A bit of context

I came up with this question after someone suggested that two fuses are better than a main one in the below configuration. The converters are SMPSs which regulate output voltage.


simulate this circuit – Schematic created using CircuitLab

Now someone justified that F1 and F2 are a better implementation through the following example: both converters are functioning within normal limits, each drawing 1A (and F1 and F2 are both designed to trip at 2A). If Converter 1 would fail open and Converter 2 would continue working normally, then during the finite instant when the aforementioned conditions start applying, as current cannot change instantly and has only one path to follow from supply to ground, the 2A will flow through Converter 2 and potentially damage it. Since F2 is there, it would protect against the over-current condition. If converters reverse roles in the same scenario, F1 will be there to open the circuit.

The justification basically says that the fuses are there to protect against an open-circuit event...instead of a short-circuit.

That's as much context as I can give. Current can't change instantly, because that's just how the real-world works. I agree it's a weak reasoning, as the "why?" is missing, but I would like to prove it wrong or right to myself (with your help). To help answerers, I would like to know both if the parasitic inductance or anything else (other than parasitic inductance) could recreate the effects described.

Main part

Now, remove the converters and the loads and place two real-world resistors between each fuse and ground, both (resistors) drawing 1A normally. Again, the total current drawn from the power supply would be 2 A.

The questions are:

1) If one removes a resistor (ceteris paribus), will the remaining other experience the 2A through for a very short, but finite amount of time? If it does, can it be due to something else, other than parasitic inductance (and can parasitic inductance also affect things)?

2) Does 1) mean, if true, that ohm's law only applies if the time-step (or time scale) is large enough?

3) If 1) is true, is it worth taking into account this effect from an engineering perspective-i.e. during that short finite time, the 2A could damage the remaining resistor?

  • \$\begingroup\$ What is the inductance of the power supply cable? Not very high, I bet. What is the input capacitance of a converter? Higher than the inductance, I bet. \$\endgroup\$
    – user253751
    Feb 3 '20 at 16:33

A resistor generally fails due to overdissipation. If it doesn't arc, a momentary pulse of current wouldn't destroy it. Semiconductors, on the other hand, can be destroyed fairly immediately by overvoltage.

Based on the limited details above, I doubt that you'll have enough inductance in the circuit to do any damage. If you're particularly concerned, you can include Zener diodes to mitigate it. Most ICs have diodes to supply and ground to protect the part, although they are small.

The second fuse will do nothing to prevent a momentary voltage spike due to current redirection. At twice its rated capacity, it's likely to survive several seconds before blowing. This is an eternity of time in terms of electrical damage.

  • \$\begingroup\$ You've answered from the point of view of the inductance, but I wonder if there's something else out there (other than inductance) which could keep current flowing (2A, assumption was that the current won't change instantly in real life and has one path remaining) through the remaining resistor for a nano-second or micro-second and potentially damage it. That would violate ohm's law , but apparently it doesn't apply for extremely small transients...what do you think? \$\endgroup\$ Feb 5 '20 at 9:00
  • 1
    \$\begingroup\$ Well, I have no idea...I'm only guessing about the inductance, because you've provided only a block diagram. Of course based on the block diagram it's possible that you have an externally powered current source, or a 10 megahenry inductor, or something. However, if your theory violates Ohm's law, I'm skeptical without further information. \$\endgroup\$ Feb 5 '20 at 13:47

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.