CMRR formula gives wrong result

[simpComp]

Hello,

If you go to this link:

http://en.wikipedia.org/wiki/Common-mode_rejection_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?

 

[berkeman]

Hello,

If you go to this link:

http://en.wikipedia.org/wiki/Common-mode_rejection_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?

 

[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?

 

[berkeman]

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...?

 

[simpComp]

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 uVThankyou for your help!

 

[berkeman]

Woot! Good job!