Question

1) Calculate the ac emitter resistance r'_e for the value of I_E = 5.52 mA. (In Ω)

2) Calculate the value of ac load resistance r_L. (In kΩ)

3) Calculate the voltage gain A_V.

4) Calculate the value of output voltage. (V_p-p)

Required information In the circuit below: Vin = 10 mVp-p, R2 = 2.2 k12, R1 = 15 k12, Vcc = 21 V, RC = 3 k12, B = 100, RE = 3602, and RL = 6.2 k12. V cc RI ww RC Cout HE B Cin RL R2 RE CE

Answer 1

( U emitter resistance VT IE re V = 25mv If = 5.52mA 25x03 re 3 .5.52x10 re =4.5801 (8) AC equivalent Circuit Capacitors are shorted sources are are shorted open circuited. hé model. DC voltage DC Carrent transistor Sources is repdated by Vout. @gmube Rc Rei + Vin Rillez < Abere E

0 Ac load resistance. Y Re //RL V = 3kull 3ku/16 6.2kn V = 3x6.2 3+6.2 ņ = 2.02 kn 3 Voltage gain Vout -gm Voe (Re/lR) that Voe Vbe = Vin We Ob Seave Vout Av -Im Vin (Re//RL) - gm (Re//RL). Ic gm Vout Vin LIE Ic VT VT E L= = d = o.gg (+1 - 3 99) (5:52) X10 gm 258103

218-59 Xios Av = gm = 218 .5.9 malu. -gm.(Rellow) Av = 441-93 v/v 3 2.02 x 10 ( 4 ) output Vol Hage Vout Av Vin Vin tomVp-p. given Vout 441-93 ro X183 Vout 4.42 Vp-p

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