Question

when rl = ∞), the voltage gain av = 0.98. when rl = 1 kω, the voltage gain av = 0.49. what are the values of gm and ro? select one: a. gm = 1.86 ma/v and ro = 100 kω b. gm = 0.98 ma/v and ro = 50 kω c. gm = 1.5 ma/v and ro = 75 kω d. gm = 0.49 ma/v and ro = 25 kω

Explanation:

Step1: Recall voltage - gain formula for source - follower

The voltage - gain formula for a source - follower is $A_v=\frac{g_mR_L}{1 + g_mR_L+\frac{R_L}{r_o}}$ when $R_L$ is finite and $A_v=\frac{g_mr_o}{1 + g_mr_o}$ when $R_L=\infty$.
When $R_L = \infty$, $A_v=\frac{g_mr_o}{1 + g_mr_o}=0.98$. Cross - multiply to get $g_mr_o=0.98 + 0.98g_mr_o$. Then $g_mr_o-0.98g_mr_o = 0.98$, so $0.02g_mr_o=0.98$, and $g_mr_o = 49$.

Step2: Use the formula for finite $R_L$

When $R_L = 1k\Omega$ and $A_v = 0.49$, we have $A_v=\frac{g_mR_L}{1 + g_mR_L+\frac{R_L}{r_o}}=0.49$. Substitute $R_L = 1000\Omega$ into the formula: $\frac{1000g_m}{1 + 1000g_m+\frac{1000}{r_o}}=0.49$.
From $g_mr_o = 49$, we have $r_o=\frac{49}{g_m}$. Substitute $r_o=\frac{49}{g_m}$ into $\frac{1000g_m}{1 + 1000g_m+\frac{1000}{\frac{49}{g_m}}}=0.49$.
Simplify the denominator: $\frac{1000g_m}{1 + 1000g_m+\frac{1000g_m}{49}}=0.49$. The denominator is $1 + 1000g_m(1+\frac{1}{49})=1 + 1000g_m\times\frac{50}{49}$.
So $\frac{1000g_m}{1+\frac{50000g_m}{49}}=0.49$. Cross - multiply: $1000g_m=0.49+ \frac{24500g_m}{49}$.
$1000g_m=0.49 + 500g_m$. Then $1000g_m-500g_m=0.49$, so $500g_m = 0.49$, and $g_m=0.98mA/V$.
Since $g_mr_o = 49$ and $g_m = 0.98mA/V$, then $r_o=\frac{49}{0.98}=50k\Omega$.

Answer:

B. $g_m = 0.98\ mA/V$ and $r_o = 50\ k\Omega$