[With regard to 'a' changing with frequency]
Not in the sim I posted.
a is the input AC voltage amplitude, which is a given. So the plot of a/v(out) is in fact a plot of v(in)/v(out), i.e. the CM gain reciprocal (which as I understand it is the CMRR). LTspice has done all the necessary division and dB calculation arithmetic
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Hi alec,
Oh i have not doubt that 'a' does not change in your simulation, it's clear that it is a defined source with constant AC amplitude. But in the calculation of CMRR we not only need to calculate the common mode gain, we also need to calculate the differential gain, and then perform the division:
DiffGain/CMGain
and DiffGain is not a constant factor over frequency but usually gets lower with increasing frequency as i am sure you know. So in effect we have:
VdiffGain(f)/CMGain(f)
showing more clearly this dependency on the frequency f.
So point 1 is that CMRR is not the input voltage divided by the common mode voltage it is the differential gain divided by the common mode gain. It is the ratio of two GAIN's.
Point 2 is that the differential gain is not equal to the input voltage times any constant factor, unless the gain is perfectly constant over all frequency and that's not very typical at all.
For example, the VcmGain (common mode gain) goes UP with frequency, while the differential gain VdiffGain goes DOWN with frequency.
For the circuit in question, the differential gain doesnt change THAT much with frequency so your estimate of a/CMGain is reasonable at least to some degree. For this circuit the gain is about 10 at f=10Hz and is about 6 at f=100kHz, so to see the two extremes we could do just two such 'a':
a/CMGain=10/CMGain
and
a/CMGain=6/CMGain
and that may be good enough, so your good idea still has merit i believe.
What we intended to do is plot the CMRR over the frequency of interest, but the only way to do that is to have a function that actually calculates differential gain over frequency, but it may also be replaceable with a curve fitted function for example. I thought the easiest way would be just to recreate the entire circuit again but change the input excitation such that we could measure diff gain as well as CM gain, then just divide.
When i first stated this i had no idea how to do this, but now it is becoming more clear. I havent had to worry about this too much in the past.