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Getting the correct wire guage for 12V/5A LED extension wires

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Siwastaja:

--- Quote from: DW1961 on July 17, 2020, 07:00:20 am ---So if the voltage drop is the same for any wire, why use 24v over 12v?

--- End quote ---

Because given the same current, at 24V, a drop of 1V is just 4.2% of the supply, when it is double that at 12V. This alone means doubling the voltage halves the losses.

Add to this that doubling the voltage allows halving the current for same power transfer (assuming the load is re-designed to do what it is doing at this higher voltage). This halves the voltage drop.

Add the two effects: the relative voltage drop, and also the power loss, is one quarter!

Increasing voltage pays off quickly.

DW1961:
Also reply to @Siwastaja


--- Quote from: blueskull on July 17, 2020, 07:09:59 am ---
--- Quote from: DW1961 on July 17, 2020, 07:00:20 am ---So if the voltage drop is the same for any wire, why use 24v over 12v?

--- End quote ---

Because to carry the same power, you need half the current at 24V than 12V, so your IR drop will be halved.

--- End quote ---

When I input the metrics into this calculator, nothing changed except voltage percent of total volts. Voltage drop in volts stays the same.
https://www.rapidtables.com/calc/wire/voltage-drop-calculator.html

So I'm confused. :)

DW1961:

--- Quote from: Siwastaja on July 17, 2020, 07:08:19 am ---
--- Quote from: DW1961 on July 17, 2020, 07:00:20 am ---So if the voltage drop is the same for any wire, why use 24v over 12v?

--- End quote ---

assuming the load is re-designed to do what it is doing at this higher voltage).

--- End quote ---

I think this is what I was missing?

Siwastaja:
Well let's give an easy numerical example.

#1, let's assume you have a magical constant current load drawing 1A.

Input voltage = 10V
Wire resistance =  1 ohm
Load current = 1A

Voltage drop on wire = 1 ohm * 1A = 1V
Input power: 10V*1A = 10W
Power at load: 9V*1A = 9W
Power loss in wire = 1V*1A = 1W
Efficiency of transfer = 9W/10W = 90%

Then up the voltage to 20V. Still,
Wire resistance = 1 ohm
Load current = 1A

Voltage drop on wire = 1 ohm * 1A = 1V
Input power: 20V*1A = 20W
Power at load: 19V*1A = 19W
Power loss in wire = 1W
Efficiency of transfer = 19W/20W = 95%

So already better, you are transferring twice the power with the same loss!

But, in reality, if you used a constant power load... For example, you used a wide input voltage switch mode converter. Or, if you used a simple resistive load, like a light bulb or a heating element, you would just change the load to one designed for the higher voltage, lower current, same power.

Let's recalculate the 20V case:

Load current = 0.5A
Voltage drop on wire = 1ohm * 0.5A = 0.5V
Input power = 20V*0.5A = 10W
Power at load = 19.5V*0.5A = 9.75W
Power loss in wire = 0.5V * 0.5A = 0.25W
Efficiency = 9.75W/10W = 97.5%

So by doubling the voltage, you dropped the wiring loss to one quarter.

Datman:
Here, in Italy, I would use a classic 3x2,5mmq (13AWG), with 2 of the three wires tied together:
14' / 2,5mmq + 14' / 5mmq = >> 35+17=52mOhm; 5A x 52mOhm = 260mV.

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