# CMRR measurement

Discussion in 'General Electronics Chat' started by DenzilPenberthy, Feb 20, 2013.

1. ### DenzilPenberthy Thread Starter New Member

May 28, 2012
20
7
Hi everyone,

I'm helping some students out with a project and we're doing a bit of head scratching...

The project is a differential amplifier which will amplify some pretty tiny differential signals in the presence of some pretty large common mode interference (it is going in an MRI scanner). I've got some questions about measuring Common Mode Gain so we can work out the Common Mode Rejection Ratio.

If we put a large common mode signal into the amp (1kHz sinewave) what we get out is not an attenuated 1kHz sinewave but messy glitchy spiky stuff like you'd get out of a SMPS.

To calculate the CM gain we know Vin but do we take Vout to be the RMS value of the glitchy spiky signal or do we just look at the 1kHz component of the output signal. i.e. does the CMRR take into account the nonlinearity and distortion introduced by the device under test?

Along the same lines; if you were to connect a spectrum analyser up to do this it would give you a graph of gain vs frequency but does it measure the return signal just at the frequency of interest or over the whole frequency range?

Cheers!

2. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,953
1,826
How large is this input signal? It sounds as if it is overwhelming the amp and the peaks are not getting attenuated but instead getting thru.

Does a smaller common mode signal (say 1V P/P) also have these peaks?

Anyway, how you calculate the CMRR depends on what is useful to you in the application: if your MRI scanner can tolerate these SMPS glitches then you can spec them out.

A spectrum analyzer doing the graph of gain vs frequency graph will have a passband of some frequency width that sweeps over the interval. You can read the response at a single frequency if desired.

Back in the day we had a line of built to spec power regulators to test, and the spec for "ripple rejection" was something like 60dB (1000:1) at 1200 Hz. (1200 Hz is full wave rectified 3 phase 400Hz). To perform the test we injected a 1200 Hz sine signal into the input power and read the output at 1200 Hz on a spectrum analyzer. The test just squeaked by at 62-63 dB: using an RMS DVM would not pass due to other noise on the output.

I don't think the test actually measured anything as we never failed any units for this test.

3. ### crutschow Expert

Mar 14, 2008
20,262
5,737
As ErnieM noted, it would appear you are exceeding the common-mode voltage range of the amp.

4. ### gootee Senior Member

Apr 24, 2007
447
51
I know that this is just for students. But CMRR varies with frequency and is typically worse for higher frequencies. If I was only going to measure at one frequency, I would try to measure at a frequency that would be critical for the application, and would also try to choose a worst-case.

If I was going to do it "right", I might use a spectrum analyzer with a built-in sweep generator (or even just a separate sweep gen), and sweep both of the inputs with the same signal while observing the spectrum plot (and recording it to Excel through a GPIB or HPIB adapter between the spec an and my computer), which should then be a plot that was proportional to the inverse of the CMRR versus frequency.

But actually, with most modern spectrum analyzers, you could first plot the sweep signal, from the amp inputs, then freeze that trace, then use a second trace to plot the output spectrum, then use trace arithmetic to subtract the input spectrum from the output spectrum (which is like dividing output / input, but for dB or dBm you subtract), and get a direct plot of the CMRR. Or, you could just capture both traces in Excel (even separately) and subtract one column from the other and plot the result. A Network Analyzer would make it even easier, I guess. But in either case, you would want to make sure that you used one that went to low-enough frequencies. A lot of them are geared toward RF and don't go below 10 MHz, or a few hundred kHz.

Last edited: Feb 20, 2013
5. ### DenzilPenberthy Thread Starter New Member

May 28, 2012
20
7
Hi all,

Thanks for the replies. Some useful stuff there. I'll pass it on,

Cheers!