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I have a self built custom made computer that is using a Ultra X3 1000 watt power supply. Over the years I have been reducing the power needs by adopting more energy efficient hardware. My computer is now using about 200 watts at any given time and can jump to a maximum of 360 or so when gaming or folding. So the PSU is overkill by 60% minimum.

I also have a 2560x1440 LG S-IPS LED monitor I built myself (63 watts max typical) using parts I sourced from around the globe and on eBay. The LG electronics that powers the monitor only needs a 24V 5A power brick for which I do not have yet. It is only $23 on ebay. However, I am wondering if I could use one of the empty +12V modular connectors on my PSU to supply the voltage/current to the monitor as well?

The PSU is a 85% efficient model with 70 amps on the +12V rail. I need to boost it to 24V, but I only need 5A according to the original power brick these come with.

Would a cheap DC-DC boost module work for this? Such as this boost converter.

The PSU I am using is connected to a very nice and conditioned power strip.

I am concerned that this boost converter says "Input current :16A (MAX) (more than 10A please strengthen heatsink)". I thought current is only determined by the load draw required? Thus if the device I am powering only needs 5A at max, then why would I be concerned with the Input current of 70A? Isn't the input current just the maximum my PSU can supply, but draw is determined by the device using it?

My goal is to try and avoid using a cheap power supply, and saving a few bucks, and since my PSU is/was very expensive I consider it a much higher quality supply.

So can a boost converter do what I need to do safely? Or is this dangerous?

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Yes. The current through the converter is determined by the load. I doubt the monitor is pulling all 5 amps from the original supply anyway. Just throw an ammeter on the ground of the pc side and measure it to be in the safe side, but it should work without modification. With 100% efficiency the input would be, at max draw, 12v 10A, for 24v 5A. In free air inside the pc case, with fans you should be alright assuming 4A load draw

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A little pedantry

Almost all power supplies for consumer electronics, including PC supplies, operate in what old-timers might call "voltage mode". This means the output current will vary in response to varying loads, in an effort to maintain a constant voltage.

This type of power supply is so common it's borderline pedantic of me to call it out as "voltage mode". Anyway.

Defensive design

Your LED monitor wants a maximum of 24V * 5A = 120W, but you say it typically only uses 63W. The following calculations will assume the worst-case scenario where the monitor requires a constant 120W (safety margin #1).

The boost converter is rated at 94% efficiency for 19V input, 16V output, but I'll assume it's at least 85% efficient at 12V in, 24V out (safety margin #2).

This means:

  • 120W / 0.85% efficient = 141.2W going into the converter
  • 141.2W input power, divided by 12V input voltage, means the input current will be 11.8A, which means you should "strengthen heatsink"

Practical advice

However, if the actual maximum power is really 63W, then you can ignore the above worst-case estimates and assume instead:

  • 63W / 0.85% efficient = 74.1W
  • 74.1W / 12V = 6.2A (no extra heatsink required)

I'd say, if the boost converter is cheaper than the monitor, AND if it fails-safe, then you could omit the extra heatsink and treat the converter as an expensive fuse. Or, add the extra heatsink and have more peace of mind.

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  • \$\begingroup\$ Isn't there a mixup between input and output in your calculations? 24V is on the output side, where the power is maximum 150W. 12V is on the input side, with increased power due to efficiency (94%). The current draw on the input is therefore 13A (< 70A so you're fine) not 7A, and on the output 6.25A (> 5A so you're fine as well). However be wary that you WILL have to "strengthen the heatsink" to do that. That's 8W to dissipate when outputting 24Vx5A. If you want the 2 MOSFETs to reach 80°C max, you'll need 2 radiators of <= 10°C/W thermal resistance. \$\endgroup\$ – Mister Mystère Nov 8 '14 at 11:47
  • \$\begingroup\$ @MisterMystère Not sure, but I revisited my answer and re-did all the calculations, so I think it's more correct now. \$\endgroup\$ – hoosierEE Nov 8 '14 at 16:23
  • \$\begingroup\$ OK, except "141.2W dissipation in the converter", the dissipation is 21.4W. Almighty heatsinks you'd have there. \$\endgroup\$ – Mister Mystère Nov 8 '14 at 16:28

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