# How does neg feedback provide stability in linear region?

Discussion in 'General Electronics Chat' started by quique123, Jul 6, 2016.

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1. ### quique123 Thread Starter Member

May 15, 2015
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Im reading this:

This stability gives the op-amp the capacity to work in its linear (active) mode, as opposed to merely being saturated fully “on” or “off” as it was when used as a comparator, with no feedback at all.

By stability we mean it wont swing to V+ or V- or whatever its upper and higher limits are, right? It means the opamp can output a variety of voltages between the rails, or between GND and V+. But that would mean that it can only due that IF we add the negative feedback with resistors, right?

2. ### Papabravo Expert

Feb 24, 2006
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The resistors are not a requirement. It would be handy to know the supply voltage, because real devices will have trouble getting that close to the rails. As drawn the device is operating in it's linear region and you might want to clarify your definition of stability. I've always liked BIBO, or bounded input equals bounded output.

3. ### crutschow Expert

Mar 14, 2008
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Stability really has nothing to do with the output voltage swing of the amp which is largely determined by the amp gain and the input voltage.
Stability means that the operation of a circuit is largely defined by the passive element values of the feedback network and not by the response of the amplifier.
This is done by trading the large open loop gain of an op-amp for a much smaller defined gain, with much better gain stability, distortion, and frequency response.
The passive feedback elements are generally linear, and are relatively precise and stable, making the circuit operation also linear, precise, and stable.

4. ### quique123 Thread Starter Member

May 15, 2015
34
2
Ok I meant why does the opamp w/ neg-fb able to operate in the linear region whereas the opamp without it (ie comparator) saturates either + or -? So whereas the comparator could only produce -5 or +5, w/neg-fb it can operate from -5 to -4 to -3...to +3 to +4 to +5 and all values in between.
My question is, does this happen only if we add resistors as voltage dividers in the neg-fb or does it happen to any opamp with neg-fb? Im thinking as long as we can regulate Vin at the non-inverting (Vp) terminal we can make the Vout go through all the values and not just saturate.
Likewise, with the neg-fb loop in place, we can regulate what Vout will be through all the values and not just saturate. right?

5. ### Papabravo Expert

Feb 24, 2006
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It also happens in the case of no resistors if you configure the opamp as a voltage follower with a gain of one(1).

6. ### crutschow Expert

Mar 14, 2008
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An op amp can theoretically operate in the linear region with no feedback but it requires a very small, precisely adjusted, and stable voltage at the input.
Normally, because of the high open-loop gain of an op amp, noise, offsets, and thermal drift make it impractical to do that.
So any small input voltage will cause the op amp output to go to either the plus or minus output rail.

7. ### BobTPH Active Member

Jun 5, 2013
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An op amp operating without negative feedback has a gain on the order of 100K or 1M.

So, taking a gain of 1M, if the inputs can change only by only 10uV to get an output range of 10V.

And the actual gain is not well controlled. Different devices will have it vary considerably, so if you tune a circuit to work, then have to replace the device, it will not work.

Bob

8. ### quique123 Thread Starter Member

May 15, 2015
34
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Ok so I'm not clear then. I'm trying to see examples comparing how the neg-fb does allow voltage steps whereas the comparator doesn't.

9. ### Papabravo Expert

Feb 24, 2006
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OK so the first part of the understanding is that negative feedback reduces the gain from an order of 1,000,000 down to something a bit smaller like say 20, so that it takes a larger differential input to produce the same output voltage. In the open loop case if 10 μV (differential) produces 10 V, then at a gain of 20, the same 10 μv produces an output of 200 μV. Now for a gain of 20 you increase the differential input in steps of 10 μV and you get the steps you were looking for. That is 20 μV(differential) →400uV (output) and so on.

10. ### BobTPH Active Member

Jun 5, 2013
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And the gain of 20 with feedback is very precise. The open loop gain might be 1,000,000 or it might by 1,500,000. The specification only gives you the minimum and it is not well controlled in the manufacturing process.

Bob

11. ### quique123 Thread Starter Member

May 15, 2015
34
2
Ok so an open loop gain is so high, that small differences in Vinputs generate such huge Voutput, that its impratical to control anything in between all on or all off.
Whereas closed loop gain is lower, such that small differences in Vinputs DONT GENERATE huge Voutputs and so its easier to control Voutput to a certain value.

How can we see the large open loop gain and small closed loop gain mathematically?
Gain = Vout/ (Vp-Vn)
In open loop, Vout is huge because there is nothing feeding it back negatively. So Gain is huge.
In closed loop, Vout is less than huge or less than infinite, so Gain is not so big?

12. ### MrAl Well-Known Member

Jun 17, 2014
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Hi,

Mathematically the open loop gain is just something like A, which is considered a constant gain even though it changes a bit, so we have:
Vout=A*(vp-vn)

That's about it for the open loop gain, sometimes indicated with Aol:
Vout=Aol*(vp-vn)

The closed loop gain depends on the feedback arrangement. However, if it is just a voltage divider then that has a ratio of input to output that is less than 1. So say we have a feedback factor of 1/2. This means that we only get 1/2 of Vout at the inverting input, so vn=Vout*1/2=Vout/2.
We apply the input voltage to vp so we have a non inverting amplifier.

Now lets see what happens...
Since vn is different now, we insert that into the first equation above and get:
Vout=Aol*(vp-Vout/2)

Not too difficult right?

Now lets solve for Vout explicitly:
Vout=Aol*vp-Aol*Vout/2
Vout+Aol*Vout/2=Aol*vp
Vout*(1+Aol*Vout/2)=Aol*vp
Vout=Aol*vp/(1+Aol/2)
or:
Vout=vp/(1/Aol+1/2)

So that's the expression for Vout with feedback factor of 1/2.
For a feedback factor of B, we have:
Vout=vp/(1/Aol+B)

Not too complicated.

To see how the open loop gain affects the closed loop gain, just do this a few times with say vp=1 volt and B set to some value like 1/9, then vary Aol from say 100000 down to something like 10. You'll see why a high open loop gain is so desirable and closed loop operation is so much better.

Last edited: Jul 7, 2016
13. ### JohnInTX Moderator

Jun 26, 2012
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