Question

For the transistor circuit shown in figure, evaluate V_E, R_B, R_E given I_C = 1 mA, V_CE = 3 V, V_BE = 0.5 V and V_CC = 12 V, β = 100.

Answer 1

Let us redraw the circuit diagram given here to solve this problem.

As we know the base current is very small. So,

I_C = I_E

R_C = 7.8 kΩ

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 + 3` = 12

`R_E = 9 - 7.8` = 1.2 kΩ

`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)/(I_C/beta + 0.085) xx 10^3`

= `10.3/(0.01 + 0.085)`  .....[Given, β = 100]

= 108 kΩ