I'm just passing through, shopping for large capacitors for sale in winnipeg actually, and found your discussion.
A cap IS a current storage device, so is a car battery.
Batteries are sold with different ampacities, but the same voltage output.
A power amp needs
CURRENT to move the diaphram via the voice coil.
All the voltage in the world will not move a speaker coil if it doesn't have the required current to finish the job.
To believe this is about keeping voltage ripple off the power bus is... well you wouldn't have to worry about voltage if the current was available. Current will drop before voltage.
Besides, any amp will have inductive chokes inside the amp to remove AC off the DC bus.
To avoid loss of needed power you need a bigger alternator for more current output, more voltage is so very bad as the amp is built to use 13.5-14VDC. That's why alternators have built in voltage regulators.
Bigger the capacitor rating, the better. Period.
The RC-time constant formula cements that fact. T = R * C
R - resistance of the load of the amp on 12VDC power bus (not speaker-ohms) at any given moment of time
C - the unit of capacitance in microFarads.
Time is in seconds.
Since R is very small under load, we need a very large C in Farads, not microFarads.
The formula tells us how long before a capacitors charge is reduced by to 37% output.
http://www.tpub.com/neets/book2/3d.htmSo, as amp draws more power,Watts are increasing,
V * I = Watts = I(squared) * R,
V is constant (or supposed to be)
I is increasing exponetially, so R is decreasing proportionatly.
R*C, where C is fixed, so when the R is decreased, T becomes smaller, the cap discharges faster, and less charge available to draw from for the amp to pull the voice coil back for the alternate cycle of the coil movement and following push-pull cycles.
But, if we increase C, that changes the formula for a longer dis-charge cycle, and more boom-boom for you.
Note that the Charge/discharge cycle is a two-way street.
The Charge resistance R, is the wire resistance from your alternator, (without consideration to any effects of inductance to an LC-time constant), forgiving the amp drawing against total available current while you are trying to charge the cap.
So, to get the benefit of a stupid-huge cap, you need either bigger wires, or shorter wires, or both, assuming you have a alternator capable of supplying more than you can use in current.
Here are illustrations of people who get this concept:
http://www.youtube.com/watch?v=RRDA64OegJYhttp://www.youtube.com/watch?v=r1tDElOuLCwQuote:
30,000 - 50,000 watt set-up with 5-super high output alternators with a bank of batteries in the back to provide the 2500-amp draw of the linears. Vehicle has 454/w nitrous. For keydown leave it in park floor it hit the amp and nitrous at the same time so the motor wont stall out and you have time for a 5-7 second keydown before you pass-out from rf burns.
70,000 watts:
http://www.youtube.com/watch?v=JTW3LmM7qBAQuote:
extremes of a hobby. puttin out 20,000 more watts than the big clear channel commercial AM stations like WOWO. I saw the le clip of this from a guy who was also cam-cording and the guy who did this tape 75 feet away when he keys down then the camcorder went blank and ejected the tape. the dude inside said his vision gets so badly effected that he can barely see when he's keyin this thing down.
Once everything else has been considered and addressed, then yes, there is the law of diminishing returns. But then I think a 1000Watt sub-woofer has already past that point.
How about instead of just "more power", try a different type of speaker like this:
http://www.youtube.com/watch?v=0sne_uI2Yq4http://www.youtube.com/watch?v=2Lr8tVC-Qx0http://www.youtube.com/watch?v=YyVTvtgm11o