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)}

Step-by-Step

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(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

Question: 11.1

(a)
\ A = \frac{a}{1 + a\frac{R_{1}}{2R_{L}...

Question: 11.2

(a) We have:
\ R = \frac{R_{1}R_{2}}{R_{1} ...

Question: 11.3

(i) From Eq. (11.17),
\ V_{n(min)} = V_{s}[...

Question: 11.4

The control signal should have a value\ V_{...

Question: 11.5

(a)
\ A = \frac{α}{1 + α(R_{1}/2R_{L})}[/la...

Question: 11.7

The control signal should have a value\ V_{...