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Question
For the transistor circuit shown in figure, evaluate VE, RB, RE given IC = 1 mA, VCE = 3 V, VBE = 0.5 V and VCC = 12 V, β = 100.
Solution
Let us redraw the circuit diagram given here to solve this problem.
As we know the base current is very small. So,
IC = IE
RC = 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 VB = VE + VBE = 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, RB = `(12 - 1.7)/(I_C/beta + 0.085) xx 10^3`
= `10.3/(0.01 + 0.085)` .....[Given, β = 100]
= 108 kΩ
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