Why is the CMRR of an op-amp defined as \$20\log(A_d/A_c)\$?

I know that the value of \$A_d\$ is far greater than \$A_c\$, so we take the log, but why are we multiplying 20 with it?

  • 5
    \$\begingroup\$ Actually it isn't, it is defined as A(d)/A(c). Only when you want to express CMRR in dB, then 20*log(CMRR) is calculated. \$\endgroup\$
    – jippie
    Commented Sep 8, 2015 at 16:32
  • 3
    \$\begingroup\$ possible duplicate of Confusion between Voltage gain & Voltage gain in decibels (dB) Though this is ostensibly about CMRR, CMRR is just a ratio of voltages, just like gain. \$\endgroup\$
    – Kaz
    Commented Sep 8, 2015 at 21:18

1 Answer 1


The \$20\log_{10}(X/Y)\$ is how we change ratios into decibels or dB.

Original, dB was used to compare ratios of power. Energy and power are often related to the square of some variable. \$1/2 mv^2\$ for kinetic energy or \$P=I^2 R\$ for power through a resistor. In this original usage, we would use \$10\log_{10}(X/Y)\$. Thus if \$X = 10 Y\$, it would be 10 dB and if \$X = 100 Y\$, it would be 20 dB. This was nice as we didn't have to deal with a bunch of decimal places to get decent comparisons.

Now, eventually we started using dB for signals such as voltage. Voltage is related to the square root of power. \$P = V^2 / R\$. Now, convert voltage to power and place it in the dB equation:

$$10\log_{10}\left(\frac{V_1^2 / R}{V_2^2 / R}\right)$$

By the rules of logarithms, we can remove the power of 2 from the inside of the \$\log_{10}\$ to a multiplier of 2 in front of the \$\log_{10}\$. This is how we get \$20\log_{10}(X/Y)\$. Many contributions to signal processing came from electrical engineering and thus the \$20\log_{10}\$ stuck.

Here are a few more links for reading.

  • \$\begingroup\$ It may be worth noting that while a power ratio will be the square of the voltage or current ratio in cases where the load impedance and phase are constant, the squaring convention is used even in cases where the power ratio has nothing to do with the voltage or current ratio (e.g. it's not uncommon for either voltage or current will be nominally zero, but have a slight offset which is not proportional to the input signal; in such cases, the "power" ratio is essentially irrelevant, but the a 10:1 change in signal would still be treated as a 100:1 change in power). \$\endgroup\$
    – supercat
    Commented Sep 8, 2015 at 17:17

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