A microphone having a DC level of zero drives a CG stage biased in saturation region at ID - 0-5 mA- If WL - 50- -n Cox - 100 - A V2- VTh - 0-5 V and VDD - 1-8 V assuming negligible channel length modulation then

I_D = 12 μ_nC_ox WL (V_GS – V_Th)^2 0.5 × 10^ 3 = 12 × 100 × 10^ 6× (50)(V_GS – 0.5)^2 V_GS – V_Th = 0.4472 V V_GS = 0.947 V For the transistor to remain in s ...

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Byju's Answer

Standard XII

Physics

Motional EMF

A microphone ...

Question

A microphone having a $D C$ level of zero drives a $C G$ stage biased in saturation region at $I_{D} = 0.5 m A$ . If $(\frac{W}{L}) = 50, μ_{n} C_{o x} = 100 μ A V^{2}, V_{T h} = 0.5 V a n d V_{D D} = 1.8 V$ assuming negligible channel length modulation then

The maximum value of voltage gain is equal to

6.06 V / V

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The maximum value of

R_{D} = 2.71 k Ω

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The minimum value of

R_{D} = 7.21 k Ω

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The value of

V_{G S} = 0.947 V

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Solution

The correct option is D The value of $V_{G S} = 0.947 V$
$I_{D} = \frac{1}{2} μ_{n} C_{o x} \frac{W}{L} (V_{G S} - V_{T h})^{2}$
$0.5 \times 10^{- 3} = \frac{1}{2} \times 100 \times 10^{- 6} \times (50) (V_{G S} - 0.5)^{2}$
$V_{G S} - V_{T h} = 0.4472 V$
$V_{G S} = 0.947 V$
For the transistor to remain in saturation
$V_{D D} - I_{D} R_{D} > V_{G S} - V_{T h}$
And hence, $R_{D} < 2.71 k Ω$
Also, the above value of $(W L)$ and $I_{D}$ yield $δ_{m} = \sqrt{μ_{n} C_{a x} (\frac{W}{L}) \times 2 I_{D}} = 0.00223 Ω$
$A_{V} \leq δ_{m} R_{D}$
$A_{V} \leq 6.06 \to (A_{V})_{m a x} = 6.06$

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A microphone having a DC level of zero drives a CG stage biased in saturation region at ID=0.5mA. If (WL)=50, μnCox=100μAV2, VTh=0.5V and VDD=1.8V assuming negligible channel length modulation then

A microphone having a $D C$ level of zero drives a $C G$ stage biased in saturation region at $I_{D} = 0.5 m A$ . If $(\frac{W}{L}) = 50, μ_{n} C_{o x} = 100 μ A V^{2}, V_{T h} = 0.5 V a n d V_{D D} = 1.8 V$ assuming negligible channel length modulation then