Question

Please do not answer if you are going to copy the answers of others. Only do it if you know how. The others are wrong.

8. Consider the circuit shown in Figure A Q-point value for Ic between a minimum of 4 mA and a maximum of 5 mA is required. Assume that resistor values are constant and that ranges from 100 to 300. It is

Answer 1

We know I_c = B I_B For minimum value of I_c & B 4 times 10^-3/100 = I_B I_B = 4 times 10^-5 I_B = 0.04 mA From the circuit above I_B = 5 - V_B E_Q/R_B R_B = 5 - 0.7/0.04 times 10^-3 = 4.3/4 times 10^-5 = 4.3/4 times 10^-4 = 10.75 times 10^4 implies 1075 times 10^2 implies 107.5 k ohm By applying KVL in given circuit 5 - I_B R_B - I_E R_E = 0 here I_E = I_B + I_C \r\n I_C = V_CC = 15/1000 = 15 mA 5 - 0.04 times 107.5 - (I_B + I_C) R_E = 0 5 - 4.3 - 15.04 R_E = 0 R_E = 0.7 times 1000/15.04 implies 700/15.04 implies 46.54 ohm (b) The above circuit can also be drawn as. This circuit also called as self Bias circuit. The above opt(a) can be drawn as where V_th = V_CC R_2/R_1 + R_2 & R_th = R_B = R_1 Vertical-bar Vertical-bar R_2 = R_1 R_2/R_1 + R_2