Question 11.6: For the four-diode gate shown in Fig 11.20(a), RL = R2 = 50 ......
For the four-diode gate shown in Fig 11.20(a),\ R_{L} = R_{2} = 50 kΩ and\ R_{1} = 1 kΩ,\ R_{f} = 25 Ω,\ V_{s} = 10 V. Calculate (a) A (b)\ V_{(min)} (c)\ V_{C(min)} (d) for\ V = V_{(min)}
(a) We have
\ R = \frac{R_{2}R_{1}}{R_{2} + R_{1}} , \ R_{3} = R + R_{f}
And\ α = \frac{R_{2}}{R_{1} + R_{2}}
\ R = \frac{50 × 1}{51} = 980 Ω
\ α = \frac{50}{51} = 0.98
\ R_{3} = 980 + 25 = 1005Ω
and\ A = α × \frac{R_{L}}{R_{L} + R_{3}/2} . Therefore,
\ A = 0.98 × \frac{50}{50 + (1.005/2)} = 0.98 × \frac{50}{50 + 0.502}A = 0.97
(b)
\ V_{min} = \frac{R_{2}}{ R_{1}} × \frac{R_{3}}{R_{3} + 2R_{L}} × V_{s}
\ = \frac{50}{1} × \frac{1.005 × 10}{(1.005 + 100)} = \frac{502.5}{101.005} = 4.97 V
(c)
\ V_{C(min)} = A V_{s} = 0.97 × 10 = 9.7 V
(d)
\ V_{n(min)} = V_{s} × \frac{R_{2}}{R_{1} + R_{2}} − V × \frac{R_{1}}{R_{1} + R_{2}}
Here,\ V = V_{(min)} = 4.97 V
Therefore,
\ V_{n(min)} = 10 × \frac{50}{50 + 1} − 4.97 × \frac{1}{50 + 1} = 9.8 − 0.97 = 9.7 V