How does the feedback work in an inverting op amp amplifier?

[crutschow]

View attachment 90929
Can someone give me a hint about why the output voltage is less than the input in the right circuit? Where should I start to analyze it? The input is a sine wave with a magnitude of 1 volt and frequency 100Hz.

Both your circuits have the capacitor directly on the output, so the response depends upon whatever the amp's output impedance is. That's not a normal or proper way to connect the capacitor so there's no good reason to try to analyze those circuits.
That's why my circuit has a resistor between the output and the capacitor (on second look I realize that R1 is not needed).
The feedback circuit is a simple LP filter that rolls off the AC feedback above its corner frequency (in this case the RC time-constant is 100F * 1kΩ = 100ks for a corner frequency of 1.59 microHz).
(I realize that 100F is not a practical capacitance but the simulator doesn't know that).
This means there is no significant AC feedback down to a below 1Hz , thus the output is the open loop AC response.

 

[Heidi]

Both your circuits have the capacitor directly on the output, so the response depends upon whatever the amp's output impedance is. That's not a normal or proper way to connect the capacitor so there's no good reason to try to analyze those circuits.
That's why my circuit has a resistor between the output and the capacitor (on second look I realize that R1 is not needed).
The feedback circuit is a simple LP filter that rolls off the AC feedback above its corner frequency (in this case the RC time-constant is 100F * 1kΩ = 100ks for a corner frequency of 1.59 microHz).
(I realize that 100F is not a practical capacitance but the simulator doesn't know that).
This means there is no significant AC feedback down to a below 1Hz , thus the output is the open loop AC response.

The output voltag magnitude in your top circuit is equal to the open-loop gain of the op amp only at very low frequencies. An op amp's gain will decrease with frequency, but for other frequencies, the output doesn't represent the op amp's gain for those frequencies. Is it correct?

 

[crutschow]

View attachment 90955
The output voltag magnitude in your top circuit is equal to the open-loop gain of the op amp only at very low frequencies. An op amp's gain will decrease with frequency, but for other frequencies, the output doesn't represent the op amp's gain for those frequencies. Is it correct?

You have it backward.
The voltage output of that circuit is the true open-loop gain at any frequency above a very low frequency (about 1 Hz for the values shown), since there is no significant negative feedback above that frequency.

If you look at the open loop gain in a real op amp's data sheet you will see a similar Bode plot.

 

[crutschow]

Double post.

 

[Heidi]

You have it backward.
The voltage output of that circuit is the true open-loop gain at any frequency above a very low frequency (about 1 Hz for the values shown), since there is no significant negative feedback above that frequency.

If you look at the open loop gain in a real op amp's data sheet you will see a similar Bode plot.

Thank you very much, crutschow. May I ask you a few more questions?

If we don't want the AC feedback to go to the inverting terminal, can we use the left hand circuit below and take its output as the op amp's open-loop gain?

As for the unit gain follower, the "gain" which is equal to -3dB at 1 MHz is refreeing to the voltage follower's gain, the open-loop gain of the op amp itself is still 0 dB at 1 MHz, is it correct? What else can the circuit tell us? The AC feedback totally go to the inverting terminal?

 

[crutschow]

........................
If we don't want the AC feedback to go to the inverting terminal, can we use the left hand circuit below and take its output as the op amp's open-loop gain?
View attachment 90966
As for the unit gain follower, the "gain" which is equal to -3dB at 1 MHz is refreeing to the voltage follower's gain, the open-loop gain of the op amp itself is still 0 dB at 1 MHz, is it correct? What else can the circuit tell us? The AC feedback totally go to the inverting terminal?
View attachment 90963

That may work in simulation with the ideal op amp model but any real op amp will saturate at one of the supply rails due to the high open-loop gain and the small input offset that all real op amps have.
Try the simulation with one of the op amp models for real devices with power supplies and you will see that it doesn't give the normal open-loop plot.

Yes, the unity gain follower output gain is obviously the gain for that configuration.
That has no effect on the open loop gain.
The circuit can tell you whatever else you might want to know about a follower (large signal AC response, transient response, etc.)
The AC and DC feedback transfer function from the output to the minus input is indeed 1. I don't understand your question mark for that.

 

[Heidi]

That may work in simulation with the ideal op amp model but any real op amp will saturate at one of the supply rails due to the high open-loop gain and the small input offset that all real op amps have.
Try the simulation with one of the op amp models for real devices with power supplies and you will see that it doesn't give the normal open-loop plot.

Oh, yes, it would saturate. What a silly question!
Thank you very much, crutschow.

 

[Heidi]

The feedback circuit is a simple LP filter that rolls off the AC feedback above its corner frequency (in this case the RC time-constant is 100F * 1kΩ = 100ks for a corner frequency of 1.59 microHz).
(I realize that 100F is not a practical capacitance but the simulator doesn't know that).
This means there is no significant AC feedback down to a below 1Hz , thus the output is the open loop AC response.

Hi, crutschow

For an op amp that its gain can be described by a low-pass RC filter, I can derive the relationship between its open-loop gain and frequency, and the result agrees with your LTspice simulation. But I'd still really like to know how I can fint Vout analytically. I only know that
Vn = Vout*[ (1/SC) /(R+1/SC) ] ---- (1)

Could you please give me some clues about how to proceed next?

 

[Heidi]

Sorry, equation (1) refers to this:

 

[crutschow]

For that circuit, except for frequencies below about a Hz, the output depends upon the open loop response of the op amp since there is no significant feedback.
In that case the open-loop response for a typical gain-of-one compensated op amp is typically a 6dB per octave roll-off (single pole) from the maximum open-loop DC gain to the unity gain frequency (GBW value) .

 

[Heidi]

For that circuit, except for frequencies below about a Hz, the output depends upon the open loop response of the op amp since there is no significant feedback.
In that case the open-loop response for a typical gain-of-one compensated op amp is typically a 6dB per octave roll-off (single pole) from the maximum open-loop DC gain to the unity gain frequency (GBW value) .

If I use a capacitor of, say 1nF, to force more feedback to appear at the inverting terminal, then that circuit will be more like a unity-gain voltage follower when frequency below perhaps 10 KHz. Is this correct?

There's one more thing I don't understand. Referring back to the original circuit with C=100 F, why is the output voltage so large? I thought it would saturate if feedback were not provided.

At first I think maybe we're using a (nearly) ideal 'opamp', but when I use a real one, for example LM308, instead, the output is still huge. Why doesn't it saturate?

 

[MrAl]

Hi,

The model might be using an arbitrary source to generate the output, rather than a full blown transistor stage. If you put even one transistor stage on the output you'll see it clamp to the supply voltage. If you put a theoretical controlled voltage source you'll see it go up to billions of voltage (ha ha).

You have to realize soon or later (if you already didnt that is) that the end analysis depends highly on the op amp model chosen. If you go with purely theoretical models, then you have to impose your own limitations, either directly with (for example) a diode clamp or implicitly by simply rejecting any solutions that go above or below the power supply. A little transistor voltage follower might do it too, or better yet an inverting transistor stage, just to get a more realistic output. Sometimes it can get tricky though, because some models will be able to put out billions of amps too, so a diode clamp wont even work in that case because it will just happily pass a million amps and therefore still put out a voltage that is too high <chuckle>. The voltage follower might even allow the emitter to pass millions of amps anyway and still allow an extremely high voltage. This is the way of world of the ideal models and the way they behave in feedback loops.

 

[Heidi]

Hello MrAl,

So the two models (opamp and LM308) both belong to the model that we have to put clamping mechanism by ourselves, is it right?

 

[crutschow]

You need to understand the difference between AC (small signal) and Transient (large signal) analysis in Spice.

Signals in AC analysis don't saturate since a linear model is assumed and that model has no voltage limitations. So even though AC analysis is considered small signal, there is no limit on the signal amplitude (other than the numerical dynamic range of the simulator). Thus AC analysis can only be used to calculate relative gains between input and output.

Prior to doing an AC analysis, Spice calculates the circuit DC operating point using the non-linear large-signal models and from that derives the small signal linear models. That's when an op amp with no feedback will saturate and the small signal model derived from that will show the saturation effects on the AC gain and frequency response.

Transient analysis uses a non-linear model which will exhibit saturation effects as determined by the circuit bias voltages.

 

[MrAl]

Hi,

Yes i agree that AC analysis is done even more unrealistically than transient. In fact, some simulators even assume that the input to the circuit is 1vac rather than what you have it set for.

It might be better to turn to a model that you create yourself so that you can see everything about how it works rather than having to guess at what the model is doing.

 

[crutschow]

Yes i agree that AC analysis is done even more unrealistically than transient. In fact, some simulators even assume that the input to the circuit is 1vac rather than what you have it set for.

You are putting words in my mouth. I don't think the AC analysis is done "unrealistically".
The Spice AC analysis uses a technique for calculating the gain and frequency response of a circuit similar to the way you would do it by hand.
You calculate the bias point of each circuit stage and from that calculate the stage gain. That gain is then assumed to be independent of signal level (even though we know the circuit will saturate at some point) for subsequent amplifier gain calculations.
It's immaterial what voltage is used for the AC analysis since it's only the relative gain from input to output that's of interest.

The transient AC response can be done, of course, if you want information on distortion and clipping levels etc. of the circuit at any one frequency.

 

[MrAl]

Hi Carl,

That's all well and good, but not always true. While the 'relative' gain is important, there are occasions where i'd like to see the actual voltage output vs frequency, not the relative gain. What i am forced to do in these cases is divide or multiply the output by some factor. For example, with 2v input i'd have to multiply the output voltage by 2, when plotting voltage of course, because the software automatically assumes 1v input even though i've got a 2vac sine source there. Maybe other simulators are different and allow for quick setting changes. We could compare.

Sorry that you dont like my terminology, but that's one valid description intended to show that there is a difference between the REAL ac output that would appear in a REAL circuit and the FAKE output shown with the simulator without some active intervention on the part of the user. Thus the word "unrealistic" seems appropriate.

 

[crutschow]

Well it's quite obvious you don't like simulators from the many negative words you use so we'll let it go at that.

 

[MrAl]

Hi,

Well not really, i guess i like simulators but there are always going to be things we dont like about software that we didnt write ourselves. I write a lot of Windows based software so i see these little pitfalls now and then in many types of software. It certainly doesnt render the software useless in most cases, but leaves something to be desired. When i write a program myself at least i can modify it anytime i want to, so i prefer that route when possible, and that goes for any software not just simulators. So i wouldnt say i dont like simulators, i just dont like certain things and i have to spend time finding a work around. Overall i'd say they are pretty good.
I was fed up with Windows Explorer a while back too, for many types of operations, so i wrote my own file manager. But i still use Windows Explorer for certain things. The nice thing about this is i can then incorporate just about any function possible on the computer in that file manager. I downloaded a couple other types that were free, but you never get the same kind of personalization as when you write it yourself.
For another example, my personal Notepad equivalent can write frequency domain or state vector differential equations just by drawing the circuit inside the program environment. My Calculator equivalent can solve 5th order differential equations. So you can sort of say i am pretty picky about my software

 

[Heidi]

Thank you very much, MrAl and crutschow.

I have decided to postpone further exploration about op amps until I study one specific internal circuitry example. I don't like very much about "general" descriptions of something in textbooks, they sometimes just leave me a lot of confusion and strange thoughts. However, I just can't help asking one last question, and I need crutschow's response please.

The top is the open loop simulation.
DC feedback is provided by R1 and R2, with C1 suppressing the AC feedback down to a very low frequency.

In the top circuit in post #95, I think that in order to obtain relationship between the open-loop gain and frequency with that circuit, you have to know the voltage at the inverting terminal will approach zero above a certain frequency when there's no significant feedback, because Aol*(Vp-Vn)=Vout, since Vp=1, Vn must be zero so that Vout represents the open-loop gain. (Vp: voltage at the non-inverting terminal, Vn: voltage at the inverting terminal)

If I understand the purpose of that circuit correctly, I'm curious, "How did you know the voltage at the inverting terminal would approach zero with no feedback in advance?"