Question

For the transistor circuit shown in figure. Find the value of $R_B\left(\text{in}k\Omega \right)$ nearly is (integer only).
[If $I_C=1\text{mA},V_{CE}=3\text{V},V_{BE}=0.5\text{V},V_{CC}=12\text{V}$ and $\beta =100$]

• A. 108.00
• B. 108.0
• C. 108

Answer 1

Answer: (A), (B), (C)

Given, $I_C=1\text{mA};V_{CE}=3\text{V}{V}_{BE}=0.5\text{V};V_{CC}=12\text{V}\beta =100$



The current gain, $\beta =\frac{I_C}{I_B}$

$\Rightarrow I_B=\frac{I_C}{\beta }=\frac{1\times {10}^{-3}}{100}=1\times {10}^{-5}\text{A}\left[\text{very small}\right]$

$\therefore I_E=I_B+I_C=I_C......\left(1\right)$

From the figure, we get,

$I_C{R}_C+I_E{R}_E+V_{CE}=V_{CC}$

$(R_E+R_C)\times I_C+V_{CE}=12[\because I_E=I_C]$

$(R_E+R_C)\times 1\times {10}^{-3}+3=12$

$\Rightarrow R_E+R_C=9\times {10}^3\Omega =9k\Omega$

$\Rightarrow R_E=9-7.8=1.2k\Omega$

Further,

$V_E=I_E{R}_E$

$=1\times {10}^{-3}\times 1.2\times {10}^3=1.2\text{V}$

Also, $V_B=V_E+V_{BE}=1.2+0.5=1.7\text{V}$

Now, $I=\frac{V_B}{20\times {10}^3}=\frac{1.7}{20\times {10}^3}=8.5\times {10}^{-5}\text{A}$

$\because V_{CC}=(I+I_B)R_B+V_B$

$\therefore R_B=\frac{V_{CC}-V_B}{I+I_B}$

$=\frac{12-1.7}{8.5\times {10}^{-5}+1\times {10}^{-5}}$

$=108421=108k\Omega$