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Question:

The transfer ratio β of a transistor is 50. The input resistance of the transistor, when used in the common emitter mode is 1kΩ. The peak value of the collector alternating current for an input peak voltage of 0.01 V is?

0.25µA

500µA

0.01µA

100µA

Solution:

The input peak voltage is Vin = 0.01V
The input resistance = Rin = 1kΩ
Hence the peak in input current = ib = Vin/Rin = 10µA
The transfer ratio is defined as ratio of output peak current to input peak current.
⇒ ic/ib = β
⇒ ic = βib = 500µA