# op-amp

Discussion in 'Homework Help' started by vvkannan, Nov 14, 2008.

1. ### vvkannan Thread Starter Active Member

Aug 9, 2008
138
11
Hello
I have a basic doubt about op-amps.By the concept of virtual ground the output produced tries to maintain the same voltage level at the two input terminals,which means the difference in input voltages is zero.
Also the output of an op-amp is proportional to the difference in inputs.
This is confusing for me.How can it produce an output if the inputs are at same potential(virtual ground).
Thank you

2. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Well the inputs are at the same potential for an ideal op-amp. For a real one this is not true because there is a small voltage difference (Vd) between the inputs enough to produce the output voltage (Vo=A*Vd). However, because the gain (A) is very large we can assume that Vd is approximately zero.

3. ### vvkannan Thread Starter Active Member

Aug 9, 2008
138
11
Op-amp would try to maintain same potential.But there would still be a very small potential difference between the two input terminals that would produce a finite output equal because of the large gain.Is this right?

If there is one thing as an ideal op-amp(though not possible) what wuld happen?
Lets say i build an op-amp for which i give input to one terminal almost equal to the gain(though not possible) . By the concept of virtual ground it should
produce large output that make the other terminal at a similar high potential.
Now there may be a small difference but how can it produce the same output.(because the gain multiplied by a number < 1 cant produce an voltage equal nearly to gain)
Though unlikely iam asking these questions just to make my understanding better.sorry if iam wrong
Thank you

4. ### steveb Senior Member

Jul 3, 2008
2,433
469
mmm..... a thought experiment. You may have a career as a theoretical physicist. Your are correct, the voltage difference would not be <1, or small compared to 1. But, what you would say is that the one volt difference on the input is small compared to the output voltage itself. So, it is comparitively small.

5. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Yes thats right.

The output is not necessary the same value as the input signal because in the feedback path there are resistors (or generally impedances) which act as a voltage divider. The output can be greater amplified or less (attenuated) depending on the value of these resistors.