I'm planning on adding multiple IoT devices in my house, and I'm looking for an elegant solution to power them all. They'll all need 5 V, a small number of them (cameras) would draw up to 2 amperes, while most of them would draw less than 500 mA.
I'm thinking of running a separate 5 V line, through the house, with 10 AWG cable (to reduce the voltage drop) and a 120 V – 5 V, 20 ampere transformer on each floor. I would add some kind of connectors, each a couple of feet, in order to be able to connect to the line easily in the future.
The main purpose of all that is to keep my electrical outlets free, save all the single transformers I'll need, and allow me to extend my IoT network in the future, without using any extra outlets / unnecessary hanging cables. Is that a viable solution, or is there a more efficient way of accomplishing my goal?
I'm contemplating whether it would be better to use a 120 V – 12 V transformer, in order to compensate any voltage drop and add buck converters / linear regulators on each device, but I would like to save the extra hardware if it's viable.
Best Answer
12 gauge cable has a resistance of ~= 1.6 Ohms per 1000 feet or per 500 "loop feet".
So you get 1.6V drop per amp for 500 loop-feet or about 3 mV per loop-foot per amp.
Option 1:
A 5V run of say 100 feet at 10 A will drop 3 mV x 100 x 10 = 3V.
That's obviously too much at 5V.
You MAY deciede that 5A max is OK - or that the mean current is half the max if ecenly distributed, and a max length of 50 feet may be acceptable.
But , making compromises will still give you drops in the 0.5 - 1.5 V range.
A few cameras at the far end are liable to 'spoil your day'.
Your network needs to either be "dumb" and able to supply max power at 5 VDC or intelligent and only supply power by negotiation. At 10A that's 5V x 10A = 50 W and probably double that to allow for drop and fusing and ... . 50 - 100W is enough to start a fire with a little ingenuity. Murphy has lots of ingenuity. It's not fatally flawed but care is needed. This is lower current and lower voltage than mains AC circuits provide - so if you take as much care with your DC network as your AC mains one it may be OK - noting the need for switches able to handle DC well at rated current (DC being substantially more difficult than AC).
Option 2:
Distribution at say 8VDC or higher (10-12VDC better) (or you can use AC) will allow freedom from voltage dropout issues. If you use linear local regulators then you lose efficincy with increasing feed voltage. If 8VDC is "just enough" you are losing (8-5)/8 ~= 40% of you power. And more again with more voltage. And you have added complexity and cost and a nonstandard system.
If you use switchmode local converters efficiency is relatively constant with feed voltage, and wiring losses drop as voltage increases due to resistive losses being proportional to current squared. But the converters add complexity and cost. POE (power over ethernet) is one version of this - and the cost per converter is liable to be substantial for off the shelf equipment, and compared to using mains powered commercial supplies.
Option 3:
A single power socket per location allows you to operate either a local 5v PSU OR a mains AC plugboard with a collection of plugpacks.
At the desk where I am typing this I have 18 x 12V powered USB hard drives on a shelf. I've considered the use of a 20A plus 12V supply, custom leads (std socket and cable from HDD to a connection system) but, so far, the 18 x 12V 1A powerpacks plus requisite mains plugboards has won through. Voltage drop is not an issue, I have excellent power supply redundancy either by providing say 2 spare 12V psus or by borrowing a supply from a less critical supply if needed.
My 5V "USB" supply needs are met by the several PCs with USB sockets, plus some of the HDDs have powered 5V outputs. And 5V "USB" psus are very very available, well priced (especially as used ones are usually reliable) and the connector is universal.
And mains AC (230 V 50 Hz in my case) is available throughout the house, or in any home or business I visit.
For me, and probably for you, option 3 makes most sense.