Background for the problem:

I rebuilt my computer in a new case where the components fit a bit better. Everything is great, and I have no issues with it. However, because of the way they engineered the front panel LEDs, I don't have an LED indicator of computer sleep status. The power light is powered off their control board, which is powered from the PSU directly, so I don't have a sleep indicator.

Now, I could easily just get an LED and extension cable for a simple power LED installation, but I got to thinking that I'd like to try something a bit more elaborate. That is, instead of a single 3.5V LED, I want to use a 12V light source (such as a LED spotlight or cold cathode) and install that to have something that could indicate sleep/power status with a bit more style. I have a mesh grill on the front of my case, so I figure something that illuminates under the mesh would be cool looking and noticeable but also not be disruptive in a dark room (like if I'm sleeping). There's some hurdles to this, though. For one, I obviously can't power one of these straight from the motherboard connector. It doesn't provide enough voltage. For another, the PSU shuts off my case lighting when in sleep mode.

Proposed solution:

So, I'd like to make something to do what I want. I'm a programmer but I have tinkered with electronics kits as a child, so I'm familiar with some (very) basics, and have assembled circuits before (at least, from pre-written designs). It has been a while though and my efforts in designing my own analog circuitry are best not to be spoken of, so I hope I can do something simple, here. :)

What I figure I would have to do is build a simple circuit that uses the power output from the power LED connection on the motherboard to power a switch (of some sort) that would provide power to the 12V light source from a battery, which could be recharged from the PSU. I believe the 3.5V connection on the motherboard just provides straight power to a LED (if one is connected). But instead of a LED, I want to make it so that while the computer is powered on, it would close the circuit for the 12V light so it would illuminate, while simultaneously charging the battery. When in standby mode, it would light the LED (my motherboard causes it to blink) by drawing power from the now-charged battery.


I think that a simple relay switch would probably be adequate for controlling the 12V circuit, but they're noisy. :) Is there a solid-state alternative? Since I'm dealing with two power sources, I'm not sure if it's possible for me to isolate them easily in a single circuit without physical separation (as a relay switch would provide). As I said, I only tinkered with electronics kits as a child. I didn't take electrical engineering in college, as I got into computer programming instead. But if it is and is simple enough to implement, I'd prefer it.

Also, I'm sure that dealing with a rechargeable battery brings with it its own set of caveats and things to worry about (like worries of overcharging). Is there anything else that I'd have to take into account?

  • \$\begingroup\$ Note that there is a 12V supply available from the ATX power supply, via the yellow wire on the molex hard drive power connectors. \$\endgroup\$
    – pjc50
    Commented Oct 8, 2012 at 9:31
  • \$\begingroup\$ @pjc50 Yes, however in standby mode, the PSU cuts power to all but the motherboard. Hence the complication. \$\endgroup\$ Commented Oct 8, 2012 at 14:16

1 Answer 1


The motherboard connection expect a LED.

Just use an optocoupler or an optotriac to drive next stages. I had once a 250W halogen lamp connected to the hard disk activity led via phototriac/triac :-)

  • \$\begingroup\$ That's cool! It still leaves the question of the battery, though. \$\endgroup\$ Commented Oct 8, 2012 at 14:21
  • 1
    \$\begingroup\$ You could just use an additional wall ac/dc adapter to provide the 12V. \$\endgroup\$
    – Jim Paris
    Commented Oct 8, 2012 at 15:57
  • \$\begingroup\$ @JimParis That actually is a good idea. Can't believe I hadn't thought of that. \$\endgroup\$ Commented Oct 8, 2012 at 17:15

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