Two op amp circuit

On this topic I highly recommend reading:

How to Identify Unstable DC Operating Points, Michael Green, IEEE Transactions on Circuits and Systems, V39, No. 10, October 1992

(Almost) Half of Any Circuit's Operating Points are Unstable, Michael Green, IEEE Transactions on Circuits and Systems, V41, No. 4, April 1994

A professor in the Czech Republic took up the study of this problem around 1996 and he has published a number of papers describing another method. Some of those papers are listed at:

http://publicationslist.org/dalibor.biolek

http://vutbr.academia.edu/DaliborBiolek

I see that he has made one available at:

http://www.valachnet.cz/biolek/articles/mwscas96.pdf

I have found others after some effort searching.
 
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Thread Starter

Jony130

Joined Feb 17, 2009
5,487
Hi everyone, thank for your comments.
Today after the work I decided breadboard this "strange" differential amplifier.
My power supply was +/-15V. And I use DC voltage as a input (both V1 and V2 where ground referenced).
R1 = R3 = 10k 5% and R2 = R4 = 22k 5%. And I set V1 = +1V and V2 = -1V.
The theory tell us that for R1 = R3 and R2 = R4 we have
Vo1 = -(V2 - V1)* R2/2R1 ; Vo2 =(V2 - V1) * R2/2R1 and Vo1 - Vo2 = (V2 - V1)*R2/R1,
So I expect this results
Vo1 = -2.2V ; Vo2 = +2.2V and Vo1 - Vo2 = -4.4V And LTspice confirms those results at lats for ideal opamp and LT1001A.
But the real world results look like this
3.PNG
As you can see this circuit "work" but not quite as I expected, no visible oscillations on my oscope.
And if I change V2 = 0V the Vo1 - Vo2 = changes to -2.2V.
It seems that the circuit has some problem with common mode voltage? And we need some kind of common mode control loop to bring DC-offset around 0V.
What do you think about those results ?
 
The circuit has two opamps. In your simulations you should be able to access the parameters of the opamp models. Try changing the gain of the opamp model for one of the opamps; make one opamp have 2 times the gain of the other and see what results you get from the simulation.
 

Thread Starter

Jony130

Joined Feb 17, 2009
5,487
I see it now, this mismatch in DC voltage is due to mismatch in op amps open loop gain. Thx The Electrician.
 

WBahn

Joined Mar 31, 2012
29,976
Hi everyone, thank for your comments.
Today after the work I decided breadboard this "strange" differential amplifier.
My power supply was +/-15V. And I use DC voltage as a input (both V1 and V2 where ground referenced).
R1 = R3 = 10k 5% and R2 = R4 = 22k 5%. And I set V1 = +1V and V2 = -1V.
The theory tell us that for R1 = R3 and R2 = R4 we have
Vo1 = -(V2 - V1)* R2/2R1 ; Vo2 =(V2 - V1) * R2/2R1 and Vo1 - Vo2 = (V2 - V1)*R2/R1,
So I expect this results
Vo1 = -2.2V ; Vo2 = +2.2V and Vo1 - Vo2 = -4.4V And LTspice confirms those results at lats for ideal opamp and LT1001A.
But the real world results look like this
View attachment 75569
As you can see this circuit "work" but not quite as I expected, no visible oscillations on my oscope.
And if I change V2 = 0V the Vo1 - Vo2 = changes to -2.2V.
It seems that the circuit has some problem with common mode voltage? And we need some kind of common mode control loop to bring DC-offset around 0V.
What do you think about those results ?
I think that those results completely validate the analysis I did way back with the following tweak: The output may, indeed, be stable but it is clearly not unique -- or at least not predictable in a practical way. The DC offset for a given circuit depends on the internals of the specific opamps and is likely determined by minor variations and bias condition.
 
I think that those results completely validate the analysis I did way back with the following tweak: The output may, indeed, be stable but it is clearly not unique -- or at least not predictable in a practical way. The DC offset for a given circuit depends on the internals of the specific opamps and is likely determined by minor variations and bias condition.
I thought you had come around to the idea that the solution is unique if the opamp gains are finite. :)

Jony130 is using opamps with finite gains, so the output should be unique.

Here's what the problems is. Using the circuit in post #1, derive symbolic results for finite amplifier gains Av1 and Av2. Designate the top opamp as OP1, the lower as OP2. Find the limit of the expressions for Vx, Vy, Vo1 and Vo2 as the gains approach infinity. If we do this and substitute numerical values for the resistors as in post #1 we get this result:

Results given as a vector (Vx, Vy, Vo1, Vo2)

With A→∞
(1/3, 1/3, -3, 3)

But, now we let the gains Av1 and Av2 approach infinity at different rates by maintaining a constant ratio between Av1 and Av2. Let Av1=2*A and Av2=A. Evaluate the expressions as A→∞, and substitute numerical values. Then we get:

With Av1=2*A and Av2=A and A→∞
(0, 0, -4, 2)

This is one of the examplse WBahn gave in post #20. All the many solutions can be found by changing the rate at which the two opamp gains approach infinity:

With Av1=.5*A and Av2=A and A→∞
(2/3, 2/3, -2, 4)

We see that relatively (as opamp gains go) small changes in the ratio of the opamp gains as they approach infinity lead to substantial changes in the voltages.

Even with finite opamp gains, the same thing occurs. For example:

With Av1=20 and Av2=10
(.121212, -.060606, -3.63636, 1.81818)

With Av1=10 and Av2=20
(.727272, .545454, -1.81818, 3.63636)

These two are unique solutions.

Compare these results to those above where the gains were infinite, but approached infinity with the gains in the same ratio as the finite gains in the immediate two previous examples. We get almost the same result as the infinite gain case.

So we should expect the various voltages to be sensitive to the actual opamp open loop gains even when ther are very low. I'll do a sensitivity analysis later, but now I have to go take care of some of the business of life.
 

MrAl

Joined Jun 17, 2014
11,389
Hi everyone, thank for your comments.
Today after the work I decided breadboard this "strange" differential amplifier.
My power supply was +/-15V. And I use DC voltage as a input (both V1 and V2 where ground referenced).
R1 = R3 = 10k 5% and R2 = R4 = 22k 5%. And I set V1 = +1V and V2 = -1V.
The theory tell us that for R1 = R3 and R2 = R4 we have
Vo1 = -(V2 - V1)* R2/2R1 ; Vo2 =(V2 - V1) * R2/2R1 and Vo1 - Vo2 = (V2 - V1)*R2/R1,
So I expect this results
Vo1 = -2.2V ; Vo2 = +2.2V and Vo1 - Vo2 = -4.4V And LTspice confirms those results at lats for ideal opamp and LT1001A.
But the real world results look like this
View attachment 75569
As you can see this circuit "work" but not quite as I expected, no visible oscillations on my oscope.
And if I change V2 = 0V the Vo1 - Vo2 = changes to -2.2V.
It seems that the circuit has some problem with common mode voltage? And we need some kind of common mode control loop to bring DC-offset around 0V.
What do you think about those results ?

Well that's not too much of a surprise:
Hi,
As a matter of fact we might even find an interesting relationship between the gains A1 and A2 and the outputs.
Try swapping the op amps next and see how the outputs change (that swaps gains A1 and A2).

It looks like the relationship between the two outputs matches the ratio of A1/A2, the gains of the two op amps. So if the gains are 10:1 then one output will be ten times the other, with opposite sign.

I also think we can prove stability and output uniquenses for this circuit, putting all doubts aside forever:)
 
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Tesla23

Joined May 10, 2009
542
Following from post#39, if the op-amp gains are finite then:
- the circuit is stable (as in no oscillations)
- the circuit has a unique operating point as long as the op-amps stay linear

The issue Jony130 shows in his post:

Hi everyone, thank for your comments.
Today after the work I decided breadboard this "strange" differential amplifier.
My power supply was +/-15V. And I use DC voltage as a input (both V1 and V2 where ground referenced).
R1 = R3 = 10k 5% and R2 = R4 = 22k 5%. And I set V1 = +1V and V2 = -1V.
The theory tell us that for R1 = R3 and R2 = R4 we have
Vo1 = -(V2 - V1)* R2/2R1 ; Vo2 =(V2 - V1) * R2/2R1 and Vo1 - Vo2 = (V2 - V1)*R2/R1,
So I expect this results
Vo1 = -2.2V ; Vo2 = +2.2V and Vo1 - Vo2 = -4.4V And LTspice confirms those results at lats for ideal opamp and LT1001A.
But the real world results look like this
View attachment 75569
As you can see this circuit "work" but not quite as I expected, no visible oscillations on my oscope.
And if I change V2 = 0V the Vo1 - Vo2 = changes to -2.2V.
It seems that the circuit has some problem with common mode voltage? And we need some kind of common mode control loop to bring DC-offset around 0V.
What do you think about those results ?
is not any sort of meta-stability, or the sort of unstable operating point described by Green in the first of his papers (I haven't read the second), but simply hyper sensitivity of the operating point to small imperfections, in this case the input offset voltages of the op amps. Including input offset voltages into the solution for Vx

\(
V_x = \frac{\beta(1+\beta+A_v)V_1 + \beta A_v V_2 + A_v^2 v_{io1} + (1+\beta+A_v)A_v v_{io2}}{(1+\beta)^2 + 2(1+\beta)A_v}
\)

Including asymmetric Av and beta doesn't add any higher order terms so this is adequate for estimating the order of magnitude of effects.

so approximately

\(
V_x = V_{x0} + \frac{A_v(v_{io1}+v_{io2})}{2(1+\beta)}
\)

so for zero input common mode voltage Vx should be zero, but the input offset voltages drive it away from there. For the LM358, typical numbers I have are for a input offset voltage of 2mV and an open loop gain or 100k, so for just one op-amp

\(
V_x = V_{x0} + \frac{A_v(v_{io1})}{2(1+\beta)} = V_{x0} + \frac{10^5(2mV)}{2(1+2)} = V_{x0} +33.3V
\)

i.e. it changes Vx by 33V. The two op-amps Jony tried must have has either offset voltages a little better than this to stay linear, or they cancelled a bit. Either way, this shows that the change in output common mode voltage is likely to be caused by variations in the input offset voltages, and not any metastability issues. It's a lousy circuit!

I don't really understand the significance of showing that you get different answers by allowing the Av's to go to a limit in fixed ratios as the Electrician was doing, is there a reason for doing this, some reference perhaps to what this shows about the circuit?
 

MrAl

Joined Jun 17, 2014
11,389
Hello again,

Using a formula similar to Mason's Gain Formula i found the four explicit equations for Vx,Vy, Vo1, and Vo2 solving four equations simultaneously. The resulting equations are:

Vx=(A1*R1*R4*V2+R2*R4*V1+A2*R2*R3*V1+R2*R3*V1)/(R2*R4+A1*R1*R4+R1*R4+A2*R2*R3+R2*R3+A2*R1*R3+A1*R1*R3+R1*R3)
Vy=(R2*R4*V2+A1*R1*R4*V2+R1*R4*V2+A2*R2*R3*V1)/(R2*R4+A1*R1*R4+R1*R4+A2*R2*R3+R2*R3+A2*R1*R3+A1*R1*R3+R1*R3)
Vo1=(A1*(R2*R4*V2+R1*R4*V2-R2*R4*V1-R2*R3*V1))/(R2*R4+A1*R1*R4+R1*R4+A2*R2*R3+R2*R3+A2*R1*R3+A1*R1*R3+R1*R3)
Vo2=-(A2*(R2*R4*V2+R1*R4*V2-R2*R4*V1-R2*R3*V1))/(R2*R4+A1*R1*R4+R1*R4+A2*R2*R3+R2*R3+A2*R1*R3+A1*R1*R3+R1*R3)

A1 is the internal gain of op amp 1, A2 is the gain of the other op amp.

Note that numerical results from these equations match the simulation results out to 5 or 6 significant digits for several different values of V1, V2, A1 and A2.
The fact that these equations resulted from an analysis using just linear algebra must say something about the validity of the circuit, even if it turns out to be impractical or there is a better way to achieve the same end result (possibly a differential amplifier with inverter for example).
Im sure they can be shown in a simpler way too.

I agree that the effects of input offset voltage should be investigated as well.
We also did not yet look at the AC characteristics using real op amps with some phase shift, etc.

Even if this circuit turns out to be impractical (as the gains wont be the same for different op amp sections without some sort of gain matching) i still think it is interesting from a purely academic standpoint.

Just to simplify a little here...

If we define dv=Vx-Vy then we get:

dv=(R2*(R4+R3)*V1-(R2+R1)*R4*V2)/((R2+(A1+1)*R1)*R4+((A2+1)*R2+(A2+A1+1)*R1)*R3)

and then we have:
Vo1=-dv*A1
Vo2=dv*A2

Assuming we have matched resistors in their respective positions R4=R2 and R3=R1, then we get:
dv=-(R2*V2-R2*V1)/(R2+(A1+A2+1)*R1)

and for the original circuit where R2=2*R1 we get:
dv=2*(V1-V2)/(A1+A2+3)

which for large A not infinite, we get:
dv=2*(V1-V2)/(A1+A2)
[which generalizes to dv=G*(V1-V2)/(A1+A2) where G is R2/R1 when R2/R1=R4/R3]

and for A2=A1=A we get:
dv=(V1-V2)/A

so we get:
Vo1=V2-V1
and
Vo2=V1-V2=-Vo1
 
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I don't really understand the significance of showing that you get different answers by allowing the Av's to go to a limit in fixed ratios as the Electrician was doing, is there a reason for doing this, some reference perhaps to what this shows about the circuit?
It provides a test for indeterminate behavior of the circuit. If a variable k is added to enforce a constant ratio between the gains of the opamps and the symbolic solutions are examined, k vanishes if the circuit is well behaved when A→∞. For example, if the technique is applied to the well known 3 opamp instrumentation amplifier, k is absent from the expressions for node voltages when A→∞. For the circuit of this thread, k is present in the symbolic node voltage expressions.

The use of the k multiplier also provides a way to find the many solutions that satisfy the circuit of this thread when A→∞.

For the circuit of this thread, it also shows that variations in opamp gain of the magnitude to be expected for real opamps can cause the observed variations in Vo1 and Vo2, including changes in common mode output, without even taking offset voltages into account.

Normally, we would expect feedback to make circuit behavior relatively immune to parameter variations in the opamps, but not for this circuit. This is a really crappy circuit, but analyzing it is edifying.
 

Tesla23

Joined May 10, 2009
542
This is a really crappy circuit, but analyzing it is edifying.
It is really crappy, but it has a remarkable characteristic - it tends to work when one of the op amps is limited to the rails.

If the circuit remains in the linear region, then analyzing the common mode output voltage shows that the contribution from the offset voltage and the unmatched gains should put the output outside the rails most times.

Looking at Jony's results, it looks in each case as if one op amp may be limiting, yet the differential output voltage is correct. Do the analysis and you find that if one amp saturates the circuit is such that it sets the output from the other amp to produce the correct differential output voltage. Now I'm not sure that it always works, obviously if one amp saturates to the positive rail and the differential output is such that the other amp should go higher then they would probably both saturate, maybe someone can show that it does in fact always work. Otherwise Jony's results are remarkable considering how unlikely they are.
 
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t_n_k

Joined Mar 6, 2009
5,455
As a further matter for consideration, I'll refer to WBahn's proposed options in post #20.

Three possibilities (of the many available) were suggested

Case 1: Vo1=-4V , Vo2=+2V
Case 2: Vo1=-1V , Vo2=+5V
Case 3: Vo1=-7V , Vo2=-1V

to which I'll add the 'obvious'

Case 4: Vo1=-3V , Vo2=+3V

In the absence of any defining parameter, suppose such an ideal physical system moves towards it lowest energy state. Which set of the above conditions represents the lowest energy state? One could base this on the resistor losses.

Suppose R1=R3=1 kOhm and R2=R4=2 kOhm ...

Then, for the above cases in the order listed, the losses will be

Case 1: 15 mW
Case 2: 15 mW
Case 3: 27 mW
Case 4: 13.33 mW

Of the four proposed, perhaps Case 4 would be the likely condition on this energy state basis ....????

Just a thought.
 

Thread Starter

Jony130

Joined Feb 17, 2009
5,487
But are you sure that the op amps gains difference is the main causes that Vo1 - 13.5V and Vo2 - 9.1V?
So how you explain this result from LTspice.
2.PNG
I'm sure that both op amp has the same open loop gain.
 
Are the opamp models identical? I would assume they are unless you made changes to them. Both input offset voltage and open loop gain differences can contribute to differences in output voltage.

The theoretical difference in Vo1 and Vo2 should be 4.4 volts, which seems to be the case. Notice that the two opamps have the same voltages (Vx and Vy) applied to their inputs, but with opposite polarity.

If the opamp models have a fixed input offset voltage it would account for your result.

If I solve your circuit with Av=100000 for each opamp, and an offset voltage of .1167 millivolts for each opamp, I get Vo1=9.4700 V and Vo2=13.8700 V.
 
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Hello again,

Using a formula similar to Mason's Gain Formula i found the four explicit equations for Vx,Vy, Vo1, and Vo2 solving four equations simultaneously. The resulting equations are:

Vx=(A1*R1*R4*V2+R2*R4*V1+A2*R2*R3*V1+R2*R3*V1)/(R2*R4+A1*R1*R4+R1*R4+A2*R2*R3+R2*R3+A2*R1*R3+A1*R1*R3+R1*R3)
Vy=(R2*R4*V2+A1*R1*R4*V2+R1*R4*V2+A2*R2*R3*V1)/(R2*R4+A1*R1*R4+R1*R4+A2*R2*R3+R2*R3+A2*R1*R3+A1*R1*R3+R1*R3)
Vo1=(A1*(R2*R4*V2+R1*R4*V2-R2*R4*V1-R2*R3*V1))/(R2*R4+A1*R1*R4+R1*R4+A2*R2*R3+R2*R3+A2*R1*R3+A1*R1*R3+R1*R3)
Vo2=-(A2*(R2*R4*V2+R1*R4*V2-R2*R4*V1-R2*R3*V1))/(R2*R4+A1*R1*R4+R1*R4+A2*R2*R3+R2*R3+A2*R1*R3+A1*R1*R3+R1*R3)

A1 is the internal gain of op amp 1, A2 is the gain of the other op amp.

Note that numerical results from these equations match the simulation results out to 5 or 6 significant digits for several different values of V1, V2, A1 and A2.
The fact that these equations resulted from an analysis using just linear algebra must say something about the validity of the circuit, even if it turns out to be impractical or there is a better way to achieve the same end result (possibly a differential amplifier with inverter for example).
Im sure they can be shown in a simpler way too.

I agree that the effects of input offset voltage should be investigated as well.
We also did not yet look at the AC characteristics using real op amps with some phase shift, etc.

Even if this circuit turns out to be impractical (as the gains wont be the same for different op amp sections without some sort of gain matching) i still think it is interesting from a purely academic standpoint.

Just to simplify a little here...

If we define dv=Vx-Vy then we get:

dv=(R2*(R4+R3)*V1-(R2+R1)*R4*V2)/((R2+(A1+1)*R1)*R4+((A2+1)*R2+(A2+A1+1)*R1)*R3)

and then we have:
Vo1=-dv*A1
Vo2=dv*A2

Assuming we have matched resistors in their respective positions R4=R2 and R3=R1, then we get:
dv=-(R2*V2-R2*V1)/(R2+(A1+A2+1)*R1)

and for the original circuit where R2=2*R1 we get:
dv=2*(V1-V2)/(A1+A2+3)

which for large A not infinite, we get:
dv=2*(V1-V2)/(A1+A2)
[which generalizes to dv=G*(V1-V2)/(A1+A2) where G is R2/R1 when R2/R1=R4/R3]

and for A2=A1=A we get:
dv=(V1-V2)/A

so we get:
Vo1=V2-V1
and
Vo2=V1-V2=-Vo1
It's not necessary to go to this much trouble to get the result that Vo1 = -Vo2

Just notice that the inputs to the two opamps are the same, except that the polarities are reversed:

Vo1 = Av1*(Vy-Vx)
Vo2 = Av2*(Vx-Vy)

So, if Av1 = Av2 then Vo1 = -Vo2

However, this is not uniquely true if Av1 and Av2 are infinite (ideal opamps).
 
Here's another way to see why the solution to the circuit is indeterminate when Av→∞. In that case (ideal opamps) Vx = Vy so that (Vx-Vy)=0. Since the output voltage of an opamp is Vo = Av*(Vx-Vy), for an ideal opamp this becomes Vo = ∞*0, an indeterminate result if no other constraints determine a value for Vo.
 

t_n_k

Joined Mar 6, 2009
5,455
Are the opamp models identical? I would assume they are unless you made changes to them. Both input offset voltage and open loop gain differences can contribute to differences in output voltage.

The theoretical difference in Vo1 and Vo2 should be 4.4 volts, which seems to be the case. Notice that the two opamps have the same voltages (Vx and Vy) applied to their inputs, but with opposite polarity.

If the opamp models have a fixed input offset voltage it would account for your result.

If I solve your circuit with Av=100000 for each opamp, and an offset voltage of .1167 millivolts for each opamp, I get Vo1=9.4700 V and Vo2=13.8700 V.
Just to be perverse, I bread boarded the circuit using several LM324 quad op-amp chips. With a maximum input offset voltage of 5mV (as noted in one data sheet) it was of little surprise to observe the outputs heading for the rails without fail.
 

Tesla23

Joined May 10, 2009
542
If I solve your circuit with Av=100000 for each opamp, and an offset voltage of .1167 millivolts for each opamp, I get Vo1=9.4700 V and Vo2=13.8700 V.
This may be what is happening, but I think it is more likely that Vo2 is the result of saturation.

Looking at an old LM358 datasheet (one with lots more useful graphs than the recent ones), the output was characterised as:



So on 15V rails you would expect the outputs to saturate at around +13.7V and -14.4V

Looking at Jony's results:


and I start to suspect that the LM358 results are from saturation rather than a happy coincidence of offset voltages that put the outputs so close to the rails.

The NE5532 datasheet shows that on +/- 15 volt supplies it typically swings +/- 13.5V. Again suspiciously close to Jony's results.

The remarkable thing about this circuit is that it tends to drive one output to saturation, but still produce the correct differential output voltage. It's easy to show that in Jony's results, that if the op-amp outputs that are close to the rails are actually the result of limiting, then the correct differential output will still be produced. I found this unexpected.
 

Attachments

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MrAl

Joined Jun 17, 2014
11,389
Here's another way to see why the solution to the circuit is indeterminate when Av→∞. In that case (ideal opamps) Vx = Vy so that (Vx-Vy)=0. Since the output voltage of an opamp is Vo = Av*(Vx-Vy), for an ideal opamp this becomes Vo = ∞*0, an indeterminate result if no other constraints determine a value for Vo.
Hi,

That is definitely not proof of anything because it does not depend on the absolute levels it depends on what happens 'near' the two extremes.
This is what allows us to take the limit for other circuits as well which always work in practice but have those same limits.

For a simple example, consider 'a' and 'b' such that:
c=a*b

Now let a=1 and b=1, we get of course c=1*1=1.
Now let a=10 and b=0.1, we get c=10*0.1=1.
Now let a=100 and b=0.01, we get c=100*0.01=1.
It's obvious that b=1/a, so we have c=a*1/a=a/a.
Now let 'a' go to infinity, we have c=1 once again.

Another quick example is:
x*ln(x) as x approaches zero from the right. This also turns into the indeterminate form 0*inf, but L'Hospital's Rule tells us how to handle that too after we turn it into a fraction.



Back on topic, here are two equations for Vo1 and Vo2 that include the input offsets of the two op amps...

Vo1=(A*(R2*R4*V2+R1*R4*V2-R2*R4*V1-R2*R3*V1+E1b*R2*R4-E1a*R2*R4+E1b*
R1*R4-E1a*R1*R4+E2b*B*R2*R3-E2a*B*R2*R3+E1b*B*R2*R3-E1a*B*R2*R3+E1b*R2*R3-E1a*R2
*R3+E2b*B*R1*R3-E2a*B*R1*R3+E1b*B*R1*R3-E1a*B*R1*R3+E1b*R1*R3-E1a*R1*R3))/(R2*
R4+A*R1*R4+R1*R4+B*R2*R3+R2*R3+B*R1*R3+A*R1*R3+R1*R3)
Vo2=-(B*(R2*R4*V2+R1*
R4*V2-R2*R4*V1-R2*R3*V1-E2b*R2*R4+E2a*R2*R4-E2b*A*R1*R4+E2a*A*R1*R4-E1b*A*R1*R4
+E1a*A*R1*R4-E2b*R1*R4+E2a*R1*R4-E2b*R2*R3+E2a*R2*R3-E2b*A*R1*R3+E2a*A*R1*R3-
E1b*A*R1*R3+E1a*A*R1*R3-E2b*R1*R3+E2a*R1*R3))/(R2*R4+A*R1*R4+R1*R4+B*R2*R3+R2*
R3+B*R1*R3+A*R1*R3+R1*R3)

where
A is the internal gain of op amp 1,
B is the internal gain of op amp 2,
E1a is the input offset for op amp 1 at the inverting terminal,
E1b is the input offset for op amp 1 at the non inverting terminal,
E2a is the input offset for op amp 2 at the inverting terminal,
E2b is the input offset for op amp 2 at the non inverting terminal.

All of the input offset voltage sources have their positive terminal connected right to the input of the offset, in series with the respective input.
To reverse any of them (flipping the source so the negative terminal connects to the op amp input) make that one negative.
Note the equations can be simplified.

Using these two equations it is not hard to see that the circuit will be very impractical in real life (unless maybe with a very low input offset op amp). That's a shame but that's life. Even with a small input offset of 0.001 on one of the inverting inputs, we see a marked change in outputs which results in an unusable circuit, at least in it's current form. I did not investigate the saturated form yet however, but it seems that if one output saturates in the same direction as the other output (because of input offset for example) then we would see the second output go to some other voltage than what it should have been, even if still bounded (for example -8v instead of -4v).

After working with this circuit however i cant help but feel that there is some significance to it, in the form of perhaps a redundant op amp where we need redundancy in case of some sort of failure. Just a passing thought though which could be hard to actually get to work right, but there may be some other significance also. For example using a sampled system we could get around the gain problem, for what it is worth.

So i'll ask Jony the question that is going to be or already is on everyones mind:

Jony:
Where did you get this circuit and was it associated with any particular application?
 
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