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 *μ*F 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.

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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 *μ*F

\ T = 2 × \frac{T}{2} = 134 *μ*s \ f = \frac{1}{ T} = \frac{1}{134 × 10^{−6}} = 7.46 kHz

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