Question

The BJT amplifier circuit in the figure below has β = 90. Use VT = 25 mV

a- Draw the DC equivalent circuit and calculate IC.

b- Draw the small signal equivalent circuit and calculate the small signal parameters.

c- Find the voltage gain

+9V 3 kΩ C2 = 1 uF HE C = 1uF + υο 10 kΩ Cz = 50 uF - . υ 9 kΩ Η. 6. ΚΩ -9 V

Answer 1

(2) For DC capacitors will act as open circuits +9V E 3KL Vc Ic Ic = BIB 9kn w VB B=90 IB VBE + TE = (1+B) 18 VE 1 E 6ұл IE - 9V DC Equivalent circuit Consider VBE = 0.7 V Applying KvL for Loop@, qIB + VBE +6 IE -9 = 0 q IB +0.7 +6(1+90) IB -9 =0 > 9-0.7 IB = - 14.955 MA 9 +6191) Ic = (90) (14.955) = = 1,346 MA TE = (92) (14,955) = 1,360 mA

(b) For small analysis will act as short ärmits signal is capacitors During small signal analysis all the active sources will be turned OFF B C t + V, 9kn h 38 44 Dam VT 33kn 10kn Vo I E Small signal Equivalent arcuit A= B Im and gm = = 1 (C) NA = VI Applying kcl at "c" node, gm Var + Vo 3)| 10 =0 > Vo = -a m VIT C 3/1 10) kn) 49m VI)( 34410) *103 - ) Vo 3 - 1.346 25 Х 30 X10 13 -124.246 VI Voltage gain, Av = -124.246 Vv