Amplitude control circuit for oscillators.

Thread Starter

cdummie

Joined Feb 6, 2015
124
I have a problem with understanding this particular circuit given above, just to make it clear, this is the circuit i am talking about (part of this circuit is amplitude control though, but still).
ампл контр.png

Now, it turns out that, when we are looking for voltage V0, the easiest way to do it is by superposition principle, now, that means that we consider one generator at the time while others are represented as short circuits (since all of them here are voltage generators), thereby, we have the following:

when we consider Vcc only V0=Vcc* (R4+R5)/(R2+R3+R4+R5)

NOTE: i know that this is not the complete expression, V0 will eventually be the same as if we had inverting opamp (in case that R2=R5 and R3=R4)

but, this expression confuses me because it implies that current is the same through all of the resistors R2,...,R4 (R1 is shortened in this case) i don't know how is that possible since we have resistor Rf and and output of OPamp, shouldn't both of them have a current flowing through, thereby, we would have different currents on resistors R2 and R5 for example. How come then this expression still happens to be correct in all textbooks, what am i missing? Any help appreciated!
 

MrAl

Joined Jun 17, 2014
8,996
Hello,

Are you calculating Vo with R2=R5 and R3=R4 and with the op amp output disconnected, and still assuming that the inverting input is at virtual ground?

If so, then the expression is right. What voltage do you get at Vo in that csse? You should be able to find out that there is no current in Rf.

Once the op amp is allowed to fully function though, then everything changes except we still would most likely assume the virtual ground still exists. If at any time the op amp comes out of the linear mode, then we have to ditch that assumption but for this circuit that may never happen.
 

Thread Starter

cdummie

Joined Feb 6, 2015
124
In case that R2=R5 and R3=R4 then Vo= - Rf/R1.

Inverting input is on the virtual ground, however, i cannot see why there's no current through Rf.
 

MrAl

Joined Jun 17, 2014
8,996
In case that R2=R5 and R3=R4 then Vo= - Rf/R1.

Inverting input is on the virtual ground, however, i cannot see why there's no current through Rf.
Hello there,

You are right if we only have Vcc, but i was assuming that the negative supply was there also all the time or it would not make sense at all because if the negative supply is NOT there then what is there, is it ground or is it open circuit.
I think the question was posed in a little strange way because the assumption is that only Vcc is present, but then they dont say how the other things are 'killed' either open or short circuit. For example, if the op amp is removed entirely (because the output is being assumed to be NOT connected) then the left side of Rf might be considered to be open circuited.

So i think better clarification of what exact circuit we are dealing with when they say "Vcc" is only there. Once we get that, we can answer this question perfectly with no doubts.

I think a better idea would be to analyze the circuit just as it is, except have an input voltage source. That would then allow us to see any response we want to with any input voltage.

Thanks for bringing that important point up. Now I'll wait for more information on this.
 

Thread Starter

cdummie

Joined Feb 6, 2015
124
As far as i know, when using superposition i am supposed to replace every single voltage generator with short circuit, now, due to the location of -Vcc generator i thought that we should actually replace it with ground, but to be honest, i am not sure if it's legitimate thing to do.

Sure, analyzing the circuit as it is can be good alternative to this.
 

MrAl

Joined Jun 17, 2014
8,996
As far as i know, when using superposition i am supposed to replace every single voltage generator with short circuit, now, due to the location of -Vcc generator i thought that we should actually replace it with ground, but to be honest, i am not sure if it's legitimate thing to do.

Sure, analyzing the circuit as it is can be good alternative to this.
Hello again,

Well in the entire scope of an analysis technique that has been proved to be successful if a certain sub analysis technique is considered successful based on the success of the overall analysis technique, then that sub analysis technique is useful and of course it works. So let us assume that the method you are using works, due to the majority of texts you said claim this to be true.
That means of course that at least for now we assume that the statement:
V0=Vcc* (R4+R5)/(R2+R3+R4+R5)

is true and works and leads to the final analysis that is correct.

Given that assumption, and given that it is not clear why they are choosing to show this method, we would need to see the entire analysis where that statement is part of that complete analysis. Without that we have to figure it out for ourselves, which although is possible, it is much easier to simply review the entire analysis method and determine if it agrees with a secondary method that proves to be correct.

So this means that you should show the entire method rather than just part of it, and that way we might quickly see what they want to achieve with this idea.
It could be as simple as they just want to show what happens with no other parts in the circuit.

If they really want to claim that statement is true though it looks like if we ground -Vcc then we also have to ground Vi (the input voltage) and we may have to treat the output of the op amp as disconnected at least temporarily. This may also mean that Rf has to be disconnected temporarily because if Vi=0 and Vo>0 then there must be current in Rf.

Try this next:
Treat the op amp section as a voltage controlled voltage source. That means the output must be grounded also during a Vcc driven partial superposition analysis (the statement above for Vo). So proceed as if you were just using a VCVS with some gain that later goes to infinity (or just use some high gain for now).

The only problem i still see here though is that usually we dont disconnect Rf. It may be that they are treating Rf as part of the dependent source though, but it would be best if we could see their entire analysis (from the texts you quote).

The analysis can be done without using superposition, but i assume that you want to do it that way. We can look at methods to do that too, but i'll wait to see if you can post the entire analysis you were reading about first.

After looking at it again i see that calculating Vo that way may make no sense unless we know what they are after, because that node is controlled by the output of the op amp not by the series connection of resistors, except for the clamping action which we could consider later.
 
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Thread Starter

cdummie

Joined Feb 6, 2015
124
Ok, here goes entire analysis:

First off, since we have a circuit with more than one generator, we can watch one generator at the time, meaning, we could use superposition theorem so we could determine Vo with respect to every generator and then sum it all up to get final solution (also, values in the circuit are assumed such that both diodes are off)

Let's say we have only Vcc generator in circuit (other generators are not mentioned here, nor circuit is drawn for this specific case), in that case we can see that Vo=Vcc*(R4+R5)/(R2+R3+R4+R5)

now, if we have a -Vcc only, then Vo=(-Vcc)*(R2+R3)/(R2+R3+R4+R5)

and now, when only Vi is present we have that Vo= -Rf/R

Thus, final expression for Vo is:

Vo= Vcc*(R4+R5)/(R2+R3+R4+R5) + Vo=(-Vcc)*(R2+R3)/(R2+R3+R4+R5) -Rf/R

and finally, if R2=R5 and R3=R4, then Vo=-Rf/R

That's all i've found about this specific circuit in textbooks.
 

MrAl

Joined Jun 17, 2014
8,996
Ok, here goes entire analysis:

First off, since we have a circuit with more than one generator, we can watch one generator at the time, meaning, we could use superposition theorem so we could determine Vo with respect to every generator and then sum it all up to get final solution (also, values in the circuit are assumed such that both diodes are off)

Let's say we have only Vcc generator in circuit (other generators are not mentioned here, nor circuit is drawn for this specific case), in that case we can see that Vo=Vcc*(R4+R5)/(R2+R3+R4+R5)

now, if we have a -Vcc only, then Vo=(-Vcc)*(R2+R3)/(R2+R3+R4+R5)

and now, when only Vi is present we have that Vo= -Rf/R

Thus, final expression for Vo is:

Vo= Vcc*(R4+R5)/(R2+R3+R4+R5) + Vo=(-Vcc)*(R2+R3)/(R2+R3+R4+R5) -Rf/R

and finally, if R2=R5 and R3=R4, then Vo=-Rf/R

That's all i've found about this specific circuit in textbooks.
Hello again,

Oh ok ha ha, that is interesting to say the least.

May i ask what book this came from?

Here's my take on this 'method'...
It looks like a misuse of the superposition theorem. It looks like a self-fulfilling method that only works for certain parameter values.

For example, what if the minus supply was not equal to the negative of the positive supply?
So what if the positive supply was 10v and the negative supply was -5v ?
Then it doesnt work.
This is also true if the upper branch resistors do not equal the lower branch resistor values.

The solution is:
Vo=-Vi*Rf/R1

and that has nothing to do with the resistors on the output.
For example, the resistor values could be 1k, 2k, 3k, and 4k and then the statement in question does not work.

Superposition works regardless what the values of the resistors are in a circuit, and regardless what the voltage supply values are. We would in general get a different answer with different value resistors yet the actual value of the output does not change with those resistor choices.

What does change then.
What does change is the clamp voltages. You might want to look at that next, if that is part of what you are looking to solve. The solutions for that does depend on the value of the resistors on the output.

I assume that the reason you asked this question is because you did not see how the statement in question could be true. It turns out that you were right to ask this question because your suspicion was correct...it does not work that way.
 
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Thread Starter

cdummie

Joined Feb 6, 2015
124
The textbook where i found this is not quite a textbook, its a material professor gave me in written form since that professor hadn't released a book yet for this particular course. However, explanation on this site relatively similar https://electronicsarea.com/nonlinear-amplitude-control-of-a-sinusoidal-oscillator/

Maybe i was not precise enough, Vcc and -Vcc do have same voltage looking by absolute value, it is a condition that has to be met in this particular case.

I know that given statement is true only for specific values of resistors R2=R5 and R3=R4 as i have said previously.
 

MrAl

Joined Jun 17, 2014
8,996
The textbook where i found this is not quite a textbook, its a material professor gave me in written form since that professor hadn't released a book yet for this particular course. However, explanation on this site relatively similar https://electronicsarea.com/nonlinear-amplitude-control-of-a-sinusoidal-oscillator/

Maybe i was not precise enough, Vcc and -Vcc do have same voltage looking by absolute value, it is a condition that has to be met in this particular case.

I know that given statement is true only for specific values of resistors R2=R5 and R3=R4 as i have said previously.
Hello again,

It is one thing to say they are using superposition, but it is another thing to judge if that is really applicable.
Superposition has no such constraints as to voltage values or resistor values. Superposition is much more general than that. Thus we can only conclude that they all made the same mistake, and it may be because they read each others work and thought it was right.

In the attachment, i show three circuits. The top circuit Circuit 1 is the same circuit we had been talking about.
Circuit 3 is a circuit similar to circuit 1 only with the op amp replaced by a new constant voltage source Vs.
Circuit 2 is also similar except it has a resistor between the new voltage source and the node Vo.

Superposition is applicable to circuit 2 but not to circuit 3 (for the node at Vo).
It is not applicable to circuit 3 because the source Vs completely controls the node Vo, so it does not matter what resistors are connected there save for a short of some kind.
It is applicable to circuit 2 because all three sources (Vs, Vcc, -Vcc) ALL have an effect on the node at Vo.

Note the difference. In circuit 2 when we short Vs we still get a response that is determined by either of the two remaining sources, while in circuit 3 when we short Vs we get absolutely NO response from EITHER of the two remaining sources. This means it makes no sense to try to do that.

These kinds of mistakes come up from time to time, there's not a lot we can do about it except try to reason out what may have happened and what should have been said instead of what was said.

Another way of stating this is that we never have to set certain voltage or resistor values just to make it look like superposition is somehow working ok. We never have to do that, and we never should have to do that because superposition does not require those kinds of specifications.

You might also want to note that the principle of superposition states that the response at any point in a linear circuit having more than one independent source can be obtained as the sum of the responses of each independent source acting alone.
In that statement you will not see any mention of specific resistor or voltage values.

I might add that i dont see them using that idea in the link you provided, although i do see them possibly using it for the two nodes A and B. Those two nodes would be acceptable for the use of superposition.
 

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Thread Starter

cdummie

Joined Feb 6, 2015
124
Ok, thanks a lot for your help, i thought of this as a relatively simple circuit, obviously, it is not. However, i think i can conclude that whenever R2=R5 and R3=R4 then V0=-Rf/R1.
 

MrAl

Joined Jun 17, 2014
8,996
Ok, thanks a lot for your help, i thought of this as a relatively simple circuit, obviously, it is not. However, i think i can conclude that whenever R2=R5 and R3=R4 then V0=-Rf/R1.
Hello again,

Sorry, but i dont understand your goal here because that conclusion does not seem to be worthwhile even if true. But only you can clear this up if you care to elaborate why you even wish to come to that conclusion.

For example, If R2=R5+1 and R3=R4 then Vo still is -Vin*Rf/R1, for most of the relative input range.

That is, Vo=-Vin*Rf/R1 until such time as one of the diodes starts to conduct. So the gain is -Rf/R1 until the diodes conduct, if they conduct.
Once the top diode conducts, the gain changes to Vo=-Vin*(Rf||R3)/R1 and so the gain goes DOWN.

To make sense of R2, R3, R4, R5, we note that we see a symmetrical gain change when R2=R5 and R3=R4, but that's the main point to be made about these four resistors...they set the point where the gain should be lowered for both positive and negative going outputs.
There is also the point that the ratios have to be constrained also, but that would be a secondary point as there would be no soft clipping if they did not follow that requirement.

I hope you can see this now as i think you will be happy to understand the circuit completely.
 

Thread Starter

cdummie

Joined Feb 6, 2015
124
Hello again,

Sorry, but i dont understand your goal here because that conclusion does not seem to be worthwhile even if true. But only you can clear this up if you care to elaborate why you even wish to come to that conclusion.

For example, If R2=R5+1 and R3=R4 then Vo still is -Vin*Rf/R1, for most of the relative input range.

That is, Vo=-Vin*Rf/R1 until such time as one of the diodes starts to conduct. So the gain is -Rf/R1 until the diodes conduct, if they conduct.
Once the top diode conducts, the gain changes to Vo=-Vin*(Rf||R3)/R1 and so the gain goes DOWN.

To make sense of R2, R3, R4, R5, we note that we see a symmetrical gain change when R2=R5 and R3=R4, but that's the main point to be made about these four resistors...they set the point where the gain should be lowered for both positive and negative going outputs.
There is also the point that the ratios have to be constrained also, but that would be a secondary point as there would be no soft clipping if they did not follow that requirement.

I hope you can see this now as i think you will be happy to understand the circuit completely.
Oh, i am sorry, when i said 'whenever' i meant as long as diodes are off, i was not considering the case when one of the diodes are on.
 

MrAl

Joined Jun 17, 2014
8,996
Oh, i am sorry, when i said 'whenever' i meant as long as diodes are off, i was not considering the case when one of the diodes are on.
Hello again,

No problem, and i am sorry too that i may not have made the point about the four resistors clear.

What i should have said (he he) was simply that it does not matter if R2=R5 and R3=R4 or any other reasonable resistor connected to the output. The output is still going to be -Vin*Rf/R1.
The requirement that R2=R5 and R3=R4 is ONLY necessary to ensure symmetrical soft clipping action. It has nothing to do with anything when the diodes do not conduct.

To put this another way, remove both diodes from the circuit completely but dont change anything else. Now the output is -Vin*Rf/R1 period, and it does not matter what the values of R2 through R5 are or their relationship to each other.
 

Thread Starter

cdummie

Joined Feb 6, 2015
124
I see, i think i understand this circuit much better now, this electronics stuff is still somehow new to me so thanks a lot for taking your time to help me out, i appreciate it!
 
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