Question

In the figure, given that $V_{BB}$ supply can vary from 0 to 5.0 V, $V_{CC}$=5 V,${\beta }_{dc}=200,R_{B}k\Omega ,R_C=1k\Omega and{V}_{BE}=1.0V,$ The minimum base current and the input voltage at which the transistor will go to saturation, will be respectively :

• A. $25\mu Aand3.5V$
• B. $25\mu Aand2.8V$
• C. $20\mu Aand3.5V$
• D. $20\mu Aand2.8V$

Answer 1

The correct option is A $25\mu Aand3.5V$

In the saturation of a transistor, we know that, $V_{CE}=0$

Now,

$V_{CE}=V_{CC}-I_C{R}_C$

$I_C=\frac{V_{CC}}{R_C}=\frac{5V}{1K\Omega }=\frac{5V}{1000\Omega }=5\times {10}^{-3}A$

We have,

${\beta }_{dc}=\frac{I_C}{I_B}$

$I_B=\frac{5\times {10}^{-3}A}{200}=25\mu A$

In the input,

$V_{BB}=I_B{R}_B+V_{BE}$

$V_{BB}=\left(25\mu A\right)\left(100k\Omega \right)+1\text{V}=\left(2.5+1\right)\text{V}=3.5\text{V}$

So, option (A) is correct.