Given B = 120, and 1E = 3.2 MA for a common emitter configuration with $v_0 = \infty $

Determine:

a] Zi

b] AV if a loud of 2k is applied.

c] Ai with the 2k load.

Ans: Consider ct configuration.

$Z_1 = r \pi$ where $r_1 = \frac{B26mV}{1_c} = \frac{26mv \times 20}{1_t}$

$r_\pi = 120 \times \frac{26mv}{3.3 MA} = 7.878 \times 120 = 945.45$

$\therefore Zi = 94545$

When load register of 2k is applied.

Av = -qm (2k) qm = $\frac{1_c}{VT}$

= -126.92 m x 2 k $= \frac{3.3m}{26mv}$

Av = -253.84 = 126.92m $\mho$

Current gain (Ai)

$A_i = \frac{1_0}{1_i}$

$A_i = \frac{gm v \pi}{(\frac{v \pi }{r \pi })}$

$A_i$ = gm r $\pi $ = B

$A_i = B = 120$