Consider a differential amplifier circuit shown in the figure below-The two transistors are exactly matched with Vt-0-5 V- -nCox-500 - A-V2 and WL-100 - Then the value of voltage Vs is equal toAssuming T1 and T2 are operating in saturation region-

The current of both the transistors are equal since they are perfectly matched. Thus, I2=12μ_nC_ox(WL)(V_GS1 V_t)^2 10×10^ 3=12×500×10^ 6×100(V_GS1 0.5)^2 ∴ V_ ...

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Standard XII

Physics

Inductor

Consider a di...

Question

Consider a differential amplifier circuit shown in the figure below:

$The two transistors are exactly matched with V_{t} = 0.5 V, μ_{n} C_{o x} = 500 μ A / V^{2} and (\frac{W}{L}) = 100 . Then the value of voltage V_{s} is equal to$
$(A s s u m i n g T_{1} a n d T_{2} a r e o p e r a t i n g i n s a t u r a t i o n r e g i o n .)$

2.413 V

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-1.124 V

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1.868 V

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3 V

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Solution

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

$Thus, \frac{I}{2} = \frac{1}{2} μ_{n} C_{o x} (\frac{W}{L}) (V_{G S 1} - V_{t})^{2}$
$10 \times 10^{- 3} = \frac{1}{2} \times 500 \times 10^{- 6} \times 100 (V_{G S 1} - 0.5)^{2}$
$∴ V_{G S 1} = V_{G S 2} = 1.132 V$
$Thus, V_{s} = V_{c m} - V_{G S 1} = 3 - 1.132 = 1.868 V$

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Consider a differential amplifier circuit shown in the figure below: The two transistors are exactly matched with Vt=0.5V,μnCox=500 μA/V2 and (WL)=100. Then the value of voltage Vs is equal to (Assuming T1 and T2 are operating in saturation region.)

The correct option is C 1.868 VThe current of both the transistors are equal since they are perfectly matched. Thus, I2=12μnCox(WL)(VGS1−Vt)2 10×10−3=12×500×10−6×100(VGS1−0.5)2 ∴VGS1=VGS2=1.132V Thus, Vs=Vcm−VGS1=3−1.132=1.868V