Op Amp, trying to approach it again.

WBahn

Joined Mar 31, 2012
29,978
You can't use static equations to determine stability because stability is an inherently dynamic phenomenon.

Again, consider the case of a marble sitting at the bottom of a bowl or balanced at the top of a ball. In both cases they are in static equilibrium. The forces are balanced and there is no reason for the ball to do anything other than sit there and look stupid.

The difference is when the ball is perturbed slightly from that equilibrium position. In one case the forces that result return the ball toward the equilibrium position and in the other the forces drive it further away. Determining which is which cannot be done by just finding the point at which the forces are balanced, which is the same result in either case.

Whether the output is negative for a positive input has nothing to do with it. Look up the circuit for the classic non-inverting opamp circuit.

Closed loop gain equations are always for circuits that exhibit negative feedback (I'm pretty sure that's a true statement, but it's good enough for the circuits you are talking about). If the circuit has positive feedback then there is no closed loop gain because the circuit is unstable.
 

WBahn

Joined Mar 31, 2012
29,978
Also what do you think of an answear I've also received about positive feedback ? : "even though for positive feedback when Vin is connected to positive input of OP Amp the Output must be positive but is negative."
Where did you receive this answer from? I can only find where YOU have stated it in this thread. That is an incomplete sentence, without context, that makes no sense. Perhaps it makes sense in the context in which it was written.
 

Thread Starter

Xenon02

Joined Feb 24, 2021
500
Yes.
And if you look at the voltage polarities you will see that it is not possible for the circuit with positive feedback to be stable with the output at other than one of the rail voltages.
That's what you are ignoring.
Ok so as you said closed loop gain can be calculated.
But what do you mean by voltage polarities ?


You can't use static equations to determine stability because stability is an inherently dynamic phenomenon.
I thought that stability can be calculated staticly.


Whether the output is negative for a positive input has nothing to do with it. Look up the circuit for the classic non-inverting opamp circuit.
Even if the output is negative for positive input when input voltage has 1V?


Closed loop gain equations are always for circuits that exhibit negative feedback (I'm pretty sure that's a true statement, but it's good enough for the circuits you are talking about). If the circuit has positive feedback then there is no closed loop gain because the circuit is unstable.
But closed loop gain for positive feedback can be calculated.

Where did you receive this answer from? I can only find where YOU have stated it in this thread. That is an incomplete sentence, without context, that makes no sense. Perhaps it makes sense in the context in which it was written.
It is not written here in this post
Here: " OK - with corrected resistor values the calculated gain is Acl=-11/9 (as given by you). Question: Can this be a realistic value? Answer: No - because the input signal acts on the non-inverting input and the gain cannot be negative. This explanation is sufficient. This is also clear because we can see that there is much more positive feedback (factor Hr+=0.5) than negative feedback (factor Hr-=-1/11)"

For this picture :

1663877085372.png

From those word I assumed that Vout cannot be negative because the Acl (closed loop gain) = Vout/Vin. so Vout can't be negative ?
 

Thread Starter

Xenon02

Joined Feb 24, 2021
500
You don't understand what polarity means? :rolleyes:

More likely where, which voltage polarities ?
Why do they have a meaning ?

Like WBahn said :


Whether the output is negative for a positive input has nothing to do with it. Look up the circuit for the classic non-inverting opamp circuit.

Now that's a Troll question.
I know that I am an idiot. You don't have to reminde me this :D Of course you didn't say that I am an idiot. I said it :).
 

WBahn

Joined Mar 31, 2012
29,978
I thought that stability can be calculated staticly.
How? Again, go back to that marble sitting perfectly still at it's equilibrium point. Is it stable? You can't tell because the fact that it is in static equilibrium at a particular position is not enough information. You need to know how the forces are going to act on it if it begins to move. The same with an opamp circuit. The fact that a particular output is in static equilibrium with a particular input does not mean that it is stable. Stability depends on which direction the output will be driven if it begins to change.

Even if the output is negative for positive input when input voltage has 1V?
All that means is that the circuit is inverting. It says nothing about stability.

But closed loop gain for positive feedback can be calculated.
So? All that means is that you can throw some numbers at an equation that doesn't apply to that situation. The result is meaningless.

It is not written here in this post
Here: " OK - with corrected resistor values the calculated gain is Acl=-11/9 (as given by you). Question: Can this be a realistic value? Answer: No - because the input signal acts on the non-inverting input and the gain cannot be negative. This explanation is sufficient. This is also clear because we can see that there is much more positive feedback (factor Hr+=0.5) than negative feedback (factor Hr-=-1/11)"
So that circuit has a positive feedback path and a negative feedback path and the result is due to the interaction between them. That is the context of that comment. It is not a general statement about positive and negative feedback, but is a statement in the context of that specific circuit.
 

Thread Starter

Xenon02

Joined Feb 24, 2021
500
All that means is that the circuit is inverting. It says nothing about stability.
What if it's negative but connected to positive input (non-inverting). Vin positive connected to positive input Op Amp.
it is pretty much related to this one :


So that circuit has a positive feedback path and a negative feedback path and the result is due to the interaction between them. That is the context of that comment. It is not a general statement about positive and negative feedback, but is a statement in the context of that specific circuit.
The quote I've mentioned was telling about why the calculated gain was incorrect. The gain was negative, but the positive feedback was "stronger" and that's why it is negative.
So the conclusion was that gain cannot be negative because Vin was connected to non inverting input.
The question is what does Vin decides the Vout there ?

You have said that Vout has nothing to do here. So why ?
 

WBahn

Joined Mar 31, 2012
29,978
What if it's negative but connected to positive input (non-inverting). Vin positive connected to positive input Op Amp.
Show the SPECIFIC circuit you are talking about, showing exactly where the Vin is being connected to. Stop talking in nebulous generalities.
 

Thread Starter

Xenon02

Joined Feb 24, 2021
500
Show the SPECIFIC circuit you are talking about, showing exactly where the Vin is being connected to. Stop talking in nebulous generalities.
1663882864760.png

Here it was the same :

1663882881237.png

Vin = 1V.

1663882892698.png

Gain was negative. Vout/Vin = -11/9.
So Vout is negaive.
But Vin is connected to non inverting input of the Op amp.

You said that Vout doesn't determine it.
 

WBahn

Joined Mar 31, 2012
29,978
You are showing two different circuits. Which one do you want to discuss. Pick one and then let's talk about just that one circuit.
 

Thread Starter

Xenon02

Joined Feb 24, 2021
500
You are showing two different circuits. Which one do you want to discuss. Pick one and then let's talk about just that one circuit.
I mean both of them have the same issue that I have stated there.
Which is the calculated closed loop gain for positive feedback and for both feedbacks it is negative for both. Also both of them have positive voltage on the non inverting (Vp like you mentioned before) input of the op amp. But the gain is negative so the Vout is negative. But it is wrong somehow.
 

crutschow

Joined Mar 14, 2008
34,281
More likely where, which voltage polarities ?
Why do they have a meaning ?
All voltage polarities.
If you don't know why voltage polarities have meaning, then you need to do more study on elementary circuit theory.

In this case it's because the relative polarity of the voltages on the two op amp inputs determines the polarity of the output signal.
(And please don't ask what "relative" means).+
 

Thread Starter

Xenon02

Joined Feb 24, 2021
500
All voltage polarities.
Okey so what is the problem here :

1663888422906.png

Gain will be negative, so Vout is also negative.
Vin is positive connected to V+ or the positive input of Op Amp or Vp or however you will call it.

I don't see the problem here why it can't work.
 

WBahn

Joined Mar 31, 2012
29,978
Okey so what is the problem here :

View attachment 276796

Gain will be negative, so Vout is also negative.
Vin is positive connected to V+ or the positive input of Op Amp or Vp or however you will call it.

I don't see the problem here why it can't work.
There IS no closed loop gain for this circuit. This circuit is a Schmitt Trigger comparator -- it is NOT a linear amplifier.

To analyze this circuit, you need to know the supply rails (or the max voltage swing of the output, which we usually assume are the rails unless we know different for the specific opamp). So let's assume the rails are +15 V and -15V. Let's also assume that the open loop gain of the opamp is one million (which is pretty typical of a lot of opamps such as the TL072).

Let's define the input differential voltage to be

Vpn = (V+ - V-)
That means that Vout is as follows:

IF
Vpn > +15 µV, THEN Vout = 15 V
ELSE IF
Vpn < -15 µV, THEN Vout = -15 V
ELSE
Vout = (10^6)·Vpn

We know that V- = 0V, but what is V+?

V+ = Vin + R1·(Vout - Vin)/(R1+R2)

V+ = Vin·[R2/(R1+R2)] + Vout·[R1/(R1+R2)]

For the sake of simplicity, let's stipulate that R1 = R2 = R. This then reduces to

V+ = Vin/2 + Vout/2

So

Vpn = V+ - V- = Vin/2 + Vout/2

What is the derivative of Vpn with respect to Vout?

dVpn/dVout = +1/2

This means that if the output voltage were to GO UP, say due to noise, by 1 µV, that the input differential voltage would GO UP by 0.5 µV and the output voltage would then be driven by 5 V. But now the input differential voltage is WAY more than +15 µV and so the output will rail at +15 V and stay there.

But what if the output voltage were to GO DOWN by 1 µV? Now the input differential voltage would Go DOWN by 0.5 µV and the output voltage would be driven down by 5 V, which will quickly exceed -15 µV need to rail the opamp output at -15 V.

That is because there is positive feedback, as evidenced by

dVpn/dVout > 0

If you swap the opamp inputs in that circuit and do the same analysis, you find that

dVpn/dVout = -1/2

and thus has negative feedback.


SO...

How does the circuit you have shown actually behave (and let's no longer assume that R1 = R2)?

Let's start with it in a condition where Vout = +15 V.

What does Vin need to be in order to make V+ equal to V-?

V+ = Vin·[R2/(R1+R2)] + Vout·[R1/(R1+R2)] = 0 V

Vin = - Vout·[R1/R2]

Let's say that R1 = 1 kΩ and R2 = 10 kΩ. That means that we need

Vin = - Vout·[1/10] = -1.5 V

Any voltage greater than this will result in V+ being greater than 0 and driving the output to +15 V.

As soon as the voltage gets below -1.5 V, the output will be driven to -15 V and will stay there. But now the voltage that we need to get V+ back to 0 V is +1.5 V.

So what will the output be when Vin = 0 V?

We don't know. It will be either +15 V or -15 V, depending on the recent history of the output. This phenomenon is therefore called "hysteresis".

When the output is +15 V, the input voltage must go below -1.5 V in order to get it to change. Taking it to 0 V will not be enough and the output will stay at + 15 V. But once you do get below -1.5 V, the output will change to -15 V and now if you take the input to 0 V it's not enough to get it to switch back and the output will stay at -15 V. In order to switch back, the input has to rise above +1.5 V. The circuit is therefore said to have 3 V of hysteresis.
 

WBahn

Joined Mar 31, 2012
29,978
Second question is for both feedbacks :

I've had from the last post all calculations I will post them here
View attachment 276747
View attachment 276748
View attachment 276749
1663908730050.png

If you do the analysis for the equilibrium output, you get:

Vout = Vin · R2(R3+R4) / (R2·R4 - R1·R3)

But this only gives you the static equilibrium output. It says nothing about whether that output is stable or not.

To get that, you need to find

dVpn / dVout and find where that is negative.

If you do that analysis, you get that

dVpn / dVout = (R1·R3 - R2·R4) / [(R1+R2)(R3+R4)]

For this to be negative, you need:

R2·R4 > R1·R3

Looking back at the equation for the static equilibrium output, this means that the output is stable only for non-inverting gains.

You can calculate the gain and the output for inverting gains all day long, but no real circuit will behave like that because it is unstable and will behave like a comparator instead.

Basically, this circuit has both positive feedback and negative feedback. It is stable only if the negative feedback is dominant, resulting in net negative feedback.

As an extra exercise, let either R1 -> 0 or R2 -> ∞ (or both) and you will see that you get the gain for the classic non-inverting amplifier circuit, which is what the circuit has become.
 
Last edited:

Ian0

Joined Aug 7, 2020
9,667
What’s the point in analysing a circuit with no practical use?
No one ever uses an op-amp with negative and positive feedback simultaneously. The closest thing is a Wien bridge, but the feedback is frequency dependent to make it oscillate.
 

WBahn

Joined Mar 31, 2012
29,978
What’s the point in analysing a circuit with no practical use?
No one ever uses an op-amp with negative and positive feedback simultaneously. The closest thing is a Wien bridge, but the feedback is frequency dependent to make it oscillate.
How many circuits that students analyze in a Circuits I course have any practical use? Some do, but the vast majority are there to develop the skills needed to analyze and/or design arbitrary circuits. For that matter, how many problems that students work in a physics, chemistry, or math class have direct practical use?
 

Jony130

Joined Feb 17, 2009
5,487
No one ever uses an op-amp with negative and positive feedback simultaneously.
This is not exactly true. In rare situations, we are using both negative and positive feedback.
feedback.PNG
https://www.ti.com/lit/an/snoa718/snoa718.pdf?ts=1663933263578&ref_url=https%3A%2F%2Fwww.google.com%2F (page 7)

https://www.researchgate.net/profil...hancement-Techniques-for-Amplifier-Design.pdf

Or to linearize the nonlinear component (PT100) in this case.

mostek_pom_ua776_1946.jpg
https://pdfserv.maximintegrated.com/en/an/AN3450.pdf

Of course in both cases, the negative feedback is stronger than the positive one. And I don't even want to mention about the negative impedance converter.
 

Thread Starter

Xenon02

Joined Feb 24, 2021
500
There IS no closed loop gain for this circuit. This circuit is a Schmitt Trigger comparator -- it is NOT a linear amplifier.
But here crutschow said it can be calculated for positive feedback if we assume that V+ = V- like for the negative feedback.

Yes.
And if you look at the voltage polarities you will see that it is not possible for the circuit with positive feedback to be stable with the output at other than one of the rail voltages.
That's what you are ignoring.
You also used the gain here :


Vin = - Vout·[R1/R2]

What is the derivative of Vpn with respect to Vout?

dVpn/dVout = +1/2

This means that if the output voltage were to GO UP, say due to noise, by 1 µV, that the input differential voltage would GO UP by 0.5 µV and the output voltage would then be driven by 5 V. But now the input differential voltage is WAY more than +15 µV and so the output will rail at +15 V and stay there.

But what if the output voltage were to GO DOWN by 1 µV? Now the input differential voltage would Go DOWN by 0.5 µV and the output voltage would be driven down by 5 V, which will quickly exceed -15 µV need to rail the opamp output at -15 V.

That is because there is positive feedback, as evidenced by

dVpn/dVout > 0

If you swap the opamp inputs in that circuit and do the same analysis, you find that

dVpn/dVout = -1/2

and thus has negative feedback.
I don't understand this part ?
I mean derivative? I haven't been taught to do it for Op amp.


SO...

How does the circuit you have shown actually behave (and let's no longer assume that R1 = R2)?

Let's start with it in a condition where Vout = +15 V.

What does Vin need to be in order to make V+ equal to V-?

V+ = Vin·[R2/(R1+R2)] + Vout·[R1/(R1+R2)] = 0 V

Vin = - Vout·[R1/R2]

Let's say that R1 = 1 kΩ and R2 = 10 kΩ. That means that we need

Vin = - Vout·[1/10] = -1.5 V

Any voltage greater than this will result in V+ being greater than 0 and driving the output to +15 V.

As soon as the voltage gets below -1.5 V, the output will be driven to -15 V and will stay there. But now the voltage that we need to get V+ back to 0 V is +1.5 V.
Why with condition that Vout must be 15V ?

For negative feedback we can also make a condition that Vout = 15V so Vin must be also specific to keep V+ =V-.


Looking back at the equation for the static equilibrium output, this means that the output is stable only for non-inverting gains.

You can calculate the gain and the output for inverting gains all day long, but no real circuit will behave like that because it is unstable and will behave like a comparator instead.

Basically, this circuit has both positive feedback and negative feedback. It is stable only if the negative feedback is dominant, resulting in net negative feedback.

As an extra exercise, let either R1 -> 0 or R2 -> ∞ (or both) and you will see that you get the gain for the classic non-inverting amplifier circuit, which is what the circuit has become.
Hmmm.
What do you think about this quote ?


" OK - with corrected resistor values the calculated gain is Acl=-11/9 (as given by you). Question: Can this be a realistic value? Answer: No - because the input signal acts on the non-inverting input and the gain cannot be negative. This explanation is sufficient. This is also clear because we can see that there is much more positive feedback (factor Hr+=0.5) than negative feedback (factor Hr-=-1/11)"
You have said that the gain is for the classic non-inverting amplifier.

But does the Vin connected to Vp or Vn (no directly but with a resistor like in here) must define the Vout ?
What do I mean is that Vin with R1 connected to Vp must give positive Vout, or it can make negative Vout ? Even if I connect Vin with R1 to Vp (also remembering that Vn is grounded).

I'm asking this because of the Acl equation which was Aol(V+ - V-) or Aol(Vp - Vn). Here Vin doesn't define the Vout. I know that you have showed me the calculation for Vout/Vin. I just mean that if always everytime Vin with R1 connected to Vp must result in positive Vout (or negative if Vin is negative). Because when I see more complicated circuits liek Jony has send then I don't know how it can be easly defined whether it will be "positive feedback" or "negative feedback".

1664060587415.png


I don't understand just why V+ = V- doesn't work in op amp for positive feedback. Even with speculation that we calculate it by saying that V+ = V- then what.

Also the part where it was said that Vout is negative so it must be positive feedback because Vin is connected to Vp so it should be Vout positive but it is negative. And this part I also don't understand. It is in the quote with the gain :


" OK - with corrected resistor values the calculated gain is Acl=-11/9 (as given by you). Question: Can this be a realistic value? Answer: No - because the input signal acts on the non-inverting input and the gain cannot be negative. This explanation is sufficient. This is also clear because we can see that there is much more positive feedback (factor Hr+=0.5) than negative feedback (factor Hr-=-1/11)"
 

WBahn

Joined Mar 31, 2012
29,978
But here crutschow said it can be calculated for positive feedback if we assume that V+ = V- like for the negative feedback.
As I said, you can calculate anything you want, that doesn't mean that it has any meaning.

What does the closed loop gain mean for a circuit in which the output is never equal to the input multiplied by that gain?

You also used the gain here :
No I didn't. I found what the input must be in order to make V+ equal to V- for a given Vout. That is VERY different than finding what Vout is for a given Vin.

I don't understand this part ?
I mean derivative? I haven't been taught to do it for Op amp.
You mean you haven't been taught derivatives yet, or don't know how to apply them to a circuit?

If you haven't been taught them, use the other approach I showed in which you assume that the output changes by a small amount and then see what change that has on the differential input voltage and then see whether or not that change will result in the output going back towards where it was, or moving even further away.

Why with condition that Vout must be 15V ?
I explained why that was used. With positive feedback, that circuit will behave like a comparator and so the output will always either be as positive as it can be or as negative as it can be. Since you didn't give any values for the supply voltages, I stipulated that I would assume that they are the usual +15 V and -15 V and that the opamp output can actually swing all the way to the rails (some opamps can, and some can't).

For negative feedback we can also make a condition that Vout = 15V so Vin must be also specific to keep V+ =V-.
Sure. So what? For the negative feedback circuit you can ask what Vin needs to be in order for Vout to be 15 V, 13.2 V, 4.15 V, 0 V, -0.05 V, -8.93 V, or any other output between the rails. But for positive feedback you can't do that because the only stable output voltages are the max output voltage (which I assumed to be +15 V in this case) and the min output voltage (assumed to be -15 V).

Hmmm.
What do you think about this quote ?
I think it's a great quote.

You have said that the gain is for the classic non-inverting amplifier.
I said: "As an extra exercise, let either R1 -> 0 or R2 -> ∞ (or both) and you will see that you get the gain for the classic non-inverting amplifier circuit, which is what the circuit has become. "

But does the Vin connected to Vp or Vn (no directly but with a resistor like in here) must define the Vout ?
What do I mean is that Vin with R1 connected to Vp must give positive Vout, or it can make negative Vout ? Even if I connect Vin with R1 to Vp (also remembering that Vn is grounded).
Do what I said. Set R1 equal to zero, which is the same as replacing it with a wire, and set R2 to infinity, which is the same as removing it from the circuit.

1664082213968.png

This IS the classic noninverting opamp circuit. Don't believe me? Google "classic noninverting opamp circuit" and see what comes up.

I'm asking this because of the Acl equation which was Aol(V+ - V-) or Aol(Vp - Vn).
Again, this is ONLY the case if the opamp is operating in its linear region.

What if V+ = 3 V, V- = 1 V, and Aol = 1,000,000? That equation says that the output would be two million volts! Do you believe that? Of course not. The output will simply be as positive as it can be, which will be somewhere close to its positive supply voltage.

Here Vin doesn't define the Vout. I know that you have showed me the calculation for Vout/Vin. I just mean that if always everytime Vin with R1 connected to Vp must result in positive Vout (or negative if Vin is negative).
Once again you are trying to draw simplistic conclusions that aren't the case in general. Just because Vin connects to Vp via a resistor does NOT mean that the output will always be the same polarity as the input. The output is a function of the entire circuit, not just a single connection.

Because when I see more complicated circuits liek Jony has send then I don't know how it can be easly defined whether it will be "positive feedback" or "negative feedback".

View attachment 276958
The closed loop gain of this is NOT negative! The closed loop gain doesn't exist! Any number you come up with for it by throwing stuff at an equation is meaningless!

I don't understand just why V+ = V- doesn't work in op amp for positive feedback. Even with speculation that we calculate it by saying that V+ = V- then what.
Because the circuit isn't stable! It will not operate at a point where V+ = V- because the positive feedback (if it is net positive in case both types are present) will immediately drive the output away from that point.

Also the part where it was said that Vout is negative so it must be positive feedback because Vin is connected to Vp so it should be Vout positive but it is negative. And this part I also don't understand. It is in the quote with the gain :
Whoever provided you with that quote was drawing some inferences and conclusions without backing them up adequately. WIthin the context of that circuit, what they said is correct, but it requires the reader to make those inferences that I don't think you are ready to be making.
 
Top