Instrumentation Amplifier Question-Help Please

[themaccabee]

Hi,

I want to know one thing about a 3 op-amp Instrumentation amplifier circuit diagram below


I understand that the CMRR of the circuit is very much dependent upon the matching of the resistors R4,R5,R6,R7 provided the XOP1 &XOP2 are matched well.
Im putting R4=R5=R6=R7 =x ,same value & Gain is introduced in the first stage only.

Now Im varying x=10K to 100K.. giving all the resistors(R4,R5,R6,R7) one value at a time.
So what will be its effect on CMMR??Will CMRR be affected..If so why?, can somebody explain..?

Thanks for any help
Regards

 

[mtwieg]

Basically there are two signal paths; once going through Xop1, R1, R2, R4, R5, and Xop3 and the other going through Xop2, R3, R2, R6, R7, and Xop3. Each of the signal paths has its own gain. Anything that causes these two gains to be different will hurt CMRR. R2 is shared between both of them, so it doesn't affect CMRR. But any of the other values will matter. If you want to derive the exact relationship, just calculate the gain from the common mode source to the output as a function of the resistances. Shouldn't be too complicated.

 

[dick_freebird]

Small signal CMRR may not be affected but beware large signal
clipping of the intermediate nodes if there is any front end gain.

 

[themaccabee]

Small signal CMRR may not be affected but beware large signal
clipping of the intermediate nodes if there is any front end gain.

Sorry i couldnt understand that..
Does the front end gain refers to the input stage gain?

 

[Milad-D]

The common mode gain is Acm= (R6/(R6+R7))*(1+R5/R4)-R5/R4. Therefore for infinite cmrr this value must be zero. About your circuit, the input is not fully differential. Because one node senses the vcm and the other one is vcm+vin. Therefore the gain of circuit is 0.5*Acm+Adm. At which frequency you check the cmrr? What about the cmrr of the second opamp itself. Maybe by changin the resistor values, the common mode pole of second amplifier become smaller and you get less cmrr.