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Close to all desktop motherboards have a pin header that is supposed to power an LED on the front of an enclosure with to indicate that the computer is on. In the documentation for my board, these pins are described as "Power LED Signal anode (+)" and "Power LED Signal cathode (-)".

This made me curious and by measuring both pins with a multimeter seperately, I found out that the anode on the board is always supplied with 5V (compared to GND) as long as the PSU is connected to AC current while the cathode is supplied with 5V normally and gets pulled down to ground potential when the PC is turned on.

What I would've expected was the cathode being connected to GND directly and the anode being pulled down normally and pulled up to 5V when the PC turned on.

Is there a good reason why the manufacturer would choose to do this? Is there any advantage in powering a LED this way over the way I described? Is this probably even common practice with desktop mainboards?

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  • \$\begingroup\$ If someone provided you with two wires that measured 5V between them, do you know how much current those wires can provide? \$\endgroup\$
    – Andy aka
    Commented May 10, 2015 at 13:50
  • \$\begingroup\$ No I don't, but why does that matter? \$\endgroup\$
    – iFreilicht
    Commented May 11, 2015 at 11:29
  • \$\begingroup\$ Because there is probably a current limiting resistor for the LED that drops the voltage to the required amount when the LED is connected. Your meter would just read the open circuit voltage as 5V. \$\endgroup\$
    – Andy aka
    Commented May 11, 2015 at 11:33

2 Answers 2

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This would be standard in all motherboards compliant with the Intel Front Panel I/O Design Guide.

Note that what you see as +5V is actually a resistor of a few hundred ohms to the +5V supply.

http://www.formfactors.org/developer%5Cspecs%5Ca2928604.pdf

The likely reasons include the practical (a short of either side to ground damages nothing) and possibly they were thinking a MOSFET drive could be used, so the higher carrier mobility of electrons over holes means an N-channel MOSFET would be preferred. I believe bipolar drivers are typically used at least on motherboards sold at retail- probably for ESD immunity reasons- no motherboard maker wants to have to take returns because an LED driver got fried by a sloppy assembler.

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  • \$\begingroup\$ Thanks for linking the I/O Design Guide, that is a very insightful document. \$\endgroup\$
    – iFreilicht
    Commented May 11, 2015 at 12:27
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Some controllers have open collector or open drain ports which allow to switch voltages larger than the supply voltage of the controller.

The following sketch shows a device supplied with 3.3V only, which can switch a load connected to 5V.

However, I have no idea if this plays a role for the design of the LED circuits of a computer mainboard. It's just one possible reason for switching the GND side.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ That's a very interesting reason, and I think it's kind of the same thing @"Sphero Pefhany" meant, but thank you very much for the schematic, that's very helpful. \$\endgroup\$
    – iFreilicht
    Commented May 11, 2015 at 11:29

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