Question

If $$\alpha =0.98$$and current through emitter$$I_E=20mA$$, the value of$$\beta$$ is
$$\begin{array}{rl}& A)4.9\\ & B)49\\ & C)96\\ & D)9.6\end{array}$$

Answer 1

Hint: Here, values of current gain and emitter current are given. To find the value of $$\beta$$(current gain factor in the common emitter configuration), we need values of collector current and base current. Collector current can be determined using the formula for $$\alpha$$(current gain in common base configuration). And we know that, emitter current is the sum of collector and base currents. Therefore, base current value can be determined using this equation and thereby we can find the value of$$\beta$$.
Formula used:
$$\alpha ={\displaystyle \frac{I_C}{I_E}}$$
$$\beta ={\displaystyle \frac{I_C}{I_B}}$$
$$I_E=I_C+I_B$$

Complete answer:
$$\beta$$ is the current gain factor in the common emitter configuration. It is the ratio of the collector current (output current) to the base current (output current).
Let us have a look at the CE configuration.

A transistor is known to consist of 3 terminals, namely, emitter (E), base (B) and collector (C). The input and output connections can be made by setting any one of these three terminals as common. For the CE configuration, the input is set in between the base and the emitter and the output is placed in between collector and the emitter. Input characteristics is defined as the variation of base current $I_B$ with base-emitter voltage$V_{BE}$ , similarly, the output characteristics is defined as the variation of collector current $I_C$with collector-emitter voltage $V_{CE}$.
$$\beta ={\displaystyle \frac{I_C}{I_B}}$$ --------- 1
Where,
$$I_C$$ is the collector current
$$I_B$$ is the base current.
We have,
$$\alpha ={\displaystyle \frac{I_C}{I_E}}$$
Where,
$$I_E$$ is the emitter current
 $$\alpha$$ is the large signal current gain in common base configuration. It is the ratio of collector current to the emitter current or it is the ratio of output current to input current in common base configuration.
Then,
$$I_C=\alpha I_E$$ --------- 2
Given that,
$$\alpha =0.98$$
$$I_E=20mA$$
Substitute the value of $$\alpha$$and $$I_E$$ in equation 2. We get,
$$I_C=0.98\times 20=19.6mA$$
We know that emitter current is the sum of collector current and base current. i.e.,
Emitter current, $$I_E=I_C+I_B$$ ----------- 3
Substitute the value of $$I_C$$and$$I_E$$in equation 3. We get,
$$20=19.6+I_B$$
$$I_B=0.4mA$$
Substituting the values in equation 1, we get,
$$\beta ={\displaystyle \frac{19.6}{0.4}}=49mA$$

Therefore, answer is option B .

Note:
Alternate method to find the value of$$\beta$$.
The relation between $$\alpha$$ and$$\beta$$are given by,
$$\beta ={\displaystyle \frac{\alpha }{1-\alpha }}$$
Given that,
$$\alpha =0.98$$
Then,
$$\beta ={\displaystyle \frac{\alpha }{1-\alpha }}={\displaystyle \frac{0.98}{1-0.98}}=49$$