## Q. 7.2

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. ## Verified Solution

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 