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Posted by: Jeffs9146 Mar 7 2008, 07:17 PM

I need to know the max load that an amp will put on a battery when turned up and all speakers loaded!

I know there are some very smart people here who might know how to figure this out, so give it your best!

PS: This is the amp http://www.kicker.com/06/tech-support/manuals/manuals/IX406manual.pdf

Thanks

Posted by: Spoke Mar 7 2008, 07:34 PM

Assuming you will power 6 channels @ 40W RMS,

Total power is 6 x 40W RMS = 240W RMS

Change to peak power: 240W x sqrt(2) = 340W peak

Assuming 100% efficiency of the amp (all power dissipated in speakers), total power input to amp is 340W peak

Watts = Volts x Amps

rearrange:

Amps = Watts/Volts = 340W/14V = 24 Amps peak

Posted by: jsayre914 Mar 7 2008, 07:48 PM

QUOTE(Spoke @ Mar 7 2008, 08:34 PM)

Posted by: Jeffs9146 Mar 7 2008, 08:12 PM

Thanks!! I had just finished a Google search on that exact question and came up with the same info. I did forget about the peak part though!

Thanks

Posted by: Scott Carlberg Mar 7 2008, 09:28 PM

Jeff,
I asked my buddy who manages & does the install at a car stereo shop here in Modesto.

Here's what he said:
The Maximum amount of Current that the Amp can draw from the Battery is EQUAL to the amount/Size of the Fuse in the Amp.

NO amp should be able to draw more current than the Fuse in the Amp.

Posted by: sww914 Mar 7 2008, 10:23 PM

20 amp fuse X 12v = 240 watts
30 amp fuse X 12v = 360 watts
The 5 channel US amps amplifier I have in my dually uses 3 30 amp fuses.
It's plenty.

Posted by: Jeffs9146 Mar 8 2008, 01:38 AM