Current Sensors

[Richard Wallace]

10 AWG is about one milliOhm per foot, while 12 AWG is about 1.6 milliOhm per foot.

Let's consider a hypothetical robot motor circuit using 2 ft. of red and 2 ft. of black to connect a Victor with a CIM motor. The CIM's maximum power point (per its 2005 data sheet) is 337 Watts and at this power its specified current draw is 67.9 Ampere. At that draw, 12 AWG wiring would dissipate 4 x 0.0006 x 67.9^2 = 11 Watts more than 10 AWG wiring. As a fraction of the 337 Watt motor output that's about 3.3%. Seen in terms of battery current, that's about one Ampere more (11W / 12V = 0.92A).

Of course your robot won't load the CIM to its maximum power point very often -- at least we all hope that it won't, because that would indicate a serious design problem that has nothing to do with wiring. However, on a intermittent basis the loading could be even higher, since the CIM's stall current is 133 Amperes -- at that loading the extra dissipation and extra battery current draw that could be avoided by using 10 AWG instead of 12 AWG would be about four times what was calculated in the example above.

Now let's consider the weight difference. 10 AWG is 0.0314 lb/ft, while 12 AWG is 0.0198 lb/ft. So the four feet used in this hypothetical circuit would weigh an extra 0.05 lb if 10 AWG is used instead of 12 AWG.

Does the difference in current draw and wasted power seem small? Is it worth the increase in weight to avoid it? Engineering is full of decisions like this one.