capacitor

Since you don't seem to listen or care what others have to say we will do dome simple math you stated you have 1000W RMS we will assume an almost unheard of efficiency of 85% which means you need to actually make 1150W to get the 1000W rated then divide that by 12.6 for the battery and you need about 92 amps just for the amp . Most vehicle alternators are rated between 85 and 105 amps which means you have between -7 and 13 amps left over to run everything else in the car. Again, capacitors do not make power only an alternator (or generator in older cars) can make power therefore adding a capacitor does nothing except give your already struggling charging system something else to charge. About the only use I can see for an alternator would be as a voltage ripple smoother but only after all other electrical upgrades are done. --->Richard2008 Scion xB

Pioneer AVIC-D3

RF 3Sixty.2 sound processor

Stock speakers (for now ;))

1. Capacitors do, of course, "work." They do exactly what any capacitor does in an electrical system, they resist changes in voltage. They are part of every amplifier power supply to help stabilize rail voltage. The point of the sarcasm is that they ONLY work as intended (to help an amplifier sustain peak outputs during short transients) they do NOT improve the capacity of your electrical system. Indeed, if the alternator is already operating near its capacity, a cap will only make matters worse, not better. 2. Many installers use them for their intended purposes, but far more install them because they look cool, or because they do not understand how they work and have bought into the marketing hype, or because people expect them, or because they can charge a lot of money from unsuspecting customers for them. 3. You need a high-output alternator. Period. Most cars do when a large car audio amplifier is installed. The above word written by DYohn in response to another member.

"...even with how good I am with math and electrical stuff, not a whole lot of it made sense. Could some of the facts that didn't make sense explain why your conclusions are incorrect? Fakepete provided a well written explaination.

"...even with how good I am with math and electrical stuff, not a whole lot of it made sense. Could some of the facts that didn't make sense explain why your conclusions are incorrect? The truth is in the numbers. Fakepete provided a well written insight: Capacitance Capacitance is a capacitors ability to store an electric charge. When a capacitor is connected to a DC voltage source the initial current is very high. Electrons from the negative side of the source pour into the negative plate of the capacitor. As a negative charge accumulates, an equal amount of electrons are repelled from the positive plate through electrostatic induction. The electrons flow out of the positive plate to the voltage source. The electrons flowing in and out of the capacitor give the illusion of current flowing between the plates even though they are not connected. As the charges continue to accumulate a voltage is developed across the plates. The voltage continues to increase until it matches the source voltage, and at this point the capacitor is fully charged and the current stops flowing. The Farad The Farad is the standard unit of measure of capacitance. One Farad is the amount of capacitance needed to store one coulomb of electric charge under the influence one volt. For those who may not be familiar with it, the coulomb is a measure of a quantity of electrons and the charge they have. The volt, amp and other units of electrical measure are based on it. One coulomb is 6.25 x 10^18 electrons, so a one Farad capacitor connected to a 12 volt source will store 7.5 x 10^19 electrons, or 12 coulombs of charge(coulomb = voltage x capacitance). This isn't very useful by itself since you can't count electrons, but it is used in the next equation, which solves for energy. Energy is measured in joules or watt-seconds. (one joule = one watt-second). The next equation is Energy = coulomb x voltage/2. Therefore, 12 x 12/2=72 watt-seconds. 72 watt-seconds is all you get out of a one Farad capacitor at 12 volts. The intention of the capacitor on the power wires of an amp is to help maintain the voltage at its highest level. If the voltage begins to drop, the energy stored in the capacitor is used to try and stabilize the voltage. If the drop lasts for one second, only 72 watts of power will come from the capacitor before its completly discharged. The power isn't spread evenly over the second, there's a rush at the beginning and it drops off quickly after that. Now the capacitor needs an additional 72 watt-seconds from the vehicle charging system to recharge, which was already strugling to maintain the voltage. The interesting thing here is that if the time of the voltage drops decreases, the power goes up. If the voltage were to drop for only 0.1 second, then 720 watts would be available over that 0.1 second before it was discharged (72 watts / 0.1 seconds).   This is using 12 volts, if using 14.6 volts, the energy goes up to 106 watt-seconds for a one Farad capacitor. Over the same 0.1 second, you now have 1065 watts available, which I believe is enough to have at least some help in maintaining the voltage. None of the calculations here take into effect the many variables such as ESR of the cap, resistance if the wire, charge and discharge curves, etc. This is only an overview of the whole thing, there are many other factors to consider, and pages of additional calculations as well, but should be enough information to form your own opinion.

Sorry about the double post

So to play a half second bass note at max volume on a 1000 watt system without any voltage drop I'd have to have a 35 farad capacitor?  Btw my system's actual usage is nowhere near 1000 watts, that's just the watt ratings of all the equipment added together.  I have an 800 watt inverter and between the EQ, the mini-amp, the other amp, and the other amp connected to it I have no idea how much it really draws but it's somewhere in the neighborhood of 400-500 watts.  Then there's my 200 watt sub amp.  So probably 700 and that's at max volume, which I usually don't play it and if you saw my system, you'd understand why :P  And then there's the fact that some equipment like the 15 band EQ is always drawing the same amount of power and the tweeters aren't very variable either and there's 15 of those!  The other 3 are subs so they're the culprits for most voltage drops cuz bass notes are intermittent.  So maybe there's 300-400W of difference between bass note and not bass note power.  So assuming a bass note lasts 500 milliseconds, I need about a 2.5 farad capacitor to cover all that?  Plus what I never understood about capacitors is how they don't just change the timing on the voltage drop?  Or do they?  Let's pretend my sound system only played 500 ms bass notes with 500 ms in between and 500ms of bass after 500ms of nothing makes it drain the capacitor until it's 90% empty.  So then in between bass notes, the capacitor recharges at, I assume, the same rate at which it drained it which draws a bunch of power from the entire circuit and drops the voltage.  In fact wouldn't it draw the same amount of power as the original bass note would have, it just delayed it half a second so now the voltage drops over and over in between bass notes instead of during them?

What brand capacitor did you have on your system when the alternator went out?

"If anyone's system is 1000W or more, they need a capacitor for like 5 different reasons no matter if their alternator is a crappy 70A one or a brand new 300A one". Do you have any facts to support such a claim?

desolator144 wrote: Having your alternator constantly re-adjust as it plays bass notes at 1000W then 0W, it will wear our your alternator. It's about as good for your alternator as turning your rear defrosters on and off once per second.

That's wrong. Your alternator is DESIGNED to do just that... AC, turn signals, headlights, windshield wipers... ALL very transient and high inrush current components. The alternator doesn't care WHAT kind of transient current it has to make, as long as the current doesn't stress it's internals, especially the diodes. An alternator is a current transformer, (NOT a voltage rtansformer) in it's simplest terms, it just needs a prime mover (the engine) to do it's job. It transforms/converts rotational energy to electrical energy - a transformer.

desolator144 wrote: I assume a 200A alternator's lowest mode has a pretty high drain on the engine compared to the lowest mode on a normal output one so even if nothing really power sucking is running, it's still destroying gas mileage.

Another wrong assumption. No load on the output side, no load on the mechanical input side. IF the rotor on the HO alternator happens to be heavier than an OEM alternator's rotor (and the rotor rarely changes - it's the stator that gets heavier for HO units) then there may be an infinitessimally larger mechanical load on the engine, but it's vanishingly small. Until the current demands go up, there is no additional energy demand on the input side. As I said before, it's a transformer.

desolator144 wrote: Hey guess what, when my couple hundred watt refrigerator turns on, the lights dimm and that's on a 3600+ watt house circuit. Having tons and tons of power available doesn't stop a large, sudden draw from dropping the voltage. Guess what would happen if I had a capacitor attached to the fridge? Yeah, it wouldn't do that.

Well, yeah, it probably still would, because you have not addressed the SERVICE DROP. The service drop to the house would be the equvalent to your alternator. If it's too small (as in wire gauge) whatever you have for load will eventually overload that service drop. I installed solar (and when I say "I installed", I mean *I INSTALLED*, I didn't hire somebody to come do it...) at my house two years ago, and for the same project, I re-wired my garage for my workshop. The #4 service drop I had exhibited the VERY SAME symptom you describe there - the one with the dimming lights - (even before the additional demands of a workshop - think "new ADDITIONAL car amplifier") but I had PG&E provide me with a new service drop of 1/0 for my new 200A main panel, and guess what? No caps attached to my components, and my light dimming stopped. It has NOTHING to do with the load, if the source is adequate.

desolator144 wrote: Hey, there's one of those mini power station thingies down the street from me that conditions the power for the surrounding area. Why does it have a bunch of capacitors in it? Maybe to even out sudden draws from a single source even though there's thousands of amps available? Yup, think so.

Again, you'd be wrong. All those caps you see are NOT for "surges". They are to keep the power phases correct, and I don't mean between phase A, B, and C. I mean the power phases, as in current or voltage lag or lead. In reactive circuits (read AC circuits) impedance based on reactive components (inductance and capacitance) cause current leads or lags, (I remember "ELI the ICE man" from my electrical classes - in an inductive circuit - "ELI" - voltage leads current, and in a capacitive circuit - "ICE" - current leads voltage) reducing DRAMATICALLY the efficiency of the transmissions. Those power stations are power factor correcting stations, adding finite, carefully selected values of capacitance and inductance to the power grid to "repair" the power curves.

desolator144 wrote: So then in between bass notes, the capacitor recharges at, I assume, the same rate at which it drained it which draws a bunch of power from the entire circuit and drops the voltage. In fact wouldn't it draw the same amount of power as the original bass note would have, it just delayed it half a second so now the voltage drops over and over in between bass notes instead of during them?

Your assumption is wrong. When a cap becomes discharged, the charging current can be THOUSANDS of times what the discharge rate was. It is determined by the charging potential and the ESR of the cap. Then apply Ohm's law, then apply DeltaT. If you have a cap with an ESR of .002 Ohms (typical for a "better" cap), and your car's voltage is 14.4V, then your charge current CAN be as high as 28000A. (Why do you think they give you a resistor to charge them?) How fast can 28000A dump one coulomb of energy into a cap? I don't know either (that's DeltaT) but it's pretty DAMN fast, and a hell of a lot shorter than your original discharge took, if connected to the power leads of a car power amplifier. Now, by the time we reach current levels like this, there is no single group 45 SLA in the world that can provide that, let alone the resistance of the power cables and connections between said SLA and the cap. Your recharge current now becomes dependent upon the charging source. The battery in the car. Once this battery provides the return charge to the cap and do you think this return charge (remember, this is all still in watt-seconds) is free? It isn't - there is a electrochemical reaction within the battery to provide it, and this electrochemical reaction WILL discharge the battery a given amount, EXACTLY the same as loading the battery with a load demanding 1/1000 the current for 1000 times as long. Now the battery is discharged a given amount. Where does it get IT'S recharge energy? That's right - the alternator. Multiply this current demand by all of the accessories in the car, and you'll see that soon the alternator will now have to power more load than it can do effectively. Forget the cap. They are bandaids on bullet wounds.It all reminds me of something that Molière once said to Guy de Maupassant at a café in Vienna: "That's nice. You should write it down."

Haem has spoken...end of argument. lol