Question

In the given figure,the $V_{BB}$ supply can be varied from $0V$ to $5V$.The silicon transistor has ${\beta }_{DC}=250$ and $R_B=100k\Omega ,R_C=1k\Omega ,V_{CC}=5V$.Assume that when the transisitor is saturated,$V_{CE}=0V$ and $V_{BE}=0.8V.$The minimum base current for which the transistor will reach saturation is:

• A. $10\mu A$
• B. $20\mu A$
• C. $30\mu A$
• D. $40\mu A$

Answer 1

Answer: (B)

$20\mu A$

Given data :

${\beta }_{DC}=250$-0.751

$R_B=100k\Omega$

$R_C=1k\Omega$

$V_{CC}=5V$

$V_{CE}=0V$

$V_{BE}=0.8V$

We know:

$V_{BB}=V_{CC}\frac{R_C}{R_B+R_C}$

$V_{BB}=5\left(\frac{1}{100+1}\right)$

$V_{BB}=\frac{5}{101}$

$V_{BB}=0.049V$

The minimum base current will be given by:

$I_B=\frac{V_{BB}-V_{CE}}{R_B+R_C(\beta +1)}$

$I_B=\frac{0.049-0.8}{100+\left(1\right)(250+1)}$

$I_B=\frac{-0.751}{351}\frac{V}{k\Omega }$

$I_B=0.002\times {10}^{-3}A$

$I_B=20\mu A$