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

A, B

Explanation

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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