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Q. Prove mathematically that the operating point does not depend on beta, in a potential divider bias circuit ?
To determine the operating point, consider the input section of the voltage divider biasing circuit.
The base current IB is very small compared to the currents in R1 and R2.
I1 ~ I2
So,
VB=R2/ (R1+R2) * VCC
The voltage across the emitter resistor RE
VE=V2-VBE
Then the emitter current is
IE= V2-VBE
And
VC=VCC-ICRC
VCE=VC-VE = (VCC-ICRC)-IERE
Since collector current is approximately equal to emitter current
VCE=VCC-(RC+RE) IC
Since in the above analysis nowhere does beta appear in any equation, which means that operating point does not depend upon the value of beta.
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