Question

Consider the circuit shown in the figure below


The value of $|V_{BE}|=0.7V\text{and}\beta =100.$ Then which of the following statement is correct?

• A. The transistor is operating in reverse active region
• B. The transistor is operating in cutoff region
• C. The transistor is operating in saturation region
• D. The transistor is operating in linear region

Answer 1

The correct option is D The transistor is operating in linear region

$V_{EC\left(sat\right)}=0.2V$

$V_{BE}=0.7V$

Assume that BJT is in saturation region
Apply KVL in outer most loop:
$-10+1k\left(I_E\right)+0.7+270k\left(I_B\right)=0$

$1k(I_B+I_C)+1k{I}_C=10-0.2=9.8$

$1k{I}_B+2k{I}_C=9.8$ ...(i)

Appky KVL in emitter base loop
$-10+1k\left(I_E\right)+0.7+270k\left(I_B\right)=0$

$\left(271k\right)I_B+1k{I}_C=9.3$ ...(ii)

From (i) and (ii)

$I_B=0.0162mA;$

$I_C=4.892mA$

$I_{Bmin}=\frac{I_{C\left(sat\right)}}{\beta }=0.04892mA$

$I_B<I_{Bmin}$

$\therefore$ Transistor is in linear or active region.