# Question 11.7: For the FET series gate shown in Fig. 11.30. Vs = ±1 V, Rs =......

For the FET series gate shown in Fig. 11.30.$\ V_{s}$ = ±1 V,$\ R_{s}$ = 50 Ω,$\ R_{L}$ = 20 kΩ. The FET has the following parameters,$\ V_{GS(OFF)max}$ = −10 V and$\ R_{D(ON)}$ = 20 Ω. Calculate the voltage levels of the control signal,$\ I_{D}$, error due to$\ R_{S}$ and error due to$\ R_{D(ON)}$.

Step-by-Step
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The control signal should have a value$\ V_{1}$ for$\ Q_{1}$ to be ON.
Therefore,$\ V_{1} = V_{s(peak)}$ = 1 V.
For$\ Q_{1}$ to be OFF, the control signal should be$\ –V_{2}$
$\ −V_{2} = −[V_{s(peak)} + V_{GS(OFF)max} + 1 V]$
$\ −V_{2} = −(1 + 10 + 1)$ = −12 V

$\ I_{D} = \frac{V_{S}}{R_{S} + R_{D(ON)}+ R_{L}} = \frac{1}{50 + 20 + 20000}$  = 49.8 μA

$\ I_{D}R_{S} = 49.8 × 10^{−6} × 0.05 × 10^{3}$ = 2.49 mV

Error due to$\ R_{S} = \frac{2.49 mV}{ 1 V}$ × 100% = 0.249%

$\ I_{D}R_{D(ON)}$ = 49.8 μA × 20 Ω = 0.996 mV

Error due to$\ R_{D(ON)} = \frac{0.996 mV}{1 V}$  × 100% = 0.099%

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: 11.2

## For the four-diode gate shown in Fig. 11.20(a), RL = R2 = 100 kΩ and R1 = 1 kΩ, Rf = 25 Ω, Vs = 20 V. Calculate (a) A (b) V(min) (c) VC(min) (d) Vn(min) for V = V(min) ...

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