Q_{1}$ is ON and$\ Q_{2}$ is OFF at$\ t = T_{1}−$

$\ V_{CN2}(T_{1}−) = V_{CC2}$ = 18 V        $\ V_{EN1}(T_{1}−) = V_{BB} − V_{σ}$ = 12 − 0.7 = 11.3 V

$\ V_{CN1}(T_{1}−) = V_{BN2}(T_{1}−) = V_{BB} − V_{σ} + V_{CE(sat)}$ = 11.3 + 0.3 = 11.6 V

$\ V_{EN2}$ exponentially falls to zero.

$\ V_{EN2}(T_{1}−) = V_{BN2}(T_{1}−) − V_{γ}$ = 11.6 − 0.5 = 11.1 V

$\ Q_{2}$ begins to conduct when$\ V_{EN2}$ falls to 11.1 V.

At$\ t = T_{1}+, Q_{2}$ conducts and$\ Q_{1}$ goes into the OFF state because of the regenerative action.$\ R_{e}$ is the parallel combination of$\ R_{e1}$ and$\ R_{e2}$. In the above circuit,

$\ R_{e1} = R_{e2}$ = 3 kΩ        $\ R_{e} = R_{e1}//R_{e2} = \frac{3 kΩ}{2}$  = 1.5 kΩ

$\ h_{FE}$ = 30,        $\ I_{C2} = \frac{V_{CC1}}{R_{e}} = \frac{15V}{1.5 kΩ}$  = 10 mA

$\ I_{B2} = \frac{I_{C2}}{h_{FE}} = \frac{10 mA}{30}$  = 0.33 mA      $\ I_{B2}R_{C1} = (0.33 mA)(0.5 kΩ)$ = 0.1655 V

$\ I_{C2}R_{C2}$ = (10 mA)(0.2 kΩ) = 2 V

We have$\ V_{γ} = 0.5V, V_{BE2}$ = 0.6 V,$\ V_{σ} = 0.7 V, V_{CE(sat)}$ = 0.3 V

$\ V_{1} = V_{CC1} − I_{B2}R_{C1} − V_{BE2} − V_{CE(sat)} + V_{γ}$ = 15 V − 0.165 V − 0.6 V − 0.3 V + 0.5 V = 14.435 V

$\ V_{CN1}(T_{1}+) = V_{CC1} − I_{B2}R_{C1}$ = 15 − 0.1655 = 14.8345 V

$\ V_{D} = V_{1} − V_{EN1}(T_{1}+)$ = 14.435 − 11.3 = 3.135 V

$\ V_{CN2}(T_{1}+) = V_{CC2} − I_{C2}R_{C2}$ = 18 − 2 = 16 V

The waveforms are shown in Fig. 7.17.

We have,
$\ R_{e2}$ = 3 kΩ, C = 0.1 μF

$\ ∴\frac{T}{2} = 3×10^{3}×0.1×10^{−6} In\frac{15}{12} = 67 × 10^{−6}s$

$\ T = 2 × \frac{T}{2}$  = 134 μs        $\ f = \frac{1}{ T} = \frac{1}{134 × 10^{−6}}$  = 7.46 kHz

Question: 7.1

## Design an astable multivibrator, assuming that silicon devices with hFE(min) = 40 are used. Also assume that VCC = 10 V, IC(sat) = 5 mA. Let the desired frequency of oscillations be 5 kHz. For transistor used, VCE(sat) = 0.2 V, VBE(sat) = Vσ = 0.7 V. ...

\ R_{c} = \frac{ V_{CC} − V_{CE(sat)}}{I_{C...
Question: 7.8

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Assume$\ Q_{1}$ is OFF and\ Q_...
Question: 7.7

## A symmetrical collector-coupled astable multivibrator has the following parameters: VCC = 10 V, RC = 1 kΩ, R = 10 kΩ, and C = 0.01μF. Silicon transistors with hFE = 50 and rbb′ = 0.2 kΩ are used. Plot the waveforms and calculate the overshoot. Also plot the waveforms if the circuit uses p-n-p ...

Given$\ V_{CC}$ = 10V,\ R_{C}[...
Question: 7.6

## In an astable multivibrator if R2 = 60 kΩ, R1 = 40 kΩ, C1 = C2 = 2.9 nF, find its frequency and duty cycle. ...

\ T_{2} = 0.69 R_{1}C_{1} = 0.69 × 40 × 10^...
Question: 7.5

## For the multivibrator shown in Fig. 7.13(a), VCC = 20 V, VBB = 10 V, R1 = R2 = R = 10 kΩ, C1 = C2 = C = 0.01μF. Find the time period and the frequency. ...

\ T = 2RC  ln \left(1 + \frac{V_{CC}}{V_{BB...
Question: 7.4

## For the astable multivibrator shown in Fig. 7.1: (a) Find the value of C to provide symmetrical oscillations if R = 10 kΩ and f = 10 kHz. (b) Determine the values of capacitors to provide a train of pulses 0.1 ms wide and at a frequency of 1 kHz, if R1 =R2 = 1 kΩ (c) Find the minimum value of RC ...

(a)   Given R = 10 kΩ and$\ f$ = 10 k...
(a)   Given$\ R_{1} = R_{2} = R$ = 10...