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Question:

For a common-emitter transistor, input current is 5µA, β= 100. Circuit is operated at load resistance of 10kΩ. The voltage across collector emitter will be?

12.5 V

10 V

7.5 V

5 V

Solution:

IB=5µA
IC=βIB
IC=100 × 5 × 10⁻⁶ A = 0.5mA
VCE=ICRL
VCE=0.5 × 10⁻³ × 10⁴
VCE=5 V