Question

In the circuit shown in figure, find the value of R_C.

Answer 1

I_E = I_C + I_B

I_C = βI_B 

I_CR_C + V_CE 6 I_ER_E = V_CC

RI_B + V_BE + I_ER_E = V_CC

From (3) Ie ≈ Ic = βI_B 

(R + βR_E) = V_CC – V_BE,

 I_B = `(V_(C C) - V_(BE))/(R + βR_E)`

= `11.5/200` mA

From (2)

R_C + R_E = `(V_(C C) - V_(CE))/I_C`

= `(V_(C C) - V_(CE))/(βI_B)`

= `2/11.5 (12 - 3)` KΩ = 1.56 KΩ

R_C = 1.56 – 1

= 0.56 KΩ