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A transistor connected in a common base configuration has the following readings IE = 2 mA and IB = 20 μA. Find the current gain α.
1. 0.95
2. 1.98
3. 0.99
4. 0.98

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Correct Answer - Option 3 : 0.99

Current amplification factor: It is defined as the ratio of the output current to the input current. In the common-base configuration, the output current is emitter current IC, whereas the input current is base current IE.

Thus, the ratio of change in collector current to the change in the emitter current is known as the current amplification factor. It is expressed by the α.

\(\alpha = \frac{{{\rm{\Delta }}{I_C}}}{{{\rm{\Delta }}{I_E}}}\)

Where, IE = IC + IB

Calculation:

Given,

IE = 2 mA

IB = 20 μA = 0.02 mA

From above concept,

IC = 2 mA - 0.02 mA = 1.98 mA

Current amplification factor is given as,

\(\alpha=\frac{I_C}{I_B}=\frac{1.98}{2}=0.99\)

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