Re: [CAD_CAM_EDM_DRO] wire for motor ?

John Dammeyer wrote:

>Hi Dave,
>
>If it were 28g solid wire then it has about 68 Ohms resistance per thousand
>feet. If you have a six foot cable then the resistance is 0.4 ohms. Your
>3V 3A motor has a DC resistance of 1 Ohm. Total resistance is therefore
>about 1.4 ohms.
>
>

Unfortunately, there's a common mistake in there. Wire resistance specs
are per conductor, and current flows through two conductors in a
properly functioning system. :)
The total resistance from the wire would actually be about 0.8 ohms per
motor phase, split into two "resistors" per phase. Assuming that the
driver can supply 3A, the power dissipated in the cable will be as high
as 2 (phases) * 3A^2 * 0.8 ohms (per phase) = 14.4W. The motor will be
using 18W, since it's 1 ohm per phase, and the driver will have to
supply nearly 6 V to get the motor to full current.

>Since your driver measures the current through the wire and the motor and
>will be set up to provide 3A it will result in 3.6W of power dissipated
>across the wire.
>
>If you made up a cable using 18g wire it's 6.3 ohms per 1000 foot. Or 0.0063
>per foot. Power dissipated across the wire is now 0.06W.
>
>John Dammeyer
>
>

I have to agree that heavier gauge wire is better. Additionally, the
AWG specs you see are usually for "plain" wire, not something that's
molded in several layers of rubber along with several other
current-carrying conductors. You need to significantly derate wire when
you expect it to get hot (which computer cables certainly are not meant
for), and when you expect it to be bundled together.

I almost never remember wire gauges or resistances, so I google for "AWG
ampacity" (I added the word "bundle" this time), and look at the charts
when I need to :)

- Steve