Question 11.3: For the four diode gate shown in Fig. 11.21(a), Vs = 20 V, R......
For the four diode gate shown in Fig. 11.21(a),\ V_{s} = 20 V,\ R_{L} = 200 kΩ ,\ R_{C} = 100 kΩ,\ R_{f} = 0.5 kΩ,\ R = R_{s} = 1 kΩ. Find\ V_{n(min)}, A and\ V_{C(min)}.
(i) From Eq. (11.17),
\ V_{n(min)} = V_{s} = 20 V (11.17)
(ii) From Eq. (11.15),
\ V_{C(min)} = V_{S} \left(2 + \frac{R_{C}}{R_{L}} \right) \left(1 + \frac{R}{4 R_{f}} \right) \thickapprox V_{S} \left(2 + \frac{R_{C}}{R_{L}} \right) (11.15)
\ V_{C(min)} = V_{S} \left(2 + \frac{R_{C}}{R_{L}} \right) \left(1 + \frac{R}{4 R_{f}} \right)\ = 20 \left(2 + \frac{100}{200} \right) \left(1 + \frac{1}{2} \right) = 20 × 2.5 × 1.5 = 75 V
(iii) From Eq. (11.16),
\ A = \frac{1}{1 + \frac{1}{R_{L}} \left(R_{S} + R_{f} + \frac{R}{4}\right) + \frac{1}{R_{C}}\left(2R_{S} + R_{f} + \frac{R}{2}\right) \left( 1 + \frac{R_{f}}{2R_{L}} \right) } (11.16)
\ A = \frac{1}{1 + \frac{1 + 0.5 + 0.25 }{200} \left( \frac{2 + 0.5 + 0.5}{100}\right) \left( 1 + \frac{0.5}{400}\right) }\ = \frac{1}{ 1 + 0.00875 + 0.03} = 0.963
A = 0.963