Electronics > Beginners

What happens if a short goes into the oscilloscope?

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Mp3:
I watched a bunch of youtube videos on oscilloscope safety. Here is one situation i am not confident regarding my understanding of what would happen in this scenario.

- The Measurement source is a battery powered device which has a fixed known resistance and is outputting about 32w of between 6-7 volts.
- The measurement source occasionally shorts out.
- The device will blow a fuse before exceeding 10A if it shorts out.

I would be putting the scope probes on the two outputs of the battery powered device i am measuring.

Can this even be safe to run into an oscilloscope, or should I stick to a DMM with logging features?

I have a Siglent 1052DL+.

atmfjstc:
A short circuit in your device will not affect the scope. Any large currents will go through the short itself, not the oscilloscope, which basically appears as a very high value resistor. The only way the oscilloscope will ever be in danger is if the short somehow causes the outputs to exceed the maximum input voltage tolerated by the scope. Although if your device is battery powered as you say, the output voltage will never exceed the battery voltage barring some super exotic situations.

tggzzz:
There will be inductance in the wires connecting carrying the 4-5A current. In the absence of better information, assume 1nH/mm of wire, i.e. 1m -> 1µH.

If that current is interrupted there will be a voltage induced across the inductor of V=Ldi/dt. If the current changes by 5A in 1ns in 1m of wire, that implies there will be an induced voltage of 1e-6 * 5 / 1e-9 = 5kV. That would be sufficient to damage a scope's input!

Obviously you can choose those figures to make any point you want :)

brucehoult:

--- Quote from: tggzzz on August 10, 2020, 03:26:46 pm ---There will be inductance in the wires connecting carrying the 4-5A current. In the absence of better information, assume 1nH/mm of wire, i.e. 1m -> 1µH.

If that current is interrupted there will be a voltage induced across the inductor of V=Ldi/dt. If the current changes by 5A in 1ns in 1m of wire, that implies there will be an induced voltage of 1e-6 * 5 / 1e-9 = 5kV. That would be sufficient to damage a scope's input!

Obviously you can choose those figures to make any point you want :)

--- End quote ---

Wouldn't the inductance itself work to prevent such a rapid change in current?

tggzzz:

--- Quote from: brucehoult on August 10, 2020, 08:38:07 pm ---
--- Quote from: tggzzz on August 10, 2020, 03:26:46 pm ---There will be inductance in the wires connecting carrying the 4-5A current. In the absence of better information, assume 1nH/mm of wire, i.e. 1m -> 1µH.

If that current is interrupted there will be a voltage induced across the inductor of V=Ldi/dt. If the current changes by 5A in 1ns in 1m of wire, that implies there will be an induced voltage of 1e-6 * 5 / 1e-9 = 5kV. That would be sufficient to damage a scope's input!

Obviously you can choose those figures to make any point you want :)

--- End quote ---

Wouldn't the inductance itself work to prevent such a rapid change in current?

--- End quote ---

Yes. The induced voltage is "an attempt" to prevent the rapid crrent change.

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