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I was reading a paper from 1977 regarding the analysis for the CMRR of a three opamp instrumentation amplifier, I got the paper through my Unviersity but I cant post the paper since its copyrighted, however here is the link to download it legally through IEEE Xplore: C.M.R.R. analysis of the 3-op-amp instrumentation amplifier

The circuit im refering to is as follows 3 OPAMP IA

Basically the short (1 page) paper describes the procedure to obtain the CMRR of an instrumentation amplifier. The formula is the following:

$$ CMRR=\frac{Ad\cdot cmrr_o}{1+Ad\cdot cmrr_o(\alpha_2-\alpha_1)}$$

Where $$Ad=\frac{(R1+R2+R3)}{R2}$$ is the differential gain of the buffer stages (both opamps at the input) $$cmrr_o \text{ is the CMRR of the differential amplifier stage}$$ and $$ \alpha_i=(cmrr_i-1)/cmrr_i $$ where $$cmrr_i $$ is the CMRR of the ith opamp

The paper follows the equation with a simple example:

An instrumentation amplifier was designed and built with the following parameters: Ad = 100, R4=R5=R6=R7= 10k (nominal value), with R5 adjustable. 741-type op-amps were used with the following measured cmrr's: $$cmrr_1= -111\times 10^3$$ $$cmrr_2= 40 \times 10^3$$ $$cmrr_3= -52.5 \times 10^3$$

R5 was used to obtain $$cmrr_o = 100 dB$$

The measured CMRR of the IA was $$-2.4 \times 10^4$$ in reasonable agreement with the previous equation which predicts a CMRR of $$CMRR=-2.9 \times 10^4$$

I can do the math and arrive at the same result as the paper, but my question is this: how is it possible to obtain negative CMRRs (when not stated in decibels) such as the values of cmrr_1 and cmrr_2, does it mean the differential gain or common mode gain is negative in that particular opamp? in that case what does it mean that an opamp has a negative differential or common mode gain?

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    \$\begingroup\$ Does it just means that it’s an inverting differential amplifier ? \$\endgroup\$ – nidhin Oct 29 '17 at 4:14
  • \$\begingroup\$ The thing is that it is the CMRR of the opamp itself that is negative \$\endgroup\$ – S.s. Oct 29 '17 at 4:33
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    \$\begingroup\$ Start with the definition, what does CMRR mean? \$\endgroup\$ – Neil_UK Oct 29 '17 at 5:11
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Neil hit things on the head. I'll try and write it up a little.

The paper you are looking at is a little aged and the terms they use and the way they use them may be a little unfamiliar. First off, take a look at the wikipedia page on the topic: Common-mode rejection ratio. There you will see the following equation:

$$V_o=A_d\left(V_{\left(+\right)} - V_{\left(-\right)}\right)+\frac{1}{2}A_{cm}\left(V_{\left(+\right)} + V_{\left(-\right)}\right)$$

Let's fabricate a somewhat fuller schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

(To get their nominal gain of 100, \$R_1=R_3=49.5\cdot R_2\$. And yes, I did take note that they adjusted the gain towards its nomimal value by making changes in \$R_5\$.)

If you take some measurements of the inputs and the output of each of the three opamps while setting \$V_{CM}=0\:\textrm{V}\$ and \$V_D=1\:\textrm{mV}\$, and do this a second time with say \$V_{CM}=100\:\textrm{mV}\$, then you can solve for both the differential gain values as well as the common mode gain values for each of the opamps. And these values will not only be different, they will also differ in sign for \$A_d\$ for each.

The following solution set to use for each opamp would be:

$$\begin{align*} A_d&=\frac{V_{O_1}\left(V_{\left(+\right)_2}+V_{\left(-\right)_2}\right) - V_{O_2}\left(V_{\left(+\right)_1}+V_{\left(-\right)_1}\right)}{\left(V_{\left(+\right)_1}+V_{\left(-\right)_1}\right)\left(V_{\left(-\right)_2}-V_{\left(+\right)_2}\right)-\left(V_{\left(+\right)_2}+V_{\left(-\right)_2}\right)\left(V_{\left(-\right)_1}-V_{\left(+\right)_1}\right)}\\\\ A_{cm} &= \frac{V_{O_2} \left(V_{\left(-\right)_1} - V_{\left(+\right)_1}\right) + V_{O_1}\left( V_{\left(+\right)_2} - V_{\left(-\right)_2}\right)}{V_{\left(-\right)_1} V_{\left(+\right)_2} - V_{\left(-\right)_2} V_{\left(+\right)_1}} \end{align*}$$

In the above pair of equations I used the subscript of '1' to indicate the first measurement with, say, \$V_{CM_1}=0\:\textrm{V}\$ and the subscript of '2' to indicate the second measurement with, say, \$V_{CM_2}=100\:\textrm{mV}\$. I suspect they made these kinds of measurements in order to arrive at their values.

Given the above solution equation set and the circuit arrangement, then it's actually true that there will be a negative value for \$A_{d_{A1}}\$ and a positive value for \$A_{d_{A2}}\$ and a negative value for \$A_{d_{A3}}\$, with this arrangement. Nothing magical here.

Yes, this doesn't comport with the usual meaning for the computation of decibels today, where the differential gain is always taken as positive. But in this case, I think they were actually making voltage measurements using a voltmeter and using the above purely mathematical solution approach to computing \$A_d\$. In that case, you can and will get negative values for \$A_d\$, given this topology.

I think that's all it means.


I don't want to try and evaluate a paper I haven't read, though. But at least I can see how they were able to arrive at values such as those you mentioned. The sign is all about the topology here.


To get the equation solutions I gave earlier, let's say we assume there exists a value \$A_d\$ known as the differential gain and that there exists a value \$A_{cm}\$ known as the common mode gain. Let's say that we want to work out what those values are.

Well, we have two unknowns so we will need two equations:

$$\begin{align*} V_{O_1}&=A_d\left(V_{\left(+\right)_1} - V_{\left(-\right)_1}\right)+\frac{1}{2}A_{cm}\left(V_{\left(+\right)_1} + V_{\left(-\right)_1}\right)\\\\ V_{O_2}&=A_d\left(V_{\left(+\right)_2} - V_{\left(-\right)_2}\right)+\frac{1}{2}A_{cm}\left(V_{\left(+\right)_2} + V_{\left(-\right)_2}\right) \end{align*}$$

If you simultaneously solve those two equations for \$A_d\$ and \$A_{cm}\$, treating everything else as measurements you made, then you will get the solution equations I provided earlier.

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  • \$\begingroup\$ Can you elaborate on how did you arrive to the solution set of Ad and Acm for each opamp? \$\endgroup\$ – S.s. Oct 29 '17 at 16:19
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    \$\begingroup\$ @A.J. The differential gain, \$A_d\$ as used in the paper isn't a "stand alone" definition regarding each opamp's sole behavior at a single common mode voltage. If that were the case, all three opamps would show a positive \$A_d\$ (of course.) It's a definition of \$A_d\$ that develops from the solution set for two different common mode values combined in this circuit. \$\endgroup\$ – jonk Oct 29 '17 at 21:02
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    \$\begingroup\$ @A.J. Perhaps this will help. Imagine that all three opamps have positive \$A_d\$ but slightly different positive values. The circuit arranges things so that the values for A1 and A3 are slightly less positive than for A2. The total circuit behaves to construct an \$A_d\$ that is a middling positive value. But the computed \$A_d\$ values used by the authors compares all three against this middling value, so that A1 and A3 are "negative" relative to it and for A2 it's positive relative to it. I know this isn't helpful. But I don't want to write out all the math here. It does work out. \$\endgroup\$ – jonk Oct 29 '17 at 21:05
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    \$\begingroup\$ @A.J. Perhaps you should wire up a sample circuit in Spice. Take measurements using Spice. Write them down. Change the common mode value. Repeat. Plug them into those equations. You will see negative and positive values for \$A_d\$ and you will see them match the sign in your paper, too. I'll write a little more about developing those equations I gave -- using Sage. If you have another program for it, fine. \$\endgroup\$ – jonk Oct 29 '17 at 21:08
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    \$\begingroup\$ @A.J. I've edited in a short note at the bottom which should allow you to use Sage (or any other program) to provide a simultaneous solution. Or you can work it out by hand. You should reach the equations I provided. Once that is achieved, get Spice running and plug in some opamps per the circuit I provided and ... take some measurements. Do so for each opamp and for two common mode voltages. Plug the results in and compute values. Report back. \$\endgroup\$ – jonk Oct 29 '17 at 21:19

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