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I am building a new PC, I just got my motherboard in the mail today. It is a ASRock X58 Extreme 3.

I am reading over the manual for it and one of the jumpers says this.

There is a 3 pin jumper labeled 1 2 3, with pins 1 and 2 shorted by default it is "+5v"

If I manually change it to short jumper pins 2 and 3 instead of 1 and 2 then it will be "+5vSB (standby) for PS/2 or USB wake up events.

So I am confused as to exactly what this is for. Can someone explain what it means, why I would want one other the other?

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2 Answers 2

Reset to default
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It sounds like it powers a circuit that detects mouse or keyboard events and pulls the system out of suspend if detected. You'd disable it if you didn't want mouse or keyboard events (wiggling or button pushing) to wake the system once put in standby.

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  • that is what I was thinking it would do. So if that is the case, how could I get the PC out of sleep mode if the keyboard/mouse was disabled from doing it?
    – JasonDavis
    Commented Apr 15, 2010 at 0:56
  • You'd have to push the power button.
    – Ignacio Vazquez-Abrams
    Commented Apr 15, 2010 at 1:04
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I don't know for sure, but I can take a guess based on the description:

If it's in +5V mode, USB/PS2 ports will have +5V when the computer is on and the CPU is running. If it's in +5VSB, then USB/PS2 ports will have +5V even when the computer is asleep, so your keyboard/whatever devices will have power so they can wake the computer.

If you want to save a tiny bit of power and won't wake your computer using a USB/PS2 device, put it in +5V mode. If you might ever want to wake your computer from a peripheral device, put it in +5VSB mode.

Of course, I've never used your motherboard, so that's kind of a guess. Try it and see. It doesn't sound like it could possibly hurt anything if it's in the wrong side.

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