I need a sanity check.

My house is wired with a Coax cable that is currently completely useless to me.

I want to use the coax wiring to send 5V DC power for small appliances (like Raspberry Pi's) throughout the house.

I want to use this PSU: https://www.amazon.ca/gp/product/B00N2RW72C/ref=ox_sc_act_title_2?smid=ASNOLMMI4SF6N&psc=1

that will feed into a DC step-down and a splitter like this: https://www.amazon.ca/gp/product/B00DIGACBU/ref=ox_sc_act_title_3?smid=A3DWYIK6Y9EEQB&psc=1

I am aware that more length = more resistance so I am wondering if I am good with slightly increasing the voltage on the source side and measuring on the output until I have an exact 5V.

It just feels like I am missing something.

EDIT: Looks like (and makes total sense) going with the 12V and dropping it at termination point is the way to go. Thank you everyone.

  • 3
    Output voltage depends on resistance (length of wire) and current. Are you sure the load current is not changing significantly ? – Warren Hill Aug 31 '16 at 11:44
  • 11
    Is it 50Ω TV antenna cable? If so watch out for splitters which have built in impedance matching resistors, and such like. They will really mess with your power distribution. – Majenko Aug 31 '16 at 12:05
  • 2
    (Typically video is 75 ohms) But definitely check to see what other devices are connected to the coax as-installed. – user2943160 Aug 31 '16 at 12:14
  • A lot of splitters blovk DC and only have DC pass on certain ports, typically to allow DC 'upstream' to a masthead amp. – D-on Aug 31 '16 at 12:21
  • Suggest measuring the port to port resistance of the splitter and then decide whether to use it or not. – scorpdaddy Aug 31 '16 at 13:30

A "better" way is to send 12 V (or more) over the coax and have local regulation to 5V at each RaPi outlet. You can use cheap buck regulators available on Ebay (a few GBP or dollars) to take the 12 V DC and efficiently convert to 5 V locally.

With 12 V being sent down the wire and with local switching buck regulators, the overall current down the coax is less than 50% of the current had you put 5 V on the line and this immediately drops less voltage and makes the whole system more viable.

You might even consider using DC-to-DC (isolating types) converters at each RaPi connection to avoid "earth" issues - they would also give a measure of protection against local (not direct) lightning strikes.

  • 11
    and with local regulation, voltage drops stop being a 'will it work?' issue, and start being an efficiency issue. – Neil_UK Aug 31 '16 at 11:54
  • @ Andy aka What happens at each "split"? Does it not also matter how many amps the device on the other end require? Let's say he has 5 raspi's in his house. According to to raspberrypi.org/documentation/hardware/raspberrypi/power/…" We have found that purchasing a 2.5A power supply...will provide you with ample power..." Theoretically he would need 2.5*5 = 12.5 Amps because each device is drawing power. Please correct me if I am wrong, but I think he will needs to worry about current. – Ted Taylor of Life Aug 31 '16 at 15:36
  • 1
    @TedTaylorofLife It's all about how to overcome the effects of volt drops by supplying a higher voltage than 5V. I used the example of 12V and, given the typical operatting power efficiency of buck regulators, if the amps needed from a 5V distribution was 12.5 amps then distributing 12V and having local buck regulation (90% power efficient) the 62.5 watts needed by the RaPis would be more like 70 watts from the 12V power source implying a current of 5.8 amps. And, for instance, if a 24V power source were used the current leaving that power source would be about 2.9 amps. – Andy aka Aug 31 '16 at 17:09
  • @TedTaylorofLife You're confusing the output current of the DC-to-DC converter (what it supplies to the device) with the input current of the DC-to-DC converter (what it draws from its supply). Read the second paragraph again. – David Schwartz Aug 31 '16 at 17:09
  • @TedTaylorofLife Having said all of that we know nothing about the loop resistance of the coax or how long it is thus, my answer is couched in somewhat general terms in order to answer the op's point that he just feels like he is missing something. It's also worth mentioning that the actual amps requirement for a RaPi is probably under 2A - see this: raspberrypi.org/help/faqs/#powerReqs – Andy aka Aug 31 '16 at 17:10

The splitter in your link is for distributing RF cable TV signals - it will seriously attenuate your DC power, if it passes DC at all.

If you are only using the coax to distribute DC (or low voltage AC), you can join several coax cables as if they were simple two-conductor cables. There is no need to worry about impedance matching and other RF complications that your suggested TV splitter deals with.

I am aware that more length = more resistance so I am wondering if I am good with slightly increasing the voltage on the source side and measuring on the output until I have an exact 5V.


(Did I say "no"?)

Current draw is very not constant with electronics. In normal running, you might have enough current to get exactly 5V, dropping 1V on the cable. At startup when it takes more current, you might be up to 2V drop on the cable so the supplied voltage will be 4V and your kit browns out. When you change the code so it goes into sleep mode, the 1V drop might go to virtually nothing, so your electronics will see a full 6V and be fried (i.e. permanently dead).

If you want to do this, I suggest using a higher voltage such as 12V for your DC supply. Each Raspberry Pi or other device must then contain its own 5V regulator. The 12V supply will wander up and down as each device draws more or less current and voltage is dropped in the cables, but you should still have more than 7V which will give you a solid supply for a 5V regulator to run on.

You also need smoothing capacitors to protect against brown-outs. And reverse-polarity protection to stop the 12V supply trying to suck charge back through the 5V regulators. A good tip is to use a regulator like an LM2940 which has this kind of protection already built in, instead of a more basic regulator like a 7805.

  • Don't even use a linear regulator -- they pretty much burn off the extra power as heat. Using a buck converter (and filtering capacitors to ensure the buck output isn't noisy) is much more efficient, especially when you're already facing the line losses. Also, what do you mean about "sucking charge back through the 5V regulators"? I've never heard of such a thing. – Doktor J Sep 1 '16 at 13:19
  • @DoktorJ Sure, SMPSUs are more efficient. They can be easier to get wrong for beginners though, where linear regulators are more fire-and-forget. If you've only got low currents and you don't care too much about efficiency (remember, these are plugging into a wall socket so it's not like we need to conserve battery life), and particularly if you're not so experienced and you're using breadboard/stripboard for your circuits, then KISS. – Graham Sep 1 '16 at 14:37
  • @DoktorJ You have a 10uF smoothing cap (or something large-ish) on the 5V reg output. Now suppose the 12V input is shorted to 0V (or something below 5V anyway). The 10uF cap will then discharge backwards through the 5V reg if it's not protected. Or simply connect two circuits to the same supply and power off the supply. The circuit which draws most current will discharge its smoothing capacitor first, and will then suck charge from the other circuit via the reg. An input diode protects against this (and supply polarity), but it can be easier to just buy a reg with protection build in. – Graham Sep 1 '16 at 14:51
  • @DoktorJ Realised I missed one point there. If discharging the smoothing cap backwards through the reg was harmless, then no worries. In fact though it's not at all good for the reg (because the peak current can potentially be high if the supply is shorted) so it's definitely worth avoiding. – Graham Sep 1 '16 at 14:55

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