[SOLVED] CMRR for a fully differential amplifier & Monte Carlo Simulations

Hi guys,
I am trying to simulate the CMRR of a fully differential telescopic amplifier. After having read quite a bit in the forum here I have come to a solution but I'm very unsure if I am really looking at the CMRR so I wanted someone who knows more than me to confirm it (see screenshots).
And actually I wanted to see the CMRR in a Monte Carlo Simulation but do I have to look at the DC value then or at a special frequency? My first results make me think I did sth. wrong cause the CMRR goes from -2,88dB to -75 dB... that can't be right...
Thanks for any help,
Winny

Hi!

The CMRR is how the output can reject the input common mode. If your OpAmp is perfect (no mismatch) then the CMRR should be very low (-400dB). But, if a mismatch occurs, for example between the size of the transistors in your input differential pair, then with a common voltage at the inputs you should get a differential voltage at the output.
For your analysis, do an AC simulation (ie from 1Hz to 10GHz). Put a common voltage at the inputs (like you did) with a DC offset and sweep the AC. (You can change the DC value to see how the CMRR can vary). And you should see the output differential mode (VoutMD = VinMD x Gd + VinMC x GMC). As VinMD is Zero, you'll know the GMC, then deduce the CMRR as you already know the gain of your OpAmp.

Before an MC sim, you can first modify the input transistors (change for example the W of the inverted input by a factor of 1%) and run the simulation. Once you get it, you can try with monte carlo.

Hope it helps!

Hi, thank you for your reply =) so what you say is, my CMRR for let's say 1 Hz would be 40 (which is my closed loop gain of the amplifier) - (-194 dB) = 234 dB without mismatch at first? Do I have to take the closed-loop or open loop gain for calculating the CMRR? Then I changed the dimensions of one input transistor as you said and my GMC looked totally different but I hope that's normal. And what frequency do I choose for the GMC value, for my calculations I just chose 1 Hz but is there a more intelligent way?

Re: CMRR for a fully differential amplifier & Monte Carlo Simulations

my CMRR for let's say 1 Hz would be 40 (which is my closed loop gain of the amplifier) - (-194 dB) = 234 dB

Click to expand...

The CMRR is the gain in differential mode of the OpAmp in OPEN LOOP divided by the common mode gain of the OpAmp. On your plot, you have only the Out2+, as it's a fully differential OpAmp, Vout = Vout+ - Vout-. If it's perfect, the common mode gain should be 0. (which correspond to -400dB, the minimum of the simulator)

Then I changed the dimensions of one input transistor as you said and my GMC looked totally different but I hope that's normal.

Click to expand...

Sure! The CMRR is greatly affected by the input mismatch!

And what frequency do I choose for the GMC value, for my calculations I just chose 1 Hz but is there a more intelligent way?

Click to expand...

Good question... it may depends on your frequency of interest. At 1Hz you have the CMRR in "static". This corresponds to the value you can read on datasheets. With an AC simulation, you have an idea on how it can degrade.

Thank you, you really helped me! Hope I will be able to go on with the simulations now