Question 7.7: A symmetrical collector-coupled astable multivibrator has th......

Given\ V_{CC} = 10V,\ R_{C} = 1 kΩ, R = 10 kΩ, C = 0.01μF,\ r_{bb^{′}} = 0.2 kΩ and\ h_{FE} = 50.
Assume\ V_{σ} = 0.7V and\ V_{CE(sat)} = 0.2V.

Let\ Q_{2} be ON and in saturation, then\ Q_{1} is OFF. After a time interval\ T_{1}, Q_{1} is ON and\ Q_{2} is OFF. Suddenly when a device switches from the OFF state into the ON state, there is an overshoot at its base. Similarly, when the device switches its state from ON to OFF there is an overshoot at its collector. To account for these overshoots the base spreading resistance is taken into account. Let\ Q_{2} switch from OFF to ON and\ Q_{1} from ON to OFF, then the equivalent circuit is shown in Fig. 7.18(a).

When\ R \gg R_{C}, then\ I_{R} \gg I^{\prime}_{B2}. Hence, R is omitted in Fig. 7.18(b).

Let\ δ^{′} and\ δ be the overshoots at the collector of\ Q_{1} and the base of\ Q_{2}.

\ V_{σ} = 0.7 V      \ V_{γ} = 0.5 V      and      \ V_{CE(sat)} = 0.2 V
\ δ = V_{B2} − V_{γ} = I^{′}_{B2}r_{bb^{′}} + V_{σ} − V_{γ} = I^{′}_{B2}r_{bb^{′}} + 0.7 − 0.5

But,\ δ = δ^{′}, since the collector\ Q_{1} and the base of\ Q_{2} are connected through a condenser C and identical changes are required to take place at these two nodes.

\ δ^{′}= V_{C1} − V_{CE(sat)} = V_{CC} − I^{′}_{B2}R_{C} − V_{CE(sat)} = I^{′}_{B2}r_{bb^{′}} + 0.2 V

9.8 V −\ I^{′}_{B2}R_{C} = I^{′}_{B2}r_{bb^{′}} + 0.2 V

\ I^{′}_{B2}(r_{bb^{′}} + R_{C}) = 9.6 V          \ I^{′}_{B2} = \frac{9.6  V}{0.2 + 1}  = \frac{9.6  V}{1.2  kΩ} = 8 mA

\ δ = I^{′}_{B2}r_{bb^{′}} + V_{σ} − V_{γ} = 8 × 10^{−3} × 0.2 × 10^{3} + 0.7  V − 0.5  V = 1.6 + 0.2 = 1.8 V      \ δ = δ^{′}

The waveforms are shown in Fig. 7.18(c). For the astable multivibrator circuit using p–n–p transistors, the waveforms are shown in Fig.7.19.