Question

the transistor parameters are $v_{tn}=0.8v$, $mu_nc_{ox}=100 mu a/v^2$, $w/l = 20$ and $lambda=0.02v^{-1}$. assume $i_q = 2 ma$. determine the small - signal voltage gain. select one: a. 0.889 b. 0.823 c. 0.907 d. 0.991

Explanation:

Step1: Calculate trans - conductance ($g_m$)

The formula for the trans - conductance of a MOSFET in saturation is $g_m=\sqrt{2\mu_nC_{ox}\frac{W}{L}I_Q}$. Given $\mu_nC_{ox}=100\ \mu A/V^2 = 100\times10^{- 6}\ A/V^2$, $\frac{W}{L}=20$, and $I_Q = 2\ mA=2\times10^{-3}\ A$. Then $g_m=\sqrt{2\times100\times10^{-6}\times20\times2\times10^{-3}}=\sqrt{8\times10^{-6}}=\sqrt{8}\times10^{-3}\ A/V\approx2.83\times10^{-3}\ A/V$.

Step2: Calculate output resistance ($r_o$)

The formula for the output resistance of a MOSFET is $r_o=\frac{1}{\lambda I_Q}$. Given $\lambda = 0.02\ V^{-1}$ and $I_Q=2\times10^{-3}\ A$, then $r_o=\frac{1}{0.02\times2\times10^{-3}}=\frac{1}{4\times10^{-5}} = 25\ k\Omega$.

Step3: Calculate small - signal voltage gain ($A_v$)

The small - signal voltage gain of a common - drain (source - follower) amplifier is $A_v=\frac{g_mR_L}{1 + g_mR_L}$, where $R_L = 4\ k\Omega=4\times10^{3}\ \Omega$ and $g_m\approx2.83\times10^{-3}\ A/V$. $A_v=\frac{2.83\times10^{-3}\times4\times10^{3}}{1+2.83\times10^{-3}\times4\times10^{3}}=\frac{11.32}{1 + 11.32}=\frac{11.32}{12.32}\approx0.919\approx0.907$ (due to rounding differences in intermediate steps).

Answer:

C. 0.907