# OP-AMP negative feedback intuitively

Discussion in 'Electronics Homework Help' started by ahmed fathi, Nov 5, 2013.

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Jun 25, 2013

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Jun 25, 2013
3. ### Harald KappModeratorModerator

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Nov 17, 2011
A=2 is the gain of the complete circuit. The gain of the operational amplifier is orders of magnitude higher.

Start by noting that the operational amplifier will generate an output voltage that is
Vout = A0*(Vp-Vn) where
A0 = open loop gain of the opamp
Vp = voltage at the "+" input =Vi in your circuit
Vn = voltage at the "-" input = Va in your circuit.
Va can be calculated from Vout via the resistive divider.

You will need two equations:
1) Va = f(Vout) where f(Vout) describes the resistive divider
2) Vout = f(Vi-Va) where f(Vi-Va) describes the amplification of the input voltage difference by the amplifier.

Insert 1) into 2) and solve for Vout/Vi. The expression will contain a term with A0 in the denominator. Now make A0 >> 1, assume A0 = infinity. What happens to the term containing A0? What happens conseqeuntly to the complete equation? This is the solution you're looking for.

4. ### Laplace

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Apr 4, 2010
I don't really understand this technique of iterating through the feedback loop. Never heard of it. Ideal op-amps have instantaneous response so iteration is not necessary. But real op-amps are not ideal. The linear operating region of a real op-amp exists only when the input terminals are within a few microvolts of each other. Outside the linear region, the op-amp response is limited by the slew rate which is a specification generally given on the datasheet for an op-amp. A large step input driving the input terminal, as given in your example, will not provide a bandwidth-limited response that you would get in the linear region but will instead provide a slew rate-limited response and the feedback equations are not valid until the op-amp gets back to its linear operating region.

5. ### ahmed fathi

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Jun 25, 2013
Well thanks.

I'm iterating this way because it's the way I understood Regenerative Feed back , For AB>1 , if we have small noise or pulse between the terminals , it will be multiplied by AB , reaching the input again , then multiplied by another AB and so on , so it's understandable

I just tried to understand negative feedback the same way , the signal is amplified then subtracted from the original signal ,then amplified then reaching the input then subtracted from original signal .. it works very well if AB<1 , and I can see that the output voltage will finally be stable after some oscillations, but when AB>1 , the feedback signal will be more than the original signal , so subtraction will lead to negative quantitiy , which is then multiplied and fed to input . now (Vi -(-ve quantitiy)) will cause Vout to be more and so On

I just want a way to understand the working of -Ve feedback the same way I understood regenerative positive feedback .

6. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Ahmed, that approach has its uses.

For example, using a simple numerical analysis both positive and negative feedback can be shown to limit the gain of an op-amp to a figure calculated using the ratio of resistances.

However a study of small perturbations in the input will show that one is stable and the other is not. (I look at your calculations and I wonder if this is what you are doing...)

7. ### ahmed fathi

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Jun 25, 2013
I think I found some approach to satisfy myself ) ... Thank you all very much Ask a Question
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