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Comment by tardo99 | Hacker News Reader

Could people with houses use legacy PSTN wiring to distribute 12V throughout the house?

POTS wires are 24 gauge [0] which can handle 3.5 amps [1] so you'd be able to transmit 12V * 3.5A = 42 Watts.

POTS often had 36V on it to power the ringer on phones so you could potentially up the power transmission by running your system at 14V. Some devices which consume a fixed amount of power would use less current, other devices would just use more power at higher voltage.

[0] https://en.wikipedia.org/wiki/Plain_old_telephone_service [1] http://www.powerstream.com/Wire_Size.htm

42 watts seems ok for a very small installation, right? I'm thinking enough to charge a phone and maybe a laptop. And, that would be driveable with a small 12V battery and a small solar panel.

As others have pointed out voltage drop is a significant problem with low voltage systems. #24 is 84 ohms per km, or 0.084 ohms per metre. Say the system is running at 13.4 V and the loads will work on as little as 11.5V before they brown out, that means we can have 13.4-11.5V voltage drop = 1.9V voltage drop. ohms law V=IR so for 3.5A and 1.9V R=0.54 ohms total resistance, and at .084 ohms per metre we can transmit 3.5A 6.46 metres before the voltage drop will be unacceptable.

You could charge a phone, but even my small laptop chargers are over 60W. The big ones are 90W.

Also don't try this in your house! pushing 3.5A through that little pinner wire it could get hot and start a fire.

Makes sense. Could you instead apply a DC bias to the AC power jack?

Wouldn't there be 4 conductors, so resistance would be half when you pair them together?

Yes.

But i forgot about the resistance from the load back to the source so above where I said 6m should actually be 3m. Or if two wires are parallel both from source to load and load back to source then we're back to 6m.

Voltage drop is one way to look at it, but since efficiency is paramount, it's better to look at losses. Losses in the conductor are I^2 * R, so when there's a current of 3.5A, losses are 6.6W at an input of 3.5*13.4=46.9W. Efficiency is 88% which is very bad. An average inverter that converts the low DC voltage to AC is much more efficient and even with taking into account the wiring loss, there is a net gain. Generally, using a low voltage DC grid isn't a good idea.

Some laptops use over 150 watts, so it would have to be an extremely tiny laptop.

The Macbook Air has a 45W charger. Assuming you're using DC (no conversion inefficiencies), you should be able to charge it.

Sorry but No. The phone line has DC to power the old-style carbon microphone and AC to ring the bell. If you try to draw appreciable current, the system thinks your phone is "off hook", so the phone is out of service.

You can draw a tiny current without triggering the "off hook" detector, but that is miniscule.

the question is about the wiring, not about keeping the PSTN working while using the wiring.

> POTS wires are 24 gauge ... transmit 12V * 3.5A = 42 Watts.

But AWG24 cable also has a resistance of 0.08Ω/m (0.024Ω/ft). With a 10m (33ft) cable this amounts to (current has to travel both ways): R=2⋅ℓ=1.6 Ohms, at 3.5A this drops 5.6V. So, in the end you'd loose RI²=19.6W in the cable and only get 22.4W (3.5A at 6.4V) in the load...

Yes, but you'd be better off using a higher voltage and switching DC-DC converters at the end.

The POTS lines would have ~50V on them in normal operation so they'd be insulated for 48V. However, V=I/R, and the lines have a constant resistance so if you can raise V and lower I you can get the same total wattage for a lower line loss. This is also why distribution networks are at 11kV and up.

Modern switching converters are pretty efficient, and you can also tailor them for the device so you don't have to double convert.

Since you have 4 conductors to work with, you could run 12V for low-amperage applications and 48V for high-amperage stuff that justifies the switchgear.

Yes, but only for very small amounts of power, as others have pointed out. A better idea would be to run DC on a select few runs of the existing 14-gauge romex from your breaker box to the rooms they serve. You'd be taking those 110v lines out of service and using them exclusively for DC, but you could pump a lot more power across them than you could across telephone wire. (You still wouldn't be able to run electric heating appliances or big motors or incandescent lights, but pretty much everything else in a modern house would work, modulo the need for switched DC-DC voltage converters.)

incredibly inefficiently, and only for loads of 5-10W max... phone wiring is not thick enough in gauge and 12V has very high voltage drop over distance for any appreciable amperage.

there is a good reason why we use 110-240VAC to distribute power in houses.

http://www.britishtelephones.com/howtele.htm

Is 16 ohms per thousand feet really that high?

Yes. Typically devices can tolerate a voltage drop of about 3%. With that resistance, a 1A load can only go 22.5 ft from the source. Considering you might have to run up a wall, across the ceiling and down another wall, you might not even get to the opposite side of a room.

The phone line is essentially "constant current". There's 48V before you draw current, but that quickly drops to zero when you lift the handset, because there's an exchange relay in series which senses when the phone is "off-hook"

No. In fact most newer solar installations are using microinverters to convert to AC at the source. 20-30amps of DC requires quite a bit of copper to go even a few meters.

20-30amps of DC requires very little copper if it's high-voltage. It's low voltage that requires fat copper, not DC vs. AC.

The same number of amps will require the same cross-sectional area of Copper, for both DC and AC.

You can use less copper for the same power with higher voltage as a reduced current is required for the same power. This is also independent of AC/DC.

(nitpicking) That's correct if you take into account the skin effect when computing the cross-section area (or if you design your transmission line to avoid it. Better solution). But at 50/60 Hz, it's negligible (the "skin depth" of copper at these frequencies is around 9mm)

Except for electric cars there is very little consumer facing HV DC. Solar panels are 12/24/36/48V. With a 20-30 meter cable run it makes sense to convert it to higher voltage at the source. Since it's going to be used in traditional AC grid system, skip a step and convert it to AC as well. Kills 2 birds.

True. And HV DC (more than 50 volts or so) is more likely to kill you than HV AC if you grab a live conductor. Just the same, if I were building a new PV-powered house today I would try very hard to wire it as a mostly-DC house with 12-gauge or bigger romex and switchable voltages at the breaker box in increments of 12, up to 48. I'd put in point-of-use inverters at the few places that needed them instead of having 100 point-of-use rectifiers like we have today. The overall efficiency of a DC house would be quite a bit higher than a typical AC home.

I live on a boat and from my experience you're better off using AC. We have 12VDC outlets and 5VDC USB outlets around the boat. Except for basic lighting they are practically useless. The 12VDC cigar adapters aren't designed for the load we put on them 80-90W (120W and they get extremely hot). And with motors you get into other concerns like the starting current can be 3-4x higher.

We have a single 3kVA inverter for the entire boat. From this we can run clothes washer, power tools, kitchen tools, and blow dryer from our batteries. Modern wall warts are incredibly efficient. You are losing, at most, 3-5% efficiency. Plus, it's really hard to find 12V anything thats any good. A 12VDC vacuum cleaner is a joke compared to a proper AC vacuum cleaner.

Trust me whatever efficiency you think you'll be saving is better off buying more solar or bigger battery bank. Keep the distribution simple, 120/220V. Point of use inverters are for charging laptops in the car.

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