The enhancement type MOSFET in the circuit below operates according to the square law. \mu_{ {n}} {C}_{ {ax}}=100 \mu {A} {/} {V}^{2}, the threshold voltage ( {V}_{ {T}}) is 500 {mV}. Ignore channel length modulation. The output voltage {V}_{\text{out }} is

GATE Electrical Engineering (EE) 2019 Solved Paper

Question : 54 of 65

Marks: +1, -0

The enhancement type MOSFET in the circuit below operates according to the square law.

µ_{ {n}} {C}_{ {ax}}=100 µ {A} {/} {V}^{2}

, the threshold voltage

( {V}_{ {T}})

500 {mV}

. Ignore channel length modulation. The output voltage

{V}_{\;\text "out "\;}

Solution:

{V}_{ {s}}=0 {V}

and

{V}_{ {D}}= {V}_{\;\text "out "\;}

⇒ {V}_{ {Ds}}= {V}_{ {out}}

{V}_{ {gs}}= {V}_{ {g}}- {V}_{ {s}}= {V}_{\;\text "out "\;}

{V}_{ {th}}=0.5

{V}_{ {ds}}( {sat}) \;\; = {V}_{ {gs}}- {V}_{ {th}}

\;\; = {V}_{0}-0.5

V_{ {ds}}> {V}_{ {ds}}

(sat)
The transistor in saturation region

{I}_{ {D}}=\;{1}/{2} µ_{x} C_{a x}(\;{W}/{L})(V_{G S}-V_{T})^{2}

5 × 10^{-6}=\;{1}/{2} × 10010^{-6}(10)( {V}_{ {GS}}- {V}_{ {T}})^{2}

5 × 2=1000( {V}_{ {GS}}- {V}_{ {T}})^{2}

{V}_{ {GS}}- {V}_{ {T}}=10^{-1} {V}

{V}_{ {GS}}= {V}_{ {T}}+0.1

=600 {mV}

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