Voltage Drop Calculation

[puppycrack]

I am trying to calculate voltage drop given a wire gauge, wire length and ampere current. I've found several calculators, but they seem to be giving different results. For example, assume:
One-way wire length: 30'
Amps: 3A
Wire: 16AWG

http://wiresizecalculator.net/calcul...oltagedrop.htm
Voltage Drop:0.9V

https://www.calculator.net/voltage-d...feet&amperes=3
Voltage Drop: 0.72V

It seems the first link is using the formula:
2 * K * L * I / Cm
K = ohms-cmil per ft - 12.9 for copper
I = current (or amperes) of load
L = length of conductor in ft. (one-way)
cmil = circular mil area of the conductor

https://www.rpc.com.au/pdf/Wire_Chart.pdf


Whereas the second is using: 2 * L * I * R / A
L = one-way length
I = amps
R = resistance of cable - 0.017 for copper
A = cross sectional area of cable in (mm²)

http://www.adamselectric.coop/wp-con...ltage-Drop.pdf


What formula is correct in the context of pixels and this hobby?

Thanks,
- pc

[DrNeutron]

Go here: http://spikerlights.com/calcpower.aspx
Works a treat and helps a ton to get a rough estimate to help plan out display.

[P. Short]

Without having looked into it in detail, I suspect that what you noticed may be due to the difference between using 60 Hz AC RMS voltages (and RMS currents) vs DC voltages.

[steve123]

You can obtain the approximate resistance per foot (meter) of AWG copper wire on the wikipedia page here: https://en.wikipedia.org/wiki/American_wire_gauge

V = I * R

Remember, your total conductor length is the length of the V+ wire plus the length of the GND wire. That is why there is a "2" multiplier in those examples you provided.

[P. Short]

Without having looked into it in detail, I suspect that what you noticed may be due to the difference between using 60 Hz AC RMS voltages (and RMS currents) vs DC voltages.

I was thinking of skin effect, which is negligible for wire of that diameter. Instead, if you look at the text for the first calculator mentioned, it's for a temperature of 75C. The resistance of copper increases with temperature, which accounts for the difference in voltage drop between the two calculators. From what I could find on that most wonderful of sources, the internet, the resistance of copper increases by 0.393% per degree C. The temperature difference is 55C, so the resistance increases by a factor of (1.00393^55) = 1.24, and 0.72V * 1.24 = 0.89V (approximately the numbers given by the two calculators).

[puppycrack]

Thank you Phil - I didn't notice that distinction. I also now see the two formulas are really the same, albeit one is imperial and the other metric.

Happy that I'll have less voltage drop to worry about in upstate NY...