Question 11.5: In the circuit shown in Fig. 11.16(a), RL = R1 = 200 kΩ, R2 ......
In the circuit shown in Fig. 11.16(a),\ R_{L} = R_{1} = 200 kΩ,\ R_{2} = 100 kΩ and the signal has a peak value of 10 V. Find (a) A (b)\ V_{C(min)} (c)\ V_{n(min)} (d)\ R_{i}
(a)
\ A = \frac{α}{1 + α(R_{1}/2R_{L})}
\ α = \frac{ R_{2}}{R_{1} + R_{2}}= \frac{100}{200 + 100} = \frac{1}{ 3} \frac{R_{1}}{2R_{L}} = \frac{200}{2 × 200} = \frac{1}{2}
Therefore,
\ A = \frac{(1/3)}{1 + ((1/3) × (1/2))} = \frac{(1/3)}{(7/6)} = \frac{1}{3} × \frac{6}{7} = \frac{2}{7} = 0.285
(b)
\ V_{C(min)} = \frac{R_{2}}{R_{1}} × \frac{R_{3}}{R_{3} + 2R_{L}} V_{s}
\ R_{3} = R + R_{f} ≈ R = 66.6 kΩ
\ V_{C(min)} = \frac{100}{200} × \frac{66.6}{66.6 + 2 × 200} × 10\ V_{C(min)} = 0.71 V
(c)
\ V_{n(min)} = \frac{−R_{2}}{R_{1}} × V_{s} = \frac{−100}{ 200} × 10 = −5 V
(d) When the diodes are ON
\ R_{i} = \frac{R_{L}R_{2}}{R_{2} + 2R_{L}} + \frac{R_{1}}{2}\ = \frac{200 × 100}{100 + 400} + \frac{ 200}{2} = \frac{ 200}{5} + 100 = 40 + 100 = 140 kΩ