# Why is the CMRR of an op-amp defined in such a way?

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?

• 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. Commented Sep 8, 2015 at 16:32
• 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.
– Kaz
Commented Sep 8, 2015 at 21:18

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.