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From the output characteristics shown in figure. Calculate the values of \(\beta _{\mathit{ac}}\)  and \(\beta _{\mathit{dc}}\)  of the transistor when \(V_{\mathit{CE}}\)  is \(10\;V\)  and \(I_C=4.0\;\mathit{mA}\)
Question: From the output characteristics shown in figure. Calculate the values of  beta _{mathit{ac}}  and  beta _{mathit{dc}}  of the transistor when  V_{mathit{CE}}  is  10;V  and  I_C=4.0;mathit{mA}

Answer

\(\beta _{\mathit{ac}}=\left(\frac{\Delta I_C}{\Delta I_B}\right)_{V_{\mathit{CE}}},\beta _{\mathit{dc}}=\frac{I_C}{I_B}\)
For determining \(\beta _{\mathit{ac}}\) and \(\beta _{\mathit{dc}}\) at the stated values of \(V_{\mathit{CE}}\)  and \(I_C\)  one can proceed as follows. Consider any two characteristics for two values
of \(I_B\)  which lie above and below the given value of \(I_C.\)  Here \(I_C=4.0\mathit{mA}\)   \((\) Choose characteristics for \(I_B=30\)  and \(20\mu \; A.\) At \(V_{\mathit{CE}}=10\) we read the two values of \(I_C\)  from the graph. Then
\(\Delta I_B=\left(30-20\right)\mu A=10\mu A,\Delta I_C=\left(4.5-3.0\right)\mathit{mA}=1.5\mathit{mA}\)
Therefore, \(\beta _{\mathit{ac}}=\dfrac{1.5\mathit{mA}}{10\mu A}=150\)
For determining \(\beta _{\mathit{dc}}\)  either estimate the value of \(I_B\)  corresponding to \(I_C=4.0\mathit{mA}\)  at \(V_{\mathit{CE}}=10V\)  or calculate the two values of \(\beta _{\mathit{dc}}\)  for the two characteristics chosen and find their mean. Therefore, for \(I_C=4.5\mathit{mA}\)  and \(I_B=30\mu A\)   \(\beta _{\mathit{dc}}=\dfrac{4.5\mathit{mA}}{30\mu A}=150\)
and for \(I_C=3.0\mathit{mA}\)  and \(I_B=20\mu A\)
\(\beta _{\mathit{dc}}=\dfrac{3.0\mathit{mA}}{20\mu A}=150\)
Hence, \(\beta _{\mathit{dc}}=\dfrac{(150+150)}{2}=150\)
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