Hello,
I followed Dave's video here:
https://youtu.be/vDe_BHvRpksAnd arrived at the following formula:
(20 * log10( RMS_diff_probe / input )) + dB
dB == 20 for 10x probe and dB == 40 for 100x probe.
I then applied it to a differential probe I got and I achieved a much higher than expected CMRR number, so I thought I'd ask here.
Attached is a screen shot of my test setup and the HT8100's (100MHz by Hantek) results. I also am including a screen shot of my PVP2350 passive probes's CMRR for comparison. I'm using an MSO5354. I'm feeding a 1MHz signal into the scope through a 50ohm termination. The differential probe is a 50X probe. Just out of curiosity, I turned on the FFT to see where the noise was coming from.
My calculation is as follows:
(20*log(0.018349/2.4508))+30 == -67.8919dB
The CMRR on the datasheet says >= 50dB at 1MHz.
NOTE: The scope is set to "high res" mode, meaning, it's doing averaging or over sampling to get a more accurate result. Therefore, it's BW is limited to 55MHz in the screen shots. If you turn off "high res" mode the RMS noise increases by about 10mV.
EDIT: The FFT is set to 70dB offset and 5dB/div. It's span is 500Hz to 1.2Mhz. My thinking was to try and see if the 1Mhz signal was leaking through (and it's not).
Am I doing the CMRR measurement right?
Thanks!