Question

For the amplifier shown in the figure, the BJT parameters are V_BE  = 0.7V, β = 200,   and thermal voltage V_T =  25mV. The voltage gain ( v₀/v_i) of the amplifier is _______.

Answer 1

V_BE = 0.7V, β = 200, VT = 25mV

DC Analysis:

V_B = 12 x 11k/(11K + 33K) = 3V

V_E = 3 - 0.7 = 2.3V

I_E = 2.3/(10 + 1K) = 2.277mA

I_B = 11.34µA

I_C = 2.26mA

r_e = 25mV/2.277mA = 10.98Ω

A_v = V₀/V_i = - βR_c/(βr_e + (1 + β)(R_s)) = (- 200  x 5k)/(200 x 10.98 + (201)10)

A_v = - 237.76