I am using a 24 PIN 875W PSU (Dell H875EF-00) for some experiments. PINs are like here (Version 2.0):

enter image description here

I am using the +12V1 DC from PIN 10 and some COM ground (I dont remember which ground is used for the halogen light and which one is used for the Arduino after cutting and resoldering them). +5VSB is connecto to VIN of an Arduino, another COM is GND for Arduino and PS_ON is shorted to a GND of the Arduino.

I am using the +12V1 DC on a Philips H7 halogen light that is 12V 55W. According to the PSU label the 12V connectors can handle up to 18A, so no issues there. I have fixed the +12V to the halogen light. When I make contact with one of the COMs it lights up neatly and all the other components attached to the PSU work as desired. When I remove the jumper cable that connects the COM to the ground of the halogen light the PSU resets (clicking noise of the relais). Standby power still works (nothing happens to the arduino) but for example the R-PI connected to the 5V active supply also resets.

Am I supposed to "switch" the "live" wire for the halogen light? I was under the impression that for DC that does not matter. It is not over current as the light shines until I disconnect the ground from it, even for several seconds. To reiterate: The PSU resets when 12V load/ground is disconnected

Does anyone know why this happens?

Connections: enter image description here

  • Switch connected --> Light on
  • Switch disconnected (after connected) --> Light off, but PSU resets
  • \$\begingroup\$ please add a diagram of your circuit \$\endgroup\$
    – jsotola
    Oct 3, 2020 at 22:31
  • \$\begingroup\$ That’s hard to say, I don’t know what the internal circuit of the PSU looks like \$\endgroup\$ Oct 3, 2020 at 22:32
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    \$\begingroup\$ @YanickSalzmann - Hi, It might be that another way of phrasing your question is: "Why do the 12 V and 5 V rails briefly drop when the only 12 V load is removed?" - yes? Or do you have other test results, which mean that re-phrasing cannot apply? I would recommend trying (for example) switching the +12 V (not Gnd) to the bulb, to see if the same behaviour is observed. I would also try using 2 such bulbs in parallel, and disconnecting only 1 of them (thereby leaving some load on the 12 V rail, instead of removing it all). The results of those tests will better characterise the behaviour. \$\endgroup\$
    – SamGibson
    Oct 3, 2020 at 22:50
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    \$\begingroup\$ I really like that phrasing. And I am going to perform some more tests on what actually causes the drop tomorrow! I thought maybe this might be some common error/misconception (but hard to find via google, as its all about some generic PSU questions) since I did (almost) 0 research past finding the pinout for the PSU, but it seems it might not be like that :-) \$\endgroup\$ Oct 3, 2020 at 22:53
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    \$\begingroup\$ Most likely the power supply does not handle a case where loading on 12V line drops suddenly from 55W to zero. It is designed to power a computer, not switched lights. \$\endgroup\$
    – Justme
    Oct 3, 2020 at 22:55

2 Answers 2


You are most likely violating ATX power supply specifications. The load current has a maximum slew rate limit of 1A change per microsecond, and the jump on 12V supply must be less than 50% of the rated maximum load, which your power supply says is rated for 18A, so the jump must be limited to 9A.

When you connect the bulb, that 55W lamp is a 4.58A load jump that is limited only by wire inductance, so the jump will be faster than 1A per 1us. However, it practice is worse than that, because the bulb cold resistance is much lower so initial current can easily exceed the 9A load change rating. A sudden large load also can make the output voltage droop until the regulation loop corrects for this. For some reason, connecting the load does not trigger overcurrent protection or undervoltage protection.

For reference, to limit 12V to have change rate of 1A/us, you would need a 12 Henry coil.

The problem is even greater when the lamp is disconnected, as the loading goes from 4.58A directly to no load at all. As the PSU output coil has inductance, the 4.58A cannot change instantly, and when load is disconnected, the current only flows into PSU output capacitor and output voltage jumps higher than intended which triggers overvoltage protection. Having extra capacitance connected would cause the voltage to rise less by capturing the current surge until it stops, and having extra load to draw some current would consume the voltage rise from the current surge faster.

  • \$\begingroup\$ You can assume a factor or 12 cold-to-hot resistance for incandescent bulbs. \$\endgroup\$
    – winny
    Oct 4, 2020 at 13:15

Problem: >4A low side dry switch on 12V causes over voltage fault, shutdown, restart. Sounds like +ve spike over voltage protection is not suppressed very well or insufficient steady load to achieve good regulation as a dummy load. or could it be a combination of both?

-If we use EE theory, the voltage produced from a dry contact switch or wire we expect a large fly back voltage from the low side V=LdI/dt. We don’t know the rise time of voltage, but the time is almost instantaneous as the open circuit capacitance may be << 1 pF. The wire and filament inductance will be ~10nH/cm and increased somewhat by the small coil of filament wire compared to its stretched out length.

Meanwhile the stored energy in the secondary coils is small, yet there will be more prior energy drive on the primary and some lag in the response to over voltage due mutually coupled load sensed on the 5V line. Regulation is accomplished by very high mutual coupling and each secondary regulated by turns ratios.

The OVP threshold may be somewhere between 5 & 10% before shutdown and retry but you have 2 conditions for over voltage and also the output impedance of the SMPS for push pull functions may not be as low for pulling down over voltage as it is for under voltage.(guess)

If you switch the high side of the lamp voltage off, this results in a negative fly back voltage snubbed by the low impedance forward diodes coupled to the secondary, mutual L and primary impedance. But a positive spike would reverse bias the diode rectifier.


You must try to reduce the rise time of turn off or use the high side or both to prevent this fault and use twisted pairs to prevent radiated crosstalk of the 4.5A*12V load for E and H field transients.


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