Do I need to include overhead when picking a power supply?

Question

I need to power 40 Raspberry Pis, each requiring a 5V input and have a maximum current draw is 2.5A.

This would mean:

5 x 2.5 = 12.5W 
12.5W x 40 = 500W

That I need a 500W PSU. I was planning on getting a 504W Meanwell SMPS, which outputs 12V, and step it down with a buck converter (LM2596) (one converter per Pi) to 5V which each Pi would be connected to.

However I am unfamiliar if there's a known 'overhead' you should include when choosing power supplies.

Am I safe to choose a 504W PSU knowing my maximum power draw would be 500W, or should I be adding for example a 10% overhead.

Answer 1

If I were doing this:

I would make a tiny PCB with a small SIP DC to DC converter module on it and a USB plug. Like this one (this one is 1A, there are also 2A versions, they can also be paralleled):

https://www.digikey.com/en/products/detail/xp-power/TR10S05/6707495

I would either measure the raspberry pi under the max processor loads that I would be using or use these power numbers.
https://www.pidramble.com/wiki/benchmarks/power-consumption

2.5A looks like it's a max draw, but average power usage doesn't exceed 1.2A (so maybe I'd plan for 1.5A. The DC DC power does need to be factored it, but don't put margin on top of margin. If you need 15% margin to not exceed total power for the converters and 20% margin for total power, then use 20% margin for the total system power and that will also cover the other margin, or if you want to be safe add 5% more for 25%. Odds are these things aren't going to be pulling a full 2.5A the whole time, so I'd think 500W would be plenty.

Also plan for system growth and mean well isn't that great of a supply.

Answer 2

It is actually a complicated question. What you DON'T want to do is pad everything and end up with 2x or 3x more power than you need. On the other hand, sometimes you have to really read the fine print on a power supply to see if it puts out its rated power 24 hours a day 7 days a week, or if the rated power is only applicable in a room kept at 25C with airflow of 10m/s over the case. So take that as a homework assignment for you. Are there any footnotes about the power output on the power supply?

Also, just as far as you calculation goes, you have to factor in the efficiency of the 12V to 5V buck. So if each Pi really needs 12.5 W, and you choose a buck with an efficiency of 90%, then you need 40 * 12.5 / 0.9 = 556W at 12V. Your calculation assumed that the bucks were infinitely efficient. Note, I just made up the number for efficiency. You should do the calculation with the real efficiency from the buck you select.

On the other hand, is 12.5W just a short-term worst case that happens rarely? If so, maybe the 504 W calculation is already generously over-sized. So the fact that you didn't allow for losses in the buck, and maybe you have to derate it even a little more because it is going to be running in a warm area would be OK.

Regarding this part of it, what would help a lot would be if you could measure the ACTUAL 5V current consumed by one Pi doing the exact job that you expect the Pi's to be doing. Then you could multiply by 40 and get the true 5V current requirement. Then calculate the 5V power. Divide by efficiency to calculate the 12V power, and see if it is enough.

So, I think I gave you more questions than answers, but this is how I would think about a system design problem like this. By the way, it is a good idea to distribute the bulk power at 12V and down-convert to 5V at the device. Even though you will have some losses from the buck, it is much better than the nightmare of fat cable you would need to run to a 500 or 600W 5V supply to carry 100 to 120 Amps! If you can find readily available 48Vin buck converters, distributing the power at 48V is even better.