In an n-p-n transistor, 1010 electrons enter the emitter in 10–6 s. 2% of the electrons are lost in the base. Calculate the current transfer ratio and the current amplification factor. Charge on electron is .


Correct option is

0.98, 49


The current is rate of flow of change (q/t). Since 1010 electrons enter the emitter in 10–6 s, the emitter-current is



            = 1.6 mA

Similarly, the base-current is 



           = 0.032 mA.

In transistor, the emitter current  is the sum of the base current  and the collector current 


            = 1.6 mA – 0.032 mA     

            = 1.568 mA.

The current transfer ratio is 


            = 0.98. 

The current amplification factor is 


            = 49.



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