I'd suggest something like a mini-360 buck converter for the 5 volt supply, powered directly from the source. The mini-360 converters are cheap, a buck or less.

Good suggestion for use with 20V signals, but probably not for 28V. The mini360's maximum input voltage is just 23V.

 

Good suggestion for use with 20V signals, but probably not for 28V. The mini360's maximum input voltage is just 23V.

My apologies. I missed that detail, but as I explained, 28 volt USB PD supplies aren't that common, and would be more expensive if purchasing one in required. If 15 volts provides enough overhead for the DC-DC converter, it will be the most-acvailable power supply.

 

My apologies. I missed that detail, but as I explained, 28 volt USB PD supplies aren't that common, and would be more expensive if purchasing one in required. If 15 volts provides enough overhead for the DC-DC converter, it will be the most-acvailable power supply.

Thanks again. If 20V works then that's great.

In another corner of the internet, somebody told me this: "If you were to try to pull both 12v and 5v from a 20v supply, you would very likely have issues with the power draw."

Is that true? This claim that 20V does not provide sufficient headroom to power the +12V, -12V and +5V rails together is what I've been trying to verify this whole time. I'm still not sure if it's accurate or misleading.

 

A 60 watt PD supplies 3 amps at 20 volts.

Per your specs some pages back:

+12v @ 1 amp = 12 watts
–12v @ 700mA = 8.4 watts
5V @ 400 mA = 2 watts

That's a total of 22.4 watts. Let's say the worst case efficiency is 70% (likely closer to 80%).

22.4 watts /0.7 = 32 watts considering efficiency. A 60 watt [20 volt] PD supply will be coasting along.

A 45 watt [15 volt] PD supply shouldn't be over-taxed at ⅔ rated load.

 

Thanks for clarifying, @Jon Chandler! Seems like the other guy's claim that 20V won't provide enough headroom is in fact misleading. You've helped me a lot today

 

You've helped me a lot today

My pleasure!