4 ms·
Office equipment doesn't really have "transformers" anymore, everything uses switched-mode power supplies. The way they work, is that they have an output buffer
by crote 3y ago
Office equipment doesn't really have "transformers" anymore, everything uses switched-mode power supplies. The way they work, is that they have an output buffer (a capacitor or inductor) and rapidly turn a switch on and off to keep that output buffer at the right level: if the output buffer is below the target voltage, the switch closes and the voltage will start to rise. Once the output buffer rises above the target voltage, the switch opens and the voltage will drop as power is being used from the buffer. This happens thousands or even millions of times per second. This principle works exactly the same for modern AC/DC power supplies as it does for DC/DC ones.
The problem with your plan is Ohm's Law, and more specifically that fact that wires aren't perfect and have some resistance (for now!). Ohm's Law gives us V=I*R, where V is the voltage in volts (V), I is the current in ampere (A), and R is the resistance in Ohm (Ω). In a wire, the resistance is constant, and V is the voltage loss across the wire. So how do we reduce the loss? We reduce I. Luckily we only actually care about the total power, which is given by P=U*I. If we want the power to stay the same and reduce the current, we have to increase the voltage.
Let's say the two wires from our central transformer to the computer are 14-gage copper, and they are 100 feet long. Their resistance is about 0.5Ω combined. We want to power a 120W computer. If we transfer that at 12V (the normal voltage computers use internally), we'd have to transfer 120W/12V=10A. The voltage loss across our wires is 10A*0.5Ω = 5V! So we put in 12V, but get out only 7V as we burned 50W in the cable itself. To get out the desired 12V we'd have to put in 17V at 10A instead, or 170W to power a 120W computer. It would also mean supplying way too high of a voltage to a computer connected with a 3-foot cable.
If we increase the voltage across the wires to 120V and down-convert that to 12V at the computer we'd only need to conduct 1A and the wire loss would be 0.5V, which at 1A is a power loss of 0.5W. That's completely acceptable, and because the computer down-converts anyways we don't really have to care about it getting 119.5V instead of 120V either.
But now we are back with a power supply at each individual computer, so in the end we didn't really gain anything. Instead of an AC/DC power supply in every computer we now have a virtually identical DC/DC power supply, so what's the point? You might have some small gains by doing the initial AC/DC conversion centrally, but in practice it probably isn't enough to care. It is only really worth it when your power comes from DC anyways, like an office with rooftop solar.
Alternatively we can use way thicker cables, but to get that same 0.5W loss at 10A would mean a wire with a resistance of 0.005Ω. To illustrate, that means using two 0000 AWG wires in parallel.
- free_energy_min 3y agoNot op but thanks for the very detailed explanation!
- Animats 3y agoYes, that's the tradeoff. High voltage for the long haul, low voltage locally is common. There are computer equipment racks where distribution within the rack is at 12 VDC. These often have big busbars in the back, and a power supply in the base. Facebook's OpenRack started at 12VDC, but a later rev is at 48 VDC. That's just within the rack; there's a power supply in the rack base running off something like 3-phase 220VAC. There are advantages to running off 3-phase power; there's always power available from at least one phase, and the capacitors needed to smooth DC are far smaller. Telephone central offices have run the whole office at 48VDC for a century, with a big battery for backup power. Big bus bars carry that around the building. (Do telco offices still do that?) Much industrial control gear runs at 24VDC. So do many military vehicles. It's a reasonable voltage to send a few meters, but not hundreds. (The extreme case is ultra-high voltage DC power transmission, where power is sent thousands of kilometers at a million volts.)
- wongarsu 3y agoFor telco gear - at least the local distribution stuff in a box on the side of a road - a common setup is AC for normal operation, but 24VDC (or is it 48VDC?) between UPS and device. Because of the telco heritage those devices come with both AC and DC power inputs, and using both protects against power supply failures and allows the UPS to last a bit longer.
- mordae 3y ago> Office equipment doesn't really have "transformers" anymore, everything uses switched-mode power supplies. The way they work, is that they have an output buffer (a capacitor or inductor) and rapidly turn a switch on and off to keep that output buffer at the right level... Overwhelming majority of mains-powered devices does, in fact, have a transformer. It is part of a switched-mode power supply, though. The only devices that do not have one are those that are fully enclosed in plastics and user can not under any circumstances come to contact with any of the conductive parts. Typical example would be a LED light bulb or wall-socket powered WiFi repeater (without RJ45 port). It is pretty hard to create a transformer-less device using plain rectifier/switch/capacitor topology. The biggest issues obviously are the high voltage before the switch and dead time when the mains voltage drops near zero 100 times per second. Most switch ICs made in the "west" cannot go up to 325V and are thus usually used with a transformer to step the voltage down below 60V. If you get a Chinese chip that can work off-line (as in directly with the 325V mains voltage), such as KP1063 <https://datasheet.lcsc.com/lcsc/2103171532_Kiwi-Instruments-KP1063CSPA_C310438.pdf https://datasheet.lcsc.com/lcsc/2103171532_Kiwi-Instruments-...>, it still needs an inductor to bridge the mains dead time. This is incidentally done by the transformer in traditional power supplies. Capacitor would have to be huge to smooth over those for any significant power draw. > If we increase the voltage across the wires to 120V and down-convert that to 12V at the computer we'd only need to conduct 1A and the wire loss would be 0.5V, which at 1A is a power loss of 0.5W. That's completely acceptable, and because the computer down-converts anyways we don't really have to care about it getting 119.5V instead of 120V either. We could use 48V, which would bring the cable losses to about 3.3W (for those 100 ft). Or maybe instead of wiring from the mains box, we would place transformers along the wall sockets. Some new installations already do that and install charging USB ports. USB-PD is now specified up to 48V / 5A. I think that we are already seeing devices abandoning the traditional DC connectors in favor of USB. A lot more people now have a large bank of charging USB-A and USB-C ports on their desks. Some vendors already integrate them in extension cords. It won't take long for LED lamps to come with an USB cord and an optional mains/USB transformer SMPS ("phone charger"). Laptops are already charging (and docking) via USB-C with PD, it might not take long for screens to follow. Stepping 48V DC down to 12V, 5V, 3.3V is way easier and can actually work the way you have described above.