Consider a differential amplifier circuit shown in the figure below:

If the two tansistors are exactly matched then the voltage $V_{s}$ is equal to
(Assuming $T_{1}$ and $T_{2}$ are operating in saturation region)
$V_{t} = 0.5 V, μ_{n} C_{o x} = 500 μ A / V^{2} a n d (\frac{W}{L}) = 100$

-1.124 V

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3 V

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2.413 V

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1.868 V

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Solution

The correct option is D 1.868 V
The current of both the transistors are equal since they are perfectly matched

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

$10 \times 10^{- 3} = \frac{1}{2} \times 500 \times 10^{- 6} \times 100 (V_{G S 1} - 0.5)^{2}$

$∴ V_{G S 1} = V_{G S 2} = 1.132 V$

Thus, $V_{s} = V_{i n} - V_{G S 1} = 3 - 1.132 = 1.868 V$

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Similar questions

The two transistors are exactly matched with V_{t} = 0.5 V, μ_{n} C_{o x} = 500 μ A / V^{2} and (\frac{W}{L}) = 100 . Then the value of voltage V_{s} is equal to

(A s s u m i n g T_{1} a n d T_{2} a r e 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 .)

T_{1}

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

(\frac{W}{L})

V_{T} = 25 m V a n d V_{B E} = 0.7 V)

If the value of R_{E} is increased, then

{(\frac{W}{L})}_{1} = 1.75

{(\frac{W}{L})}_{3} = 27.8

{(\frac{W}{L})}_{2} =

μ_{n} C_{o x} = 86 μ A / V^{2} a n d V_{T N} = 1 V

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