3
\$\begingroup\$

I am trying to convert an ATX power supply to a lab bench supply following this Instructable, but with one change: Since I only had one red plug on hand, I connected the +5V and +12V rails to a SPDT switch with the common lead then connected to the load resistor and the (1.5A fused) output:

Image

The problem I am getting is that when the 12V rail is selected, the PSU turns off and does not supply power, i.e., neither the 5V or the 12V rails are "hot" and the fan spins down. When 5V is selected, it works fine (and both the 5V and 12V rails are "hot.").

Why is this happening?

\$\endgroup\$
1
  • \$\begingroup\$ Ideally, both 5v and 12v would have individual load resistors. \$\endgroup\$
    – Passerby
    May 25, 2014 at 15:12

1 Answer 1

4
\$\begingroup\$

The PSU's output feedback is based on the +5V rail. You need a load connected always to that rail. I suggest you leave the +12V rail without load (if it works).

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
5
  • \$\begingroup\$ I agree but I'd go with a bigger resistor. That's 2.5W of pure electronic heat on that poor \$R_1\$ \$\endgroup\$ May 25, 2014 at 17:32
  • \$\begingroup\$ @VladimirCravero I would have said the same, but I've seen a TL494 based PSU whose maximum power (current) on the 12 V rail was directly dependent on the load of the 5 V rail. \$\endgroup\$
    – Cornelius
    May 25, 2014 at 17:35
  • 1
    \$\begingroup\$ @VladimirCravero: If you don't have a resistor that can handle that much power, use a few larger ones in parallel, so that the total current remains unchanged. \$\endgroup\$
    – Ben Voigt
    May 25, 2014 at 18:47
  • \$\begingroup\$ @BenVoigt well thanks, but I am not the OP. And I am lucky, I've got my own bench PSU ;) \$\endgroup\$ May 25, 2014 at 18:48
  • \$\begingroup\$ It works without a resistor on the 12V rail, just on the 5V one, and it doesn't get hot, so I feel like it's okay! Thank you all! \$\endgroup\$
    – Asa Graf
    May 26, 2014 at 18:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.