Question 3.12: Find IB , IC and vo in the transistor circuit of Fig. 3.41. ......

For the input loop, KVL gives

  -4 + I_{B} ( 20 × 10 ³) + V_{BE} = 0

Since V_{BE} = 0.7 V   in the active mode,

I_{B} = \frac{4 – 0.7}{ 20 × 10³}= 165 μ A

But

I_{C} = β I_{B} = 50 × 165  μA = 8.25 mA

For the output loop, KVL gives

-v_{o} – 100 I_{C} + 6 = 0

or

v_{o} = 6 – 100 I_{C} = 6 – 0.825 = 5.175 V

Note that v_{o} = V_{CE}  in this case.