Question

The output characteristics of a transistor are shown in the figure. When $V_{CE}$ is $10V$ and $I_c=4.0mA$, then the value of ${\beta }_{ac}$ is ________.

Answer 1

Step 1: Given data

Value of $V_{CE}$=$10V$

Current $I_C=4.0mA$

Step 2: Formula used

The ${\beta }_{ac}$ which is a current gain can be computed using the following formula:

${\beta }_{ac}=\frac{∆I_c}{∆I_b}where{\beta }_{ac}=currentgain,∆I_c=outputcurrentand∆I_b=inputcurrent$

Step 3: Compute the ${\beta }_{ac}$

To determine the $∆I_b$, choose any two values of input current from the graph. We select $I_{B1}=20\mu A$ and $I_{B2}=30\mu A$. So,

$\begin{array}{rcl}∆I_B& =& 30-20\\ & =& 10\mu A\end{array}$

And the corresponding output currents are $I_{C1}=3mA$ and $I_{C2}=4.5mA$ respectively. So,

$\begin{array}{rcl}∆I_C& =& 4.5-3\\ & =& 1.5mA\end{array}$

Therefore, the ${\beta }_{ac}$ is,

$\begin{array}{rcl}{\beta }_{ac}& =& \frac{∆I_C}{∆I_B}\\ & =& \frac{1.5\times {10}^{-3}}{10\times {10}^{-6}}\\ & =& 150\end{array}$

Hence, the value of ${\beta }_{ac}$ will be $150$.