Question 11.2: For the four-diode gate shown in Fig. 11.20(a), RL = R2 = 10......
For the four-diode gate shown in Fig. 11.20(a),\ R_{L} = R_{2} = 100 kΩ and\ R_{1} = 1 kΩ,\ R_{f} = 25 Ω,\ V_{s} = 20 V. Calculate (a) A (b)\ V_{(min)} (c)\ V_{C(min)} (d)\ V_{n(min)} for\ V = V_{(min)}
(a) We have:
\ R = \frac{R_{1}R_{2}}{R_{1} + R_{2}} , \ R_{3} = R + R_{f}
\ R = \frac{100 × 1}{101} = 990 Ω
\ R_{3} = 990 + 25 = 1015 Ω
and
\ A = \frac{R_{2}}{R_{1} + R_{2}} × \frac{R_{L}}{R_{L} + (R_{3}/2)} = \frac{ 100}{100 + 1} × \frac{100}{100 + (1.015/2)} \ = 0.99 × \frac{100}{100 + (1.015/2)} = 0.99 × \frac{100}{100 + 0.507}
A = 0.985
(b)
\ V_{min} = \frac{R_{2}}{R_{1}} × \frac{R_{3}}{R_{3} + 2R_{L}} × V_{s}\ = \frac{100}{1} × \frac{1.015 × 20}{(1.015 + 200)} = \frac{101.5 × 20}{201.015} = 10.1 V
(c)
\ V_{C(min)} = AV_{s} = 0.985 × 20 = 19.7 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)}= 10.1 V Therefore,
\ V_{n(min)} = 20 × \frac{100}{100 + 1} − 10.1 × \frac{1}{100 + 1} = 19.80 − 0.1 = 19.70 V