I got a new digital oscilloscope as a gift. As a training course, I observed some waveforms of my test 12V full bridge circuit that is driven by TL494 and two IR2110 operate the IRF3710 100V mosfets. During my tests I was running it off a limited 12V 5A floating power supply.
My goal is to observe waveforms generated by possible parasitic elements of my pcb layout that are described in an-978.pdf, in section 5 "How to deal with negative transients on the Vs pin". I observed Vs-COM undershoot successfully.
However I was curious how high goes high side and what's it's rise time. So I hooked oscilloscope GND to Vs and put the probe to HO. I observed waveform, but mosfet it operated died. I did not noticed that immediately because I was running off a limited supply and input current increased only twice. I noticed that only after poweroff, when I touched heatsink, it was warm. And that high side mosfet is died with short.
It's a silly thing what I did? How measure that point safely?
During that test, scope GND was hooked only to Vs and only one channel was used. Scope was not grounded (my building does not have PE wire).
Best Answer
So, without a PE wire, the live and neutral, (via capacitors inside the scope's power supply) will "force" the scope's ground to about half the AC voltage of the supply. The scope uses capacitors because it will filter the supply for pretty normal sensible reasons.
You then apply that ground connection (now at maybe 60VAC or 120VAC) to a sensitive point on your circuit.
You shouldn't be surprised at what happened. Think about how this pans out: -
The above picture shows an "inline" AC EMI filter that uses a safety ground. Now your scope will have an earth and you, in effect, have it disconnected. Now, the two capacitors marked Cv will form an AC potential divider and the outgoing earth is at 60VAC or 120VAC. Yes it won't be enough to kill someone but it will be enough to puncture through a gate on a MOSFET.