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Wire size (200A)

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I would just like to confirm wire sizing for my battery.  I am used to aviation specs where the minimum wire size for 200A would be 110mm^2...  But I see the inverter manual says 55mm^2 and the SANS spec seems to say this is OK. But you are still looking at 50W loss into the cable at max draw.

55mm^2 is obviously a lot easier (readily available cable and lugs) - and obviously guaranteed to fit in the inverter cable glands.

What is the general feeling about battery cable sizes?

On 2020/08/13 at 12:39 PM, JustinSchoeman said:

Thanks.  May be best to wait for the inverter, and see what the max size cable is I can fit through the glands and/or fit on the terminals...  55 really does seem a little small.

Which inverter is it?  Also 200amps sounds like max draw...  For a 5kW unit @ 48volts, 50mm should be ok, 70mm ideal if you spend a lot of time at high power.

On 2020/08/13 at 9:51 AM, JustinSchoeman said:

What is the general feeling about battery cable sizes?

Ideally you want around a 1% voltage drop, but iirc the recommendation is <3%.

So on a 50V battery, that's 0.5V. Generally the cable runs are fairly short from the battery to the inverter. If we assume 10 meters total length (5 meters on either side). 50mm^2 cable has a resistance of 0.35Ω/km, so for 10 meters that's 0.0035Ω. 0.5/0.0035 = 142A. That's for ten meters, which I deliberately chose to be on the long side. Put the inverter closer, and 50mm^2 should be good for 200A or 10kW.

If you need to go bigger than that, then use 35mm^2 and use two cables (double up).

  • Author

Total cable length (positive and negative together) is around 2m. So I suppose 50mm^2 is actually more than sufficient.  Although I would still want to use 95mm^2 for joining the two layers of the battery pack (to reduce voltage errors to the balancer).  But I suppose it makes sense to take the end-plates off and add extra terminal bolts and do 2x 50mm^2 - that way I get half the contact resistance too...

12 hours ago, plonkster said:

50mm^2 cable has a resistance of 0.35Ω/km

I think this information (and how to get it) may be useful for some other people, so I thought that this is as good a place as any to teach others how to find the resistance of their cables.

A 1mm^2 copper wire with a length of 1m has a resistance of 17mΩ (i.e. 0.017Ω). This is the only number you need to remember.

For any cable, of any cross-sectional area and length, you simply divide the 17mΩ by your cable's area and multiply the result by the length in m.

So to use @plonkster's example above for 10m of 50mm^2 cable. (0.017Ω / 50mm^2) * 10m = 0.0034Ω

Just now, JustinSchoeman said:

So much easier with aviation wiring... Take the fuse rating, look it up in the table, and get the minimum wire diameter. No room for choice/acceptable losses/etc.

I disagree. The 17mΩ rule is much easier to remember . Of course tables are available for this sort of thing too, just google for it. There are tables for both copper and aluminium. This is usually what I do, I just google it. Which is how come I'm 0.01Ω off on my anwer above.

@Stanley's rule is an awesome little tool that is definitely going into my bag of tricks, along with the 345 rule (Pythagoras) and my Afrikaans Sints/Cosas/Tanta trick to remember how the trig functions work 🙂

 

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