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maninnet said:get stuck on a fundamental question on feedback, looks simple, but just can not figure out what goes wrong, please refer to the attachment
A.Anand Srinivasan said:your beta is wrong... the output is sampled and fed back to input as current... here Vo is across R2 and hence -(Vo/R2)*R1 is the feedback and hence beta is -R2/R1... the negative sign is due to the polarity of Vo...
aryajur said:Actually this circuit has been analysed both as a voltage-voltage feedback and voltage current feedback circuit, your β would change in both cases.
Fawad Elahi said:think why we need negative feedback.
the one reason is to reduce the output to a acceptable value.
but there is also a second reason.
leg1234
V(-)=R2/(R1+R2)*Vin+R1/(R1+R2)*Vo
Vamsi Mocherla
The analysis is pretty easy here. Firstly, I do accept that the feedback factor is still R1/R1+R2, with R2 being the feeback resistor. But this is with respect to the virtual ground, since the positive node is connected to ground. Hence assuming that the gain is high enough, the voltage at this node is Vin* (R2/R1+R2) using superposition and Vout = 0. Now if you multiply this with the inverse of the feedback factor, you will get Vout = Vin * (-R2/R1). The minus sign comes because of the negative input put forth.