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It’s early so I may not have the following details exactly correct:
Cable length = 20 feet
Cable is a loop so total cable length = cable length * 2 = 40 feet
Cable resistance = cable resistance per foot * 40
Voltage drop = cable current * cable resistance (E=IR)
Cable current = 1A
Therefore:
Voltage drop = 1 * total cable resistance
Copper cable resistance is dependant only on the wire gauge.
This table gives the ohms per length versus gauge.
Don’t have time to look into to it fully now (I’m just a visitor here not an employee), but I think you’ll need cable that is 19AWG or larger (e.g. 18AWG) to get < 400 milliohms @ 20 feet
I’ve spent some of my personal time to try to help and provided way more information than you should have needed if you work in electronics or even were an electrician.
Perhaps when DigiKey employees show up they will help further but I’ve done all I’m willing to do for free at this point.
The resistance per 1000 ft you’re looking to target is 20 Ohms or less, which isn’t difficult to achieve, looking at random 24AWG copper is 16 Ohms per 1000ft.
Since I’ve had it memorized for my work of 45+ years, I’m pretty sure 22AWG cable is ~16 ohms/1000 feet. We sell 500 foot 22 AWG optional sensor cables for our analog instruments and custom calibrate them with 16 ohms in series (500’ in a loop = 1000’)
Charts I have say 24 AWG wire is 25.67 milliohms/foot = 25.67 ohms / 1000 feet one way, 51.34 ohms for a 1000 foot loop.
The OP asked for 0.4 ohms maximum for 20 feet one way.
Maximum loop = 40 feet so they want <0.4 ohms for 40 feet. 0.4/40 = 0.01 ohm/foot.
So they want < 10 milliohm/foot
20AWG = 10.15milliohm/foot which is just barely too high (0.406 ohms)
The smallest standard gauge that works is 19 AWG = 8.051 milliohm/foot.