# Op Amp Negative Feedback

Discussion in 'General Electronics Chat' started by vindicate, Dec 15, 2009.

1. ### vindicate Thread Starter Active Member

Jul 9, 2009
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I'm trying to understand the concept of negative feedback in an op-amp. I've read the wikipedia article and the allboutcircuits article and have a couple of questions.

If you have 2 inputs, and inverting and non-inverting, the ouput will be the polarity of whichever one has a higher voltage(if no negative feedback is used). Basically if the non-inverting has a higher voltage the output is positive and the out voltage will be whatever v+ is. This happens because the gain is so high it saturates easily.

Hopefully that is right.

Then it gets muddy when it comes to negative feedback. Why does the output "follow" the input? If both the inverting and non inverting inputs are the same voltage shouldn't the output be 0V?

The way I see it. If the non-inverting input is 5v and Vin+ is 15v. The output will be 15V and than negative feed back into the system making the non-inverting ouput higher and switch Vout to Vin-.

I just need a good explanation of how the negative voltage works.

2. ### Duane P Wetick Active Member

Apr 23, 2009
408
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Probably the simplest explanation is the thermostat in your home. Interestingly, the US Patent office rejected the first patent application on negative feedback saying that it was impossible.

Cheers, DPW [ Spent years making heaters out of op-amps]

Last edited: Dec 15, 2009

Apr 14, 2005
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Jul 9, 2009
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5. ### vindicate Thread Starter Active Member

Jul 9, 2009
158
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Can anyone offer some insight

6. ### Wendy Moderator

Mar 24, 2008
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Negative feedback is self correcting, if it overshoots it compensates. Basically it is trying to make both inputs the same exact voltage. Positive feedback does exactly the opposite, it causes the output to go to maximum values and stay there. Both have uses with op amps, and both look very similar, the differences are subtle. What is it you need help understanding?

7. ### eblc1388 Senior Member

Nov 28, 2008
1,542
102
In real life they are never exactly the same but differ by an extremely small amount. But practically they appear to be the same to an user.

If you want to know about the magnitude of this difference, you can just divide the opamp output(whatever voltage) by the open loop gain of the opamp.

8. ### vindicate Thread Starter Active Member

Jul 9, 2009
158
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So what happens with negative feedback, is that it somehow feedsback just the right amount of voltage to the inverting input that the difference between it and the non-inverting input x Gain = the original voltage of the non-inverting input? And then the output voltage will be the the same voltage as the noninverting input, even if V+'s voltage changes.

How is that possible?

9. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Draw a classic op amp circuit and follow what happens with three conditions, too high, too low, and just right. While eblc1388 is correct, for beginners assume the gain is infinite (an ideal op amp).

1. The negative input is lower (more negative voltage) than the positive input. The output swings positive.

2. The negative input is higher (more positive) than the positive input, the output swings negative.

3. The only balanced state is where the two inputs are the same, if there is any variation conditions 1 and 2 apply. The output assumes whatever voltage is required to make the inputs the same.

10. ### eblc1388 Senior Member

Nov 28, 2008
1,542
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It is possible because it uses negative feedback technique.

Say your non-inverting input is at 5V and the opamp output at say 7V, because of this condition, the inverting input(which is connected to the opamp output in a voltage follower) is also 7V and thus the opamp will actively reduces its output voltage because voltage at IN- is higher than IN+.

But when will it stop reducing its output? It stops when the voltage difference between IN+ and IN- multiplied by the loop gain equals to the IN+ voltage. But as this difference is extremely small in real life, we usually say it stops when the output is equal to the IN+ voltage.

If the opamp output voltage drops, for example someone connects a load to the opamp output, then IN+ is now higher than IN- and the opamp will increases its output to compensate.

11. ### vindicate Thread Starter Active Member

Jul 9, 2009
158
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In your example. In+ is more positive so output is 7V. It feedsback the 7V to In-. In- is now more positive, so wouldn't Output then be whatever V- is? Probably -7 V or 0V?

12. ### eblc1388 Senior Member

Nov 28, 2008
1,542
102
Voltage on IN+ is determined by external applied voltage and is not in anyway affects by the opamp output performance.

13. ### Ron H AAC Fanatic!

Apr 14, 2005
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But if the output went more negative than the input, the negative feedback would force it back in the positive direction, to equilibrium.
I think you are imagining an oscillatory state. This can happen if the op amp has delay between input and output, but op amps are internally compensated so that oscillation won't occur.

14. ### vindicate Thread Starter Active Member

Jul 9, 2009
158
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Yes that is what I'm imagining. So there is something internal to the opamp that will keep my scenario from happening?

15. ### Ron H AAC Fanatic!

Apr 14, 2005
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Yes. Keep in mind that the negative feedback provides 180° phase shift around the feedback loop, due to the inversion. If the internal delay of the op amp provides another 180° phase shift at some frequency, the op amp will oscillate if the gain is greater than 1 at that frequency. The internal compensation guarantees that the open loop gain will always be less than 1 at frequencies where the internal delay can cause 180° phase shift.

16. ### vindicate Thread Starter Active Member

Jul 9, 2009
158
0
Thank you everyone for your help. I think I understand it...enough anyways haha.