brendonius

Hey guys...sorry for the odd title..couldn't think of something good for this.
I have a rockford fosgate p325.1 for a bass amp and i was just looking over how i could make it a little bit safer (ie. make sure it is properly fused in both front and back of vehicle.) The amp has 2 onboard fuses of 30A a piece. Now my question is...does that mean this draws 60A total? or is that just a double safeguard in case the first one doesn't give out in time. Any way to figure this out?

2k1Toaster

P=V*I
Power = Volts * Amps

So if it is a 1380W amp, and the car runs at 13.8v, then it draws 100A at most. The problem comes from that wattage. A $10 4000W amp isnt going to draw 4000W, it is just marketing. But looking at the manual you should be able to figure out actual draw.

__________________

brendonius

Going by the performance verification sheet then...

240W RMS @ 4 ohms
1179W Max

therefore,

1179 = 12 x I
I = 1179/12
I = 98.25A MAX

or

I = 240/12
I = 20A rms

Going by this...i'm not quite sure if i understand the purpose of 2x30A fuses. Obviously maxing the amp out at 1179W isn't going to happen because both fuses would blow. Should i replace them with 2x50A to leave some overhead in case it ever did max out?

brendonius

****...just realized its not 12V exactly...assuming my car puts out 14.4V its still around 80A max for draw. Still over and above the fuse rating...yet i've never blown a fuse.

h3rk

another way to find the amps is to substitute ohms law for voltage in the power equation.

since P = I * E

and E = I * R

P = I * I * R = I^2 * R ( I squared R)

then you can find the current based solely on the load resistance and rated power.
that way you don't have to make rough estimates on voltage.

EDIT: But I still cant get it to wrk out to need two parallel fuses @30A, maybe series?

even using 1180 as a RMS value, therefore dividing it by 0.707 then dividing that dividend by 4 ohms, taking the sqrt of the next dividend (28 amps now) even a 125% margin doesn't get you to 60. but if they were in series, that would make more sense.

pokki

Just fuse at the front (battery) as that is where the power comes from. No point fusing at the back unless you run a battery in the boot.


Yes, current draw will be around 60amps when running at peak output.


No. Never ever replace factory fuses with anything other that what the originals are rated for. Doing so will void your waranty, increase the damage to the amplifier, and create a fire risk.

I can't be bothered following the power calculations but heres' how I do it.

Voltage x Amps x Efficiencey (90% class d or 60% class b) = Max continous RMS output (approximation only).

So with your amp the best case scenario is:
14.4 volts (car on) x 60 amps x 90% = 777 watts RMS

This is above the 600 RMS at 2ohm that RF state, so it should meet it's specs.


Fuses don't blow at exactly their rating. It's normally 2x their rating for instant blow, or 1.x times over a long period of time. Hence how the max peak output can be higher than the max rms.

brendonius

Thanks. Definately a different way of looking at it. Using 4ohms as R and 1179W as P, i arrived at 17.17A? Maybe i messed up flipping the equation... I = sqrt(P/R) ?
hmm..anyway
Even if they were in series, seems like the first one will still always blow at 30A no matter what is on the other side of it. Parallel would kind of make sense, equation wise though...i dont understand it. Still seems very odd...maybe i'll email rf tech support and get their explanation. Interesting...

pokki

Followign on from my edit.

If they instant blow at 2x amps, then if wired correctly 2x30 amp fuses, one would blow when total current = 90 amps.

If they used a singe 60 amps fuse, instant blowing wouldn't occur until 120amps of current draw, which is likely to late to pretect the amps electronics.

Either way what the amp draws isn't overly relevant to fusing the wire from battery to amp. As this fuse is there to protect your car from catchign fire incase your power cables short.

Fuse to the cable size.
http://www.the12volt.com/info/recwirsz.asp

brendonius

Ahhh thanks Pokki...definately totally left out efficiency in the equation. Performance verification sheet says 65.546%...damn thats not very efficient.

So...
P = V*I*(efficiency)
P = (14.4)(60)(0.6546)
P = 565.57= 566W RMS.

According to rap sheet:
240W RMS @ 4ohms
393W RMS @ 2ohms

why such a big discrepancy?

pokki

Is the birt cert/rap sheet done at 12 volt?


That will be at minimum load. Maybe the amp is happy down to 1ohm.

brendonius

B+ Test Voltage: 14.4 VDC

h3rk

here's another way to figure it out with a calculator

http://www.bcae1.com/images/swfs/sys...nassistant.swf
be sure to select the class inside the amp, in adition to entering the efficiency.

It provides a bunch of other useful info like wire size. The link at the lower right will show some more of the theory behind it.

I totally spaced out efficiency. Good explanation on the 2 fuse thing.

Mylt1

you guys are going about this the wrong way. the fuse at the batt is to protect the car. you fuse the wire, not the device. so if your running say a 4awg wire max fuse at the batt should be 125 amps. go up a size to 2awg and the max fuse rating is 200 amps. so the man fuse should be fusing just the wire where as the fuses on the amp actually protect the amp itself.

h3rk

you're right, lets just assume a wire and fuse it.

I think the gist of the previous calcs was for a deeper understanding of the device fuses (which was a question of his in the second post), and a means for a confident determination of the wire size to the amp based on it's current draw, which would be protected as you have so eloquently and kindly stated.

If you look through pokki's posts again, he has already stated that, and it's well covered in the link I provided.

No one likes a guy coming in, reading half of the thread, missing the relevant info, and telling everyone they're "wrong". Have a nice day

2k1Toaster

Ok, very simple...

1179W x 65.46% = 771.77W actual.

Most cars run at 13.8v. Some run at 14.4 but usually those are very sepcial cases.

So 771.77W / 13.8V = 55.93A maximum draw. So 2 fuses at 30A in series make perfect sense.

__________________