# Two op amp circuit

Discussion in 'Homework Help' started by Jony130, Nov 10, 2014.

1. ### Jony130 Thread Starter AAC Fanatic!

Feb 17, 2009
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Hi, I would like to ask if there is a simple way to solve this op amp circuit.

Without involving a lot of math?

My way:

First I write nodal equations for Vx and Vy
For Vx node

(V1 - Vx)/R1 = (Vx - Vo1)/R2 (1)

For Vy

(V2 - Vx)/R3 = (Vx - Vo2)/R4 (2)

Next
Vo1 = (Vy - Vx)*Ao (3)
Vo2 = (Vx - Vy)*Ao (4)

Next I substitute Vo1 and Vo2 into 1 and 2 and solve for Vx and Vy.
Then I use 3 and 4 to find Vo1 and Vo2 for Ao--->00

$Vo1 = \frac {(R1 + R2) R4 V2 - R2 (R3 + R4) V1}{R2 R3 + R1 (2 R3 + R4)}$

$Vo2 = \frac { R2 (R3 + R4) V1 - (R1 + R2) R4 V2 }{R2 R3 + R1 (2 R3 + R4)}$

Dose anyone knows a simpler way?

2. ### MrAl Well-Known Member

Jun 17, 2014
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432
Hi,

I dont know if it will come out simpler or not, but you could try using the properties of op amps that have infinite gain, which means the gain of each section is -Rf/Rin, and Vy=Vx.

Using this idea the output of the top and bottom sections for example could be stated as:
Vo1=-(V1-Vy)*R2/R1+Vy
Vo2=-(V2-Vx)*R4/R3+Vx

I am pretty sure this can lead to the output differential Vo1-Vo2 but you can take it from here to see what you can get from this.

Last edited: Nov 10, 2014
3. ### WBahn Moderator

Mar 31, 2012
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I don't think these are correct. Let's consider the limiting case of when all of the resistors are the same, namely R.

$Vo1 = \frac {V2 - V1}{2}$
$Vo2 = \frac {V1 - V2}{2}$

Now, if V1=V2, this says that V01 = V02 = 0V

But does this make sense?

If we apply the same voltage to V1 and V2, and if we just require that R1=R3 and R2=R4, then all we would know about the output is that V01 = V02, but they could take on ANY value and, by symmetry, Vx and Vy will be the same.

I suspect that the circuit is not solvable and is probably not stable.

Keep in mind that these are just my immediate thoughts and I may well be missing something.

4. ### crutschow Expert

Mar 14, 2008
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3,068
I simulated the circuit and got some strange results with the output saturating for 1/2 of a sine-wave input. So I agree that the circuit seems kind of flaky and don't think it represents a circuit that does anything useful. But I'm open to edification.

5. ### MrAl Well-Known Member

Jun 17, 2014
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432
Hello again,

It looks like a differential input to differential output amplifier. If it works in real life or not would probably have to be proved on a breadboard. It could be one of those op amp circuits that looks good in theory but then it latches up or oscillates or something. The troubling part is that one op amp output changes the others input and vice versa, and it looks like positive feedback. Not all positive feedback is bad however so we'd have to look at this closer i guess or just do a quick breadboard.

To make it simpler, i make R4=R2 and R3=R1, and get results:
Vo2=-(V2-V1)*R2/(2*R1)
Vo1=(V2-V1)*R2/(2*R1)

This shows that one output is the negative of the other when we match resistors in their respective positions.

When i do a simulation i get results that match these results.
For example:
V1=1v, V2=0v, Vo1=-1v, Vo2=+1v
V1=2v, V2=1v, Vo1=-1v, Vo2=+1v
V1=3v, V2=2v, Vo1=-1v, Vo2=+1v

The models use limited gain so the results are not that exact, for example 1v might come out 1.0008v.

I use op amp models that are voltage controlled current sources with low output impedance for the simulations.
I use voltage controlled voltage sources for the nodal analysis.

6. ### Jony130 Thread Starter AAC Fanatic!

Feb 17, 2009
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Yes, indeed this circuit looks very interesting. But MrAl I still don't see any way how I can solve this
Vo1=-(V1-Vy)*R2/R1+Vy
Vo2=-(V2-Vx)*R4/R3+Vx
Because we have only two equations and four unknowns. I must have some kind of blackout today.

7. ### joeyd999 AAC Fanatic!

Jun 6, 2011
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If the circuit operates closed loop, then Vx = Vy.

But I've been unable, since last night, to develop a transfer function of the (expected) form:

(Vo1-Vo2)=(V1-V2) * X

Where X is an algebraic combination of R1, R2, R3, and R4.

I don't like not being able to find the general equation for (simple) circuits like this, so I think I'll keep trying for a while.

8. ### joeyd999 AAC Fanatic!

Jun 6, 2011
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BTW, here is my starting position:

Assuming Vx=Vy=v, then:

v = (Vo1-V1)(R1/(R1+R2))+V1

and

v = (Vo2-V2)(R3/(R3+R4))+V2

so

(Vo1-V1)(R1/(R1+R2))+V1 = (Vo2-V2)(R3/(R3+R4))+V2

There are no unknowns, so this should resolve to the proper form with algebraic manipulation.

9. ### joeyd999 AAC Fanatic!

Jun 6, 2011
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And, this is where I get stuck:

Vo1(R1R3+R1R4) - Vo2(R1R3+R2R3) = V2(R1R4 + R2R4) - V1(R2R3 + R2R4)

Again, trying to resolve to the form:

Vo1 - Vo2 = X (V1 - V2)

where X is an algebraic combination of R1 through R4.

10. ### joeyd999 AAC Fanatic!

Jun 6, 2011
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If I set R1=R3 and R2=R4, I get:

Vo1-Vo2 = (V2 - V1)((R2^2 + R1R2)/(R1^2 + R1R2))

Haven't check it, but this should be the answer for the special case as illustrated.

Need a better mathematician for the general solution.

11. ### joeyd999 AAC Fanatic!

Jun 6, 2011
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Rearranging to a (more) proper form:

Vo1 - Vo2 = (V1 - V2) (-((R2^2 + R1R2)/(R1^2 + R1R2)))

So, the differential gain for the original example is -2, which agrees with MrAl's simulation.

12. ### Jony130 Thread Starter AAC Fanatic!

Feb 17, 2009
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And this is what I got

$Vo1 - Vo2 = \frac{-2 R2 (R3 + R4)}{R2 R3 + R1 (2 R3 + R4)}*V1 +\frac{2 (R1 + R2) R4 }{R2 R3 + R1 (2 R3 + R4)}*V2$

13. ### joeyd999 AAC Fanatic!

Jun 6, 2011
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Without checking for correctness, this is a cumbersome form to work with, as it tells you nothing about the differential output with respect to the differential input.

14. ### Jony130 Thread Starter AAC Fanatic!

Feb 17, 2009
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I'm sure that this is a correct solution. So please tell me what to do next?

15. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Rather curiously, I have ...

$V_{o1} - V_{o2 }=(V_2-V_1) \frac{(R_2+R_4)}{(R_1+R_3)}$

.....????

Jun 6, 2011
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17. ### WBahn Moderator

Mar 31, 2012
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I don't think there is a unique solution for a given set of input. Consider a load line analysis. Pick a ratio for R2/R1 and then plot Vo1 vs Vy for a series of inputs values for V1. In essence, you are pretending the Vy is a reference signal supplied by a separate circuit (which is exactly what it is). Now pick a ratio for V4/R3 and plot Vo2 vs Vy on the same set of axes. Now focus on the curve for a particular value of V1 and a particular value of V2 and ignore the rest. These are the load lines for the circuit. Since they were plotted using the same axis for Vx and Vy, ANY vertical line that you care to draw satisfies the requirement that Vx=Vy and, as you can see, each choice of Vx=Vy results in a pair of values for Vo1 and Vo2 that are suitable. But any other choice produces a different pair of values that are equally suitable. So you don't have a unique solution.

18. ### t_n_k AAC Fanatic!

Mar 6, 2009
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My solution only applies with certain constraints.
Such as - where the source is a floating (not ground referenced) input connected between the V1 & V2 terminals. In addition R1-R4 values may be arbitrarily selected.
Other conditions (including ground referenced inputs) are possible but the solution is not universal - which suggests Bill Bahn is correct.
It may also be the case that my solution is simply a reduced form of the equation posted by Jony where the aforementioned constraints are applied.

Last edited: Nov 12, 2014
19. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Are you considering the case where both V1 and V2 are ground referenced, or the case solved by t_n_k where there is a floating voltage source between V1 and V2?

Taking the case where V1 and V2 are ground referenced, and letting V1=2 volts, V2=-1 volts in both cases, can you exhibit two sets of values for Vx, Vy (Vx=Vy, of course), Vo1 and Vo2 such that KCL is satisfied at Vx and Vy for both sets of values (using the resistors in post #1)?

20. ### WBahn Moderator

Mar 31, 2012
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I was considering the case when both are ground referenced.

Trivially.

Would you agree that

Vo1(V1, Vy) = Vy·[1 + (R2/R1)] - V1·(R2/R1)
Vo2(V2, Vx) = Vx·[1 + (R4/R3)] - V2·(R4/R3)

At this point it should be obvious that if you specify V1 and V2 (along with all of the resistor values) that I can trivially find pairs of output values Vo1 and Vo2 that are consistent with it just by choosing different values of Vx=Vy.

So, let's throw your numbers at it. Using the specific resistances given in the original schematic, we have R2/R1 = R4/R3 = 2, making our equations reduce to:

Vo1(V1, Vy) = 3Vy - 2V1
Vo2(V2, Vx) = 3Vx - 2V2

So here's a pair:

Vo1(2V, 0V) = -4V
Vo2(-1V, 0V) = 2V

Here's another pair:

Vo1(2V, 1V) = -1V
Vo2(-1V, 1V) = 5V

Vo1(2V, -1V) = -7V
Vo2(-1V, -1V) = -1V

Note that, for the case where the two resistor ratios are matches, the output difference, (Vo2-Vo1), is independent of Vx,Vx which only serve to adjust the output offset. But in general this is not the case.

I'll leave it to you to show that there is no problem with KCL. As to whether there are stability issues, I suspect there are and that the system will rail, though to which rail is probably unpredictable. When you think about it, you have two interconnected negative feedback loops and, as such, the overall loop has positive feedback.