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I'm looking at a very simple method for overvoltage protection of a load connected to a standard external switching power supply. The load is powered by a 12V power supply, but this could be mistaken and exchanged with a higher voltage power supply instead, i.e. a 15V power supply. I was planning to simply connect a diode in parallel with the load. Is this a bad idea? My point is that the diode should have an activation voltage bigger than 12V (we can say 13V for safety reasons) and, since the power supply itself has overcurrent protection, it will automatically shut down once the diode is ON. Here it comes the circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

How should the diode be rated? The load is a 150W resistive/inductive type. Should I use a zener diode instead? What are the differences between the two types of diodes?

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    \$\begingroup\$ Regular diodes all have an "activation voltage" (as you say - but it's actually called forward voltage) lower than 1V. However, zener diodes can have reverse breakdown voltages in the range you're looking for. So this is what you should use, indeed, but they must then be placed the other way around (cathode towards +). I would also add a fuse, because, since the supply can be changed, relying on the presence of its overcurrent protection is a bit risky. \$\endgroup\$
    – dim
    Jan 4, 2017 at 16:20
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    \$\begingroup\$ Also, don't use a 12V zener diode. You need to allow for some headroom so it doesn't leak a lot of current in the nominal case, or if the supply voltage is a bit higher than nominal. Use at least 13V. \$\endgroup\$
    – dim
    Jan 4, 2017 at 16:25
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    \$\begingroup\$ This info is usually given in the datasheet. But it is very low, and the diode will have to dissipate a lot, which is why you really need a fuse: to relieve the diode before it burns. You can't choose something too small. The required power rating of the diode will also depend on the time the fuse takes to melt. Overall, it is not actually that easy to size well, unless you really oversize the diode to make it safe. \$\endgroup\$
    – dim
    Jan 4, 2017 at 16:34
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    \$\begingroup\$ If you just limit the current through just the zener, then it won't protect the circuit any more. Your trouble is going to be that any supply capable of delivering 150W at 12V will fry a zener pretty much instantly. \$\endgroup\$
    – Simon B
    Jan 4, 2017 at 17:21
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    \$\begingroup\$ Unfortunately, I think @Simon is right. I didn't actually realize there was about 12A nominal flowing from the supply. So, given the size of the fuse you need, the zener, whatever its size, will never be able to survive from an overvoltage before the fuse blows. The fact it will keep at least 12V across it makes it dissipate way too much. The SCR crowbar solution may be better, since the voltage across the SCR will be much lower once triggered. So it will have to dissipate much less. See this \$\endgroup\$
    – dim
    Jan 4, 2017 at 20:42

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This type of circuit is often known as a crowbar. The diode is replaced with an SCR and the gate of the SCR is controlled by a 12.5V zener diode and a resistor. (resistor pulls the gate down to 0V and zener is connected from gate to 12V)

It combines the best features of both devices, the high voltage switching of the zener and the current handling and robustness of the SCR.

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    \$\begingroup\$ I am aware of the crowbar-type circuits but I wonder if I could find a much simpler and lower cost solution. Why my circuit would not work? \$\endgroup\$
    – Francesco
    Jan 4, 2017 at 16:13
  • \$\begingroup\$ Because the device you are proposing does not exist the crowbar gets closest to this at minimum cost. \$\endgroup\$
    – RoyC
    Jan 4, 2017 at 16:14
  • \$\begingroup\$ I am just proposing a diode in parallel with the load! The diode will activates when the voltage is bigger than 12V. \$\endgroup\$
    – Francesco
    Jan 4, 2017 at 16:15
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    \$\begingroup\$ Yes but what if that 12V source can supply an (almost) infinite amount of current. Your (zener)diode/SCR would simply explode. You need to add a fuse in series with the supply to stop that current. Indeed you could rely on the supply's overcurrent protection but what if that was broken ? \$\endgroup\$
    – Bimpelrekkie
    Jan 4, 2017 at 16:19
  • \$\begingroup\$ Fair comment question did state supply has overcurrent protection and I was assuming that would take care of it. The addition of a fuse would make it safer. \$\endgroup\$
    – RoyC
    Jan 4, 2017 at 16:20

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