Question

The figure shown below represents a MOS transistor with

$K_n=\frac{{\mu }_n{C}_{ox}W}{2L}=10mA/V^2$. The figure is constructed for the purpose of the small signal analysis. If the MOSFET is biased at a D.C. current $I_D$ of 10 mA and a small signal input voltage of $V_{sig}=1mV$ is applied, then the small signal output $\left(V_0\right)$ will be equal to _______ V.
(Assume that the effect of channel length modulation is negligible)

• A. -0.25
• B. -0.15
• C. -0.20
• D. -0.20

Answer 1

The correct option is D -0.20

To calculate the small signal equivalent of the MOSFET, we need to find the value of $g_m$
Now, $g_m=2\sqrt{K_D{I}_D}$

Given, $K_n=10mA/V^2$

$\therefore$ $g_m=2\times \sqrt{10\times {10}^{-3}\times 10\times {10}^{-3}}$

$g_m=20mA/V$
Therefore, the small signal equivalent,

$\therefore V_0=-g_m{R}_D{V}_{gs}$

but $V_{gs}=V_{sg}$

$\therefore$ $V_0=-g_m{R}_D\times V_{sig}$

$=-20\times {10}^{-5}\times 10\times {10}^3\times 1\times {10}^{-3}V$

\(=-200\,mV\] )

thus, $V_0=-0.20V$