# How does negative feedback in an op-amp work conceptually.

1. Mar 7, 2012

### glenn21

Consider an op-amp in which there is a DC voltage of 100mV for example applied to the non-inverting (positive) terminal of the op-amp and the negative terminal of the op-amp is connected to the output directly. Also take the voltage supplied to power the op-amp, Vcc, to be 15V and the gain, A, to be 10000 for example.

Now from my understanding, excluding common-mode signals, the output voltage of the op-amp is given by: vo=A(vp-vn)

Initially vo=10000(0.1-0) = 1000V (theoretically) (But because Vcc is 15V the voltage will be clipped to 15V)

Then this is fed back to vn and so vn=15V.

vo=10000(0.1-15) = -149000V (again it will be clipped but to -15V this time).

Now, vo=10000(0.1+15) = 151000V (clipped to 15V)

So I don't understand how this works as a voltage follower, to me it seems the output is constantly oscillating between positive and negative saturation? Any help on what is going on would be great.

I understand from mathematics derivation that you argue that because Vo/A = 0 if A is large then vp=vn=vo so we have unity gain. But then that doesn't exactly make sense because if negative feedback isn't applied vp!=vn always. But from a practical standpoint I don't understand how unity gain is achieved (I don't like just accepting things).

2. Mar 7, 2012

### Staff: Mentor

As the output voltage (and hence the - input voltage) passes through the + input voltage, what happens...?

3. Mar 7, 2012

### glenn21

As the output voltage passes through the positive input voltage there will be a voltage (appearing across the input resistor Rin) causing a new output voltage vo. If for example the positive voltage is 10mV and the gain is 1000 there will be 10V at the negative terminal. Then as the voltage passes through there will be a larger differential input voltage appearing across the terminals, but this time the positive terminal is less than the negative terminal so the output will be a negative value.

I'm referring to the model presented under the title Ideal op-amps.
http://en.wikipedia.org/wiki/Operational_amplifier

Last edited: Mar 7, 2012
4. Mar 8, 2012

### jim hardy

We all have trouble when op-amps are first introduced.
It seems counterintuitive that something with such high gain could behave itself.

Try reversing your thinking. I find it helpful to anthropomorphize and imagine myself inside that little triangle.
It is the duty of the designer to surround the op-amp with a circuit that allows it to force its inputs equal. To that end, usually he gives the opamp contol over its inverting input.

So - when Mr Opamp's inverting input is tied to output, does he have control over it? Of course.

In your example, if Mr Opamp sets his output to 100 mv has he forced his inputs equal?
Of course.
Can he do that ? sure.

Now to your point - he has gain of let's say a million.
So to produce output of 0.1 volt requires that he set inputs to 0.1 microvolt differential.

So, instead of 0.1 volt output he will settle at 0.099999 volt out , providing that 0.1 microvolt differential input.
We usually ignore that small differential of 1/(open-loop-gain). Observe how close is 0.099999 to 0.10000 , only an accountant could complain.

Now - your oscillation scenario is sure possible if there's a delay in getting the inputs balanced and you'll study that soon enough when get to "phase margins".

But today it's more important for you to get your mind working the op-amp as a device that wants to make his inuts equal.
It's the job of designer to surround him with a circuit that allows that.

5. Mar 10, 2012

### glenn21

Hi Jim, thanks for your response.

I know that an op-amp is a device that wants to make it's inputs equal. I just don't understand how it forces it's inputs to be equal, and that is what I want to understand.

Maybe my misunderstanding has got to do with me thinking that before the positive input voltage is applied to the op-amp the negative terminal is at 0V? That's the only thing I can think of. Because even if you start applying a small voltage say 1microvolt to the positive input you will get 1V fed-back to the negative terminal resulting in a larger differential input then before, and hence the circuit eventually goes into the oscillation state.

This is my current thought process. If for example 100mV is applied to the non-inverting input, there is almost zero current passing through Rin and so there is almost no voltage drop across it, so vn is almost 100mV (with slight losses say a nanovolt or so) then the op-amp output becomes 100mV, so we say if A is large then the non-inverting and inverting inputs are approx equal. I've shown pictorially below what I mean using the model I mentioned in a previous post.

Is what I'm describing correct now?

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Last edited: Mar 11, 2012
6. Mar 11, 2012

### willem2

Opamp inputs have a very high impedance, and the small input current isn't what makes the inputs approximately equal.
What does make the inputs equal is the feedback network, but this only works because the response of the output of the opamp isn't immediate.
If you put 100 mV on the non-inverting input and the inverting input and the output starts out at 0 V (because everything was At 0V before you put 100mV on the non-inverting input), the output won't immediately go to the supply voltage, but will rise with a finite speed. Because of this, the non-inverting input will be at 99.9999 mV at some point, and then the output of the opamp will be stable at 100mV.

Overshoot of this voltage, and even oscillation is indeed possible, if the gain of the opamp is large enough at high frequencies. Normally you want to avoid that, see

http://en.wikipedia.org/wiki/Frequency_compensation

7. Mar 11, 2012

### glenn21

@willem2

I didn't even think about the finite rise time, that makes perfect sense, thank you so much!