# 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$\ V = V_{(min)}$

Step-by-Step
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(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

Question: 11.1

## In the circuit shown in Fig. 11.16(a), RL = R1 = 100 kΩ , R2 = 50 kΩ and the signal has a peak value of 20 V. Find (a) A (b) VC(min) (c) Vn(min) (d) Ri when the diodes are ON ...

(a) \ A = \frac{a}{1 + a\frac{R_{1}}{2R_{L}...
Question: 11.3

## For the four diode gate shown in Fig. 11.21(a), Vs = 20 V, RL = 200 kΩ , RC = 100 kΩ, Rf = 0.5 kΩ, R = Rs = 1 kΩ. Find Vn(min), A and VC(min). ...

(i) From Eq. (11.17), \ V_{n(min)} = V_{s}[...
Question: 11.4

## Consider the following situation in the FET series gate, Vs = ± 2 V, Rs = 100 Ω, RL = 10 kΩ. The FET has the following parameters: VGS(OFF)max = −10 V and RD(ON) = 20 Ω. Calculate the voltage levels of the control signal, ID, error due to RS and error due to RD(ON). ...

The control signal should have a value\ V_{...
Question: 11.5

## In the circuit shown in Fig. 11.16(a), RL = R1 = 200 kΩ, R2 = 100 kΩ and the signal has a peak value of 10 V. Find (a) A (b) VC(min) (c) Vn(min) (d) Ri ...

(a) \ A = \frac{α}{1 + α(R_{1}/2R_{L})}[/la...
Question: 11.6

## For the four-diode gate shown in Fig 11.20(a), RL = R2 = 50 kΩ and R1 = 1 kΩ, Rf = 25 Ω, Vs = 10 V. Calculate (a) A (b) V(min) (c) VC(min) (d) for V = V(min) ...

(a) We have \ R = \frac{R_{2}R_{1}}{R_{2} +...
The control signal should have a value\ V_{...