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Old 06-03-2006, 04:27 PM
sm1lodon sm1lodon is offline
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Location: Hawaii
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Default Importance of Capacitors. Pack a lunch, this is a long one!

HOW IMPORTANT CAPACITORS ARE TO THE FUNCTION OF THE AMPLIFIER.

To start off with, there are some basic electrical laws on which amplifiers operate. Electricity is made up of electrons. If these electrons are not moving, it is called "Static Electricity". If the electrons ARE moving, then it is called "electrical current"

Now, to convince electrons to move from point A to point B requires that there be some kind of pressure on one side of the electrons, and/or some kind of suction on the other side. What pushes electrons away is more electrons in the same amount of space. What pulls them close is fewer electrons in the same amount of space.

Wherever there is this electrical pressure, we call it voltage. Which is why static electricity can turn into current suddenly if it has a vent for the built up static charge to escape to, namely, a place where the charge is lesser.

So. We have electrons. And we have the pressure, voltage, that influences them to move, which makes current.

Now, a coulomb of electrons is a big, fat heap of electrons. When 1 coulomb of electrons flows past any one spot, in 1 second, it is called an Ampere, after some French guy.

It is like a gallon of water flowing past a spot in one hour is measured as 1 gallon per hour, or gph. In electrical terms, the coulomb is the gallon, and seconds are the time slot used, so Coulombs per second is amperes, or amps. This has NOTHING TO DO WITH THE WORD AMPLIFIER.

Okay. So, the ampere, or amp is the basic measure of electrical current flow.

Electricity would flow everywhere all the time with so much ease if it weren't for things that resist that flow. Resistance is the basic measure of opposition to current flow.

Resistance is like having a pipe that is bent, twisted and rusty on the inside to flow water. The more bent, twisted and rusty the inside of the pipe, the more pressure it takes to push the same amount of gallons per hour through the pipe. Same if the pipe is smaller diameter. And same if the pipe is longer.

If you have 1 volt pushing through 1 Ohm (the measure of resistance to electrical current flow), then you have 1 amp of current flow as a result.

Volts = Amps X Ohms

So, if you have at any given moment, 200 amps flowing through a 4 ohm load, then you have, obviously Volts=200X4, which is 800 volts. Yes, it takes a lot of voltage to push that much current through that little resistance.

Now, power is a measure of how many amps, times how many volts. Power is measured in watts or kilowatts, depending on how big the application.

Watts = Volts X Amps

So, a 12 volt system pushing 200 amps consumes 2400 watts.

These are the basics of direct current. More in next post.

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Loud? This is beyond LOUD. Now we're talking about LETHAL RADIUS!

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Posts: 20 | From: USA | Registered: Dec 2002 | IP: Logged |

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posted December 28, 2002 05:27 AM
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Importance of Capacitors 2

Okay, having said that in my previous post, let's look under a microscope, as it were, just HOW a car amplifier works. We will be dealing with ONE full cycle of a bass note only, to start with.

The deck sends a signal to the amplifier, telling it to hit a 24Hz bass note.

So, we want the amplifier to make our speakers move out and all the way back in again, 24 times in one second. We will examine what happens in 1/24th of a second when the speaker moves out and back in again ONE TIME.

The amplifier is a 1200 watt (peak) amplifier. The system is a 12 volt system. What a smashing coincidence!

The amplifier, when doing nothing, is ideally consuming very little power, just enough to keep the circuit board awake, as it were.

But, when the beginning of this bass note comes, something changes. The amplifier, which has NOT been allowing ANY current to flow through itself, has to change to allow current to flow through itself, because those little electrons are what the voltage uses to push the power through the amplifier.

So, since we have the amplifier turned up all the way, we know that we are going to be peaking at 1200 watts at 12 volts, or 100 amps peak current flow.

Watts=Volts times Amps

1200 watts = 12 volts times 100 amps

So, the deck instructs the amplifier to start its change from infinite resistance, allowing NO current flow, to lower and lower resistance until it is at 0.12 ohm,

Ohms=Volts divided by Amps

.12 = 12/100

.12 ohms being its lowest resistance. at this point, the maximum current of 100 amperes is flowing into our amplifier, and since Watts are Volts times amps, then 1200 watts of power is being produced to slam our subwoofer cone forward. But, wait, now that it is approaching peak, and pushed that subwoofer cone out as far as it can, the resistance of the amplifier is being told by the deck to go up again, reducing the amount of current, which reduces the amount of power, which allows that subwoofer cone to return back to its starting point...

Then, the amplifier again drops its resistance smoothly to 0.12 ohm again, allowing 100 amperes of current in, which, since it is at 12 volts, produces 1200 watts of power when it finally reaches the low level of 0.12 ohm of resistance, and the amp switches the direction of the current going to the speaker, increasing it smoothly in the other direction and sucking the subwoofer cone backward into the housing as far as it can at that level of power.

Then, again, the deck instructs the amp to smoothly return the resistance back to infinite from 0.12 ohm, blocking current from flowing into the amplifier, and the speaker, no longer being provided with as much power starts to return once again to its neutral position, and when current flow reduces to zero, there is no more power being applied to the speaker by the amp, to pull it backwards into the speaker basket.

This process repeats 24 times a second with a 24Hz bass note.

Now, here is where it gets more interesting, and to we car stereo electrical buffs, absolutely fascinating!

You see, the voltage pushing that current through that amplifier doesn't ONLY have that amplifier to deal with. It has several obstacles along the way that are reducing the pressure that those electrons are feeling on their way TO the amplifier. It is like once the electron gets shot out of the battery, it has an uphill climbing run through all the wiring and fuses and connectors before it even gets to the amplifier.

And, the more electrons try to run down that cable, the more of their pressure is lost, sort of like a bunch of those M and Ms trying to jog down the same tunnel: The more there are, the less pressure they have by the time they pop out the other end... at the amplifier.

If I have 12 volts way down at the front of the car where my battery and alternator are, and a bunch of restrictions for my electrons between the battery/alternator and amplifiers, such as several feet of wire, connectors, fuses, fuse blocks, and terminals on the actual amplifier, then I have a veritable obstacle course for my electrons to hurdle before they arrive, panting and tired, at my amplifier. And the more electrons there are, the more tired they will be when they get there. Further, the more they have to stop and start, the more energy is consumed, such as when the amp draws current to move the speaker to the front, then drops back to zero draw, then back up again to push the speaker to the rear, etc. How do I reduce the effect of this? How do I make it easy for my amplifier to receive closer to the original 12 volts than I used to?

So, we have our 12V supply pushing, at peak amplifier power (and lowest amplifier resistance) through 2 ohms in the cabling, and fuses and fuse holders and terminals on the amplifier, for a total of 2.12 ohms of total resistance to current flow between the battery/alternator and the amplifier.

Since we started way back there at the battery/alternator with 12 volts, we can now find out just how much current is getting to that amplifier:

12volts divided by 2.12 ohms = 5.66 amps. Gee, whiz that is a piddling little amount of current to supply to that amplifier, when it is thirsting for 100 amps! That only allows it to have... let's see...

Since the voltage started out as 12, but the current went through 2 ohms of resistance to get there, the voltage at the amplifier guts is now

Volts = Amps X Ohms

Volts = 5.66 amps X 2 Ohms

Volts = 11.32V!

And since we have only 5.66 amps at 11.32 volts, we have...

Watts = Volts X Amps

64 Watts = 11.32 Volts X 5.66 Amps!!!

What do we do??? How do we keep from starving our amplifier for those needed amps?

Well, first of all, you get a wire size that is appropriate for your current draw on your amplifier. If you are a fanatic, like me, then you will be searching the internet for sites like http://www.colemancable.com/catalog/Portable17a.htm where you can find truly mondo wire, such as 500 MCM copper (which is like anchor cable for the USS Nimitz compared to most, and to most people seems like total overkill. But I'm just like that!) to minimize the resistance and thus voltage drop to the amplifier. But, maybe you aren't quite that fanatical. I am!

Secondly, make sure all your connections are very secure and not corroded.

Thirdly, measure the resistance from one end of your system to the other with an Ohm-meter that measures to two or three decimal places .01 or .001 ohms, for example. If you notice that the resistance is too high, like near one ohm or more, then check individual legs of your system to see which parts are providing the excessive resistance.

The typical car stereo power wiring scheme doesn't have 2 ugly ohms of resistance in it. But, it has SOME resistance, and the more you can minimize it, the happier and healthier that amplifier is going to be. And it won't be clipping the peaks off your sound waves, which can be a damaging thing for speakers.

Another factor is inductive load. Namely, how much energy it takes to stop and start the electrons moving down the cable from the battery. The greater the current, the greater the inductive load. The smaller the cable, the greater the inductive load, even if the cable has no resistance, because the same electrons have to move faster back and forth. The longer the cable, the greater the inductive load. So, for loads that vary rapidly in their current demands, it becomes all that much more important to have a capacitor system right by said load.

So, having said all that, what is the importance of capacitors? Well, let me tell you an anecdotal reference. I had an ammeter set up on my Power Wagon; needle analog type. When my system hit a bass note, the ammeter needle would turn into a pie-slice shaped fog on the meter face, it was traveling so fast back and forth between normal and load. (at the same frequency (x2) as the bass note).

I added a 1 Farad capacitor to my system right by the amplifier.

(Remember, a coulomb is a big pile of electrons? Well, 1 Farad is when a device can store 1 coulomb when charged with 1 volt. So a 1 Farad capacitor can store 12 coulombs of electrons at 12 volts.) If it can dump all 12 of those coulombs into a load in one second, then since Amperes = coulombs/second, then it would be flowing 12 amps during that second.)

So, my amplifier was a 450 watt amplifier. Which means its peak load, with all inefficiencies and heat and resistance and internal goofing around thrown in, was closer to 900 watts.

So, when I hit a bass note, my amplifier was drawing 62.5 amps.

900Watts = 14.4 volts X 62.5 amps

But that was only at the peaks of the bass notes, so it was drawing, during a bass note of 30 Hz, 62.5 amps 60 times a second. (once per second to push the speaker out, and once per second to pull the speaker back), which was causing my ammeter to wiggle back and forth at 60 Hz.

But, after the advent of the capacitor, when I hit a bass note, the ammeter moved to the left, stayed rock steady there for the entire bass note, then went back to center.

Why???

Because the alternator output (which was going to the battery, which was supplying the amplifier) was only seeing a drop in voltage that was smoothed out by the capacitor.

Why did the capacitor smooth it out?

Because the capacitor was storing 14.4 coulombs of electrons.

In the 1/60th of a second that the amplifier was sending the speaker cone forward, the amplifier was drawing, on average, (I don't want to explain the math for this one, I would have to show you a graph) 36 amps from the system for 1/60th of a second.

During that 1/60th of a second, the amplifier's internal resistance dropped from (functionally) infinite to 0.2304 ohms.

Ohms = Volts/Amps
.2304 = 14.4/62.5

That was lowest resistance, peak power and peak current.

Average resistance during that 1/60th of a second was 0.4 ohms. So, for that 1/60th of a second, 14.4 volts was pushing through a 0.4 ohm load. So 36 amps ON AVERAGE flowed into the amplifier for the duration of that bass note.

And, since Amperes are Coulombs divided by seconds, then
36 = Coulombs divided by 1/60th
36 = Coulombs / 0.0166666666666666666
36 = Coulombs X 60
36 = .6 X 60

So, during that 1/60th of a second, only 0.6 coulombs of electrons passed through that amplifier.
An Optima battery has an internal resistance of 0.0028 ohms. Added to this is the resistance of the entire electrical system up to the amplifier from the battery, let's say 0.2 ohms. so we have 0.2028 ohms total resistance from the battery to the amplifier guts. Okay, so if the amplifier drops on average over 1/60th of a second, to an average of 0.4 ohms. Add this to the total of .2028 ohms of the entire electrical supply to the amplifier, and you have 0.6028 ohms.

14.4 volts divided by 0.6028 ohms = 23.88 amps

Since that 23.88 amps went through a 0.2028 ohm load before it even GOT to the amp, it dropped in voltage by 4.84 volts!

Volts = amps X ohms
Volts = 23.88 X .2028
Volts = 4.84V

14.4-4.84 = 9.557 volts available at the amplifier guts!

So. We want to be much closer to 14.4 volts at the amplifier guts, so we introduce that capacitor.

The capacitor has about 0.0008 ohms of internal resistance. It stores 14.4 coulombs of electrons within itself. It is kept at 14.4 volts by the battery/charging system during non-peak times, when the amplifier is drawing NO current.

So, during that time when the amplifier resistance drops to .4 ohms (average), we now see that since electricity takes ALL paths, and takes paths in proportion to the resistance OF that path, that the amplifier now has TWO sources of electron flow.

A) the battery at 0.2028 ohms, all inclusive, and
B) the capacitor at .0008 ohms, all inclusive.

Now, since this 1 Farad capacitor stores one coulomb of electrons for every 1 volt of electrical charge it is at, then if this capacitor gives up 1 coulomb of electrons, it is because the circuit it was attached to dropped in voltage by 1 volt. If you use a 2 Farad capacitor, it will only take ½ of a volt drop to convince the capacitor to give up one coulomb. 5 Farads, 1/5th of a volt drop, etc.

If you have an infinite Farad capacitor, you have an inexhaustible supply of electricity that is ONLY limited in how much current it flows by the internal resistance of the capacitor and the resistance of the load to which it is attached. It will NEVER experience a voltage drop at its terminals other than that caused by its own internal resistance, because when the voltage TRIES to drop at its terminals, new electrons, in infinite supply, rush in to fill the void back up.

So. The amplifier drops in resistance to 0.4 ohms, ON AVERAGE. for the duration of the bass note.

14.4 volts at the battery and at the capacitor, when there is no current flow. But, now, since the resistance of the circuit has dropped to 0.4 ohms, and if there is 14.4 volts present, for each 1/60th of a second 0.6 coulombs flows from the circuit through the amplifier.

So, in slower motion yet, the amplifier hits. resistance in the amplifier drops smoothly to 0.2304 ohms peak, not average. At that ohm level, and at 14.4 volts, the capacitor, which has .0008 ohms of internal resistance, and the battery/charger which has .2028 ohms internal resistance will both be drawn on by the amplifier. Since electricity flows to places in inverse proportion to the resistance, then the ratio of electricity flowing to the amplifier from the capacitor compared to the electricity flowing from the battery/charging system will be:
.2028/.0008, which is a ratio of 253.5 to 1. So, roughly speaking, an average of 0.597642436 coulombs will flow from the capacitor through the amplifier, and only .0002357564 coulombs will flow from the charging system and battery through the amplifier.

Having a capacitor is like having a pressurized tank of electrons that is finite, with a big outflow pipe, and big inflow pipe, while having a battery/charging system is like having a pressurized tank of electrons that is infinite, but has a bit smaller spout, and is a bit further away from the amplifier and capacitor.

When the amplifier drops its resistance rapidly, the voltage across the amplifier power terminals also drops slightly, as current flows through the amplifier. The voltage of the whole system drops, nowhere as low as it is right at the power terminals of the amplifier. The capacitor continues to allow current through the amplifier as long as its voltage is higher than that of the amplifier.

When the capacitor's voltage drops below whatever voltage is available at the terminals where the battery/charging system joins the capacitor, the battery/charging system flows current into the capacitor. If the amplifier is at the same voltage, the battery/charging system also supply it with current, in inverse proportion to its internal resistance compared to the capacitor.

When the amplifier raises its resistance, the voltage at its terminals rises again so that no current flows from the capacitor to the amplifier, and the electricity in the conductors between the amplifier and the capacitor becomes static electricity.

Since the amplifier is no longer drawing current, the voltage of the entire system starts to rise, refilling the capacitor with electrons, until the capacitor, amplifier, and battery are all at the same voltage, and the electricity is once again static. Since the capacitor fills up so fast, due to small internal resistance, it is ready to dump its charge into the amplifier again during the next sound wave.

So, to sum up:

A) Get yourself some nice fat cables, and/or more of the same size ones, but they must be exactly or almost exactly the same length.

B) Low-resistance connectors

C) Low-resistance circuit breakers or fuses

D) Make nice short paths between your capacitors and your amplifiers,

and you will be maximizing your sound system's potential. And, if you can, take the wire that reads system voltage, thus controlling how much the alternator puts out, and attach it to the central point of your batteries, near your amplifiers, but with the capacitors still between the alternator feed and the amplifiers.

Since the capacitor is close enough and has low enough internal resistance to pour in most of the necessary current, it does, while the battery/charging system pour in some. By being closer to the amplifier than the battery and charging system AND having lower internal resistance, the capacitor levels off the peaks that the battery and charging system used to see, and fills in the valleys that the battery and charging system used to see.

The reason capacitors charge so fast is as follows: their only speed limit is their internal resistance, which is typically much lower than batteries. So, if a capacitor is missing 1 coulomb of electrons, due to a recent bass hit, it doesn't take hardly ANY voltage difference at all to refill those missing coulombs. In fact, due to its exceedingly low internal resistance, the capacitor is refilling almost exactly as fast as voltage is rising.

Instead of wildly fluctuating current draws, what the battery and charging system see is a smooth draw of current, while what the capacitor sees is the wildly fluctuating current draw that the battery and charging system used to be subjected to, but the capacitor is exactly suited to filling and emptying itself of electrical charge repeatedly and rapidly, while batteries aren't really made for that. Also, since it has about 1/500th the internal resistance of the battery, it provides electrons with less voltage drop than the battery and charging system can, ESPECIALLY for short bursts of high current demand.

For any increase in current, at the same resistance, the voltage drop will increase.

To get more current flow, you either need more voltage, or less resistance.

Bigger electrical loads have LOWER resistance, not higher, than smaller ones.

Capacitors are important for the electrical system and sound quality because they have a readily-available source of electrical current that is responsive enough to meet the need of the high-amplitude demands of a modern bass amplifier.

And, that is why capacitors are important!
When the amplifier goes back to functionally infinite resistance at the valleys of each bass note, the battery and charging system "refill" the capacitor(s), ready for the next hit, which may be only 1/60 of a second or less away.

This is why, when my sound system was run straight from my battery and charging system, my ammeter was a blur between zero and somewhere on the discharge side of the scale every bass note I hit, but when my sound system was run from the battery and charging system with a capacitor not 12 inches of thick copper cable away from my amplifier, when the bass notes hit, the ammeter needle would move left, pause there for the duration of the entire bass note, then move back to center.

A capacitor smooths the peaks and valleys that the charging system and battery "see" that happen at double the Hz rate of whatever major bass note is playing.

So, if you have a one-second-long 30Hz bass note, the capacitor discharges and charges 60 times, while the battery/charging system sees a relatively smooth dip for one full second, before returning to normal.

This is also why in the ideal charging system, ALL the electrical devices, including the amplifiers, would be fed from one central point, and THAT point is where the system voltage would be read that signals the alternator that it is time to pump out some more current.

If you have the voltage regulation for the alternator measured from the voltage at the battery, it can be as much as 50% off of what the immediate needs of the electrical system are at the amplifier, depending on the load.

Since voltage drops along a conductor are reduced by:
A) having lower-resistance conductor materials
B) having shorter conductors, and
C) having conductors of larger cross-sectional (csa) area, or more conductors of the same csa, or better yet, more conductors of larger csa!

Then, it makes sense that the IDEAL electrical system for a very powerful vehicular stereo system would have the following elements: (These are IDEALS. They may not be possible, but they clarify just what one is shooting for when one wires up a power system for a car stereo)

A) No resistance to current flow

B) Power production capacity equal to or greater than all electrical loads of all systems of entire vehicle combined, at their peaks

C) Instantaneous response to any current demand with the exact amount of current needed, so NO voltage sags due to inadequate swiftness of response from the electrical power generation system on the vehicle ever occur.

D) Voltage that never fluctuates, but is rock-steady at 14.4 volts, no matter what the load, current, conditions, etc.

Of course the battery/charging system needs to be able to supply the system with whatever draw it needs. But. With enough capacitance, close enough to the amplifier, the once-per-0.016666666 second peaks and valleys that the alternator and battery are subjected (during a 30-Hz note) to are reduced or eliminated.

And, one can theorize all one wants, but, as I said before when I added a capacitor to my system right by my amplifier, the ammeter on my truck went from being, as I said "a pie-slice shaped fog" as it went from peak draw to nothing and back again at about 60Hz or more, depending on if the bass note was 30Hz or more, to merely moving to the left, staying there, then moving back to center after the bass note was finished.

Now, my ammeter had nothing to sell. No agenda. No hidden purpose. It wasn't trying to convince me that I was right or wrong. It merely showed that when the capacitor WAS installed, the load seen by the battery and alternator was no longer oscillating at twice the bass note's speed, but was a smooth draw that lasted the entire length of the bass note. This indicated to me that a capacitor mounted as mine was, right next to the amplifier with some heavy-gauge wire, will smooth out the load that the charging system and battery see from the amplifier.

When it comes to this stuff, you can't just average the measurement of current out over 1 second, or three days, one has to pay attention to instantaneous current flow at the valley (near zero) of the draw and at the peak of the draw.

So, to further elaborate, it is NOT the case that the amplifier ONLY draws 62.5 amps at peak load because of one thing: The voltage at the amplifier is in constant flux due to the resistance of the electrical system.

Many amplifiers have power supplies that will take (WITHIN LIMITS) whatever voltage is present momentarily at the amplifier power terminals and keep dropping the resistance until that voltage supplies the amplifier with however much current it needs to produce (in my example case) 900 watts.

With modern switching power supplies, the amplifier doesn't just sit there as a dumb unit and clip off the top of every wave that the voltage present at its power terminals dictates. Instead, it has built-in power conditioning that drops the resistance of the power-supply side of the amplifier lower and lower until it has the power it needs to produce the music at the volume it is currently being told to play it by the head unit.

So, if the current draw through the system momentarily drops the voltage at the supply side of the amplifier's circuitry to 10 volts, and what the amplifier is being instructed to do at that moment is produce 900 watts, then the amplifier will drop its resistance down to meet that demand.

Watts = volts X amps

900 = 10volts X 90 amps

So, the current flow rises to 90 amps at that moment to meet the demand. No wonder my old 1200 W RMS PPI amplifier had twin 90 Amp fuses on it. Because the smart power supply would drop the resistance low enough to feed the amplifier power, even when it wasn't in the "ideal" 14.4 volt world.

And the resistance in this case, since
Ohms = volts/amps
ohms = 10/90
ohms = 0.1111111111111111111 ohms

What the amplifier in effect does is act like its own charging system. If it needs more power than is available at its present state, it drops its resistance lower until the amount of current passing through it at whatever voltage it is at at that moment provides enough power to feed the amplifier's internally regulated power supply. Notice, this all happens in the space of fractions of a second, and is happening continually, and always changing. And that is one of the reasons amplifiers have plenty of capacitors internally, so their response to changing current draws of the amplifier is rapid, and not stymied by the voltage available at that given moment at the power input terminals of the amplifier.

The ONLY point of a capacitor in the power supply side of the audio system is to provide a low-resistance, fast-response supply of stored electrical energy that is capable of supplying immediate transient loads (not just averaged theoretical ones) with the current they need to power the amplifier with more responsiveness than the battery and alternator are capable of delivering.

Again, I must fall back to the real world here, because theory that doesn't line up with reality is not theory any longer, it is error. My ammeter on my truck shows this to be the case. When the load wasn't buffered by the capacitor, the ammeter wagged back and forth at the same frequency as the load demand at the amplifier terminals. With capacitor installed, the ammeter moved once, per bass note, to a certain level of current flow, and stayed there the duration of the bass note. When the bass note ended, the ammeter quickly returned right back to center. You can theorize all you want, but the fact is THAT is what happened, so the capacitor, for all our math and mistakes, whatever they may be, feeds the wildly spiking amplifier load with its low-resistance store of electrons, while the battery and alternator power the whole thing by keeping the capacitor juiced up.

Another real-world proof of this is the one time I forgot my battery was disconnected (my engine was turned off) and I turned on my stereo. The stored energy in the capacitor ran my system for a couple of seconds, bass and all, until it ran out of power, when I realized that my battery was disconnected and had discharged my capacitor. Now, if it was as some people's theories state that a capacitor has NO effect on the system whatsoever, what the heck was it doing powering my stereo system? If it can do it for two seconds, then surely it can do it while being constantly recharged by the battery, neh?

I know the battery and charging system are the actual power source of the amplifier. I know this. I know capacitors don't SUPPLY power. They just store it, like a low-capacity, fast-responding battery that is constantly being refilled by a high-capacity, slower-responding power source.

And, not ALL capacitors are created equal. Total capacitance is NOT the only measure of the efficacy of a capacitor. There are one-farad capacitors the size of a stack of ten quarters. But, their resistance is so high that it would take seconds to fully discharge, not fractions of a second, if they were shorted out. These capacitors are what are beginning to be looked at to be used to store energy for cellular phones, and other stuff.

So, less effective "power caps" have higher internal resistance, while better ones have exceedingly low internal resistance.

If a capacitor had NO internal resistance, it would be instantaneous in its response to current demands. Which is back to the "Ideal Electrical System" again.

But, if you have a choice of using four 2.4 Farad capacitors with 9.6 Farads total capacitance to get .001 ohms total internal resistance, or using one 2-Farad capacitor to get .001 ohms total internal resistance, and it is the same price for the four as it is for the one, then, heck, get the four. More capacitance at the same resistance for the same price? Why not?

Briefly, for best system performance, use capacitors, but check their E.S.R. Equivalent Series Resistance.
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Old 06-03-2006, 04:35 PM
sm1lodon sm1lodon is offline
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Join Date: May 2006
Location: Hawaii
Posts: 19
Default Capacitors in a nutshell

The ideal capacitor:

1) Zero internal resistance
2) Infinite capacity
3) Already charged up to your preferred voltage
4) Connected to the audio system by zero-resistance conductors
5) Zero leakage


This would, if it were to exist:
A) eliminate any need for a battery or charging system
B) eliminate any and all fluctuations of voltage at the terminals of whatever load it is supplying, which is part of the whole point of capacitors.

The less-than-ideal-capacitor, but humanly possible:

1) Exceedingly low internal resistance (easily achievable by placing several capacitors in parallel)
2) Large capacity (Easily achievable by placing several capacitors in parallel)
3) Charged by your electrical system in your car, which has enough output to handle any combination of loads your car/system can throw at it. Any car electrical system with 900 watts of peak power output will not be able to sustain a system using 1000 watts, even if there is 100 megafarads of capacitance available.
4) Connected to the audio system and charging system by the lowest possible resistance conductors, ie: shortest, fattest cables you can afford/care to install.
5) Extremely low leakage

Function: Makes the electrical energy for the audio system available to the amplifiers/processors/head unit(s)/video displays as consistent and spike-and-sag-free as possible.

Does not:
A) add power
B) increase voltage
C) provide current
D) reduce amount of overall power required by the audio system

Does:
A) reduce fluctuations in voltage downstream from the capacitors
B) reduce fluctuations in current flow upstream from the capacitors

Capacitors in a nutshell.


Another reason more capacitance is better: the capacitors are being sustained at 14.4 V by your charging system.

The battery is at 12V. So, the more capacitance you have, the more power can be drawn from the capacitors and charging system before your system voltage will even drop to the point of having to draw on the battery.

The battery is not putting out any energy until the system voltage drops below 12V. A 12V battery does not energize a 14.4V system, folks. That is like 12 PSI air expanding into container that has 14.4 PSI in it. It will never happen.

So, the more capacitance you have, the less often system voltage will dip, due to demand beyond the charging system's immediate capacity, to the point of having to draw on the battery.

When the car is being started, the charging system is not turning the engine over. It is the battery, and to some exceedingly small extent, the capacitors, that are supplying the power for that. When the vehicle is running, however, the battery is recharged, the capacitors are charged back up to 14.4V, and the charging system, which runs at 14.4V, is where the vehicle gets its electrical power. Now, if batteries operated at 14.4V, then capacitors would probably be far less useful. But most batteries operate at 12V. So the capacitor acts like a an EXTREMELY low-capacity battery that the charging system is always trying to keep charged to 14.4V.
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Old 06-05-2006, 01:04 AM
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cageman cageman is offline
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Good post, sounds like you know what your talking about.

I have a capacitor that I have never hooked up, why, because I dont know how, after reading your post, I think I will.

It is a blue one that says on the side
sprague
compulytic
32dx6824
120000 uf 15 vdc
8417c
-10% +75%

It has two posts on the top to hook the wire to, and one of them has a + by it.
How would I tie it into my power wire for my amp, I never did know, and thats why I have never used it, I bet I have owned it for 10 years, but now I want too. Do you just wire it in line with the power wire?
Thanks
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Old 08-05-2009, 11:59 PM
sm1lodon sm1lodon is offline
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Just place the capacitor as close as possible to your amplifier with the positive power wire going to the amp going in and out of the positive terminal of the capacitor.

The ground wire from the amp, capacitor, and anything else you have nearby can all be bolted to the chassis in one spot.
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Old 08-06-2009, 12:01 AM
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Dick Dick is offline
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Biggrin

Your reply is 3 years late.
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Old 05-31-2010, 08:33 AM
sm1lodon sm1lodon is offline
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Oh. Sorry about that!
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Old 05-31-2010, 11:53 AM
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Biggrin

And another 9 months to react.
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Old 06-01-2010, 01:31 AM
sm1lodon sm1lodon is offline
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Ah, but I talked to a guy who has a system and I dug up the post using Google, and found, voila! There it is!

Thanks for a great message board, BTW!
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Old 03-19-2021, 09:01 PM
sm1lodon sm1lodon is offline
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Addendum: supercapacitors, which are, IIRC tantalum slugs, can have monstrous storage capacity in a small space, but their ESR (equivalent series resistance) is pretty high.

I would think that a combination of some nice, big, one- or two-farad capacitors in parallel with a 5-, 10-, or whatever-farad supercapacitor would be able to supply one's audio system quite handily. In fact, one could mount the supercapacitor centrally, with shortest-possible wires running to the capacitors mounted right next to each amplifier.
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