Question 11.1: In the circuit shown in Fig. 11.16(a), RL = R1 = 100 kΩ , R2......
In the circuit shown in Fig. 11.16(a),\ R_{L} = R_{1} = 100 kΩ ,\ R_{2} = 50 kΩ and the signal has a peak value of 20 V.
Find (a) A (b)\ V_{C(min)} (c)\ V_{n(min)} (d)\ R_{i} when the diodes are ON
(a)
\ A = \frac{a}{1 + a\frac{R_{1}}{2R_{L}} } \ α = \frac{R_{2}}{R_{1} + R_{2}} = \frac{50}{100 + 50 } = \frac{50}{150} = \frac{1}{3}
\frac{R_{1}}{2R_{L}} = \frac{100}{ 2 × 100} = \frac{1}{2}Therefore,
\ A = \frac{\frac{1}{3} }{1 + \left(\frac{1}{3} \times \frac{1}{2} \right) } = \frac{\frac{1}{3} }{\frac{7}{6} }= \frac{2}{7} = 0.285
(b)
\ V_{C(min)} = \frac{R_{2}}{ R_{1}} × \frac{ R_{3}}{R_{3} + 2R_{L}}V_{s} \ R = \frac{R_{2}R_{1}}{R_{2} + R_{1}} = \frac{100 × 50}{150} = \frac{100}{ 3} = \frac{100}{3} kΩ
\ R_{3} = R + R_{f} ≈ R = 33.3 kΩ
\ V_{C(min)} = \frac{50}{100} × \frac{33.3}{33.3 + 2 × 100} × 20 = \frac{333}{233.3} = 1.43 V
(c)
\ V_{n(min)} = \frac{−R_{2}}{R_{1}} × V_{s} = \frac{−50}{100} × 20 = −10 V
(d) When the diodes are ON
\ R_{i} = \frac{R_{L}R_{2}}{R_{2} + 2R_{L}} +\frac{R_{1}}{2} \ = \frac{100 × 50}{50 + 200} + \frac{100}{2}
= 20 + 50 = 70 kΩ