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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.
For the circuit shown in Figure, use KCL and KVL to determine i 1 , i 2 , v bd and v x . Also, find v eb .
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