Consider the circuit shown in the figure below:

The MOSFET is biased in saturation region having the minimum value of $V_{D D}$ for which the MOSFET will remain in saturation region is _____ V
$V_{T} = 2.5 V a n d (\frac{μ_{n} C_{o x} W}{L}) = 0.5 m A / V$

2

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Solution

The correct option is A 2
For the transistor

$V_{s} = V_{B} = V_{A}$

due to virtual ground,

thus, $V_{s} = 0 V$

Hence, $I_{D} = \frac{0 - (- 10)}{10 \times 10^{5}} = 1 m A$

$I_{D} = \frac{μ_{n} C_{o x} W}{2 L} (V_{G S} - V_{T})^{2}$

$∴ V_{G S} - V_{T} = \sqrt{\frac{I_{D}}{\frac{μ_{n} C_{o x} W}{2 L}}}$

$V_{G S} - V_{T} = \sqrt{\frac{1 \times 10^{- 3}}{\frac{0.5 \times 10^{- 3}}{2}}}$

$V_{G S} - V_{T} = 2 V$

For the MOSFET to be in saturation region

$V_{D S} \geq V_{G S} - V_{T}$

$∴$ at the edge of saturation

$V_{D S} = V_{G S} - V_{T} = 2 V$

$∵ V_{s} = 0$

$∴ V_{D} = V_{G} - V_{T}$

$\Rightarrow V_{D D} = 2 V$

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Similar questions

V_{D D} = 2 V, μ_{n} C_{o x} = 100 μ A / V^{2}

V_{T h} = 0.5 V

V_{G S}

g_{m} = 5 m A / V .

R_{i n}

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},

(V_{G S}),

(I_{D})

V_{D S} = 4 V a n d 2.5 m A f o r V_{D S} = 8 V .

λ

V^{- 1}

λ

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