Electronic – ATX power supply safe on

You can open the supply and look for the TL431 chip. There will be two resistors forming a voltage divider from the 5V rail to the 2.5V that the TL431 compares against. If you calculate the right resistor values, you can set this divider to output half the voltage it currently does. Then you can get the output up so that the +5V rail becomes +10V and the +12V rail becomes +24V. You will probably need to install new output filter capacitors of higher voltage rating as well. And if you modify an ATX supply this way, please cut off the ATX connector and install something else, so it won't plug in to a motherboard.

You'll most likely have to limit the current yourself. It's under that condition that the voltage is specified. Going higher than 200\$\mu\$A will probably cause the voltage to sag below 2.4V, which is the minimum for a high level in TTL. (I've also seen 2.7V as the minimum, I guess it depends on the TTL subfamily.)

Note that the same table says that "signal type = +5V TTL compatible", and that a low level is specified as < 0.4V. That output level is 0.4V less than the maximum TTL input level for a logic 0. And the 2.4V is 0.4V higher than the minimum for a high input level. This gives a 0.4V noise margin.

If you want to control a MOSFET with the PWR_OK signal, like OP, you'll need a logic level MOSFET, which draws enough current at a low \$V_{GS}\$. The BSG103 may be a good choice; it has an \$I_D\$ of 750mA at a \$V_{GS}\$ as low as 1.5V.

edit
On second thought the BSH103 may be too good. It has a \$V_{GSth}\$ of 0.4V, which means that worst case you'll have a drain current of 1mA with PWR_OK low. Even FETs with a \$V_{GSth}\$ of 1V typical indicate 0.4V as a minimum value. Can be fixed by using a resistor divider to lower the output voltage from PWR_OK. A 15k\$\Omega\$ + 25k\$\Omega\$ gives you a minimum gate voltage of 1.5V, while the current is maximum 125\$\mu\$A.