Calculate the value of IB if IE is 1 milli amps, IC is 0.95 milli amps in a common base transistor configuration a.0.04b.-0.07c.0.07d.0.05e.-0.04
Question
Calculate the value of IB if IE is 1 milli amps, IC is 0.95 milli amps in a common base transistor configuration a.0.04b.-0.07c.0.07d.0.05e.-0.04
Solution
In a common base transistor configuration, the emitter current (IE) is the sum of the base current (IB) and the collector current (IC). This relationship can be expressed by the equation:
IE = IB + IC
We can rearrange this equation to solve for IB:
IB = IE - IC
Given that IE is 1 milliamp (mA) and IC is 0.95 mA, we can substitute these values into the equation:
IB = 1 mA - 0.95 mA = 0.05 mA
So, the value of IB is 0.05 mA. Therefore, the correct answer is d. 0.05.
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