Question

Define $\alpha$ and $\beta$ for a transistor. Derive relation between them. In a $CE$ transistor circuit if $\beta =100$ and $I_B=50\mu A$. Compute the values of $\alpha ,I_E$ and $I_C$.

Answer 1

$\alpha$ and $\beta$ are two important parameters in the transistor which define the current gains in the transistor.
$\alpha =\frac{I_C}{I_E}$
$\beta =\frac{I_C}{I_B}$
Since $I_C>I_B$; $\beta$ is very large and its value lies between 15 and 50.
Since $I_C<I_e$; so $\alpha <1$;
Its value lies between 0.95 and 0.99.
Relation between $\alpha$ and $\beta$:
$I_E=I_B+I_C$
Divide by $I_C$, $\frac{I_E}{I_C}=\frac{I_B}{I_C}+1$
$\frac{I_C}{I_E}=\alpha ,\frac{I_C}{I_B}=\beta$
So, $\frac{1}{\alpha }=\frac{1}{\beta }+1$ or $\beta =\frac{\alpha }{1-\alpha }$
$\beta =100$
$100=\frac{\alpha }{1-\alpha }$
$\alpha =\frac{100}{101}$
Also, $\beta =\frac{I_C}{I_B}=100$
$\Rightarrow I_C=I_B\times 100=50\times {10}^{-6}\times 100$
$=5\times {10}^{-3}=5mA$
$\Rightarrow I_E=5\times {10}^{-3}+0.05\times {10}^{-3}$
$=5.05\times {10}^{-3}$ Ampere or, 5.05 mA.