I need some assistance on the derivation of the formula for AV(cm) depicted in the figure below. the +/-2 delta_R/R and +/-4 delta_R/R are giving me hard time deriving. enter image description here

The book probably provided enough information to the reader for them to derive it but I'll be honest I still cannot proceed on going about it. enter image description here

  • \$\begingroup\$ Try to use superposition theorem \$\endgroup\$
    – G36
    Jun 17, 2020 at 17:38
  • \$\begingroup\$ I know about that part. But +/-2 delta_R/R and +/-4delta_R/R are not clear to me. \$\endgroup\$
    – Paumdrdo
    Jun 17, 2020 at 17:45
  • \$\begingroup\$ So the book does not mention that if you have perfected matched resistors and ideal opamp the Ac_cm is 0V/V (as any ideal differential amplifier will have)? \$\endgroup\$
    – G36
    Jun 17, 2020 at 17:59
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    \$\begingroup\$ @Paumdrdo I'll add an answer, soon. But I've a meeting coming up, shortly. So it will have to wait until after that. \$\endgroup\$
    – jonk
    Jun 17, 2020 at 20:16
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    \$\begingroup\$ @Paumdrdo There is a pretty gross error in the problem statement, which may very well be confusing you. I don't have time to discuss it, right now. But it boils down to this: they cannot be talking about \$A_{v_\text{CM}}\$ but instead about the following sensitivity equation:$$\frac{\frac{\text{d}\,A_{v_\text{CM}}}{A_{v_\text{CM}}}}{\frac{\text{d}\,R}{R}}$$In short, the % variation of \$A_{v_\text{CM}}\$ vs the % variation of \$R\$ (in the first case with \$R=R_1=R_2\$.) Or, put another way$$\frac{\text{d}\,A_{v_\text{CM}}}{A_{v_\text{CM}}}=\pm 2\,\frac{\text{d}\,R}{R}$$That may help. \$\endgroup\$
    – jonk
    Jun 17, 2020 at 20:31

2 Answers 2


In order to null common-mode gain we want \$A_{v(CM)}=0\$

enter image description here

Recall inverting voltage gain is the ratio =-k and non-inverting gain is 1+k. for R ratio k=R2/R1

We know \$A_{v(DM)}=\frac{R2}{R1}=\frac{R2'}{R1}\$

thus \$A_{v(CM)}= (\frac{R_2}{R_1}+1)(\frac{R_2'}{R_1'+R_2'})-(\frac{R_2}{R_1}) = 0 \$ for null CM gain

\$A_{v(CM)}= ({\frac{R_2+R_1}{R_1}})({\frac{R_2'}{R_2'+R_1'}})-(\frac{R_2}{R_1}) = 0 \$

to balance input Z let \$R_1=R_1'=R, ~~~R_2=R_2' \$
to null bias current DC offset voltage.

If all parts have the same polarity of tolerance error, it cancels out. Now compute error for any one part with a tolerance error of ΔR/R (%) the resulting CM gain error is the same % for k=1 but 2x for large k due to ANY 1 part tolerance error.

(For practical reasons Laser trimmed R's inside IC's are better or R Arrays with 0.01% tolerance ratios.) But even for perfect INA's with 120 dB CMRR, the imbalance of each wire in % causes CMRR error.

I'll let @jonk do the better math.

  • \$\begingroup\$ Stewart, Where do I substitute R1 and R2? If I substitute to Avdm I got k for the result. Can you give me more hint? \$\endgroup\$
    – Paumdrdo
    Jun 17, 2020 at 18:17
  • \$\begingroup\$ actually I disagree with the result in the question. e.g. 1% change of any 4 parts results in 1% Av(cm) from 0 Compute for all values then let one part differ by ΔR/R \$\endgroup\$ Jun 17, 2020 at 18:32
  • \$\begingroup\$ I'm more confused now. :( \$\endgroup\$
    – Paumdrdo
    Jun 17, 2020 at 18:36
  • \$\begingroup\$ Put 18-6,7 in 18-5 then compute Av cm with tolerance in 1 part with Avdm = k R \$\endgroup\$ Jun 17, 2020 at 18:40
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    \$\begingroup\$ @Paumdrdo I'll write something up. \$\endgroup\$
    – jonk
    Jun 17, 2020 at 19:48

I'm going to write up a simplified starting point for just the first part of the question (the slightly easier part.)

You are supposed to be able to perform the addition indicated in 18-5 and find this simplified form:

$$\begin{align*} A_{v_\text{CM}}&=\frac{R_1\,R_2^{'}-R_2\,R_1^{'}}{R_1\left(R_1^{'}+R_2^{'}\right)} \end{align*}$$

This was the simplified form I kept pressing you to achieve. It's just algebra, so you need to be able to achieve this much, given where you are at right now.

Once you have this much, what you are being asked to achieve with the first question, where \$R=R_1=R_2\$, is for you to think about the meaning of \$\frac{\Delta\,R}{R}\$. (In calculus, this is \$\frac{\text{d}\,R}{R}\$.) That is just a percentage, really. Right? So, let's say that \$R_1\$ varies by its allowed variation in one direction and that \$R_2\$ varies by its allowed variation in the exact opposite direction. Wouldn't that lead us to the worst case situation?

If so, then here's the result of that kind of thinking. We substitite \$R\left(1\pm\frac{\text{d}\,R}{R}\right)\$ for \$R_1\$ and substitute \$R\left(1\mp\frac{\text{d}\,R}{R}\right)\$ for \$R_2\$ (note the opposite sign arrangements here.) As \$R=R_1=R_2\$, it follows:

$$\begin{align*} A_{v_\text{CM}}&=\frac{R\left(1\pm\frac{\text{d}\,R}{R}\right)\,R_2^{'}-R\left(1\mp\frac{\text{d}\,R}{R}\right)\,R_1^{'}}{R\left(1\pm\frac{\text{d}\,R}{R}\right)\left(R_1^{'}+R_2^{'}\right)}\\\\ &=\frac{\left(1\pm\frac{\text{d}\,R}{R}\right)\,R_2^{'}-\left(1\mp\frac{\text{d}\,R}{R}\right)\,R_1^{'}}{\left(1\pm\frac{\text{d}\,R}{R}\right)\left(R_1^{'}+R_2^{'}\right)}\\\\ &=\frac{R_2^{'}}{R_1^{'}+R_2^{'}}-\frac{R_1^{'}}{R_1^{'}+R_2^{'}}\cdot\left[\frac{1\mp\frac{\text{d}\,R}{R}}{1\pm\frac{\text{d}\,R}{R}}\right] \end{align*}$$

I want you to consider this, first, and see if you feel I've made any conceptual mistakes while proceeding, above. I also want to draw your attention to the bracketed factor of the second term, above.

I'd like to wait, now, and see if you feel any of the above is productive or triggers any thoughts.

  • \$\begingroup\$ Is \$R\left(1\pm\frac{\text{d}\,R}{R}\right)\$ equivalent to \$R\pm\Delta\text{R}\$?. I can see now how the algebra works. Is the next step taking the partial derivative of Avcm with respect to R? I read a little about the sensitivity analysis formula. How do we get to the form given in equation 18-8? please bear with as it has been a long time since I did some calc and algebra. \$\endgroup\$
    – Paumdrdo
    Jun 18, 2020 at 18:12

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