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Old 14th September 2007, 06:17 AM   #1
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Default worst CMRR?

An op amp differential amplifier as shown in picture is built using four
identical resistors, each having a tolerance of ¡À 5 %. Calculate the worst possible

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Old 14th September 2007, 04:52 PM   #2
kubeek is offline kubeek  Czech Republic
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You just calculate the worst error it produces at the both borders of the resistor´s tolerance. They should be identical, but when the + side has a bit higher gain than the - side you get the CM error. Plus add CMMR of the opamp if not ideal.
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Old 15th September 2007, 06:46 AM   #3
AndrewT is offline AndrewT  Scotland
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would someone be good enough to show us the calculation or at least the equations?
regards Andrew T.
Sent from my desktop computer using a keyboard
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Old 15th September 2007, 05:50 PM   #4
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sorry, I don't have the equations at hand and actually I'm too lazy to do the math myself
That's the approach I would try:
-set V1=V2=Vi
-calculate the voltages at opamp-inputs
V3 as a subject of Vi, R1, Rf, Vout
V4 as a subject of Vi, R2, Rg
- assume V3=V4 and use this to get a large equation for Vout with Vi, R1, R2, Rg, Rf
- set Vi to 1 and the resistors to 1,05 resp. 0,95 for worst mismatch
- calculate Vout, this is now equivalent to common-mode amplification
- use logarithms to get to the cmmr
Have fun!
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Old 16th September 2007, 08:25 PM   #5
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this sounds like a homework question (someone concerned with CMRR would have chosen 1% parts).

in anycase, this problem is simple because of the linear nature, which will place the minimum CMRR at a worst case resistance value.

step 1. break the system up using superposition. this gives a positive gain circuit and a negative gain circuit. for equal resistors the gains are:
(Rg/(Rg+R2))*((Rf+R1)/R1) = 1
-Rf/R1 = -1
Giving a total gain of
(Rg/(Rg+R2))*(Rf/(Rf+R1)) + -Rf/R1 = 1 - 1 = 0

Normally, to find the worst case would mean finding the global maximum of the above on the surface of interest, which would mean taking partial derivitives. But in this case we're pretty confident that the worst case will occur at a corner.

step 2.
Slove (Rg/(Rg+R2))*((Rf+R1)/Rf) + -Rf/R1 for the 16 cases of interest, eg, there are 16 combinations of Rg,R1,R2,and Rf with each being either 1.05R or 0.95R.

Step 3.
This will give the common mode gain for all cases. note that the common mode gain will be 0 in the cases where all resistors are equal. Find the maximum absolute common mode gain, eg the maximum of |x|. CMRR will be 1/|x|

I get 4.9875, which is around +14dB
if only it could be used for good, not evil...
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Old 21st September 2007, 04:58 AM   #6
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firstly, ya this is a homework, but it could also be a useful info for those opamps

CMRR is defined as Ad/Acm, where

Ad is the differential gain
Acm is the common mode gain

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R3=0.95R, R3'=1.05R, R4=1.05R, R4'=0.95R
R3=1.05R, R3'=0.95R, R4=0.95R, R4'=1.05R

could the value be negative, since there is a decrease in CMRR?
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Old 21st September 2007, 05:20 AM   #7
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before asking a new question, her majesty would consider a kind remark or short nod to the poor servants which tried to help with the first one (question)?
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