5 volt power supply, 60 Amps (too much?)

Each device would be fine if connected.   You can't force the 60 amps into something that doesn't need it.

My concern I would have is the 60 amps would require a quite large wire for distribution, and if you ever have a short or failure of a device, you would have an instant heater.  Use ohms law.  A short anywhere which would pull the distribution wiring resistance down to a couple of ohms.

Try Amps squared times resistance using 60 amps and 2 ohms.     60 * 60 * 2 = 7,200 Watts

That a pretty big heater if you have a fault.  Most likely a fire.

Posted by: @whitneydesignlabs

If I understand Ohm's law correctly, no device will draw more than it needs?

That's correct. Of course, a short-circuit will consume all 60 amps, until the wire melts. So fusing is probably a good idea.

In my youth (back when the dinosaurs still roamed the Earth and there were only 4 TV networks) I worked as a technician in a place that manufactured power supplies. One time one of our assemblers forgot to install a mica insulator on one of the power rectifiers in a 100-amp power supply. When I hooked it up for testing it was pretty spectacular, not only did we get a room full of "magic smoke", it also vaporized the wires on my multimeter! The test leads looked perfectly fine, but they were now just empty insulated tubes - the wire inside was completely gone.

That's a nice find, I envy you!

😎

Bill

Thanks for the feedback, guys. I added a couple pics of the scrapyard score.

Good idea about fusing. I think an automotive 40A ATC blade fuse would do nicely. I could even fuse it at 10-15amps and use smaller wiring. That would be equivalent of about 4 or 5 high-demand cell phones all charging at full tilt, and/or a raspberry pi doing some heavy processing.  Don't know for sure, but the idea seems pretty future proof. It would be nice to do away with all the wall warts everywhere and USB seems to be a pretty good standard by now.  

Side note: the power supply also has a 12v output. That could be handy in the shop.

A large shared power supply where all the loads are 'local' can make sense, but unfortunately there are a number of issues.

As discussed above, high current sources can be destructive under fault conditions. A vaporised wire is 'good outcome' compared to a fire, which is the more common result.

In addition, voltage drop across wires of appreciable length tend to be excessive in low voltage systems.

I am sorry if the following looks pessimistic, but I would like you to carefully design your system and not get hurt.

Please note, I am only sharing my thoughts in good faith; I take no responsibility for omissions or errors on my part.

Overcurrent Protection

As already mentioned a high current supply can become very destructive if a short circuit fault occurs. The general rule is that ALL of the wire in EVERY circuit must be able to carry the 'maximum case' current, with the fuse/protection circuit tripping before the wire is damaged.

Hence, if you are charging a mobile phone, a short could occur at the phone, and your distribution system must protect the USB cable to the phone (as well as the rest of the wiring). I haven't checked the exact specifications, but clearly USB cables are quite thin, so sustained currents of more than say 2A-3A would be concerning. I think the exact maximum current specifications vary with USB variant 2.0, 3.0, C, etc., but the better quality multi port hubs (which supply power) include a protection device for each outlet which is specifically designed for the task.

Voltage drop

Expanding the physical range of a low voltage circuit can be more problematic, because the wire has resistance, so that the voltage 'seen' at the load may be too small for reliable operation. Basically, I recommend doing some simple Ohm's law calculations.

I offer the following as a simple illustration of the maths - it is NOT a recommended electrical approach - please adapt the calculation approach to your own design.

1. You wish to mount a multiport USB in different room from the power supply, with a maximum total demand of 10A. (e.g. the hub will have 5 USB ports each drawing 2A).
2. 10 metres of twin (+5V and 0V return conductors) cable are needed to connect the hub to the power supply.
3. You find 2.5mm2 "mains cable", rated for mains currents 13A-19A+ is readily available, which you might protect with a 10A fuse at the power supply end.
4. However, the resistance of 2.5mm2 conductor is about 0.0074 ohm/metre ( http://www.farnell.com/datasheets/79497.pdf)
   
5. Total conductor path length between power supply and hub is 20 metres (+5V and 0V return combined length)
6. Wire path resistance = 20 (m) * 0.0074 (Ohm/m) = 0.148 Ohm
7. Total voltage drop = 10 (A) * 0.148 (Ohm) = 1.48 V   (Ohm's Law V=IR)

Unfortunately, USB voltage is only 5V, so the voltage appearing at the USB hub = 5 - 1.48 = 3.52 V

Clearly, 3.52V is unlikely to be enough for your loads (e.g. phones) to work efficiently.

And of course, I have ignored voltage drops of fuses, connectors, etc., plus wire resistance increases with temperature, all of which will incease total voltage losses. 

I hope the above is at least informative and interesting. You will not be alone if you find the result frustrating ... many power distribution engineers face the same fundamental limitations every day.

ps Sorry, I am sure you have guessed, but '2.5mm2' is '2.5 square millimetres' .. I haven't managed to work out how to superscript the last '2'.

60A is a large amount of current, perhaps requiring very thick wire.