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For the transistor circuit shown in figure , evaluate `V_E , R_B and R_E`. Given `I_C = 1 mA`, ` V_(CE) = 3V , V_(BE) = 0.5 V , V_(CC )= 12 V and beta = 1000` |
Answer» Correct Answer - A::B Consider the figure given here to solve this problem ` I_C = I_E ` [As base current is very small] ` R_C = 7.8 K Omega` From the figure, ` I_C(R_C + R_E) + V_(CE) = 12` ` (R_E + R_C) xx 1 xx 10^(-3) + 3 = 12 ` ` R_E + R_C = 9 xx 10^(3) = 9 k Omega ` ` R_E = 9 -7.8 = 1.2 k Omega ` ` V_E = I_E xx R_E ` ` = 1 xx 10^(-3) xx 1.2 xx 10^(3) = 1.2 V ` Voltage , ` V_B = V_E + V_(BE) = 1.2 + 0.5 = 1.7 V ` Current, `I = V_B / (20 xx 10^(3)) = 1.7 /( 20 xx 10^(3)) ` ` = 0.085 mA ` Resistance , `R_B = (12-1.7) / (1_C / beta) + 0.085 = 10.3 / (0.01 + 0.085) ` [ Given , ` beta = 100`] ` = 108 k Omega ` |
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