The channel resistance when [latex]V_{GS} = −3 [/latex] V is (a) 360 Ω (b) 917 Ω (c) 1000 Ω (d) 3000 Ω

The depletion width of the channel on one side is directly propostional to the square-root of sum of the built-in voltage of the [latex]P^+N[/latex] junction and gate-source voltage. Therefore, [latex]\begin{aligned}W & \propto \sqrt{V}\\\frac{W_2}{W_1}&=\sqrt{\frac{V_{GS 1}+V_{bj 1}}{V_{GS 2}+V_{bj 2}}}\\\frac{W_2}{W_1}&=\sqrt{4}=2\end{aligned}[/latex] [latex]\begin{aligned}W_2 &=2 \mu m \\t_{ch 2}&=[10-2(2)] \mu m=6 \mu m\end{aligned}[/latex] As, [latex]R \propto \frac{1}{t_{ ch }}[/latex] Therefore, channel resistance [latex]R_2 \text { at } V_{G S}=-3 V \text { is }[/latex] [latex]R_2=600 \times \frac{\left(10 \times 10^{-6}\right)}{\left(6 \times 10^{-6}\right)}=1000 \Omega[/latex] Ans. (c)

Question 10.SG.Q.27: The channel resistance when VGS = −3 V is (a) 360 Ω (b) 917 ......

Wiley Acing the Gate Electronics and Communication Engineering [1079457]

The channel resistance when $V_{GS} = −3$ V is

(a) 360 Ω (b) 917 Ω

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The depletion width of the channel on one side is directly propostional to the square-root of sum of the built-in voltage of the $P^+N$ junction and gate-source voltage. Therefore,

$\begin{aligned}W & \propto \sqrt{V}\\\frac{W_2}{W_1}&=\sqrt{\frac{V_{GS 1}+V_{bj 1}}{V_{GS 2}+V_{bj 2}}}\\\frac{W_2}{W_1}&=\sqrt{4}=2\end{aligned}$

$\begin{aligned}W_2 &=2 \mu m \\t_{ch 2}&=[10-2(2)] \mu m=6 \mu m\end{aligned}$

As, $R \propto \frac{1}{t_{ ch }}$

Therefore, channel resistance $R_2 \text { at } V_{G S}=-3 V \text { is }$

$R_2=600 \times \frac{\left(10 \times 10^{-6}\right)}{\left(6 \times 10^{-6}\right)}=1000 \Omega$

Ans. (c)