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how to: 2 270 H.O. alts on 1 battery bank ?


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HHR Ed, The mist is beginning to clear lol. So what i think ive learned is watts needed by electrical should be equal or less then watts made by batts and to achevie the 1:1 ratio the alt needs to produce equal or greater amps than the amp hrs of the batts... Taking the alts. efficiency of 80% into consideration. So assuming the correct requirement is 6375w, one 300 amp dc power alt, one d3400 under the hood and one d3400 in the back would give me a 1:1 or better ratio and alow me to run without discharging the batts and therefore causing a drop in voltage, correct?

Also the thing i didnt understand is you said...."Taking the efficiency of maybe 80%, then you need to produce 4375w to get the amp to full power." ... how would this be achieved by running lower than 1 ohm?

Btw thanx again for the guidance.

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So what? Im trying to go down the road of doing things by the rules, now if the moderators allow personal attacks, then im not responsible.

Nothing that has been said here is a personal attack... It was stated that you talk out your ass that is not a personal attack...

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HHR Ed, The mist is beginning to clear lol. So what i think ive learned is watts needed by electrical should be equal or less then watts made by batts and to achevie the 1:1 ratio the alt needs to produce equal or greater amps than the amp hrs of the batts... Taking the alts. efficiency of 80% into consideration. So assuming the correct requirement is 6375w, one 300 amp dc power alt, one d3400 under the hood and one d3400 in the back would give me a 1:1 or better ratio and alow me to run without discharging the batts and therefore causing a drop in voltage, correct?

Also the thing i didnt understand is you said...."Taking the efficiency of maybe 80%, then you need to produce 4375w to get the amp to full power." ... how would this be achieved by running lower than 1 ohm?

Btw thanx again for the guidance.

Amplifier efficiency in never a certain number and it varies depending on many factors. So look at the method I used to figure out what you needed. Then research you brand amp to find a realistic efficiency rating for it. The lower the ohm load, the less the efficiency. So more overhead is always suggested. And using a low efficiency number will give you that overhead.

So a 3500w amp that is 80% efficient at 1 ohm would consume 4375w from the electrical system to be able to put that 3500w to the woofers.

That may be an ideal estimate. If you go below 1 ohm and for example, have 4000w at 0.5ohms at 60% efficiency, then it will be consuming 6666.666 watts from the electrical.

So it depends on your amp. Just take the power it is rated or calculated to have (3500w) and divide by the efficiency percentage (80% = 0.80) so you have 3500/0.8 = 4375

So you must know the efficiency rating or try and calculate on the low end.

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Thanks a million HHR ED. I'm pretty sure Ive got it down. I got mixed up about the amp needing 4375w to produce 3500w but I understand now. the only question I have left now is I want to make sure I have the basics down so would this statement be correct....... so what I think I've learned is it the watts needed by the electrical should be equal or less then the watts produced by the batteries and that to achieve the 1 to 1 ratio or better the alt needs to produce equel or greater amps than the total amp hours of the batteries...

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Thanks a million HHR ED. I'm pretty sure Ive got it down. I got mixed up about the amp needing 4375w to produce 3500w but I understand now. the only question I have left now is I want to make sure I have the basics down so would this statement be correct....... so what I think I've learned is it the watts needed by the electrical should be equal or less then the watts produced by the batteries and that to achieve the 1 to 1 ratio or better the alt needs to produce equel or greater amps than the total amp hours of the batteries...

I would word it more like this.

When designing the electrical system of a car audio installation

-The wattage output capabilities of the batteries must exceed the wattage output of all amplifiers combined. (by 20% or more depending on amplifier efficiency)

-The output of the alternator must also exceed the amp hours of all batteries combined by 20% or more in order to achieve a charge time ratio of 1:1 (1 minute of charge for every 1 minute of full power play)

Ed Lester

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Showtime Electronics Video Marketing

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http://www.stevemead...08/#entry511451

http://www.youtube.com/showtimespl



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Last ride 2007 HHR, current dB 153.5 and bass race 149.4 dB. 153.0 dB on music

New Ride, 2008 HHR SS. Build under way.
Loudest score ever = 171dB
2009 dB Drag Racing, North American Points Champion

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Awesome. thank you for saving me from myself lol. Hopefully someone saves this info, three systems to date and you would think i would have learned this rule of thumb by now it will save plenty-o-headaches for people im sure.

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  • 3 months later...

Does any one no if 2 C&D technologies ups 12-490mr will do me any good on a 220 alt and a 4000wt amp (American Bass 400.1). I guess what i realy want to no is what is the W on this battery per amp hr?

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I will state this fact, because I can prove it in my truck and was told the same thing before hand by MECHMAN, so of course I had to test it, they were right.

I run 4 alts, all same all connected to front battery(s) and then to rear back all equal lengths of cable and same power and grounds per alt.

2 external regs, 1 for each pair, I can turn 2 alts off, run either 2 or 4, nice feature. BUT

If I set the two external regs at different voltage, even a half of a volt (.5) what do you think happens?

Example one pair is running at 14.5 and on, second pair is set to 15.0 and off, what happens once I turn second pair on to where all 4 are running?

Food for thought....

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