EEVblog Electronics Community Forum
Products => Test Equipment => Topic started by: JohnPi on November 21, 2021, 10:33:01 pm
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I am testing some power circuits which fail under some overloads. As it fails, the damage spreads and damages other components. Therefore I want to shutdown the system as soon as the initial failure occurs -- basically when a current exceeds a threshold -- and shutdown within < 10 us by shorting the external power supply with a v. large MOSFET.
I am thinking of using an OWON VDS1022I driven by a current clamp, and using the Trigger Out signal (maybe buffered with an NPN) to effect the shutdown. I don't care about the other 'scope functions, but I can't find detailed information on the Trigger Out function. Specifically:
Is this a 50 Ω output ?
What is the amplitude ? Is it 0 V until the trigger event and then 3-5 V ?
If I set a trigger in the middle of the screen, will the Trigger Out go high as soon (< 1 us) as the trigger occurs ?
how long does it stay high for ? The remainder of the trace ? So if I set it to slow (10 ms/div ?), I could get a pulse that is 10's of ms long
(the lab power supply will OCP after that).
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This is what the trigger output looks like. Channel 1 is connected to the 1khz signal from the oscilloscope, and channel 2 is connected to the trigger output. It seems it's delayed about 1.2µs. It does stay high for the remainder of the trace(third screenshot set at 10ms/division).
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Thanks. It looks like about a 4 V signal which is OK for me.
If you used 100 Hz, would it still be delayed ~ 1 us ?
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This is at 200khz and at 100hz, seems like it's pretty constant the 1.2µs delay
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Thank you very much for the measurements. That delay is acceptable for my use.
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Thank you very much for the measurements. That delay is acceptable for my use.
:) happy to help!