Question

Define $\alpha \text{ and }\beta$ for a transistor. Derive relation between them. In a CE transistor circuit if $\beta =100\text{ and }{{I}_{B}}=50\mu A$. Compute the values of $\alpha, {{I}_{E}}\text{ and }{{I}_{C}}$.

Answer 1

Define how a transistor works and remember the basic relations to solve this type of question. Define $\alpha$ and $\beta$ in terms of the collector current, base current and emitter current. Define mathematical expression giving relation between $\alpha$ and $\beta$. The emitter current is equal to the addition of collector current and base current. Using these relations, find the required answer.

Formula Used :
$~\alpha =\dfrac{{{I}_{C}}}{{{I}_{E}}}$
$\beta =\dfrac{{{I}_{C}}}{{{I}_{B}}}$
$\begin{aligned} & \beta =\dfrac{\alpha }{1-\alpha } \\\ & Or\text{ }\alpha \text{=}\dfrac{\beta }{1+\beta } \\\ \end{aligned}$
${{I}_{E}}={{I}_{B}}+{{I}_{C}}$

Complete step-by-step answer :
$\alpha \text{ and }\beta$ are the parameters for a transistor which defines the current gain in a transistor.
$\alpha$ is defined as the ratio of the collector current to the emitter current.
$~\alpha =\dfrac{{{I}_{C}}}{{{I}_{E}}}$
$\beta$ is defined as the current gain which is given by the ratio of the collector current to the base current.
$\beta =\dfrac{{{I}_{C}}}{{{I}_{B}}}$
In the above equation, ${{I}_{E}}$ is the emitter current, ${{I}_{B}}$ is the base current and ${{I}_{C}}$ is the collector current.
For a transistor we always have,
$\begin{aligned} & {{I}_{E}}={{I}_{B}}+{{I}_{C}} \\\ & \text{Dividing both side by }{{I}_{C}} \\\ & \dfrac{{{I}_{E}}}{{{I}_{C}}}=\dfrac{{{I}_{B}}}{{{I}_{C}}}+1 \\\ & \text{putting the value of }\alpha \text{ and }\beta \text{,} \\\ & \dfrac{1}{\alpha }=\dfrac{1}{\beta }+1 \\\ & \beta =\dfrac{\alpha }{1-\alpha } \\\ & Or\text{ }\alpha \text{=}\dfrac{\beta }{1+\beta } \\\ \end{aligned}$
This is the relation between $\alpha \text{ and }\beta$.
Now, for the E transistor, we have
$\beta =100\text{ and }{{I}_{B}}=50\mu A$
Putting this value on the above equations
$\begin{aligned} & \alpha \text{=}\dfrac{\beta }{1+\beta } \\\ & \alpha \text{=}\dfrac{100}{1+100} \\\ & \alpha =\dfrac{100}{101} \\\ \end{aligned}$
Value of $\alpha$ is $\dfrac{100}{101}$
Again, to find the value of ${{I}_{C}}$,
$\begin{aligned} & \beta =\dfrac{{{I}_{C}}}{{{I}_{B}}} \\\ & {{I}_{C}}=\beta {{I}_{B}} \\\ & {{I}_{C}}=100\times 50\times {{10}^{-6}}A \\\ & {{I}_{C}}=5\times {{10}^{-3}}A \\\ & {{I}_{C}}=5mA \\\ \end{aligned}$
Also, to find the value of ${{I}_{E}}$
$\begin{aligned} & {{I}_{E}}={{I}_{B}}+{{I}_{C}} \\\ & {{I}_{E}}=50\mu A+5mA \\\ & {{I}_{E}}=.05mA+5mA \\\ & {{I}_{E}}=5.05mA \\\ \end{aligned}$
$\text{Value of }\alpha \text{ will be }\dfrac{100}{101},\text{ value of }{{\text{I}}_{C}}\text{ is 5mA and value of }{{\text{I}}_{E}}\text{ is 5}\text{.05mA}\text{.}$

Additional Information :Transistor is a n-p-n or p-n-p junction device. The central part is called the base and the other parts are called collector and emitter. The transistor can be made in such a way that either collector or emitter or base is common to both the input and output.
This gives three common configurations called common emitter, common collector and common base transistor.
For a transistor value of $\alpha$ will be always less than 1, because then collector current is always less than emitter current. Again, the value of $\beta$ is always greater than 1, because the value of collector current is always greater than base current.

Note :
Try to imagine which components are given in the question and what are to be found. If you can relate them then finding the solution will be very easy.
In the above question from the value of $\beta$ we have found the value of $\alpha$. We have proceeded in the same manner to find the other quantities.