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I am confused about the circuit of an instrumentation amplifier. In class we solved the following example:

schematic

simulate this circuit – Schematic created using CircuitLab

No big deal; with negative feedback we say that V+ = V- on the op-amps and find the voltage using KCL. However, the professor waved his hands and said due to symmetry, it is equivalent to the following circuit:

schematic

simulate this circuit

The difference is that that ground is removed and the two resistors are connected. I don't understand how this would be equivalent. Current would flow from node A to node B or vice versa. It would completely change the calculations. I've been working through the equations and I get an ugly mess of equations, not the elegant solution: $$\frac{R_4}{R_3}\left(1+\frac{R_2}{R_1/2}\right)(V_2-V_1)$$

My question is, why does this work? Why doesn't connecting the two resistors change everything like I think it should?

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  • \$\begingroup\$ at 2nd glance (2*R2/R1 +1)*R4/R3 so if R1 is open gain is just R4/R3 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 20 '16 at 0:29
  • \$\begingroup\$ I have the answer already (we did it in class), but I want to know why the circuits are equivalent. \$\endgroup\$ – Klik Sep 20 '16 at 0:29
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    \$\begingroup\$ Any voltage applied to midpoint 0 or Vcc/2 or ? within common mode range becomes a common mode output and thus 0 diff. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 20 '16 at 0:42
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    \$\begingroup\$ Despite the fact that they may have the same output formula, they are not "equivalent". For example, mismatch in R1/2 will kill your CMRR in the top circuit, but will just cause gain to be off in the bottom circuit (which is why its a "better" circuit for many purposes). \$\endgroup\$ – Scott Seidman Sep 20 '16 at 12:59
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    \$\begingroup\$ The version with one single R1 (I called it "balanced" in my first comment but probably IA is more correct) would behave differently. OA1 and OA2 outputs will copy input common mode at 2V, not amplifying it. Then we superimpose differential \$\pm100\,\text{mV}\times 10=\pm 1\,\text{V}\$ setting OA1 and OA2 outputs at 2V-1V=1V and 2V+1V=3V. So now they are not saturated and OA3 can do his job. IMHO this is most important advantage of the IA above. Cheers \$\endgroup\$ – carloc Sep 20 '16 at 17:02
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schematic

simulate this circuit – Schematic created using CircuitLab

So this is the easy version: $$U_{o1} = U_1 + I_1 * R_2 \qquad,\qquad U_{o2} = U_2 + I_2 * R_2$$

with $$I_1 = 2\frac{U_1}{R_1} \qquad I_2 = 2\frac{U_2}{R_1}$$

this yields $$U_{o1} = U_1 (1 + 2\frac{R_2}{R_1}) \qquad,\qquad U_{o2} = U_2(1 + 2\frac{R_2}{R_1})$$ and

\$U_{o2}-U_{o1} = (U_2-U_1)(1+2\frac{R_2}{R_1})\$.

As this is pretty straight forward I won't go into it any more. Now we want to show that connecting the two half-resistors will yield the same result:

schematic

simulate this circuit

In this one, finding the voltage across \$R_1\$ (combined \$2*\frac{R_1}{2}\$) is very easy as well. it's just \$U_2 - U_1\$. By this, we can calculate \$I_0 = \frac{U_2-U_1}{R_1}\$. Because no current flows in or out of the inputs, \$I_0\$ goes through both \$R_2\$ resistors equally. Now we can calculate the output voltages:

$$U_{o1} = U_1 - I_0 * R_2 \qquad, \qquad U_{o2} = U_2 + I_0 * R_2$$ $$U_{o2} - U_{o1} = U_2 - U_1 + 2*I_0*R_2 = U_2 - U_1 + 2*(U_2 - U_1)\frac{R_2}{R_1}$$

\$ = (U_2 - U_1)(1 + 2 \frac{R_2}{R_1})\$

Which is the same solution as for the first circuit. So you're right. There is a current flowing, but it's proportional to the difference between \$U_2\$ and \$U_1\$.

Edit: As it's coming up in the comments, the voltage between the two halves of R1 in the second circuit is not 0V.

schematic

simulate this circuit

As we can see, the potential between the two negative inputs splits in half on the resistors. Both potentials are \$\frac{U_2-U_1}{2}\$. If we want to calculate the absolute voltage in the middle you can go from either side: $$U_M = U_1 + \frac{U_2-U_1}{2} = U_2 - \frac{U_2-U_1}{2}$$

$$= \frac{U_1+U_2}{2}$$

which is the average of the input voltages. Also note that \$U_{o1}\$ for a given input voltage is different between connecting the resistors together and grounding both. It's just the differential output that's the same.

Remember we calculated \$U_{o1}\$ and \$U_{o2}\$ for the grounded resistors version in the beginning and they were only dependent on the respective input voltage. However, with the connected resistors we get:

$$ U_{o1} = U_1 - I_0 * R_2 = U_1 - (U_2 - U_1) \frac{R_2}{R_1} $$ $$ U_{o2} = U_2 + I_0 * R_2 = U_2 + (U_2 - U_1) \frac{R_2}{R_1} $$

So while \$U_{o2}-U_{o1}\$ is the same in both circuits, the connected one has output voltages in the first stage that are dependent on both input voltages. The very important advantage is that only the difference between the signals gets amplified in the first stage. Since real opamps have rise times and especially supply rails that can be hit even with a small differential voltage if both voltages are relatively high. Here is a plot of the two different circuits at 1V differential voltage and U1 sweeped from 0V to 10V. As you can see, the grounded circuit hits 30V and more which could easily be above the supply rail while the differential circuit is well balanced.

enter image description here

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  • \$\begingroup\$ If we take the second circuit and replace R1 with two resistors of halve the value connected in series, it should be possible to proove that the voltage at the mid point of those resistors is zero. But when this voltage is zero, there is no difference when this point is connected to ground or not, because no current will flow in this connection. I tried this proove but I failed. \$\endgroup\$ – Uwe Sep 20 '16 at 13:01
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    \$\begingroup\$ There is no reason why the voltage at the midpoint of the two resistors should be zero. It should be midpoint between v1 and v2. \$\endgroup\$ – Scott Seidman Sep 20 '16 at 17:11
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    \$\begingroup\$ Exactly, Scott. I have edited my answer with an explanation of why the circuits are indeed very different even though they provide the same differential outputs. \$\endgroup\$ – Felix S Sep 20 '16 at 17:18
  • \$\begingroup\$ @FelixS This is an excellent explanation. I'm not familiar with CMM terminology so I really appreciate that your answer was simple enough for me to understand. Fantastic job. Thank you for the time. \$\endgroup\$ – Klik Sep 21 '16 at 3:14
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It is not equivalent, with ideal components the final outputs are equivalent but the internal signals are not. With real components the second circuit is much better.

To make the analysis tractable lets start with the assumption that all the components are ideal. We can consider the affect of non-idealities once we understand the basic behaviour.


We can analyse these circuits is by superposition. We can consider our input to be made up of a common mode component and a differential mode component. The overall response of the circuit is made up of the sum of the responses to the common mode component and the differential mode component.

For a purely differential input (\$V_1=-V_2\$), both circuits behave the same. We can see this through symmetry, the voltages in the top half of the first stage are equal and opposite to those in the bottom half of the first stage. So the node connecting the two resistors must be at zero.

For a purely common mode input (\$V_1=-V_2\$), the internal behaviour is somewhat different. In the top circuit the two inputs are amplified separately by the first stage. In the bottom circuit we can see that the voltages in the top and bottom halves are the same and therefore there is no current in the gain resistor and therefore the first stage has a common mode gain of unity.

In the ideal case this change in response of the first stage to common mode inputs has no affect on the final result, because the second stage removes all the common mode anyway.


Now we understand the ideal case lets get back to reality and understand why the second version is so much better. Lets assume that our goal is to use a high gain (say g=1000) to detect a small differential signal (say 1mV) on top of a large common mode signal (say 1V). Lets also assume that as per normal instrumentation amp practice we put our gain in the first stage and have a second stage gain of unity.

The first reason is saturation. In the top circuit any gain in the first stage is applied to both the differential mode an the common mode. So to avoid saturation our gain in the first stage is limited to 10 or so. In the bottom circuit we have a common mode gain of 1, so our op-amps can easily avoid saturation.

The second reason is that in the top circuit any resistor value inaccuracies in the first stage will cause gain imblance between the two amplifier circuits which will in turn common mode signal into differential mode signal. The bottom circuit doesn't have this problem, regardless of resistor values it will essentially float on the common mode and (assuming the op-amps are linear) it will not turn common mode into differential mode.

The third reason is that in the bottom circuit the first stage essentially amplifies the common mode rejection of the second stage, since the first stage amplifies differential mode signals but has unity gain for common mode signals.

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