I'm going to be installing for a buddy with a 2003 Max.. and I'm trying to talk him into springing for a upgraded alt..

Install will be a 40x4 class A/B, and a 2400W class D amp.. expecting about 250A peaks..

So.. what do you guys think?

 

Holy crap!!! 250 amps!! Man that is a good 80 more amps of current than I am drawing. I have an Optima red top and a Phoenix Gold 1.2F cap and I'm having no problems what so ever. But 250 amps is a ton of current.

If the guy doesn't want to spring for the upgraded alt, at least a second battery would be in order. If he listened to the stereo with the car off, the battery would be dead in no time!

Hope this helps!

 

Yeah, thats incredible, I've got 120 amps and a 1F cap and all is well, but 250, that's a ton. With a yellow top and a Stinger pro 3cap, you'd probably still draw too much on the stock. I upgraded to a 185 amp alternator in my Tbird when I couldn't get the headlights to not pulse with the bass

 

thx for the replies..

is the new 3.5 alternator rated 110 like the previous gen?

 

i have a couple of questions...1) 160 watts for mids and highs, and 2400 watts for lows? something seems to be unbalanced here. i doubt u will hear anything but bass with that setup....2) 250A? thats incorrect..at full gain and full volume u are pulling 185A...the formula for this is wattage divided by system voltage....2560watts/13.8 volts = 185 amperes....ur alternator is probably listed for 130 or so amps..with a system that would pull 185 amps, u would kill the output diodes on the alternator, so if u are seriously going to do that setup (i seriously think u need to balance power to mids, highs, and lows bettter)u would need a very high output alternator....also, adding capacitors wouldnt solve ur problem at all because ur system will not be able to charge the battery and caps correctly causeing all sorts of problems...any other questions get back to me.

 

I'll agree with the previous post for sure. I'm running 75x2 to the fronts, 35x2 to the rears and 350x2 to the subs and with a good HU you can set the sub controls to decrease or increase the bass from your sub amp, but 2400w in the low end is serious overkill. I would run some speaks in the kicks to add more mid and tweet if I wanted that much boom. Oh, and I like the ability to hear as well. That damage isn't repairable.

 

agreed, but that's what he wants.. it's not my car..

you're on the right track, but you're forgetting efficiency. class D amps rarely reach 75% efficiency, and class A/B amps come in closer to 60%.

so, 2400=.75*draw, 2400/.75=draw, draw=3200w
and, 160=.60*draw, 160/.60=draw, draw=266.67w

and (3200+266.67)/13.8= 251.2 amps.

 

actually, class D amps are usually over 90% efficient if u want to be technical, where the 10% or less is power lost to heat...so that would be 2400watts*.90(acoustical power)...class A/B are only about 50% efficient...so that means 160watts*.5

calculations:

2400 x .9 = 2160
160 x .5 = 80

2160 + 80 = 2240 total watts
2240 / 13.8 = 162 amps

im not sure how u measured in efficiency and came out with higher wattage

 

I'm sorry, but you are mistaken..

how can an amplifier system draw 2240W, and output 2560? you have things backwards with your efficiency calculations, even if you want to use 90% efficiency.

DEI 1100D: 73%Crossfire BMF 1000D: 73.6%
Hollywood Sound Labs RHV 2000D: 65%Lanzar Vibe 1200d: 67%

I can keep going if you like, but I think you get the picture.

 

i dont think u understand what efficiency means bro. if it is 100% efficient then all wattage is outputted as it is rated. how can an amp output more than it is rated for if it is only 70% efficient? im an electrical engineering major, i think im sure of what im saying

 

another thing, wattage isnt drawn, it is outputted, and the links u sent didnt prove anything that u said

 

where the hell did u come up with 2560?

 

ohm's law states:

I = P/E

I = current
P= power
E = voltage


im wrong?

innefficient things do not output at rated power, and check ur damn calculations

 

there's a reason i graduated college...so i didn't have to read sh*t like this...i swear i think we're gonna get quizzed tomorow.

i wonder whose friendhasmax's friend is....

jerk....

 

you are correct with the formula, but wrong with your assumption on power. An amp that puts out 1000 watts puts out 1000 watts. If it was 50% efficient it would draw 2000 watts from the charging system and lose 1000 to heat.

 

once again, an amp does not draw wattage

 

2560* .9 = 2240

that means that the charging system will be supplying 2560 watts and through the loss of power due to efficiency it will output 2240.

 

It draws current at a certain voltage
P=iv if i need to remind you.

 

holy sh*t....an alternator does not output watts, ur confused about efficiency. u need to understand how it is calculated

 

input power - power lost to heat = output power

 

It is considered a current source that is at a 14.4 V due to a voltage regulator built in

 

tell me how amplifier efficiency is calculated

 

I would consider other majors.

Maybe you misunderstand how amplifiers are rated.. They are rated on output, not current draw. so if an amp puts out 2400 watts at 70% efficiency, it is drawing 3200 watts, losing 800 of those watts to heat.
false, an amp drawing 100A @ 12V is drawing 1200 watts. basic.
again, simple. 2400+(40x4)=2560
actually, that's not Ohms law. Ohm's law states: V=IR, aka R=V/I, aka I=V/R.

but that formula is correct for calculating power with voltage & current.

 

actually that is ohm's law stated in other terms

 

there is an www.ohmslaw.com if you need a refresher..

 

The amp has two stages and loses efficiency in between. say the power supply is able to supply a constant 14V and the amplifier draws 100A of current supplied by the alternator.
P=iv 1400=100*14
This is the input power... 1400 watts.
say it is 50% efficient, the output power will be 700 watts, and 700 will be lost to heat.

 

guess what...u have current, voltage, and resistance right? but current , voltage, and resistance can be expressed as other forms includind power (wattage)

 

yes, but as this website outlines nicely, power was not an original part of Ohm's findings..

http://www.csgnetwork.com/ohmslaw2.html

Now I'm in an ME program, and it's kinda depressing that I know this, and you don't.

 

wow a history lesson....does this change the laws? NOPE
its still part of ohm's laws

 

amplifier output wattage is based on mathematical formulas determined by amplifier circuitry. efficiency is based on actual output versus mathematical ideal...dont believe it? there is a reason why there is a Class A, B, A/B, and D. mostly determined by voltage drops across certain transistor designs.

 

this is where you are misunderstanding. amplifier manufaucturers do not rate their amps based on mathematical ideals. and as a customer, I'm damn glad they don't.

amplifier classes are based solely on circuit design, nothing to do with voltage drop. different power losses (efficiency) is a bi-product of the design used, and the specifics of the circuit..

 

haha, what the **** was i even thinking, my bad bro....i was all ****ed up on that ....lol...sorry about that...but u were wrong on the isolator question on the other thread

 

LOL!

oops, huh.

and I dunno, I'm not as familiar with isolators as I am with power ratings, but I do know that the batteries should drain simultaniously, unless there is a greater resistance between one and the drain.

but I'm not going to argue anymore tonight.. peace.

 

peace

 

Graduated hell this is the reason i dropped out!!!!!!!!
I'm oing to bed my head hurts and i aint sure if 500mg of Tylenol is more efficent than 4000mg of Excederine.

 

Once in a blue moon, I was a ME major. Boy am I glad I decided to change my major to Accounting & Finance!!! I just finished my MBA and I begin law school in the fall. These debates over stuff like this about drove me nuts. It takes a dedicated person to stick with any engineering major. I wasn't that dedicated at the time. I guess things have changed. If I didn't, my fiance would crawl up my ****!!!

C.D.

 

such a good conversation that can be solved by one word....




Rockford....

 

Why, you are an electrical engineer and already think you know everything.

 

Remember that 250 amp peak is just that. A peak. Unless you need it to be capable of holding that level more than a few seconds proper wireing and capacitance can do wonders with 110-120 amps of current. I don't know of an alt larger than 180-200 amps that will fit in the stock location. And the only alt I have tested is fairly high maintenance. I have probably replaced the belt 4-6 times this year alone. Not to mention three idler pulleys in the past two years. If you truly need 250 amps of current at 14.4 volts you need to look into finding a shop prepared to install a second alt. The only easy way to do it is to remove the AC compressor.