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I have a PCB that is equipped with short circuit protection inform of an inline fuse at the main 40V DC input (5x20 glass fuse) and some other components that I am not too familiar with.

My aim is to "apply a short-circuit to the PCB traces downstream of the input terminals and protection device to test the DUT protection circuit" (Thanks Transistor for helping to clarify)

My power supply is a varible voltage and current 60V DC 6A PSU. (This is a 5A version, I have a 6A with similar specs)

enter image description here

Can someone assist me on how to go about testing the DUT?

THE DUT CAN BREAK, I HAVE ABOUT 4 OF THESE DUT, SO I CAN DAMAGE A FEW.

enter image description here

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    \$\begingroup\$ A simple solution would be to use a power supply that can handle the abuse. You could use a standard lab power supply with current limiting, then it's no issue if you short the output. \$\endgroup\$
    – Klas-Kenny
    Commented Dec 30, 2021 at 9:17
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    \$\begingroup\$ This doesn't make sense. Can you show a bit more of a schematic including where you will put the short? It seems to me that shorting the power supply cannot possibly damage the DUT. But maybe I am not clear on where you are putting the short. \$\endgroup\$
    – user57037
    Commented Dec 30, 2021 at 9:19
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    \$\begingroup\$ Shorting the PSU output presents 0V to the DUT. I don't think this is what you want to test... \$\endgroup\$
    – TypeIA
    Commented Dec 30, 2021 at 9:24
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    \$\begingroup\$ Joey, you are not thinking straight. As pointed out in the comments already shorting the supply at the input to the DUT you will not test the DUT. You will simply be switching off its supply. \$\endgroup\$
    – Transistor
    Commented Dec 30, 2021 at 10:33
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    \$\begingroup\$ It's not possible to provide a detailed solution with such a vague question. My best guess is that you want to apply a short-circuit to the PCB traces downstream of the input terminals and protection device to test the DUT protection circuit but that's not what your question says. Your question is missing details on the power supply too. Hit the edit link ... \$\endgroup\$
    – Transistor
    Commented Dec 30, 2021 at 14:06

2 Answers 2

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Your bench power supply has adjustable voltage and current limits. That makes it pretty much indestructible in this application.

Your post suggests (but isn't clear) that the DUT has a 5 A fuse. If you check the fuse datasheet you will find a curve showing the time the fuse will survive at a given current. A 5 A fuse may not blow for quite some time at 50% overload.

You can set the PSU to 40 V and maximum current, connect to the device and apply a short-circuit to the PCB. This is often achieved by making a jig with "pogo-pins" to allow rapid insertion and connection between the test circuit - a shorting switch in your case - and the board.

On short-circuit the PSU will immediately go into current limit and I the voltage will collapse to a fraction of a volt - mostly due to voltage drop along the wires and the PCB traces. You'll have to determine if 5 A is enough to trip the protective device on the DUT.

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Your magic box would be an electronic circuit breaker, with different current limits if you want. You may want to consider the following video https://m.youtube.com/watch?v=fqeUpATJlZY which presents the basics of a current limit circuit that can be applied to any circuit. The part2 and part3, although having a different name (look in the video description for their links), builds on it with the final solution presented in part 3. You will have to change the pass transistor to a big one that can accept a higher current and still have a saturation Vce of 0.7V or less. The 2N3055 (or MJ2955 for the part 2 video circuit) can be driven at Ic=4A with an Ib=200mA. The 4A will be below the 5A DUT capacity and still allow for a short circuit test on the DUT.

EDIT: You are right SamGibson, I have linked the video because I find it very instructional, and I apologize for that.

Anyway here are the schematics of my answer: enter image description here RB needs to have a current value that keeps the VCE of Q1 below 0.7V. For that you can use the transistor base current, collector voltage like this one: enter image description here RT needs to be such that will keep the VCE of Q2 also below 0.7V so there is not much heat dissipation in it. For that you use its base current, collector voltage like this one: enter image description here And here are the results of an ngspice test with RB=250R and RT=10K:

Before the circuit triggers:

Node Voltage

vin 40

vb1 39.1217

vb2 39.3187

vout 39.2447

vin#branch -4.08096

After the circuit triggers:

Node Voltage

vin 40

vb1 39.6644

vb2 39.1545

vout 0.0352077

vin#branch -0.16257

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  • \$\begingroup\$ VictorTito - Hi, What is your association with that video? Our site rules here require that if you are associated with anything you promote (via links or otherwise), you must disclose that affiliation. You have linked to that video a few times here and on other sites, so I guess you are associated with the video. If so, please edit each place where you have linked to it on Stack Exchange, and refer to it as "my video" (or whatever wording is correct) to disclose your affiliation. \$\endgroup\$
    – SamGibson ♦
    Commented Dec 30, 2021 at 15:12
  • \$\begingroup\$ [continued] Also, links to other sites / videos / etc. are only allowed if they add supplemental information. Answers should, wherever possible, be self-contained and not rely on external links, as those links may become unavailable in the future. So as a minimum, I recommend you include the relevant schematic(s) and any further relevant analysis in your actual answers, instead of only including links to the video. Thanks. \$\endgroup\$
    – SamGibson ♦
    Commented Dec 30, 2021 at 15:12

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