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What is the best CMRR definition:

24Volts

Mar 21, 2010
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Hello,

I have seen many definitions in many articles throughout the Internet... here are some of the definitions I found:

CMRR stands for common mode rejection ratio. It is a
measure of the ability of a test instrument to reject
interference that is common to both of its measurement
input terminals. It is expressed in decibels and it is the
ratio of the actual or common signal level appearing on the
two input terminals together to the measured level.

CMRR whichis defined as the ability to reject the common
gain.it is also defined as the ratio between the
differential mode gain to the common mode gain.


The ability of a differential amplifier to not pass (reject) the portion of the signal common to both the + and - inputs.

and so on....

But my question is, for example, for a given differential op amp circuit, if we have a CMMR of:

CMRR = Ad./Acm
CMRR = 2/1

Can we say that in this case the CMRR:

"is the ability of rejecting 1 volt of common voltage for every 2 volts of differential voltage"

I am looking for a definition of CMRR that involves a reference to its numbers in its ratio!

Thanks all
 
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(*steve*)

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Look for a CMMR specified in dB. This is the ratio.

So, if it was 3db, it would be 2/1

If it were 6dB it would be 10^(6/10) / 1 == 4 / 1

If it were 87dB it would be 10^(87/10) / 1 = 500,000,000 / 1

A 741 has a CMMR of 90dB which is twice as good as that!

Now a lot of that is due to the fact that the Acm is not going to be 1. The same datasheet shows us the typical gain is 200,000. So the common mode gain must be 200,000 / 1,000,000,000 = 2 / 10,000.

Perhaps it's easier to say that Ad = 53dB and Acm = -37dB

The simpler way is to have the differential gain expressed in dB, and subtract the CMMR from this to find the differential gain in dB.

In this case, a differential gain of 200,000 is 53db, the CMMR is 90dB, so Acm = 53 - 90 = -37dB

All you need to know is the formula for CMMR, some simple math, and have a datasheet handy. :)
 

24Volts

Mar 21, 2010
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Hello (*Steve*),

I go line by line if I may!

bof!!! I definitely could be wrong but I read everywhere that:
CMRR = (Ad/Acm)

and that:

CMR = 20 Log(10)(Ad/Acm)

where if Ad/Acm = 2/1, then

CMR = 20 Log(10)(2)
CMR = 6.02 dB

therefore:

So, if it was 3db, it would be 2/1

shouldn't it be:
So, if it was 6db, it would be 2/1 ???

ref link: http://www.ecircuitcenter.com/OpModels/CMR/Op3_CMR.htm
 
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(*steve*)

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Sorry, I was using 10Log rather than 20Log.

Just recalculate all the figures. I leave that as an exercise for the student :D
 

24Volts

Mar 21, 2010
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A 741 has a CMMR of 90dB which is twice as good as that!

Now a lot of that is due to the fact that the Acm is not going to be 1. The same datasheet shows us the typical gain is 200,000. So the common mode gain must be 200,000 / 1,000,000,000 = 2 / 10,000.

Perhaps it's easier to say that Ad = 53dB and Acm = -37dB

The simpler way is to have the differential gain expressed in dB, and subtract the CMMR from this to find the differential gain in dB.

In this case, a differential gain of 200,000 is 53db, the CMMR is 90dB, so Acm = 53 - 90 = -37dB

cool!

thanks for this insight!
 

24Volts

Mar 21, 2010
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Hello (*Steve*),

I have three questions!


Q#1)
So the common mode gain must be 200,000 / 1,000,000,000 = 2 / 10,000.

I could be wrong, but isn't the CMRR: 1,000,000,000
See math at attachment "CMRR1"
The math checks out when (0.014142136^2) = Acm = 0.0002 and so CMRR = (200,000/0.0002) = 1,000,000,000 right? :(


Q#2)
Sorry, I was using 10Log rather than 20Log.

So when do we use 10 log(Ad/Acm)^2 and when do we use 20 log(Ad/Acm) ? .... I see they both give the same results... See math in CMRR2 attachment!
:cool:


Q#3)
Why do we have to take the square root of the open loop gain and the Acm in order to calculate dB's....

For example if we take the op amp we have been discussing "741" which has an open loop gain of 200,000 and a CMR of 90db's, and if we know that the Acm is 0.0002, then
why do we have to square root everything so that the CMR formula gives 90 dB's?

If I try it without squaring anything (see "CMRR3" attachment) we don't get 90 dB's, instead I get 180dB ?? :eek:

:confused:

sorry for all the questions!
Thanks
 

Attachments

  • CMRR1.jpg
    CMRR1.jpg
    42 KB · Views: 147
  • CMRR2.jpg
    CMRR2.jpg
    27.7 KB · Views: 189
  • CMRR3.jpg
    CMRR3.jpg
    12 KB · Views: 173
Last edited:

(*steve*)

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I have three questions!

Can I deal with Q2 first?
Q#2)


So when do we use 10 log(Ad/Acm)^2 and when do we use 20 log(Ad/Acm) ? .... I see they both give the same results... See math in CMRR2 attachment!
:cool:

10Log is used for power. 20Log is used for voltage because power is proportional to the square of the voltage. A factor of 2 applied to the log is that squaring.

However, Look at the second line of your math (in both cases). That looks like a mathematical error. You seem to be intending to take the constant from one side to the other of the equation, but you are only doing half a job.

I see no reason to square the gain figures as long as you're using 20Log. because 10*log(x*x) = 20*Log(x)
 

(*steve*)

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Let me repeat my math using 20Log...

Look for a CMMR specified in dB. This is the ratio.

So, if it was 6db, it would be 2/1 (20Log(2/1) = 20Log(2) - 20Log(1) = 6 - 0 = 6)

If it were 12dB it would be 10^(12/20) / 1 == 4 / 1

If it were 87dB it would be 10^(87/20) / 1 = 22,500 / 1

A 741 has a CMMR of 90dB which is 1.4142136 times as good as that! (31,500)

Now a lot of that is due to the fact that the Acm is not going to be 1. The same datasheet shows us the typical gain is 200,000. So the common mode gain must be 200,000 / 31,500 = 6.5

Perhaps it's easier to say that Ad = 106dB and Acm = 16dB

The simpler way is to have the differential gain expressed in dB, and subtract the CMMR from this to find the differential gain in dB.

In this case, a differential gain of 200,000 is 106dB, the CMMR is 90dB, so Acm = 106 - 90 = 16dB

All you need to know is the formula for CMMR, some simple math, and have a datasheet handy.

Let me add to that that you also need to use the correct dB calculation. :D

In retrospect, these figures seem a bit more believable.

Here is a tutorial that shows you how to measure this for a real OP-amp. By my estimation, you'd see about 60mV difference in Vout for a 741.
 

24Volts

Mar 21, 2010
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Hi (*Steve*),

However, Look at the second line of your math (in both cases). That looks like a mathematical error. You seem to be intending to take the constant from one side to the other of the equation, but you are only doing half a job.

I dunno... :D
when we do the anti-log on the left, it usually cancels the log on the
right.... bof ... that's the way I learned it ... :p

The rest seems clear now.... it all works with 20 log and I get the same
results with 10 log as long as we square the values. Althought I
prefer using 20 log....

I think now I am okay .... thank you for your help. :)
 

(*steve*)

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Look at the first 2 lines

90 = 20 log...

then

90/20 = 20 Log...

It should have been

90/20 = Log...

(And as I said, 20Log(x) = 10Log(x*x) so it's not unexpected that you get the same result.)
 

24Volts

Mar 21, 2010
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It should have been

90/20 = Log...

yeah that's me being lazy with the equation editor !!!! :D

I will correct that in my personal notes!!!
thanks
 
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