Question

The common emitter forward current gain of the transistor is equal to $\beta =100.$

If the forward voltage drop across the base emitter terminal $V_{BE}=-0.7V,$ then the transistor is operating in

• A. Saturated region
• B.
  Active region
• C.
  
  Cut-off region
• D.
  Reverse active region.

Answer 1

The correct option is B

Active region
$I_B=\frac{10-0.7}{(250+101)\times {10}^3}=26.57\mu A$

Now, assuming the transistor to be in forward region,

$I_C=\beta I_B=2.657mAand{I}_E=\frac{I_c}{\alpha }=\frac{101}{100}\times 2.657=2.68mA\therefore V_{EC}=10-1\times {10}^3\times 2.657\times {10}^{-3}-1\times {10}^3\times 2.68\times {10}^{-3}$
$V_{EC}=4.663V$

Thus, the transistor is operating in linear region.