I had a hair-brained idea the other day and put together the attached circuit.

The problem: the instant I push the switch to place a voltage across the inductor, the power supply shuts down. Clearly this is some sort of emergency power-down because I can start it right back up, but I'm not understanding why such a feature would get triggered? Is it just the voltage drop is sudden enough to flip off such a power supply? I would have thought that the series resistance would be enough to prevent something like this.


simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Smells XY problem. Forget the inductor and tell us what your real end goal is. \$\endgroup\$ – winny Sep 6 '18 at 15:14
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    \$\begingroup\$ The circuit you've drawn shouldn't be able to short the supply - so probably a construction error somewhere. It also won't do what you want it to - but that's a different story. \$\endgroup\$ – brhans Sep 6 '18 at 15:37
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    \$\begingroup\$ Interesting... 2 answers and no one addresses an elephant in the room. Yeah, that very elephant @brhans hinted at above, I suspect. Like the current in open circuit \$\endgroup\$ – Maple Sep 6 '18 at 16:09
  • \$\begingroup\$ @winny What is XY problem? The inductor was the whole point of me setting up the circuit... I'm sure it won't work as it's set up here (there can't possibly be enough power to drive the gate I think), but I was planning on lowering R1 and R2 to see if I could get it there. But none of that is the point of the question. I am just curious to know why my power supply keeps shutting down with this circuit. \$\endgroup\$ – TrivialCase Sep 6 '18 at 17:14
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    \$\begingroup\$ What exactly are you trying to achieve? i.e. what is the hair-brained idea intended to do? \$\endgroup\$ – Chu Sep 6 '18 at 21:38

The above circuit in of itself would not cause the 5V ATX supply to drop out, as the max amount of power that this circuit could draw would be ~12mA. The circuit may have been built something a little different than the one listed above and may be shorting out, test it with a benchtop supply before using it on an ATX supply.

The ATX power supply probably has overcurrent, and/or undervoltage protection. Undervoltage protection is the most likely, if the supply dips below a given voltage (say 4.7V) for an extended amount of time, the supply shuts down.

If you're looking to dim the LED slowly and turn it on slowly, a mosfet with an RC for the input might be best.

Like this:

enter image description here

  • \$\begingroup\$ Where did "dimming" came in to the picture? The circuit idea as stated in the question was to utilize back EMF to drive FET gate. The question itself (which you have answered) was why PCU shuts down. \$\endgroup\$ – Maple Sep 6 '18 at 18:29
  • \$\begingroup\$ I believe the OP posed this question "Is it just the voltage drop is sudden enough to flip off such a power supply?" and this one "but I'm not understanding why such a feature would get triggered?" Meaning the power supply shutting down. I answered that part of the question, in the first part of my post. As a bonus, I suggested that the op could make the LED dim with this type of circuit listed above. I would suggest to stop scanning questions and answers and read them in their entirety. \$\endgroup\$ – Voltage Spike Sep 6 '18 at 18:35
  • \$\begingroup\$ Did you just scan my comment? "The question itself (which you have answered) was why PCU shuts down". I only wondered about the "bonus" part, thinking I might have missed that somewhere. \$\endgroup\$ – Maple Sep 6 '18 at 20:12
  • \$\begingroup\$ Thanks! While my circuit was misguided and ridiculous, this answered my question and turned out to be the solution. The voltage drop was large enough to shut down the PSU. Although I'm no closer to understanding how the PSU knew the voltage had dropped enough to shut down. \$\endgroup\$ – TrivialCase Sep 7 '18 at 14:09
  • \$\begingroup\$ @TrivialCase wrong question. Of course PSU "knew" the voltage dropped. It could not have worked at all without measuring the voltage it puts out. The right question is "what exactly did I connect wrong to make such a huge voltage drop?" Because the very first line of the answer tells you that circuit assembled following your schematics should not have done this. \$\endgroup\$ – Maple Sep 8 '18 at 5:36

The only reason you would use an inductive Back EMF pulse to drive an LED is called a Boost Regulator.

A boost SMPS basically uses a low side switch ( N PN or Nch) pulsed repetitively at high f and draws a current ramp dI/dt=V/L for the duration of the pulse. THen when turned off, it releases a high voltage in the opposite voltage polarity (rising +) but at the same current and direction for a duration of L/R=T thru a reversed diode, ( thus current is switched from transistor to diode)

The diode current pulse is usually clamped to some voltage or storage cap and in this case the load is a string of higher voltage LEDs.

Thus you have a DC-DC inductive boost or "Flyback" Switched Mode Power Supply or SMPS.

Of course DC would just saturate an Inductor core and current limit the supply or shutdown, since the ramp-up in current is fast.dI/dt=V/L= 12V/100mH= 0.12A/ms. But, Relay coils have resistance that limits the current with fine wire and is always specified for each coil volt for ~ constant power.


Of course for giggles, and amuse your friends, if you have a big old car SPDT relay with the normally closed (NC) contact in series with the coil using a 9V battery or 12V to drive a relay. You now have an HV arcing buzzer. buzzzzz. ouch. Frequency and heat increase with voltage so 5V might last longer but just as intense high voltage.

Often an RF suppression cap across the supply near the relay is used. But for a few minutes of buzzing, depending on the contact gap, you can generate many kV or about 2kV/mm arcs with two wires . Just don't put the current thru your heart between two hands. It may hurt. But very easy corona and you can spread the contacts for more arcs for a few minutes until the contacts burn out.

  • \$\begingroup\$ Thanks! I have seen this idea of a flyback power supply before, although I didn't quite get it. Now I see it better. So the short of it is that while I might get large backwards voltage, the current will be so limited by the physical parts that I will not have enough power to drive the FET? It's hard to imagine why protection diodes are necessary then? My thought was to include a working version of this (if it existed) in designs where I might want to know that the flyback protection was used. \$\endgroup\$ – TrivialCase Sep 6 '18 at 17:19
  • \$\begingroup\$ The diodes protect from over voltage but must handle the same high currents but may have a larger voltage drop but for a shorter time, so these "steering or clamp" diodes are always necessary unless the slew rate is low then the analog signal just oscillates about V+, up to full scale (depending on current and impedance) so that can double Vpp. But here it steers the current above V+ or below ground if using a high side switch. \$\endgroup\$ – Tony Stewart EE75 Sep 6 '18 at 17:44
  • \$\begingroup\$ Umm... guys, what current are you talking about? When switch is opened there would be no current whatsoever, except maybe short arcing \$\endgroup\$ – Maple Sep 6 '18 at 18:12
  • \$\begingroup\$ Umm Maple you now the current thru an inductor cannot be switched off instantaneously. It has to be redirected or self-dissipated at a rate of V/L \$\endgroup\$ – Tony Stewart EE75 Sep 6 '18 at 18:15
  • \$\begingroup\$ @TonyEErocketscientist Yep. redirected through what? \$\endgroup\$ – Maple Sep 6 '18 at 18:16

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