Question

A transistor has \[\alpha = 0.95\]. If the emitter current is \[10\,{\text{mA}}\], then the collector current will be
A. \[9.5\,{\text{mA}}\]
B. \[10\,{\text{mA}}\]
C. \[0.95\,{\text{mA}}\]
D. \[95\,{\text{mA}}\]

Answer 1

Hint: Use the formula for the current amplification factor. This formula gives the relation between the current amplification factor of the transistor, collector current and emitter current. Substitute the values of the current amplification factor of the given transistor and emitter current in the formula and determine the collector current.

Formula used:
The current amplification factor \[\alpha \] is given by
\[\alpha = \dfrac{{{I_C}}}{{{I_e}}}\] …… (1)
Here, \[{I_C}\] is the collector current and \[{I_e}\] is the emitter current.

Complete step by step answer:
We have given that the current amplification factor of the transistor is 0.95.
\[\alpha = 0.95\]
The emitter current is \[10\,{\text{mA}}\].
\[{I_e} = 10\,{\text{mA}}\]
We have asked to determine the collector current which can be determined using equation (1).

Rearrange the equation (1) for the collector current.
\[{I_C} = \alpha {I_e}\]
Substitute \[0.95\] for \[\alpha \] and \[10\,{\text{mA}}\] for \[{I_e}\] in the above equation.
\[{I_C} = \left( {0.95} \right)\left( {10\,{\text{mA}}} \right)\]
\[ \therefore {I_C} = 9.5\,{\text{mA}}\]

Therefore, the collector current will be \[9.5\,{\text{mA}}\]. Hence, the correct option is A.

Additional information:
The current amplification factor of a transistor is the ratio of collector current (current at collector terminal of transistor) to the emitter current (current at the emitter terminal of the transistor). Ideally, the current amplification factor can be 1 if the collector current and the emitter current has the same value. But practically, the value of current amplification factor ranges from 0.9 to 0.99.

Note: The students may convert the unit of the emitter current in the SI system of units. But there is no need for the unit conversion as the unit of the collector current will also be in the same unit which is milliampere. Hence, the unit conversion of the emitter current term will be a waste of time.