Question

Consider the circuit shown in the figure below:

$\text{The value of}\frac{{\mu }_{\pi }{C}_{ox}W}{L}=1\text{mA/V,}{V}_T=2V\text{and}\lambda =0.\text{Then, the value of small signal voltage gain}{A}_v\text{is equal to}$

• A. -12.65 V/V
• B. -8.94 V/V
• C. -16.31 V/V
• D. -21.22 V/V

Answer 1

The correct option is B -8.94 V/V

$\text{Now,}{g}_m=\sqrt{2{\mu }_{\pi }{C}_{o}x\frac{W}{L}{I}_{DS}}=\sqrt{2\times 1\times {10}^{-3}\times 0.4\times {10}^{-3}}$

$=0.894\text{mA/V}$

$\text{Thus,}{A}_v=-g_m{R}_D=-0.894\times 10=-8.94\text{V/V}$