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The problem with low voltage power is power loss during transmission. Power in a circuit is voltage times current: Watts = Volts * Amps Power lost in transmi
by todd8 12y ago
The problem with low voltage power is power loss during transmission. Power in a circuit is voltage times current:
Watts = Volts * Amps
Power lost in transmission is current through the wire times the voltage drop across the wire. The voltage drop is easy to figure out with Ohms law:
Volts = Amps * Ohms
Rewriting these with V = voltage, I = current in amps, R = resistance in ohms and P = power in Watts and simplifying one gets:
P = V * I
P = (I * R) * I
P = I**2 * R
So power losses are proportional to the square of the current flowing through any given piece of wire (that will have a resistance independent of the voltage or current).
My house has 20 Amp circuits at 120 Volts. The same power transmitted at 5 volts needs 24 times the current (i.e. 480 Amps) and 24^2 times the copper in the transmission wire to have the same power losses. So I would basically need a wire with 576 times as much copper to have the same power loss in that one low voltage circuit.
This is why it isn't practical to run the whole house on low voltage. High efficiency transformers can step up or down a voltage while losing only a few percent of the power. When their output voltage is 10 times higher their output current will be 10 times smaller and can be transmitted over the same wire with only 1 percent of the original voltage's transmission power loss.
This is the reason that power transmission lines run at 120,000 volts--a million times lower transmission line losses. (It actually is a bit more complicated due to the alternating current interactions, but it's the basic idea.) Power transformers can be 99% efficient. I have no idea of the power losses that would occur between DC current from solar panels to 120 VAC and then back to 5 VDC for USB, but you definitely don't want to transmit 5VDC over long distances.
- deleted 12y ago[deleted]
- CucumberLime 12y agoI think the idea is use these only for low wattage applications, way below 2400 Watts. 48V at, say, 10A is plenty of power for many uses. Also the transmission distances should be relatively short if coming from a battery in the home.
- jws 12y agoThe cross over is surprisingly early. I care for a cabin that uses a 24VDC battery for its solar power storage. I run that in #6 copper to the cabin (pencil thick conductors). I regularly see a >2 volt drop on that run when the tiny high efficiency fridge kicks on, on top of the ~40watts that the telemetry and comm gear takes. You wouldn't notice that on a 120v line, but that's 10% of my power gone to heating the earth. Given a do-over on infrastructure, with today's equipment costs, it would save up front capital and power to use a small inverter at the batteries and run those items off of AC using smaller copper lines for the long run. Keep in mind that you still need DC-DC converters at the loads, since nothing is going to want to run off of your varying DC socket voltage. (Batteries closer to the cabin is not a win, the solar panels are off that direction and those lines need to be kept short too. I do have a large inverter in the battery shed and send 120v to the cabin for big loads, but its idle current is too high to keep up for the base loads. The most expensive part of this spring's wind turbine installation (and heaviest) is the copper wire. Low voltage DC is not an answer.)
- rootbear 12y agoThanks, I'm getting a great understanding of all of this from the comments here. I suspect that USB is going to end up being the universal low-voltage power source, but I still think that's a shame, as something more like 24 to 48V would be better, especially if there was a well designed, universal connector. I'm still amazed that they can get 100W out of those tiny USB wires...
- wlesieutre 12y agoThe 100W delivery over USB actually bumps it up to 5 amps at 20V. The higher power profiles aren't available by default until the devices negotiate the higher voltage and confirm that the cable can support it. With how much USB cables get beat up, staying on the lower end of voltage is probably a good idea. I've definitely seen some old ones that ended up with wires exposed. But 20V is definitely more capable than the old 5V. You'd need a lot of copper to run 100W at 5V. More info: http://www.usb.org/developers/powerdelivery/PD_1.0_Introduction.pdf http://www.usb.org/developers/powerdelivery/PD_1.0_Introduct...