# Op-Amp Feedback Basics

Discussion in 'General Electronics Chat' started by ActivePower, Feb 23, 2014.

1. ### ActivePower Thread Starter Active Member

Mar 15, 2012
155
23
I am revisiting operational amplifiers and negative feedback and having some trouble getting it well. I'd appreciate it if anyone could verify/correct my understanding at the moment.

The way I understand it, negative feedback serves as an adjusting/controlling connection which ensures the the opamp output voltage settles to a definite voltage instead of just hitting the extremes (set by adjusting the fraction of voltage sent from the output to the input using a voltage divider network).

To this end, it drives the voltage difference between the terminals to a very low value (almost but not exactly zero) which is why both the terminals appear to be shorted together in negative-feedback connections.

On the other hand, positive feedback drives the opamp harder in the direction it is already going.

So far so good. To test my understanding I drew up four basic opamp configurations in an attempt to predict the output voltage in each case.

CircuitLab Schematic

According to what I presently understand, I figured the output voltages should be:
(a) The negative feedback would try to minimize the input volt difference which gives, $V_{out}$ + $V_{2}$ = $V_{1}$.
Thus, output would probably be $V_{1}$-$V_{2}$.

(c) I figured the output would be driven to saturation because of the positive feedback. Whether it goes to positive or negative saturation is decided by which is higher $V_{1}$ or $V_{2}$. Output voltage goes to positive saturation if $V_{1}$ is higher, negative otherwise.

(b) The same as (a) above goes for this where $V_{2}$ would be set to zero giving us the familiar voltage buffer topology.

(d) This one is still a conundrum for me. I guessed it would depend on what the output voltage is at any point of time. Say the output voltage is higher than the voltage at the negative terminal it would be driven to positive saturation and negative otherwise.

Last edited: Feb 23, 2014
2. ### LvW Active Member

Jun 13, 2013
674
100
Hi,

a) There is no negative feedback at all because V2 produces a short circuit for the feedback signal. Applying the superposition rule the output voltage is identical to the voltage at the inv. terminal Vout=V2=1V.
(V1 at the pos. input has no effect)

b) The classical unity gain feedback amplifier (buffer). Vout=1V.
(Why do you see "positive feedback" for this circuit?)

c) and d) Both circuits with positive dc feedback, Hence, no linear operating point. No gain at all.

3. ### ActivePower Thread Starter Active Member

Mar 15, 2012
155
23
Sorry I wrongly labelled the circuit parts (b) and (c) while drawing them from my notebook. I've edited the same.

I am pretty sure I am missing some pretty basic things here, please bear with me.
Why do you say that there is no negative feedback in (a) but there is positive feedback in (c) and (d)?
I understand that the short between the -ve terminal of the opamp and the output would clamp the output to the voltage fixed by the stiff voltage source but doesn't the same thing hold for (c) and (d)?
What do you mean when you say that there is 'no gain' in (c) and (d)? What would be the output? Does this mean the output would hit the extremes or is it unpredictable?

Last edited: Feb 23, 2014
4. ### LvW Active Member

Jun 13, 2013
674
100
@ a) As I have mentioned, the output cannot cause any feedback signal at the inv. input because V2 (internal source resistance zero) is connected to this input. More than that, the voltage V2 is also connected directly at the output node.

@ c) Yes - correct. It is no positive feedback. Rather, it is no feedback at all and the input source is feeding directly the output (as in a).

An opamp circuit has "gain" as long as it is operated in the linear region of its transfer characteristic. This is not the case in c) and d).

5. ### crutschow Expert

Mar 14, 2008
16,571
4,476
b) is a normal follower circuit with a gain of +1, as noted.

a) and c) are not viable op amp connections and their outputs are the same whether the amp is there or not since the input voltages (V2 for a and V3 for c) override the op amp outputs (assuming ideal voltage sources).

The positive feedback in (d) means the output will go to one of the rail voltages and stay there.

6. ### MrChips Moderator

Oct 2, 2009
14,520
4,282
You cannot connect two voltage sources to each other.
While you can do this in practice, you cannot solve this in theory unless you take into account all resistances.

Remember that the output of the opamp is a voltage source.

To resolve this, place a resistor between the two voltage sources.

May 11, 2009
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8. ### crutschow Expert

Mar 14, 2008
16,571
4,476
A basic op amp connection you should look at is the inverting configuration with the (+) input to ground, an input resistor to the (-) input, and a feedback resistor from the (-) input to the output. As an exercise, predict the output for a given input for varying ratios of the input to feedback resistors.

9. ### ActivePower Thread Starter Active Member

Mar 15, 2012
155
23
Thanks for all the useful replies. I was never really any good at analog and I'm trying to iron out all the gaps in understanding that have accumulated over years of neglect.
I'm just attempting to gain some intuitive insights, not dependent entirely on equations, in order to 'think' like a circuit designer (and maybe build some).
(I may seem ignorant - it is probably because I am)

@MrChips: Indeed I seem to have forgotten that. Thank you for pointing it out.

@t06: That is a really great reference. I've read the first couple of pages into it now. Seems excellent for a resource that is free.

@crutschow: I did do as you suggested. Could you verify my understanding? Here goes:

The feedback resistor added along the path from the output to the negative input terminal would drop a certain amount of output voltage and would result in a certain net sum of the applied (source) and the output voltage appearing at the input terminal (which should approach zero as the non-inverting terminal is grounded).

Thus, adjusting the value of the feedback resistor would adjust the fraction of output voltage that must be 'sent' to the inverting input.

Let us begin at the moment we apply the source voltage to the inverting terminal. Say we apply a +5 V on the negative end with the source and feedback resistors set at 1 kΩ and 3 kΩ respectively. In addition, let us assume that there exists negligible output voltage at the moment we switch on the circuit (I don't know if this is a reasonable assumption).
The voltage divider network would split the applied voltage with most potential drop occurring across the feedback resistor. The voltage at the inverting terminal (-) which is equal to the voltage across the feedback resistor at this moment ($V_{out}$ = 0) isn't zero. This drives the high-gain amp to pump a very large momentary voltage to the output.

This large momentary voltage attempts to nullify the voltage at the inverting terminal. If the source resistance is low and the feedback resistance is higher (as in our example) this would mean that the voltage at the inverting point would fall quicker. After some oscillations the voltage at the -ve terminal does manage to reach a stable settling point where the input error voltage is almost zero at which point the output voltage is left undisturbed.

If the ratio of source to feedback resistance is lower, the output voltage would have to fall lower and lower to get the input error voltage to reach zero volt.

I admit this explanation seems far too stretched out even bordering on idiotic but I'd appreciate it if I could get an intuitive view of the whole process.

Thanks.

10. ### LvW Active Member

Jun 13, 2013
674
100
ActivePower, I`ve got the impression that you didn`t consider the fact that a second resistor between the input source and the inv. opamp terminal is required.
Otherwise it is not possible to feed back a certain portion of the output signal. Without such a resistor, your circuit acts as a current-to-voltage converter.

11. ### crutschow Expert

Mar 14, 2008
16,571
4,476
Let's start with the ideal op amp in the inverting configuration that has infinite open loop gain and an input current of zero. This means the voltage across the two inputs is zero when the op amp is operating in the linear region. (Most op amps have such a high open loop gain that this can often be assumed for many op amp circuit designs).

Now look at the circuit from a current viewpoint, not a voltage---

For the inverting configuration the input current would be Vin /Rin since the (-) input is at 0V (due to the feedback and the (+) input being at ground).

Then current through Rfb must be Vo/Rfb, again since the (-) input is at zero volts.

Therefore Vin/Rin = -Vo/Rfb. The reason for the minus sign is that a positive current from Vin must be balanced by a negative current from Vo if the (-) input is to stay at 0V. (This works because at the (-) input any small positive input voltage will cause the output to go negative and vice versa).

Rearranging terms gives a circuit gain of Vo/Vin = -Rfb/Rin. The minus sign means the circuit is a inverter with a 180° phase shift between input and output.

Note that with a typical op amp the oscillations you mentioned at startup or for a change in input voltage do not occur. This is because the op amps are internally frequency compensated to be stable with feedback and just go smoothly to the new output value with no significant overshoot or oscillations.

Last edited: Feb 24, 2014

Jan 3, 2012
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13. ### LvW Active Member

Jun 13, 2013
674
100
ActivePower, I think that I can imagine what you mean.
You try to describe the process of finding a fixed bias point (dc output voltage) resulting from a voltage step of the source voltage (starting at 0 volts), correct?
Your explanation is based on the assumption that in the very first moment the feedback isn`t active yet and the output would jump imediately to the maximum possible voltage (power rail).
However, this is not the case because the opamp has an inherent delay - mostly caused by the compensation capacity that is responsible for the limited slew rate of the device.
Hence, there remains always "enough time" for the feedback network to "feel" the rising of the output voltage and to reduce the opamp´s effective differental input voltage and, thus, finding a point of equilibrium without "up and downs" (I wouldn`t say: oscillations) around this final steady-state point.

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14. ### ActivePower Thread Starter Active Member

Mar 15, 2012
155
23
@crutschow: I do understand the math and the quantitative analysis. I was able to derive the formulae behind the basic opamp topologies - non-inverting, inverting, summing, integrating, differentiating amplifiers and the like.

What I am after is a qualitative understanding which I was not able to gain from reading the 'standard' texts (I haven't checked Art of Electronics yet). Our college course on the subject was mainly dry and quantitative with a very little (if any) design component. I'd like to be able to explain a circuit or at least gain some measure of understanding about it without delving head-first into equations.

@LvW: That indeed was my thinking. On second thoughts, I think I could have phrased my previous answer better. If I understand correctly, the limitation of a real opamp would be the lack of an instant response to an applied input. This delay would negate the fluctuations in finding the final settling point. Correct?

Thanks for all the replies.

15. ### LvW Active Member

Jun 13, 2013
674
100
Yes - I think, this is what I wanted to express - however, with other words.
As something like a counter-example, we could imagine that the opamp has no delay and no slew rate limitation, but the feedback has a certain inherent delay.
I think, in this case - caused by an input voltage step - the opamp`s output would perhaps toggle several times between both power rails until it finds the equilibrium point. Perhaps tomorrow I can try to proof this expactation via simulation.

EDIT: No - I don`t think the "counter-example" will work as expected. Hence, forget the last tree sentences.

Last edited: Feb 25, 2014
16. ### crutschow Expert

Mar 14, 2008
16,571
4,476
If you had a perfect op amp with infinite bandwidth and infinite gain then the output would instantly follow the input and there would be no fluctuations in the output. In a real op amp, due to finite bandwidth (and the consequent phase shift at high frequencies which causes positive high frequency feedback), a compensation circuit is added internally to insure that the output is stable with negative feedback and has no significant overshoot or ringing.