I have two components in parallel on a DC circuit driven by a single momentary switch. Each component is individually fused.

I need to detect failure of either fuse and in that event power on a warning indicator.

The most direct approach is to wire a NC relay in series with each fuse, but the specs of the circuit aren't tolerant to the voltage drop as power output from each component must be maintained.

I'd like to solve with common cheap components as this is a hobby implementation.

Any ideas?

  • 1
    \$\begingroup\$ Any answers that involve looking at the voltage across the fuse could be deemed to be interfering with the operation of that fuse. Take note that fuses have a rupture current rating that must be observed in some applications and those components across the broken fuse might just render a design unsafe. I'm being picky of course so just make your own mind up! \$\endgroup\$ – Andy aka Sep 17 '15 at 12:46
  • \$\begingroup\$ Right you are, @Andyaka. But for low powered, non-safety relevant applications you should be OK. Wouldn't want to see something like that in a piece of medical equipment, for example. \$\endgroup\$ – JRE Sep 17 '15 at 13:06
  • \$\begingroup\$ I do not know the details of your implementation but you might look into already commercially available automotive fuses that already have a built-in LED and resistance, that basically light up once the Fuse is blown. \$\endgroup\$ – Kvegaoro Sep 17 '15 at 19:39
  • \$\begingroup\$ If the fuses are integrated into the loads, it seems you are left with no trivial options. There are nontrivial options, like Hall sensors and stuff, but these might be too nontrivial for your case. \$\endgroup\$ – Eugene Ryabtsev Sep 18 '15 at 5:42

You could connect an LED and its ballast resistor in parallel with each load, as shown below.

The LED would stay on while the fuse wasn't blown, and you could use a high-efficiency LED to minimize its impact on the fuse.

enter image description here

  • \$\begingroup\$ Very good answer. Other than the other answers, this design is safe because it does not shortcut the fuse - and let the fuse do what it should: cut the wire in case of a dangerous situation. +1 for this. \$\endgroup\$ – Matt Sep 17 '15 at 14:53
  • \$\begingroup\$ +1 for the out of the box thinking of the LED turning OFF with the fuse blowing instead of turning on to notify of a fault. \$\endgroup\$ – Cort Ammon Sep 17 '15 at 20:17

Put a resistor and LED in series, and put that series combination in parallel with the fuse. If the fuse is intact, it's a short, and no current flows through the LED. If the fuse is open, the only current path is through the resistor and LED. Size the resistor properly for the voltages involved, and the LED will turn on.

If you need an indicator used to drive a circuit rather than just an LED, put the diode of an optocoupler in place of the LED.

Of course, this is somewhat dependent on the nature of the voltage and the load. There are imaginable circumstances where this would not work.

  • \$\begingroup\$ The fuse is integrated into the components, so there's no way to wire in parallel with the fuse without wiring in parallel with the component/load. The supply is 12V, component power is 28W giving an R of ~5ohm. Based on that load, I don't think that this would work? \$\endgroup\$ – MilitaryCoo Sep 17 '15 at 14:31

Assuming there is a voltage drop after the fuse blows, a simple P-ch MOSFET (or PNP) with gate connected to the top of the component would work. When voltage drops after fuse blows, it lights an LED. Something like this:

enter image description here


enter image description here


You might try something like the attached circuit.
Whether it will light or not will depend on the circuit being powered. If it has a relatively low resistance to ground, then the LED will light. You might have to monkey with the value of the resistor.


simulate this circuit – Schematic created using CircuitLab


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