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Derivation for Drain Current Equation

The drain of ...

Question

The drain of an n-channel MOSFET is shorted to the gate so that $V_{G S} = V_{D S}$ . The threshold voltage $(V_{T})$ of MOSFET is $1 V$ . If the drain current $(I_{D})$ is $1 m A$ for $V_{G S} = 2 V$ then, for $V_{G S} = 3 V, I_{D}$ is

9 m A

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3 m A

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2 m A

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4 m A

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Solution

The correct option is D $4 m A$
Ans : (d)

$I_{D} = K (V_{G S} - V_{T})^{2}$
$1 m A = K (2 - 1)^{2}$
$K = 1 m A / V^{2}$
Again, $I_{D} = K (V_{G S} - V_{T})^{2} = 4 m A$

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(V_{G S}),

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A n i d e a l n - c h a n n e l M O S F E T h a s t h e f o l l o w i n g p a r a m e t e r s, t h r e s h o l d v o l t a g e V_{T} = 0.65 V a n d o x i d e t h i c k n e s s i s 0.4 n m . I t i s f o u n d t h a t t h e M O S F E T o p e r a t i n g i n s a t u r a t i o n r e g i o n w i t h V_{D S} = 6 V . T h e n t h e p o w e r d i s s i p a t i o n i n t h e M O S F E T w i t h V_{G S} = 4 V i s μ_{n} C_{a x} \frac{W}{L} = 1.5 \times 10^{- 3} A / V^{2},

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