You visited us 1 times! Enjoying our articles? Unlock Full Access!

Discharging of Capacitors

Consider the ...

Question

Consider the MOS transistor circuit shown in the figure below

The two N-MOS transistors are identical with threshold voltage $V_{T} = 1 V$ and assuming the value of $λ = 0$ , $\frac{μ_{n} C_{o x} W}{L} = 2 m A / V^{2}$ , the value of drain to source voltage for transistor $T_{2} (V_{D S 2})$ is equal to ____V

5

Open in App

Solution

The correct option is A 5
For saturation region the condition is,

$V_{D S} \geq V_{G S} - V_{T}$ ...(i)

$V_{G S 1} = 5 - V_{1}$

$V_{D S 1} = 10 - V_{1}$

$∴$ $10 - V_{1} > (5 - V_{1}) - 1$

$10 - V_{1} > 4 - V_{1}$

This indicates $T_{1}$ is in saturation,
$V_{D S 2} \geq V_{G S 2} - V_{2}$

$V_{1} - V_{2} > - V_{2} - 1$

From equation (i)
$I_{D 1} = I_{D 2}$

The inequality satisfies, so $T_{1}$ and $T_{2}$ are in saturation
$I_{D 1} = I_{D 2}$

$\frac{1}{2} μ_{n} C_{o x} \frac{W}{L} {(V_{G S 1} - V_{T})}^{2} = \frac{1}{2} μ_{n} C_{o x} \frac{W}{L} {(V_{G S 1} - V_{T})}^{2}$

$V_{G S 1} = V_{G S 2}$

$5 - V_{1} = - V_{2}$

$V_{1} - V_{2} = 5 = V_{D S 2}$

Suggest Corrections

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

V_{T} = 1 V a n d μ_{n} C_{o x} (\frac{ω}{L}) = 2 m A / V^{2} .

(V_{D S})

c o n s i d e r t h e c i r c u i t s h o w n i n t h e f i g u r e b e l o w . t h e v a l u e o f C_{1}, C_{2} a n d C_{c} i s v e r y l a r g e a n d t h e t r a n s i s t o r i s b i a s e d s u c h t h a t t h e v a l u e o f g_{m} = 20 m A / V λ = 0.01 V^{- 1} I f t h e M O S t r a n s i s t o r h a s t h e n t h e v a l u e o f v o l t a g e g a i n \frac{V_{0}}{V_{s}} i s e q u a l t o . . . . . V / V .

Q_{1}

Q_{2}

I_{C} 1 = 10 I_{C 2}

V_{T} = 26

V_{1} = 25

V_{2}

T_{1}

μ_{n} C_{o x} = 50 \frac{m A}{V^{2}} a n d V_{t} = 1 V

(\frac{W}{L})

Join BYJU'S Learning Program