Question

In an n-p-n transistor 10¹⁰ electrons enter the emitter in 10^-6 s. If 2% of the electrons are lost in the base. What is the current transfer ratio and the current amplification factor?

Options

• α = 0.98, β = 49

• α = 0.98, β = 0.98

• α = 49, β = 0.98

• None of these

Answer 1

α = 0.98, β = 49

Explanation:

`"I"_"e" = "q"/"t" = ("n"_"e" xx "e")/"t"`

`"I"_"C" = ("n"_"e" xx "e")/"t" = ((98/100) "n"_"e" xx "e")/"t" = 98/100 = 0.98`

and current amplification factor

`beta = alpha/(1 - alpha)  0.98/(1 - 0.98) = 0.98/0.02 = 49`