Cascaded Op Amp circuit .. #2

Thread Starter

The Electrician

Joined Oct 9, 2007
2,914
I decided to test this circuit on the breadboard using LM358 and dual supply +/-12V

I use this schematic (DC voltage measured in a real circuit )
View attachment 243614

And the output voltage on the scope for Vin = 5VDC + 1Vp (1Khz)

View attachment 243615



The similar things we can see on LTspice
View attachment 243616





Maybe it is a DC stable now. But on the bench, after swapping the inputs circuit start to oscillate (Fosc = 14Khz).
As we can see on the scope:

View attachment 243617
The LM358 is not a rail-to-rail opamp, so when we see that U3 has very different inputs such that it's output should be saturated, we can reasonably take 10.63 volts as the max positive output with +-12 volt supplies. That being the case, the +10.28 output of U1 suggests that it may also be saturated. But the real circuit and LTspice simulation show that the AC input is producing some AC output. The LTspice waveform is clipped on the positive portion of the waveform. I take the fact that the U1 output voltage is apparently at the rail, and the clipped AC output waveform as indicating that U1 is saturated. So, it would seem that the whole circuit is latched up.

Something else that is probably affecting the various DC voltages is the fact that real opamps like the LM358 can draw current at the inputs when the opamp is saturated, and also when the input common mode range is exceeded.
 

Jony130

Joined Feb 17, 2009
5,435
@The Electrician I'm willing to accept the fact that we see the AC voltage at the output due to the fact that the opamp is saturated. Any comment on the "stable DC operating point" for the "swapped input" version?
 

LvW

Joined Jun 13, 2013
1,588
One of those other things is the positive feedback. However, not all circuits with positive feedback oscillate. In fact, some circuits are designed with intended positive feedback in order to speed up the response or increase the input impedance. There are of course limitations to these techniques, but it's enough to know that there are some circuits with positive feedback that work fine so we can not simply state that any given circuit latches up because it has positive feedback. It has to be analyzed, and analyzed very carefully.
Hi MrAl, with your words in your post #18 above (" ....some people feel the need to try to limit other people's ideas about a topic. " ) you probably were referring to my last post (" It is really a weird circuit - and, actually, it does not deserve so many comments ").
If this is true, I like to mention that I took part in this discussion over 45 contributions. And - if I am not wrong - at first it was me who has pointed to the fact that the circuit as shown in the 1st post cannot work. And in the following discussion I have tried to justify and explain the problem . So - I cannot follow your reproach.

With reference to the above quoted text I feel it necessary to mention that there is a fundamental difference between (1) Signal feedback (ac) and (2) DC feedback. This should never be mixed.
Of course, you are right that there are many applications (active filters, oscillators) which contain positive signal feedback.
But my concern about the circuit under discussion was related to positive DC feedback.
And yes - when a circuit intended for linear operation (other than two-valued hysteresis circuits) shows overall positive DC feedback we can "simply state that ......it latches up".
In such a case, a simple visual inspection is sufficient and it is not necessary to analyze it "very carefully".

Regards
LvW
 

MrAl

Joined Jun 17, 2014
9,638
Hi MrAl, with your words in your post #18 above (" ....some people feel the need to try to limit other people's ideas about a topic. " ) you probably were referring to my last post (" It is really a weird circuit - and, actually, it does not deserve so many comments ").
If this is true, I like to mention that I took part in this discussion over 45 contributions. And - if I am not wrong - at first it was me who has pointed to the fact that the circuit as shown in the 1st post cannot work. And in the following discussion I have tried to justify and explain the problem . So - I cannot follow your reproach.

With reference to the above quoted text I feel it necessary to mention that there is a fundamental difference between (1) Signal feedback (ac) and (2) DC feedback. This should never be mixed.
Of course, you are right that there are many applications (active filters, oscillators) which contain positive signal feedback.
But my concern about the circuit under discussion was related to positive DC feedback.
And yes - when a circuit intended for linear operation (other than two-valued hysteresis circuits) shows overall positive DC feedback we can "simply state that ......it latches up".
In such a case, a simple visual inspection is sufficient and it is not necessary to analyze it "very carefully".

Regards
LvW
Oh i hope i didnt offend you that was not my intention i was really just stating an opinion. I value your replies as i think of you in the 'upper class' category of members who talk here so your input is valuable to me.
We may not agree on everything but that does not change anything overall about my opinion about your replies.

Now back to the circuit(s) at hand...

Sorry to disagree but DC positive feedback is possible without latching up. Perhaps you can explain your position on that. Positive DC feedback can be used to increase input RESISTANCE also. But remember that positive feedback has a factor of it's own too, and it has to be right which usually means not too high.
So in practice this means the DC positive feedback follows the input voltage, but the factor is such that it never overpowers it. In this way the input starts to look like a 'differential' signal caused by the DC positive feedback. So the positive feedback creates less of a difference across the input resistor which means less input current which equates to higher input resistance (and thus higher impedance also).

I didnt want to dismiss this circuit too quickly that's why i was of the opinion that it may be worth looking into. To this end i have done some analysis and here are the conclusions so far...
1. There may be DC operating points that hold steady. This may be possible when we consider absolutely ideal components including the op amp, which for this analysis are modeled as voltage controlled voltage sources. This may not be preferred for real world analysis but because so many of the circuits we see on here are purely theoretical, i find it interesting to look into this mode of operation.
One thing i think i can say with certainty, and that is that the circuit is 'algebraically' stable for any DC input.
2. The circuit as mentioned by others here probably is unstable using real op amps, and if those op amps act as integrators (a rough approximation of an op amp that is informative), it is unstable. This can be proved by simply doing an analysis and looking at the transfer function and even converting that to the time domain.
What we end up with is a factor in the denominator that is a subtraction of a constant from a frequency term, and what that implies is that for some frequency that factor will go to zero, and because it is a factor that means the expression evaluates to infinity for that frequency (and there is no pole/zero cancelation). In the time domain, we end up with a term that is exponential with a positive exponent, so after some time the response exponentially ramps to infinity. Both of these tell me that the circuit is unstable. We can make it stable with a compensator design but that's not really the question at hand.
 
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Thread Starter

The Electrician

Joined Oct 9, 2007
2,914
@The Electrician I'm willing to accept the fact that we see the AC voltage at the output due to the fact that the opamp is saturated. Any comment on the "stable DC operating point" for the "swapped input" version?
The mathematical analyses I did only apply to "ideal" opamps. The LM358 is an older, not particularly high performance opamp, so I'm not surprised that its behavior is not expected. :( However the oscillations look like what we would expect from a relaxation oscillator, not a harmonic oscillator. The oscillations are apparently causing slew rate limiting in the transitions. Did you try to run a simulation of this configuration?
 

LvW

Joined Jun 13, 2013
1,588
Sorry to disagree but DC positive feedback is possible without latching up.
..................................
So in practice this means the DC positive feedback follows the input voltage, but the factor is such that it never overpowers it. In this way the input starts to look like a 'differential' signal caused by the DC positive feedback. So the positive feedback creates less of a difference across the input resistor which means less input current which equates to higher input resistance (and thus higher impedance also).
Hello MrAl,
As I am very interested in novel and versatile circuit concepts, I kindly ask you to show me an example where positive DC feedback is used without latching up.
As you know, it is a common and well established practice to use negative DC feedback for setting/fixing the bias point of "linear" active circuits.
Therefore, we can guess how and in which direction any positive DC feeback would influence the bias conditions.
That is the background of my question.
Can you please present an example which shows the benfits of such feedback ?

In the second part of your answer you are speaking about "less input current" and "higher input resistance" of a circuit.
Which kind of "input" ? Are you really referring to positive DC feedback ?
Thank you
LvW
 
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MrAl

Joined Jun 17, 2014
9,638
Hello MrAl,
As I am very interested in novel and versatile circuit concepts, I kindly ask you to show me an example where positive DC feedback is used without latching up.
As you know, it is a common and well established practice to use negative DC feedback for setting/fixing the bias point of "linear" active circuits.
Therefore, we can guess how and in which direction any positive DC feeback would influence the bias conditions.
That is the background of my question.
Can you please present an example which shows the benfits of such feedback ?

In the second part of your answer you are speaking about "less input current" and "higher input resistance" of a circuit.
Which kind of "input" ? Are you really referring to positive DC feedback ?
Thank you
LvW
Hi again,

Ok give me a little time i just woke up :)
I'll try to get back with an example a bit later. I'll simulate it first too just to make sure it works right.

Oh hey for now, first set up a differential amplifier with gain of 1 and no positive feedback. Use 100 ohm resistors on the non inverting terminal (a voltage divider with two 100 ohm resistors forming a divide by 2 voltage divider). Use higher values like 10k on the inverting terminal (also forms a divide by 2 from output to inverting input. Ground the input with the 10k's, apply 2v to the 100 ohm input (it's actually going to be a 200 ohm input though because two 100's in series equals 200 ohms).
Measure the output voltage and the input resistance to the non inverting input into the input resistor which is 100 Ohms.
Because of the gain=1 we get 2v on the output and because 2v/200ohms is 10ma, the input resistance looks like 2/10ma=2*100=200 ohms as expected. Measure the slew rate, say 0.5v/us (an LM358).

Next, add some positive feedback. Connect a 150 Ohm resistor from output of the op amp to non inverting input (positive feedback). Again with 2v at the input we measure 3v on the output and about 5ma on the input. Because we have 2v in and 5ma in, the input resistance is 2/0.005=400 Ohms. Because the output is 3v the gain increased to 1.5 up from 1.0 so it is higher also. Measure the slew rate, it will still be about 0.5v/us.

Comparing the first experiment with the second we see the gain increased and the input resistance (and thus impedance) increased although the slew rate did not change much.

I think the reason this does not latch up is because the divider ratio from output to non inverting input is less than the divider ratio from output to inverting input so the inverting input matches the non inverting input at some definite point and that's where the output stops rising. If this is true (and i think it is) we will eventually see latchup when the positive feedback resistor gets close to causing the same voltage divider ratio as the negative feedback resistors form although the actual input voltage will make a difference too so we cant go too high on the input voltage with divider ratios that are close to each other (but still the positive one should be less than the negative feedback one). So it's a matter of how low we can go on the positive feedback resistor combined with how high we can go on the input before latchup will occur.


However, since you are so sure of the latch up problem could it be possible that we can not simulate this effectively with a simulator or calculate the outcome accurately even with less than infinite internal op amp gain? I'll look into this more later.
 
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BobTPH

Joined Jun 5, 2013
6,092
I think the two of you are talking past each other.

When @LvW says positve feedback he means net, i.e. the difference between the feedback on the two inputs.

When @MrAl says positive feedback, he means only feedback ti the non-inverting input.

@MrAl : your differential amplifier of gain 1 already has negative feedback, to which you are adding a smaller amount if positive feedback. The net is still negative.

Bob
 

LvW

Joined Jun 13, 2013
1,588
I think the two of you are talking past each other.

When @LvW says positve feedback he means net, i.e. the difference between the feedback on the two inputs.

When @MrAl says positive feedback, he means only feedback ti the non-inverting input.

@MrAl : your differential amplifier of gain 1 already has negative feedback, to which you are adding a smaller amount if positive feedback. The net is still negative.

Bob
Hi Bob, thank you for your arttempt to clarify things.

When I use the term "positive feedback" - I am strictly following the definition of feedback terms (using, of course, the "loop gain concept"). How could it be otherwise?
 
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LvW

Joined Jun 13, 2013
1,588
However, since you are so sure of the latch up problem could it be possible that we can not simulate this effectively with a simulator or calculate the outcome accurately even with less than infinite internal op amp gain? I'll look into this more later.
I think, it is absolutely not necessary to simulate/calculate something.
I remember the words of somebody else: "Before discussing technical matters, clarify terms !"
As it seems, this is the core of some misunderstandings.

I have to say that I actually feel pretty confident when it comes to the terms "positive/negative feedback".
Of course, there are circuits where there is - in addition to the dominant negative-feedback - also a positive feedback loop. However, the stability criterion naturally requires that the combined resulting net feedback factor be negative as a matter of course.
In the case of a simple opamp, for example, an additional positive feedback loop has the advantage that both, loop gain and closed-loop gain, can be chosen independently (important for opamps that are not unity-gain stable). But negative feedback must always be dominant.

One comment to your (relatively complicated) example.
Quote: Comparing the first experiment with the second we see the gain increased and the input resistance (and thus impedance) increased although the slew rate did not change much.)
Did I (again) misunderstand something? I think, for increasing the gain and the input resistance of an opamp amplifier, I would select another pair of resistors. The slew rate is a property of the opamp alone and has nothing to do with the surrounding circuitry.
 
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Tesla23

Joined May 10, 2009
524
I haven't worked through everyone else's analyses, so apologies if I am rehashing, but I get that the circuit is stable with ideal op-amps if

\[\frac{R_3}{R_3+R_4} \gt \frac{R_1}{R_1+R_2} \frac{R_6}{R_5+R_6}\]

with the resistors labelled as below.

So for the given resistors it is unstable, but choosing to change \(R_2\), we find it is stable if \( R_2 > 1.4R_1 \) or \( R_2 > 7k \)

Here is an LTSpice simulation with \( R_2 = 7.01k \), on the edge of instability, where I have arranged a 50Hz ripple of 200mV p-p on the input, this leads to 400V ripple on the output with a 10kV offset (the op-amps are idealish - LTSpice default - limited Aol and GBW as shown on the schematic, but no limit on voltage output). Keep increasing R2 and at about 50k the output stays under 15V, I haven't investigated the interior nodes. With \( R_2 < 7k \) LTSpice initially converges, but then diverges on the first edge on the source.

1626469212970.png
 
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MrAl

Joined Jun 17, 2014
9,638
I think the two of you are talking past each other.

When @LvW says positve feedback he means net, i.e. the difference between the feedback on the two inputs.

When @MrAl says positive feedback, he means only feedback ti the non-inverting input.

@MrAl : your differential amplifier of gain 1 already has negative feedback, to which you are adding a smaller amount if positive feedback. The net is still negative.

Bob
Hello Bob,

Well the statement he made was about this original circuit which also has negative feedback and it is not immediately apparent that it has any predominate positive feedback yet the statement that "we see right away that is has positive feedback so it must latch up" was made about this circuit also.
The new experimental circuit i will post next.

Network-02.gif
 

MrAl

Joined Jun 17, 2014
9,638
I think, it is absolutely not necessary to simulate/calculate something.
I remember the words of somebody else: "Before discussing technical matters, clarify terms !"
As it seems, this is the core of some misunderstandings.

I have to say that I actually feel pretty confident when it comes to the terms "positive/negative feedback".
Of course, there are circuits where there is - in addition to the dominant negative-feedback - also a positive feedback loop. However, the stability criterion naturally requires that the combined resulting net feedback factor be negative as a matter of course.
In the case of a simple opamp, for example, an additional positive feedback loop has the advantage that both, loop gain and closed-loop gain, can be chosen independently (important for opamps that are not unity-gain stable). But negative feedback must always be dominant.

One comment to your (relatively complicated) example.
Quote: Comparing the first experiment with the second we see the gain increased and the input resistance (and thus impedance) increased although the slew rate did not change much.)
Did I (again) misunderstand something? I think, for increasing the gain and the input resistance of an opamp amplifier, I would select another pair of resistors. The slew rate is a property of the opamp alone and has nothing to do with the surrounding circuitry.
Yeah but you are shifting the paradigm which is sometimes referred to as "moving the goal posts".
Your original comment was about a circuit that ALSO had negative feedback in addition to the positive feedback yet you referred to it as "positive feedback". Perhaps you can explain that.

For another example that you may not realize you did this the circuit with the low value input divider you said we can increase the two resistors. Not only do we NOT do that when we are setting up an example for how positive feedback works, we may not get away with that anyway if the circuit needs those low value resistors for another reason such as bandwidth. But this practical concern is till secondary to the idea that we are trying to show what positive feedback does we dont want to change resistors so we dont have to add positive feedback because then we cant show how positive feedback works without coming up with a completely new circuit.

I mentioned the slew rate just for illustration i dont really care about that, but the bandwidth may in fact increase with positive feedback. Do we buy a faster op amp instead :)
 

LvW

Joined Jun 13, 2013
1,588
Hi Tesla23 - yes, you are right. No surprise.
In principle, this fact was revealed already some posts ago: When the loop gain is small enough (below unity), we are below the critical limit (Barkhausen condition). In this case, the overall DC loop gain would be 0<LG<1.
In this case, the negative feedback faktor k=R3/(R3+R4) is made sufficienty large (gain of the first stage smaller) ; this is a case of marginal stability.
However, for the given (original) dimensioning the positive DC loop gain is above unity (LG=+1,7, as I have mentioned in an earlier post)
 
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LvW

Joined Jun 13, 2013
1,588
Yeah but you are shifting the paradigm which is sometimes referred to as "moving the goal posts".
Your original comment was about a circuit that ALSO had negative feedback in addition to the positive feedback yet you referred to it as "positive feedback". Perhaps you can explain that.
MrAl - sorry to say, but you are wrong. I really cannot understand your position - I do not "shift" anything.........
A simple visual inspection of the original circuit shows the following:
* There are two two main active blocks - two non-inverting amplifiers (both get their input signals at the non-inv. inputs)
* In between is a simple voltage divider which must be carefully dimensioned (in the given circuit, it is not) to keep the positive loop gain small enough.
* This series connection of non-inverting blocks is equipped with a resistive feedback loop.
* Question to you: Where do you see any "negative feedback in addition to the positive feedback" ?
Perhaps you can explain that?
 
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Tesla23

Joined May 10, 2009
524
To help anyone that can't see the positive feedback, the way I analyzed the circuit was to note that if the circuit had ideal op-amps and was operating in a linear mode, then the effect of the second op amp is simply to buffer the voltage at the non-inverting input to the inverting input. Thus it may be replaced by a buffer, and the equivalent circuit is:

1626510603726.png

This is much easier to analyze, and the two feedback paths - positive and negative - are clear.

Breaking the loop at X you can derive the loop gain as:

\[LG = A_0 ( \frac{R_1}{R_1+R_2} \frac{R_6}{R_5+R_6} - \frac{R_3}{R_3+R_4} ) \]

where \(A_0\) is the op-amp gain, and you can clearly see the positive and negative feedback contributions.

My statement before was simply a statement that the negative feedback should dominate. The more esoteric situation, where positive feedback dominates but the circuit is still stable, I'll leave to others.
 

LvW

Joined Jun 13, 2013
1,588
Tesla, I must admit that I cannot follow your considerations.
It is not true that opamp 2 works simply as a unity gain amplifier.
This opamp has a non-unity feedback factor (determined by R1, R2 and R3).
Did you insert numbers into your LG expression?
What happens for ideal opamp parameters? Loop gain of infinite?
For finding the overall loop gain it is best to insert the testsignal directly into the non-inv. terminal of opamp 1 and to consider both amplifiers as fixed positive gain units.
 

Tesla23

Joined May 10, 2009
524
Denoting the first op-amp as U1 and the second as U2:

If the amplifiers are ideal and operating in a linear mode, you can solve the circuit by solving the feedback around U1 first. This is because if U2 is ideal and operating linearily, then it forces the voltage at the inverting input to equal the voltage at the non-inverting input. So we can replace U2 with a buffer (UX below) that achieves the same thing, and this will allow us to solve for all voltages up to and including the inputs to U2. This gives me the schematic below:

1626517268609.png

(This has better labelling that my previous schematic where, instead of using a new name for the buffer, I re-used the name U2, sure to create confusion.)

Solving this allows you to see the feedback paths, and calculate the current through R2. You can then easily then compute the output voltage from the original network using the current through R2, the voltage at the R5,R6 junction, and the feedback resistor on U2. I haven't done that, but I don't think you have to in order to understand the stability of the network (if the op amps are ideal).
 
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