Question

Consider the source-follower circuit shown below:

$\text{The circuit is biased so that}{g}_m=2\text{mS and the value of output resistance for MOS transistor}{r}_o=100\text{k}\Omega \text{, then the value of small signal voltage gain}\frac{V_0}{V_i}\text{is equal to}$

• A. 0.8
• B. 10
• C. -0.9
• D. -0.8

Answer 1

The correct option is A 0.8

The value of small signal equivalent is shown in the figure below:

$\text{Now,}{V}_o=g_m{V}_{gs}\left(4\text{k}\Omega ||r_0\right)...\left(1\right)$

$\text{Now,}{V}_i=V_{gs}+V_o$

$V_i=V_{gs}+g_m{V}_{gs}\left(4\text{k}\Omega ||r_o\right)$

$V_{g}s=\frac{V_i}{1+g_m\left(4\text{k}\Omega ||r_o\right)}...\left(2\right)$

$\therefore \text{Substituting (2) in (1) we get,}\frac{V_o}{V_i}=\frac{g_m\left(4\text{k}\Omega ||r_o\right)}{1+g_m\left(4\text{k}\Omega ||r_o\right)}=0.8$