I just made a circuit consisting of a number of digital components including two microcontrollers and I will be adding a home-made radio unit to it.

I will make a few other miniature circuits which are mostly lights that connect to the main circuit through cables.

Due to space and my design limitations, I have considered eliminating some parts of my circuit to allow the battery to barely fit, but that is not possible, so now, my only option is to run two 3 foot wires just for batteries.

But in terms of everything, both analog and digital, I feel I need to add other components to make the circuit function with the battery source connected so far away.

So far, on the main circuit board where the battery connections are, I connected the connections together through a 10uF capacitor and I plan to connect another capacitor right next to the battery (so that I have a 10uF at each end of the 3 foot wire pair).

Is there anything else I should take into consideration when running power (+5VDC and ground) through long wires? These long wires will deliver power to a radio unit and the digital components and I don't want them to suddenly stop working, just because I increased the length of the power wires.

And also, what is the maximum AWG (thinnest wire) I'm allowed to use to make the circuit run efficiently?

  • \$\begingroup\$ Why do you think having long wires will make the unit "suddenly stop working"? The effect of long wires is adding resistance and adding inductance. What current are you looking at? \$\endgroup\$
    – Puffafish
    Dec 2, 2016 at 16:06
  • \$\begingroup\$ 2m of wire is not usually considered "long". We don't know the resistance/impedance of the wires, or the current you need, or the shape and frequency of current consumption, but for a few 100mA with short peaks of a few mA the 10µF in the device should be plenty. \$\endgroup\$
    – JimmyB
    Dec 2, 2016 at 17:05
  • \$\begingroup\$ As to the AWG vs. efficiency: See en.wikipedia.org/wiki/… . It has figures for resistance per meter. Figure out what max. current you need and what maximum voltage drop you can accept, then choose any wire with less than that max. resistance per 2m. Example: If you need max. 100mA transferred with max. 0.5% loss of power, that'd be (0.5%*5V)/0.1A = 0.25Ohm, so for 2m total you could get away with AWG25 (about 0.1Ohm/m) or better. \$\endgroup\$
    – JimmyB
    Dec 2, 2016 at 17:12
  • \$\begingroup\$ For the most part, I wouldn't need more than one amp. Worst case scenario will be a small vibrator motor running for about one second. After that, all I'm driving is an LCD display. a ISD1700 sound chip and some LED's with series resistors of about 200 ohms. \$\endgroup\$
    – user116345
    Dec 2, 2016 at 18:33
  • \$\begingroup\$ I just left a similar comment on another post: why not regulate the voltage at the device, instead of sending regulated voltage across the wires? This helps to eliminate issues of voltage sag due to added resistance of long wires, and helps guarantee a constant voltage at the device. It can be easily accomplished with a 7805 or buck converter, depending on the current requirements. Also, this gives you added flexibility to power the device with a wide input voltage range. \$\endgroup\$ Jan 2, 2017 at 2:52

2 Answers 2


The issues with long wires is added inductance and added resistance.

First off, resistance: easier to work out the effect of this. The datasheet for the wire will tell you the resistance per metre (or at least the information to work it out). You know the current draw of your system. Applying V=IR will tell you the volt drop seen at your system end, choose a wire where this drop is low enough for you. Don't forget you'll also have the wires getting hotter if it is a big drop, so will need to factor that in.

Secondly, the added inductance: this is only an issue if you have a large switching load drawing gulps of current. In this case, it is probably best for you to simulate your system (using LTSpice or similar) and use that to see how the difference inductances of different wires will effect the power available to your system. You may find this site useful to see the inductive value of your wires.

(All this assumes that you are doing this system at home for fun, and not making a saleable product with all the regulations etc that is required for one of them).


I've found the concept of "local battery" to be the solution, to avoid transient problems. By installing resistors (often a bad idea) or Ferrite Beads in Power Leads, and placing substantial capacitors right by the circuits demanding fast-changes in current, these filters [L&C] prevent the transient charges being supplied from distant capacitors/supplies/batteries thru long wires. And the EMI is low, because fast edges are not permitted in the antennas of long wires.

Over the years, I've had many circuits oscillate, for various reasons. The "local battery", with resistors or Inductors/beads in the wiring, has been a cure for many problems.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.