Consider the MOS transistor shown in the figure below-Both transistors M1 and M2 have VTN-0-8 V and -nCox-30 - A-V2 - If the value of V0-2-91 V- then the ratio of WL1WL2 is equal to

Ans : (a) Given, V_GS2=V_0=2.91 V ∴ V_GS1=5 2.91=2.09 V ∴ I_D1=I_D2 thus, μ_nC_ox2(WL)_1(V_GS1 V_T)^2=μ_nC_ox2(WL)_2(V_GS2 V_T)^2 ∴(WL)_1(WL)_2=(2.91 0.8)^2(2. ...

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Standard IX

Physics

Direct Current vs. Alternating Current

Consider the ...

Question

Consider the MOS transistor shown in the figure below:

$Both transistors M_{1} and M_{2} have V_{T N} = 0.8 V and μ_{n} C_{o x} = 30 μ A / V^{2} . If the value of V_{0} = 2.91 V, then the ratio of \frac{{(\frac{W}{L})}_{1}}{{(\frac{W}{L})}_{2}} is equal to$

\frac{1}{8}

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None of these

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\frac{8}{3}

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\frac{14}{9}

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Solution

The correct option is C $\frac{8}{3}$
Ans : (a)

$Given, V_{G S 2} = V_{0} = 2.91 V$
$∴ V_{G S 1} = 5 - 2.91 = 2.09 V$
$∴ I_{D 1} = I_{D 2}$
thus,
$\frac{μ_{n} C_{o x}}{2} {(\frac{W}{L})}_{1} (V_{G S 1} - V_{T})^{2} = \frac{μ_{n} C_{o x}}{2} {(\frac{W}{L})}_{2} (V_{G S 2} - V_{T})^{2}$
$∴ \frac{{(\frac{W}{L})}_{1}}{{(\frac{W}{L})}_{2}} = \frac{(2.91 - 0.8)^{2}}{(2.09 - 0.8)^{2}}$

$\frac{{(\frac{W}{L})}_{1}}{{(\frac{W}{L})}_{2}} = \frac{8}{3}$

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Consider the MOS transistor shown in the figure below: Both transistors M1 and M2 have VTN=0.8 V and μnCox=30μA/V2 . If the value of V0=2.91V, then the ratio of (WL)1(WL)2 is equal to

The correct option is C 83Ans : (a) Given,VGS2=V0=2.91V ∴VGS1=5−2.91=2.09V ∴ID1=ID2 thus, μnCox2(WL)1(VGS1−VT)2=μnCox2(WL)2(VGS2−VT)2 ∴(WL)1(WL)2=(2.91−0.8)2(2.09−0.8)2 (WL)1(WL)2=83

Consider the MOS transistor shown in the figure below:

$Both transistors M_{1} and M_{2} have V_{T N} = 0.8 V and μ_{n} C_{o x} = 30 μ A / V^{2} . If the value of V_{0} = 2.91 V, then the ratio of \frac{{(\frac{W}{L})}_{1}}{{(\frac{W}{L})}_{2}} is equal to$