Hi,
You do not need to label BOTH input sources, only one, because they are both exactly the same as to amplitude and phase, at least for the CM gain calculation. So say you label the source of the non inverting input A1, then label the very output of the last stage A2. Then you can calculate the CM gain by using V(A2)/V(A1)).
Next, you need to recreate the ENTIRE circuit, that means ALL op amps and resistors and current mirrors, but this time make the second input source equal to something else like zero, label the non inverting input of that circuit A3, and label the output of that circuit A4. Now the differential gain can be calculated by V(A4)/V(A3).
Next, the CMRR is (V(A4)/V(A3)/(V(A2)/V(A1)), which is also equal to (V(A1)*V(A4))/(V(A2)*V(A3), and if A1 and A3 are the same (which they very well can and should be made to be) then this reduces to:
V(A4)/V(A3).
So the CMRR is the ratio of those two outputs.
The only hard part is you have to copy the entire circuit including all parts to do it this way, but you can probably use copy and paste. With the two input amps, two buffer amps, four current mirrors, and one output amp per circuit, that means the double circuit will take 10 op amps and 8 current mirrors as well as all the resistors.
It will be hard to say though how well this calculation or any calculation of the CMRR using spice models will relate to the real life circuit.
You do not need to label BOTH input sources, only one, because they are both exactly the same as to amplitude and phase, at least for the CM gain calculation. So say you label the source of the non inverting input A1, then label the very output of the last stage A2. Then you can calculate the CM gain by using V(A2)/V(A1)).
Next, you need to recreate the ENTIRE circuit, that means ALL op amps and resistors and current mirrors, but this time make the second input source equal to something else like zero, label the non inverting input of that circuit A3, and label the output of that circuit A4. Now the differential gain can be calculated by V(A4)/V(A3).
Next, the CMRR is (V(A4)/V(A3)/(V(A2)/V(A1)), which is also equal to (V(A1)*V(A4))/(V(A2)*V(A3), and if A1 and A3 are the same (which they very well can and should be made to be) then this reduces to:
V(A4)/V(A3).
So the CMRR is the ratio of those two outputs.
The only hard part is you have to copy the entire circuit including all parts to do it this way, but you can probably use copy and paste. With the two input amps, two buffer amps, four current mirrors, and one output amp per circuit, that means the double circuit will take 10 op amps and 8 current mirrors as well as all the resistors.
It will be hard to say though how well this calculation or any calculation of the CMRR using spice models will relate to the real life circuit.
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