1
\$\begingroup\$

I'm probing a point with 230V RMS voltage inside a machine (don't worry, I'm using a 100:1 CAT 3 probe rated at 600V), and would like to trigger my oscilloscope at the point when this signal is interrupted. That is, the sine wave ends and becomes a flat line.

I have a Rigol DS1054Z oscilloscope. Is there some clever way to arrange such a trigger? I thought perhaps I could use the slope trigger, but I didn't get it to work.

\$\endgroup\$
  • \$\begingroup\$ Normal trigger mode \$\endgroup\$ – JonRB Jul 31 '15 at 21:05
  • \$\begingroup\$ Do you mean edge trigger? It won't work, it'll trigger once every period. It'll be just luck if the last trigger point happens so close to the actual end of the train that the end ends up in memory. \$\endgroup\$ – avl_sweden Jul 31 '15 at 21:10
  • \$\begingroup\$ You're right that normal edge trigger would work if it was possible to get it to trigger at every single edge. But I don't think it's possible to configure the scope so that the trigger is active during the sweep. Or is it? \$\endgroup\$ – avl_sweden Jul 31 '15 at 21:11
  • 1
    \$\begingroup\$ While the scope itself may not have this capability, you can rectify the input, put the rectified voltage to a second input and trigger on negative going edge. \$\endgroup\$ – Pentium100 Jul 31 '15 at 21:18
  • \$\begingroup\$ Pentium100: Yeah, that would work of course. I was hoping for something not requiring external hardware. :-). \$\endgroup\$ – avl_sweden Jul 31 '15 at 21:48
3
\$\begingroup\$

In the trigger menu (right side below the trigger level knob: "menu") you can, if you have it (I think Timeout might be in the upgrades? You could hack it if you don't have it anyway) you can select Timeout.

You then set slope up and down and a timeout of 75% of the sine period, in the case of 50Hz AC that would be 15ms. It will then trigger when no slope has been detected after 15ms. (So you will get 15ms of samples of 0V, but if you adjust the trigger to the right you can move a lot of signal into your sample depth.

\$\endgroup\$
  • 1
    \$\begingroup\$ Excellent! Exactly what I was looking for! Thank you so much! \$\endgroup\$ – avl_sweden Jul 31 '15 at 22:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.