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In a domotics installation I will have several devices, Arduino's by example, that typically powers with 5V/1A. These devices are distributed in the space, up to tens of meters between them.

The obvious solution to a continuous power of these devices is use a adapter 230 AC/5 DC before each device. However, I like to reduce number of elements.

A single shared 5V line doesn't seems a good idea, too much voltage fall due to wires impedance that will made them unusable and inefficient. Thus, I'm thinking in two possibilities:

  1. A common 12v line, with DC-DC converters before each device.

  2. Instead of distribute DC current, use a common line of AC current, around 5.5Vrms AC, that is converted to DC before each devices using a diode bridge. However, voltages in the common line near to target ones are easy to fail due to wire resistance.

  3. As combination of the previous two, a common line of 12Vrms AC that is converted to DC and reduced to 5V before each device.

Knows someone which is/are the usual solutions to this subject, and the facts to be taken into account ?

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  • \$\begingroup\$ Have you considered PoE (power over Ethernet)? Even if you don’t actually use/need Ethernet, you can use standard switches, injectors, splitters and cables. \$\endgroup\$
    – jcaron
    Oct 22 at 11:52
  • \$\begingroup\$ Tenths of meters or tens of meters? What is the actual current draw, do they really use 1A each? \$\endgroup\$
    – user253751
    Oct 22 at 13:27
  • \$\begingroup\$ @user253751: tens, fixed. We can assume each device using from 200mA to 1A. \$\endgroup\$ Oct 22 at 14:46
  • \$\begingroup\$ @jcaron: The possible problem for PoE is that PoE is around 50V on the line. That means a final conversion from 50V to 5V on each device, that is bigger that valid input for most lineal or switched DC-DC converters. \$\endgroup\$ Oct 22 at 14:53
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Due to voltage drop in wires it is a good idea to use a voltage regulator in each device.

I'm going to do something similar: I will distribute 5V, and have 3V3 local LDOs because everything runs at 3V3. If you absolutely need 5V you could distribute 6V and use local 5V LDOs.

Distributing AC requires rectifiers and large capacitors in each module, which increases bulk and cost.

If you use 5V, distributing 12V instead of 6V pretty much mandates switching converters, which increases cost and EMI. 12V would only be justified if you need high power for lighting or actuators, and in this case 24V is better anyway.

If you have a large number of devices it becomes important to optimize power consumption. Besides using sleep modes, that means selecting boards and modules that have LDOs (not high dropout regulators) and low idle power, which especially means to avoid LM317 variants with 10mA idle current.

So basically, there are two scenarios:

  1. You have low power devices like arduinos, which require a couple tens of mA current. Since most of the interesting peripherals for these either require 3V3 and don't work on 5V, or work both on 3V3 and 5V, then the easiest solution is to distribute 5V and run the arduinos on 3V3 with local cheap LDOs.

  2. There are high power devices using more than a couple hundred milliamps at 5V like a large backlit display, or some actuators, or LED lights, etc. In this case the loads will decide the voltage and wire gauge, for example 24V for LED strips. 5V loads can be powered by local buck regulators, for exemple these cheap ones at 2-2.50€ each would do the trick nicely. If there is no load that requires a specific voltage (like a 24V LED strip) then a recycled 19V laptop power supply from the junk bin would be a good option too.

under no circumstances should you use the counterfeit "LM2596" modules.

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    \$\begingroup\$ 5V common line to power 3.3V devices gives you a margin of around 1.5V. Power 5 devices of 1A is 5A total, with a line AWS20 of 33ohms/1000m means a maximum length of 1.5/(5*0.033) = 9 meters maximum length. \$\endgroup\$ Oct 22 at 9:57
  • \$\begingroup\$ @pasabaporaqui Yes, I said if you need high power (like 1A) then you should be using 24V and point-of-load buck converters. For a bunch of devices that will use a couple tens of mA, like arduino, microcontrollers... or sensors that will use a couple mA or less... 5V down to 3V3 is more economical. \$\endgroup\$
    – bobflux
    Oct 22 at 12:17
  • \$\begingroup\$ 'under no circumstances should you use the counterfeit "LM2596" modules.' : could you clarify ? \$\endgroup\$ Oct 22 at 19:17
  • \$\begingroup\$ Check this answer and the links inside ; i've tested a few of these and they are definitely not LM2596, also they use super low quality capacitors that will die quickly, output ripple is huge, etc. \$\endgroup\$
    – bobflux
    Oct 22 at 19:22
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The definitive answer to this question depends on more parameters than what you specified. For example, if you can accomodate wiring your modules with 10AWG cable (5mm²), distributing 5V DC could be acceptable: for a 10m x2 (supply and ground) wire, at 1A, that will amount to 0.12V of drop, which probably won't be a problem. (voltage drop can be easily calculated: there are even some online calculators for this: e.g. https://bluerobotics.com/learn/voltage-drop-calculator)

If you want to use Cat5 cable (practical because it is cheap, and you have multiple wires, so you can use the same cable for both power and data), which is typically 24 AWG, you will obviously have a problem: the drop will be 3.28V at 1A so you'll only have 1.72V left at your device.

But if you use 12V and use DC-DC converters, at say 80% efficiency, that amounts to 520mA on the wire, and the drop will be 1.71V (power loss ~1W). This could be acceptable. If you can parallel multiple wires from the Cat5 bundle, it makes it even easier.

Now, if you distribute 12V AC through AWG24 and use bridges and linear converters, you will need 1A on the cable, the losses and drop will be greater on the cable, but you won't care because it will still be able to regulate 5V at your device side. But the total loss will be huge (7W, for a device that consumes 5W). So you will need a much bigger main supply, which may be a problem. You'll also probalby need a heatsink at each module.

So, here is the design procedure I would follow:

  • Start by deciding the kind of cable(s) you want to use for power distribution.

  • Decide on a few reasonable options for power distribution: 5V, 12V, 24V, DC-DC converters or linear regulators on the devices side, etc...

  • Then for each option:

    • deduce the current on the line, then the voltage drop, and the power losses using some online calculator.
    • Depending on the number of modules, estimate the main supply needs. You'll see that it may seem much bigger than wat you'd expect. This may have an impact on your decision.
    • Find the appropriate solution for the device-side supplies and the main supply. Keep in mind that there are integrated DC-DC solutions available around, which is much easier, and probably cheaper, than making your own (e.g. https://www.digikey.fr/product-detail/fr/cui-inc/P78E05-1000/102-5018-ND/9649654). Those aren't more difficult to use than a linear converter.
    • That's a good idea to check the total cost of solution: wires, DC-DC or linear regulators on the devices, main supply.
    • You'll also probably want to check what physical size each solution takes, both on the device and the main supply side.

At that point, you'll know what is the best option for you. Here is what you'll probably find out:

  • Distributing 5V is probably impossible, or makes the cabling too impractical.
  • 12V is probably a good choice, but using DC-DC converters. AC-DC and/or linear regulators at each module will make the solution bigger and more expensive: Each device will be bigger, and the main supply will be much bigger too.
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    \$\begingroup\$ Or if 48VDC is sent on the main cable, it can be very lightweight due to reduced current. \$\endgroup\$
    – rdtsc
    Oct 22 at 12:28
  • \$\begingroup\$ Thanks for your answer. Probably the best alternative is to use the maximum DC voltage that usual DC-DC (linear, or switching) accepts. That could mean 24V. \$\endgroup\$ Oct 22 at 15:04
  • \$\begingroup\$ @rdtsc: 48V could be bigger than usual DC-DC maximum allowed input. \$\endgroup\$ Oct 22 at 15:06
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    \$\begingroup\$ @pasabaporaqui Indeed, that is the way to minimize the power losses in the cable, and use the cheapest possible cables. At some point, however, you'll start to have trouble on the devices side to step down the voltage conveniently (DC-DC converter will be more expensive and less efficient), and you need to be careful to not exceed the voltage rating of the cable wires insulation. But 24V seems very reasonable. \$\endgroup\$
    – dim
    Oct 22 at 15:10
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    \$\begingroup\$ " for a 10m x2": note the calculator says "Enter the one-way cable length or distance", so, probably it is not need multiply length by two. In the example you say (10 meters from main supply to device), drop will be 0.06V \$\endgroup\$ Oct 22 at 19:33
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In my opinion, the most simple way to solve you problem is use 12 V common line, and use linear regulator (such is LM1117-5) before each device you need to power.

So all you need is three components before each devices (LM1117-5 and two capacitors). If you would power Arduino, most of the boards has internal linear regulator with max current ~500 mA, so for small currents you don't need any additional components.

Another advantage of this solution is electrical safety if you use a good 12 V power supply.

Disadvantage of this method it's heating linear regulators at high current, but in case you need more current you can use DC-DC converters with higher efficiency.

Voltage drop on shared wires at high current can be very big, but if you use linear regulator it's not a problem - max dropout voltage ~1.2 V, so you need a voltage more than ~6.2 V on you common line, and voltage drop on this line can reach 5.8 V, witch too much, and you can use very long lines and not too thick wires.

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  • \$\begingroup\$ Thanks for your answer. A LM117 with Vin=12V and Vout=5v will dissipate internally more power than the one provided to the load. Its efficiency will be less than 50%. \$\endgroup\$ Oct 22 at 18:37

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