# CMRR formula gives wrong result!

by simpComp
Tags: cmrr, formula, result
P: 44
Hello,

If you go to this link:

http://en.wikipedia.org/wiki/Common-...ejection_ratio

and scroll all the way down to the bottom where they show the:

"Example: operational amplifiers"

section.... we have:

 So for example, an op-amp with 90dB CMRR operating with 10V of common-mode will have an output error of ±316uV.
I get +/-222mv ????

Am I doing the math wrong?
Mentor
P: 39,599
 Quote by simpComp Hello, If you go to this link: http://en.wikipedia.org/wiki/Common-...ejection_ratio and scroll all the way down to the bottom where they show the: "Example: operational amplifiers" section.... we have: I get +/-222mv ???? Am I doing the math wrong?
I get the 316mV number. Can you show how you are typing the numbers into your calculator?
 P: 44 90/10 then the result is multiplied by 10. then 10/the result above hence: 10/((90/20)10) = 0.222??? and they get 316 micro volts?
Mentor
P: 39,599

## CMRR formula gives wrong result!

 Quote by simpComp 90/10 then the result is multiplied by 10. then 10/the result above hence: 10/((90/20)10) = 0.222??? and they get 316 micro volts?
I think there are several issues with the way you are trying to solve this. First remember that for voltage ratios, the equation is dB = 20log(ratio).

So for this problem, you start with 90dB = 20log(10/x).

To solve that equation, you divide both sides by 20, and then take 10^ for both sides.

Can you take it from there...?
 P: 44 Hi berkeman, You talking to a very slow guy here!!! so I try: I went back to my algebra notes of 20 years ago and saw the example: 2 = log(base10)(100) then I tried doing the reverse: 10^2 = 100 So then I understood that its 10 to the power of (90/20)... So I did the same while following your instructions.... 90dB = 20log(10/x) 90dB/20 = (20log(10/x))/20 4.5dB = log(10/x) Antilog = 10^4.5 so: 10^4.5 = 10/x which becomes: 10/31622.8 = x 316 uV Thankyou for your help!
 Mentor P: 39,599 Woot! Good job!

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