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!
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!
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
and
-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|
Solution:
I get 4.9875, which is around +14dB
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
and
-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|
Solution:
I get 4.9875, which is around +14dB
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
R3=0.95R, R3'=1.05R, R4=1.05R, R4'=0.95R
or
R3=1.05R, R3'=0.95R, R4=0.95R, R4'=1.05R
could the value be negative, since there is a decrease in CMRR?
CMRR is defined as Ad/Acm, where
Ad is the differential gain
Acm is the common mode gain
An externally hosted image should be here but it was not working when we last tested it.
An externally hosted image should be here but it was not working when we last tested it.
R3=0.95R, R3'=1.05R, R4=1.05R, R4'=0.95R
or
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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