Question

Please just a walk through guide with formulas and steps.

Problem 1. For the differential amplifier of Figure P-1, it is given that RA-RB-602, R2-302, RSA-RsB 3kS2, R 6kS2. To calculate bias currents, you may assume that K, = and β = oo . Also I'm = ν,,-IV (DC). You may make reasonable approximations to obtain answers accurate to 10% . Find the bias current Ic3 b) Find the bias voltages at VoUT, VE1, and Vcs c) Find the small-signal voltage gain- d) Now assume that a resistor is added to the circuit and is connected between the nodes ve and v. The value of the resistor is 8k2. Find the smal-signal voltage gain , +8V A SE Q1 EI C3 CS Q4 Q3 Q5 -8V Figure P-

Answer 1

KVL from node B_2 to to GND = -8 V OV - i_5 (6 k ohm) - 0.7 = -8 V Implies i_5 = 8 - 0.7/6 k = 1.267 mA No input current to transistor, i_input = 0: V_BE = 0.7 V Given circuit is a current mirror current through 25, is I_5, = 1.383 mA as V_BE of theta_5 = V_BE theta_4 = V_BE theta_3, same current flows through them Implies i_3 = i_4 = i_5 = 1.267 mA Implies I_c_3 = 1.268 mA As i_3 = i_4 rightarrow No current flows through R_2, \r\n As beta = infinity: I_c = I_E: I_B = 0 for all transistor Therefore i_4 flows through R_5 B Therefore V_out = 8 - i_4 (R_5 theta) = 8 - (1.267 m) (3 k) = 4.199 V V_E_1 = V_B_1 - V_BE = 1 - 0.7 = 0.3 V (at theta_1 transistor Implies V_E 1 = 0.3 V V_C 5 = 0 - i_5 R_4 = -(1.267 m) (6 k) = 7.602 Implies V_C 5 = 7.602 V \r\n Equivalent AC circuit: This is a differential amplifier whose gain is given by A_v = -R_C/R_C