# How would you compartmentalize this circuit?

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1. Feb 16, 2016

### kostoglotov

This is not directly a homework question, as I found the answer. The question did not specify that compartmentalization be used, but it was in the section of Cascaded Op Amps, and the general idea of the section is to compartmentalize such complex looking circuits into cascaded stages of inverting, non-inverting, summing, or difference op amps.

I took the long way round and got the answer using Kirchoff's Laws and systems of equations.

However, can this circuit be compartmentalized into op amp stages, and if so, how? What are the stages?

2. Feb 16, 2016

### Averagesupernova

Need some more details here. There are no resistor values.

3. Feb 16, 2016

### davenn

it's a cascading 2 stage inverting op-amp arrangement with addition feedback provided by R3 and R6

D

4. Feb 16, 2016

### kostoglotov

There's not meant to be. The only assumption is that none of them are zero or infinite.

5. Feb 16, 2016

### kostoglotov

Right, how do I handle that feedback? Can I wire it up outside the inverting op amp stage, or is it crossing into the stages? Because R3 seems to alter the internal feedback of the inverting op amps, and R3 is wired up to in between the voltage source and the first inverting op amp, doesn't this cause a problem for compartmentalization?

6. Feb 16, 2016

### Averagesupernova

I am curious to know what your answer was to whatever question was asked which I assume was what is the total gain.

7. Feb 17, 2016

### kostoglotov

Just an algebraic expression in terms of all the various R's.

8. Feb 17, 2016

### Averagesupernova

Based on the diagram you posted I would assume the answer to the question is a specific voltage gain. In the case of all resistors being the same I would think the gain would be zero.

9. Feb 18, 2016

### Tom.G

Does this work for you? It treats each stage seperately.

1) Label the output of the first stage as "Vx"
2) Vo= -(R4/R5)*Vx -(R4/R6)*Vi
3) Vx= -(R2/R1)*Vi -(R2/R3)*Vo

Then substitute Eq.2 into Eq.1

p.s. note that R3 supplies positive feedback

10. Feb 18, 2016

### kostoglotov

No :), I found the answer, and it matched the answer given in the back of the textbook, and that answer was an algebraic expression in terms of the resistances.

11. Feb 18, 2016

### donpacino

You can think of the output of an op amp as a dependant voltage source.
With that being said the two stage are isolated.
You can see that the second op amp is in an inverting/summing configuration (the two source being summed are the output of the first op amp and vi).
The first op amp is in an inverting configuration with addition feedback coming from the output of the second op amps (through R3).

you can solve this by writing two KCLs, one at the negative nodes of the op amps. You cannot completely isolate the two signals. Because of the feedback, they are dependent in some way.

note. R6 is not a feedback resistor. It will not effect the gain of the first op amp, due to the fact that it is tied to an ideal voltage source.

12. Feb 18, 2016

### kostoglotov

Right, so one does need to apply KCL equations, you can't just box the two up and use the standard formulas for the two different op amp types.

Great explanation, thanks! :)

13. Feb 18, 2016

### davenn

so what is R6 doing ? ...it's function in this circuit ?

Dave

14. Feb 18, 2016

### donpacino

its a feed forward path for vi to the second op amp. In a non ideal circuit (vi having output impedance, wire inductance/resistance, etc) r6 would have some effect on the gain of the first op amp. Any feedback signal is mitigated by the ideal source.

Last edited: Feb 18, 2016
15. Feb 18, 2016

### Averagesupernova

Here is the way I look a it assuming all resistors have the same value. Simplify the circuit by eliminating R6 and R3. Now we have 2 inverting stages each with a gain of -1. Simple enough. Now reinsert R6. When Vin goes positive, the output of the first stage goes negative. But, the input also is driving R6 which pulls in the opposite direction at the summing junction of R5 and R6. So this cancels and the output of the second stage will be 0. Inserting R3 will make no difference at this point since we are adding 0 back to the summing junction of the first stage. In the real word this is a rather pointless circuit in my opinion. Any tolerance will cause positive feedback and the output will rail. Its a good exercise though.

16. Feb 18, 2016

### donpacino

why do all the resistors have to have the same value?

17. Feb 18, 2016

### Averagesupernova

They don't. But you have to start somewhere.

18. Feb 18, 2016

### donpacino

whats wrong with leaving them as variables? knowing the generic gain of a circuit without the parameters is very very valuable for design and analysis

19. Feb 18, 2016

### Averagesupernova

Never said it wasn't. My point is that the thing will remain linear until you actually expect an output other than zero. So it looks to me like we have built a needlessly complex Schmitt trigger. That doesn't really make sense, but the circuit doesn't either, so...

20. Feb 19, 2016

### donpacino

well the gain of the circuit is

vo/vi= (G2G6 - G1G5) / (G3G5-G2G4)

Gx=1/Rx fyi

so you can size the resistors such that the circuit will have either an inverting gain, or non-inverting gain, as well as attenuate or amplify. This is actually a really cool circuit!

21. Feb 19, 2016

### Averagesupernova

I plugged some numbers into your equation and came up with a gain of -3 when using 1 for R1 and R5 and using 2 for R2, R4, R3 and R6. Then using KCL I came up with a gain of 3 ignoring R3. So you are telling me that inserting positive feedback by inserting R3 will cause the output to swing 180 degrees? KCL has never failed me yet. Am I doing something wrong?

22. Feb 19, 2016

### donpacino

no i don't think you're doing something wrong, I get the same results. Like I said this is an interesting circuit.

It seems with the values you chose, and varying R3 something interesting happens.
as R3 increases from 0 to 4 the gain goes from 0 to - inf.
then R3 from 4 - inf the gain moves from inf to three.
weird

Last edited: Feb 19, 2016
23. Feb 19, 2016

### donpacino

i'm sure you can use this circuit layout in combination with a pot to capture a range of transfer functions.
you would have to choose your resistor values carefully, but it could be done.

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24. Feb 19, 2016

### Averagesupernova

I will not claim to know how you came with your equation but I cannot see how R3 being zero ohms can cause the gain to be zero when the other resistor values are as I set them. You do agree that R3 in any resistance value causes nothing except positive feedback correct? I can see that considering everything ideal with absolutely zero offset error and no thermal noise that with no input voltage the output will be zero even with R3 being zero. But then assuming a 1 volt input the current will ALL get shunted away from the inverting input of the first stage by R3. It would end there except R6 is providing a signal to the second stage and once the second stage has any voltage on its output it will hit the rail since there is ALOT of positive feedback with R3 being so low. This is interesting. Hoped this would not come to a p!ssing match. I will admit I was very sleepy when making my last post in this thread. I still say it seriously resembles a Schmitt trigger.

Last edited: Feb 19, 2016
25. Feb 19, 2016

### donpacino

the math doesnt lie. it doesnt change simply because you don't belive it. there might be feedback in r6 BUT there is also negative feedback in R2. Thats why this circuit is so interesting.

if you don't believe me solve the circuit with R6 in the circuit (I'm not sure why you took it out at all).