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Question:

In the circuit shown in the figure, the input voltage vi is 20V, VBE=0 and VCE=0. The values of IB, IC and β are given by.

IB=20µA,IC=5mA,β=250

IB=40µA,IC=5mA,β=125

IB=40µA,IC=10mA,β=250

IB=25µA,IC=5mA,β=200

Solution:

Vi=VBE+RBiB ———(1)
20=500×103×iB
iB=40µA
Vo=VCE−RCiC ———(2)
20=4×103iC
iC=5mA
β=IC/IB=5mA/40µA=125