Can (4) 12 gauge wires in parallel handle 320A at 48 volts on a BMS?

Rider

Photon Sorcerer
So, if those 4 - 12 gauge wires are kept very short, you will have very little voltage drop and very little power loss.
This may be true, but I sure as hell would want to bench test it before trying it for real. 2 #12 wires, 1 foot long, and something to pull an increasing amount of current. Find the point when the wires exceed safe temps.
 

KenDan

Solar Enthusiast
The temperature rise on very short wires can also be drastically reduced by heat sinking effects of that they are connected to. That is why I cut mine back to about 2 inches. The 4 wires were originally about 6 inches long on each side.
 

Zwy

Solar Addict
If it were possible to make 4 #12 wires carry 320A, then all of our discussions on getting power from the solar panels to the SCC are rendered moot and if that simple, why didn't we do this before?
You forget the length of the wire affects resistance. I have yet to see a charge controller 6 inches from each solar panel.
 

KenDan

Solar Enthusiast
You forget the length of the wire affects resistance. I have yet to see a charge controller 6 inches from each solar panel.
Yes, this is precisely why you need heavy wires for long runs - such as between panels and the charge controller.
The wires on the BMS in question in this thread can be much lighter if they are kept very short.
 

Obie1Kanobe

New Member
So much misinformation here!

Ampacity and voltage drop are two DIFFERENT issues! Ampacity is the critical one here.

Ampacity is defined as the maximum current, in amperes, that a conductor can carry continuously under the conditions of use without exceeding its temperature rating. Voltage drop is self-explanatory.

See the ampacity tables at: Cerrowire.com 12 gauge wire is rated around 20 amps
 
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Supervstech

Administrator
Staff member
Moderator
So much misinformation here!

Ampacity and voltage drop are two DIFFERENT issues! Ampacity is the critical one here.

Ampacity is defined as the maximum current, in amperes, that a conductor can carry continuously under the conditions of use without exceeding its temperature rating. Voltage drop is self-explanatory.

See the ampacity tables at: Cerrowire.com 12 gauge wire is rated around 20 amps
Yes, but ampacity is for a conductor run, and designed for ac power.
It allows any length conductor, and uses derating for conditions such as number of insulation layers, pathway shielding, number of conductors in a pathway etc.
It is an insurance protection metric.

when calculating short run, or specific runs, ampacity charts don’t really have any use. What matters is voltage drop, and ALLOWED voltage drop.

short runs under a meter in length can be used to calculate at voltage drop and use acceptable loss percentage for the result.
Most consider 10% loss extreme, 5% a weak acceptable figure, 3% ok for most uses, and 1% or less as the ideal.
If you size a cable at specified length in a calculator program, and specify 1% voltage loss... the wire size will be a safe figure to use. And #12 can handle 30 amps at 10%, 25 Amps at 3%, and 20 Amps at 1%.
So, x4 for the capacity, and you have a maximum safe load of 120A... with a voltage loss of less than 2 volts... around 1.2V
Shorter runs equal less loss, and more amps capability, but in no way Will they handle 320A for more than the time it takes the solder on the board to vaporize... likely around 30 seconds...
 

Gazoo

Dumb Dumb
The BMS has two pairs of 12awg wire. The effective awg of each pair is 9awg. I used this voltage drop calculator to calculate the voltage drop over a 1 foot length. I used a couple of others as well. Is it wrong?

 

Maast

Compulsive Tinkerer
The BMS has two pairs of 12awg wire. The effective awg of each pair is 9awg. I used this voltage drop calculator to calculate the voltage drop over a 1 foot length. I used a couple of others as well. Is it wrong?

That calculator is strictly for voltage drop. It doesnt address heat generation at all.
 

Maast

Compulsive Tinkerer
True.
But heat isn't generated if you stick to the 1% or 3% scale...
I'm afraid I have to disagree, at 3% voltage drop of a 320a current you're looking at 460 watts of heat and I doubt a meter length of insulated wire could cool itself enough to avoid melting the insulation. Plus it's going to cook whatever its attached to.
 

Supervstech

Administrator
Staff member
Moderator
I'm afraid I have to disagree, at 3% voltage drop of a 320a current you're looking at 460 watts of heat and I doubt a meter length of insulated wire could cool itself enough to avoid melting the insulation. Plus it's going to cook whatever its attached to.
Really? 460W! Wow, I can't imagine it would be that much.
How did you calculate the wattage?
 

Maast

Compulsive Tinkerer
Really? 460W! Wow, I can't imagine it would be that much.
How did you calculate the wattage?
I did it two ways: The easy way: 3% of 48v = 1.44v times 320A which gave me 460W

Then I did it the hard way: 1 meter of 9AWG @ 48v and 320a results in .00518 ohms resistance and a voltage drop of 1.66v (3.4% drop)
Plug that resistance into a watts calculator @ 320A and you get 532.5 watts. Technically its 532.5 Joules but since the definition of a joule is 1 watt per second I just converted it.
 

Supervstech

Administrator
Staff member
Moderator
I did it two ways: The easy way: 3% of 48v = 1.44v times 320A which gave me 460W

Then I did it the hard way: 1 meter of 9AWG @ 48v and 320a results in .00518 ohms resistance and a voltage drop of 1.66v (3.4% drop)
Plug that resistance into a watts calculator @ 320A and you get 532.5 watts. Technically its 532.5 Joules but since the definition of a joule is 1 watt per second I just converted it.
Whoa! I didn’t see anybody say 320A was ok on 9AWG... I said 3% would be 40A... and no way would 320A be ok... I see you are just using another method to show why 320A is outrageous and weren’t really disagreeing with me.
 

Supervstech

Administrator
Staff member
Moderator
I did it two ways: The easy way: 3% of 48v = 1.44v times 320A which gave me 460W

Then I did it the hard way: 1 meter of 9AWG @ 48v and 320a results in .00518 ohms resistance and a voltage drop of 1.66v (3.4% drop)
Plug that resistance into a watts calculator @ 320A and you get 532.5 watts. Technically its 532.5 Joules but since the definition of a joule is 1 watt per second I just converted it.
Ok, I dug in a bit further... yeah, that calculator is crazy! I cannot fathom 1 foot of 9AWG loaded with 48V at 320A would have a voltage drop of 1.06%...

I meant with my statement that using a conductor chart allowing a 3% loss, showing max amps on a conductor is safe... at 40A... no freaking way can it handle 320A...

That calculator seems to think it would... odd.
Now, I gotta get some #12 and a 48v loaded to 320A... yikes... that is a 15,000W load... ok... maybe drop that to 3V... um... 320A... uhhh... 1000W umm... nope, I’m not trying that test... it would be fun to video I bet...


SSSSSSSSSMOKIN!
 
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