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Transistor: Working

Consider the ...

Question

Consider the circuit shown in the figure below:

Assume drain current of 1 mA, voltage gain
$A_{v} = 0.8, μ_{u} C_{o x} = 100 μ A / V^{2}, V_{T h} = 0.5 V, λ = 0, V_{D D} = 1.8 V and R_{G} = 50 k Ω,$ then

the value of

A_{V}

doubles if

R_{G}

is doubled

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the value of

R_{S} = 867 Ω

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the value of

R_{i n} = \infty

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the value of

V_{d s} = 0.933 V

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Solution

The correct option is D the value of $V_{d s} = 0.933 V$
$A_{v} = \frac{g_{m} R_{s}}{1 + g_{m} R_{s}} = \frac{R_{s}}{\frac{1}{g_{m}} + R_{s}}$
Now, $g_{m} = \frac{2 I_{D}}{V_{g} s - V_{T h}}$
$A_{V} = \frac{2 I_{D} R_{s}}{V_{g} s - V_{T h} + 2 I_{D} R_{s}}$
$R_{s} = \frac{V_{D D} - V_{T h}}{I_{D}} . \frac{A_{V}}{2 - A_{V}}$ $[∵ V_{g} s = V_{d s} = V_{D D} - I_{D} . R_{S}]$
$= 867$
and $V_{d s} = V_{D D} - I_{D} R_{s}$
$= V_{D D} - (V_{D D} - V_{T h}) \frac{A_{V}}{2 - A_{V}}$
$= 0.933 V$

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Similar questions

μ_{n} C_{o x} = 2.5 μ A / V^{2}, V_{T h} = 1 V a n d \frac{W}{L} = 3

I_{D}

μ A

β = 99, V_{A} = 100 V

V_{T} = 25 m V,

A_{v} = \frac{v_{0}}{v_{s}}

V_{0}

V_{T} = 1 V

λ = 0

\frac{μ_{n} C_{o x} W}{L} = 2 m A / V^{2}

T_{2} (V_{D S 2})

R_{B} (in k Ω)

I_{C} = 1 mA, V_{C E} = 3 V, V_{B E} = 0.5 V, V_{C C} = 12 V

β = 100

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