Transistor

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Question

In a n-p-n transistor 10¹⁰ electrons enter the emitter in 10^-6 s. 4% of the electrons are lost in base. The current transfer ratio will be

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Solution

Current transfer ratio is $α = \frac{I_{c}}{I_{E}} = \frac{I_{E} - I_{B}}{I_{E}} = \frac{100 - 4}{100} = 0.96$

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