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Science Vs Fiction as it Applies to Charging System Design

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Lately, I've seen lots of video and photos of 10, 20, and 30,000 "watt" builds.  While 30 runs of "oversized" 1/0 AWG power and ground cables may look the part, it's absolutely not necessary.

First and foremost, the Laws of Physics are in fact Laws, not just really good ideas.  They work in your favor as well as against it.

Let's say that we design a system capable of 30,000 Watts RMS.  As in - they system is cable of delivering 30kW of unclipped power to our speakers.  The source of power is infinite (no matter how much juice we need, it can maintain 13.8 Volts - wouldn't that be nice?)  Show me:

1 - the math on determining just how much current it would require to play music at 30kW

2 - the math to determine how many runs of 1/0 AWG copper cable it would require with the source of power located 15 foot away

If you don't know how to answer, then your time will be far better served by learning how than comparing prices on cable for your next build.

First person to get both right gets a FREE TORK-2 shipped to your front door (US only).  This isn't designed to be a freebie.  You'll have to earn it.

Tony Candela - SMD Sales & Marketing
Email me at [email protected] to learn about becoming an SMD Partner!


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I'm probably wrong because the conversion from DC to AC, but just for shiggles and gits:

1. I = P/V (30,000W ÷ 13.8VDC = 2,174A)

2. 1/0awg is should be able to carry 300A at 15ft, -10% for voltage drop puts you at 270, so 2174 ÷ 270 = 8 runs of 1/0awg.


That's the most basic answer I thought of. I'll have to re-read your books and I'm sure the actual math/answers are in there. 

Edited by reedal
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Let me take a stab at this. I like these fun little challenges

I write this with a couple assumptions:
1. The amplifiers used are likely class D with an efficiency of 70%
2. The wire used is standard OFC 0 gauge. No over sizing.
3. Wires used adhere to AWG standards, which as per the chart, will be spec's to have an equivalent cross sectional area to the solid copper
4. With above, the wire used will have a resistance of 0.00009827 ohms/foot
5. Wiring for this scenario requires use of a ground path. As frame/chassis of vehicles have a large variance, I will assume that a ground will also need to be wired, giving a total circuit length of 30 feet.
6. Due to variability in termination techniques and quality, all connections will be treated as zero resistance. This includes fusing.
7. Wiring used will be assumed in a chassis configuration - i.e. wired individually rather than an insulated bundle.
8. The able above lists current capacities for wires. Those will not be used. The code lists for an AC system, and there are some difference in how voltage drop works in AC vs DC (read up on power factor if you want to know more). Instead, I'll be referencing an ampacity chart provided by lincoln electric which establishes that 0ga <50ft at a 60% duty cycle(pretty close to music eh?) to be 350a.
9. While to my knowledge, there are no published tolerances for a 12v electrical system, I am opting to use electrical code NEC 210.19, which states a maximum allowable voltage drop for feeders+ branch circuits to be no more than 5%. Interpreted as: Total drop for the circuit can be no more than 5%. So I'll use that.
10. I am interpreting playing music at 30kw as a 50% load cycle, with an average draw that needs to be supplied as 15kw. Music is dynamic. It has periods where it's loud, pauses, kicks, varying bass lines, and varying frequencies. It's a rhythm. You're not playing a sine wave, which would be full output at all times. i.e. a burp scenario. Now, of course, some songs will be different for others. A country song might be closer to a 10-20% duty cycle, while some bass boosted edit thing might be closer to 70%. But as an overall average, I'm treating it as 50%.
11. For all intents and purposes, we're treating the 30kw/15kw number as regulated power, I'm not going to get into differences in current draw in response to voltage droop across lines. It would make this long and complicated.

30kw power at 70% efficiency requires 42,857w of input power
15kw of power at the same requires 21,429w of input power

13.8v at 5% drop is 13.11v, and 0.69v drop

So for 30kw as our maximum, 42,857w/13.11v= 3,269a
And for 15kw, 21,429w/13.11v=1,635a

Now for the wiring:
0 gauge at 30ft gives 0.00009827ohm/ft * 30ft = 0.0029481ohms/wire

Target resistance:
30kw: 0.69v/3,269a = 0.000211ohms
15kw: 0.69v/1,635a = 0.000422 ohms

Coupling that with our wire, which is simplified by X=wire/target, where X is runs of wire, target is our allowable resistance, and wire is the resistance/wire
30kw: 0.0029481/0.000211 = 13.97, or rounded to 14 runs
15kw: 0.0029481/0.000422 = 6.98, or rounded to 7 runs.

14 runs yields 3269a/14 = 234a/wire. Below the 350a limit set above
7 runs yields the same, below the 350a limit set above.

Bearing in mind the (numerous) assumptions above, such a system could be handled with 7 runs on music(50%), or 14 runs at burp(full power)

Edit: Added note on duty cycle for music. See #10

Edited by SnowDrifter
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2 hours ago, strangeduck said:

you mind letting us know what type of amps we are using? i'd like some efficiency numbers. 

Your math can take into account either or both ...

Tony Candela - SMD Sales & Marketing
Email me at [email protected] to learn about becoming an SMD Partner!


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I’m not even going to try, but I’m definitely tuned in ...

Kenwood / HELIX / Linear Power (For The Love Of Music) / Brutal Sounds / OverKill Electric Co 

Questions About Sound Quality ?? Try Here ... Sound Quality, What does it REALLY mean ?? 

SMD SOTM Winner "White Lightning" 1997 GMT400 Chevy Silverado   

"The Green Dickle" 1994 GMT400 Chevy "Phantom Dually"   

Randal's 2007 Chevy Avalanche (we haven't named this one yet)

Dylan's "Brutal" 17 Chevy Cruze RS Hatch                         

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