Question

Determine IB,IC,and VCE for a base biased transistor circuit with the following values: ẞ= 90, Vcc = 12V, RB= 22kΩ and RC= 100Ω

Answer 1

• For input circuit, we have the equation-

I_B = $\frac{V_{CC}-0.7}{R_B}$

Here, V_CC is also base biasing voltage.

$\frac{I_C}{\beta}$ = $\frac{12-0.7}{22000}$

Given $\beta$ = 90

I_C = $\frac{101.7}{2200}$

I_C = 46.23 mA ...(1)

• For output circuit, we have the equation-

12 = I_CR_C + V_CE

12 = (0.04623)(100) + V_CE

V_CE = 7.38 V

• Also, I_B = $\frac{46.23}{90}$

I_B = 0.514 mA