Hi,
I studied in opamps that I both the input terminal of the opamp will be at the same potential.i got explanation for that in the forum .but if both terminals are at the same potential, then (V1-V2)=0.hence Vo=0. But there is an output not equal to zero.how,is it possible?
even if suppose both input pins are shorted to each other, the opamp dissimilarity brings some deviation and perhaps that is why opamps have offset null adjustment to minimize this error.
Jayanthy,
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Let us say that the output is indeed 0.1V (not zero). The gain of the opamp, Vout/(V1-V2), is usually very high as 100,000 for example (open loop).
(V1-V2) = Vout / 100000 = 0.1/100000 = 0.000001 V (1µV)
So most of the time we accept that V1=V2 :wink:
Kerim
Edited:
Even if the output becomes 5V, V1-V2 in our example would be 0.00005V (50uV).
It won't be a crime if we write during the calculation that V1=V2
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that's fine for the open loop.suppose i choose a closed loop negetive feedback,then A=-Rf/Rin.if A=10,and Vi=1V,the output is 10V => (V1-V2)=1V
and hence the approximation fails.but still we take a virtual ground and solve.How is this possible?
About the open loop,i made this up from all your ideas:as the gain is large and as the difference is amplified,if the difference is large the opamp reaches saturation.hence the difference has to be small.hence equal.please conform this,and about the closed loop.
that's fine for the open loop.suppose i choose a closed loop negetive feedback,then A=-Rf/Rin.if A=10,and Vi=1V,the output is 10V => (V1-V2)=1V
and hence the approximation fails.but still we take a virtual ground and solve.How is this possible?
About the open loop,i made this up from all your ideas:as the gain is large and as the difference is amplified,if the difference is large the opamp reaches saturation.hence the difference has to be small.hence equal.please conform this,and about the closed loop.
And about the feedback, is it from output to input or is it other way round as you told.
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one thing I noticed is,reduction in gain is actually useful in the case of opamp unlike transistors.
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