How would you compartmentalize this circuit?

[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?

imgur link: http://i.imgur.com/kYrp99R.png

 

[Averagesupernova]

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

 

[davenn]

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

it's a cascading 2 stage inverting op-amp arrangement with addition feedback provided by R3 and R6D

 

[kostoglotov]

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

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

 

[kostoglotov]

it's a cascading 2 stage inverting op-amp arrangement with addition feedback provided by R3 and R6D

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?

 

[Averagesupernova]

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

 

[kostoglotov]

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

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

 

[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.

 

[Tom.G]

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

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

 

[kostoglotov]

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.

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.

 

[donpacino]

You can think of the output of an op amp as a dependent 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.

 

[kostoglotov]

You can think of the output of an op amp as a dependent 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.

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! :)

 

[davenn]

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.

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

 

[donpacino]

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

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.

 

[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.

 

[donpacino]

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.

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

 

[Averagesupernova]

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

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

 

[donpacino]

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

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

 

[Averagesupernova]

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

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...

 

[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!

 

[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?

 

[donpacino]

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?

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

 

[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.

 

[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 a lot 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.

 

[donpacino]

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 a lot 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.

the math doesn't lie. it doesn't change simply because you don't believe 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).

 

[donpacino]

if you analyze the circuit with variables instead of numbers you'll be able to evaluate the transfer function easier and better.

 

[Averagesupernova]

I am not sure we are arguing the same point. I removed R6 for simplification in post #15. After that it has never been removed. I have never said there is feedback of any kind in R6. The fact that you have R3 feeding from the output to the input of 2 inverting op amps chained together guarantees positive feedback. Nothing you can do with R6 will change that.

 

[donpacino]

I am not sure we are arguing the same point. I removed R6 for simplification in post #15. After that it has never been removed. I have never said there is feedback of any kind in R6. The fact that you have R3 feeding from the output to the input of 2 inverting op amps chained together guarantees positive feedback. Nothing you can do with R6 will change that.

I was tired last night and replaced R3 with R6.

the math doesn't lie. it doesn't change simply because you don't believe it. there might be feedback in r3 BUT there is also negative feedback in R2. Thats why this circuit is so interesting. you can have some positive feedback if you have negative feedback as well

if you don't believe me solve the circuit with R3 in the circuit (I'm not sure why you took it out at all).
if you analyze the circuit with variables instead of numbers you'll be able to evaluate the transfer function easier and better.\

if you know how to use LTspice use that.
there is also a path completely independent of any level of positive feedback.

 

[Averagesupernova]

Well you are certainly correct in that intuition does not trump math but with all due respect I don't trust the way you came up with it, however that may be. Using my resistor values and the famous opamp rule about inputs tracking let's step through this. Feed +1 volt into R1. Assuming the output of the second stage is zero volts we know that the output of the first stage will be -2 volts since there will be no current through R3 that will affect the first stage which is actually a summing amp. The second stage is also a summing amp which will output +3 volts due to the gain of -2 concerning currents through R5 and a gain of -1 concerning currents through R6 which has a voltage of +1 which is the same as the input to the whole circuit. So just after the instant we applied +1 volt to the input we now have +3 on the output but this will lower the voltage on the output of the first stage by 3 volts due to the feedback through R3 as the gain here is -1 due to the value we chose for R3 and as you said R2 supplies negative feedback. So that stage once again satisfies the rule about inputs to an opamp tracking each other. Now we have -5 volts on the output of the first stage. This feeds into the second stage which has a gain of -2 due to the values I chose for R5 and R4. So now we have +9 volts on the output and again have satisfied the rule about inputs tracking each other. We go back around again and again and again until we run out of headroom. The circuit will not settle in on a voltage on its output somewhere between the supply rails.
-
Now of course there is a catch to this. When I supplied + volt to the input I said we could assume that the output of this circuit was zero. This may not be the case. But it is irrelevant since no matter what we arbitrarily decide it could/would be the result will be the same. You can also pick a different resistor for R3 that is say 10 or larger but the result will be the same. The output will settle at one of the supply rails.

 

[jim hardy]

I like Tom G 's idea

Adding dots at the nodes shows there's only 3 of them with nonzero voltage
so i labeled the middle one v_m
and labeled the amplifiers a1 and a2

at a1's summing junction

v_m/R2 + v_i/R1 + v_o/R3 = 0

at a2's summing junction

v_m/R5 + v_i/R6 +v_o/R4 = 0

normalizing v_m in both gives two equations

(1) v_m + v_iR2/R1 +v_oR2/R3 = 0
and
(2) v_m + v_iR5/R6 + v_oR5/R4 = 0

subtract (1) from (2)

v_i X (R5/R6 - R2/R1) + v_o X (R5/R4 - R2/R3) = 0

will try to align numerator and denominator in next lines,

...(R5/R6 - R2/R1)
vo/vi = ------------------------ ( latex won't do it, had to use periods to align them)
...(R5/R4 - R2/R3)

so this circuit has the interesting properties that
if all resistors are equal gain is indeterminate , zero / zero
if R5/R6 = R2/R1 gain is zero
R5/R4 = R2/R3 makes gain approach infinite

if my algebra is okay ( I'm prone to algebra mistakes)