yeah that's a great idea. And alot of people don't even know about it. I'll surely do it for my next system in a newer car.

 

What exactly is the big 3?

I'm assuming it was battery and alternator... but was the third wiring? Or am I way off?

 

This is how to do the big 3. We should really get this stickied somewhere, it seems to come up once a week.

Here is another explanation.

 

Also I find upgrading your battery terminals will help out.

 

Whoever wrote this only has a basic grasp of electrical theory.

1- He talks about reactance, reactance is an AC theory we all have 12VDC power systems.

2- 13% loss due to heat!?!?!?! No way jose, Caps are much more efficient than that, there are light bulbs more efficient than that.

3- He talks about ESR, EQUIVILANT SERIES RESISTANCE. In our systems the cap is wired in Parallel it doesn't present any resistance to the system. The only resistance I can think of it generating is the resistance over the line caused by the extra draw to recharge the cap but it would be much lower than even 0.17ohms and the downside would be negated by the up side of having extra juice to begin with.

4- In his 9volt battery test he never tested maximum output, Its much higher on the cap (a cap can output all its energy in less than a seconde) the battery would never put out as much and would just get hot trying. not just that if he put the cap and battery it would last just as long as the battery alone as caps don't use energy they only store it. (they shed very little, the battery would be long dead before the cap sheding electrons would become a factor) and a light bulb is a constant draw not a varied draw so this experiment means nothing.

5- What??? is he somehow going to cycle the cap to do that??? where does this 50hz number come from???

6- He says he can see the cap recharging on his graph but he doesn't see any gain from having the cap. Well if he can see the cap recharging he has to be able to see it discharging. Its math it has to happen. In fact caps don't charge or discharge at a constant rate a full cap will discharge fast and first and then slope off as it depletes, the opposite when it charges it will charge fast at first then level off, anyone whos charged their cap with the light bulb they give you has seen this. So with that in mind if he took proper messurements and saw the cap charging he would HAVE to see it discharging. If he said that he saw nothing at all I would have an easier time believing it.

7- He finishes by stating that the cap cannot maintain 14volts. Thats not what its supposed to do its supposed to lighten the blow on the electrical system caused by the varied draw of the amp. If you had a constant load the cap would be useless but amps are varied they never draw a constant current. Open up your amp you'll see tons of caps in there. Hmmmm wonder what those are for? maintaining rail voltage maybe.

IMO caps are a last resort. Its like bying heat sheids and TB spacers for your car. They're expensive and give little gains but the gains are there. If you already have a big 3 and all the bells and whistles I recomend a CAP if not you have allot of work to do. You'll never maintain 14v with 1000's of watts of power unless you live in some kind of magical fairy land.

I'm an Automation and fire alarm technician and I've gone to college for electrical Technology, and I have other varied college courses like home theater and sound.

 

hell yes!

 

haha pwned that article



caps arent supposed to be like a second battery there only supposed to smoothen out the voltage draw to reducce strain on your alternator and electrical system. if your lights dim with the bass then a capacitor could be the solution....

 

It could be argued that adding a capacitor will only strain the electrical system further. When the capacitor is discharged it is only another load on the alternator and battery. If your electrical system is struggling already, adding capacitors will just compound the problem and shorten the lifespan of your charging system.

 

Not really, a constant draw of power is the way to go. The voltage on the cap will never drop below or above what the system is putting out. If your altenator is struguling at 13 volts the cap will stay at 13 volts untill the system goes back to its original voltage. Also because the cap never fully discharges it stays more than half charged it will charge at a slower rate than it discharges at.

 

While I do agree that the writer's arguement is somewhat invalid, I
believe it is because he made some electrical assumptions simply because
they supported his argument rather than trying to present an accurate
model. His math, as well as the electrical theory behind it however is
very solid.

You say reactance is an AC theory, but this is completely untrue.
Reactance comes into play any time there is voltage or current
variations. Granted, reactance is more pronounced in an oscillating
system, but every time there is a voltage drop, reactance exists. By
looking at the Fourier representation for capacitance of 1/jwC (or 1/sC
if you prefer working in LaPlace domain) it is apparant that any change
in the system voltage will result in a reactive impedence being
generated in a capacitor. If a 12 VDC system was ALWAYS 12V you would be
right, but then we wouldn't even be discussing capacitors anyways since
there would be no need for them.

As for a 13% loss due to heat, that is based on his ESR calculation
which, if anything, I consider to be a bit conservative. Capacitors
designed for low voltage, high current discharge are always very
inefficient. I never recommend shunt capacitor systems to be used on
systems below 480V for this very reason, and even a ballast regulation
capacitor in a CFL operating at 120V is only about 80% efficient.
Lightbulbs with a 13% percent loss due to heat (87% efficient) also do
not exist; standard efficiency for a non-ballasted fixture is under 20%,
and only a very modern LED would break 87% efficiency.

Regarding using an AC formula to calculate the figure of 0.017 ohms, he
never once stated that this was a reactive impedence figure. In fact his
analogy with the "unobtanium" explains the concept perfectly; a
capacitor will have both AC impedence (reactive) based on its
capacitance and a DC impedence (resistive) based mostly on the
dielectric permittivity of the material in between the plates, and also
slightly on the connective wires and the plates themselves. The resistive component is in series with the capacitance. The formula
for the capacitor's reactive impedence can only be represented in the
time-domain using derivatives (or integrals depending on what parameters
are used) which is why the Fourier representation of 1/jwC is much more
common. The fact that it is in parallel is also moot in this analysis
since as long as there is voltage across a resistance, there will be a
loss of power as heat. If he were to include the decrease in resistance,
it would actually show more power being drawn from the battery since in
paralell, P=V^2/R, and lower R results in more power.

While his 9V battery test is fine conceptually, I am not sure what he is
trying to prove there. Everyone knows that a capacitor is not meant to
function as a battery and I think the writer is just playing dumb there
to support his argument with a useless test.

The "cycling" of the cap he mentioned will be dependent on the frequency
being played. Since each sine wave cycle has a positive and negative
peak, and each of these peaks is supplied with a DC power supply, the
capacitor should be cycling at about twice as fast as the frequency
being played. This is why the capacitors regulating BJT current mirrors
in a DC rectifier are cited at cycling at 120 Hz when they are made for
a 60 Hz system. While his number of 50Hz is somewhat arbitrary and I do
not agree with it, the concept of capacitors refreshing at a given cycle
on a DC system with voltage drop is well-documented and accepted.

I cannot see his graphs so I cannot comment on them.

While the writer's argument seemed biased in that he went in to prove a
theory and chose his values to support it, I agree with pretty much all
of his points. People seem love citing that bigger capacitances are
better, without realizing that more capacitance there is, the longer it
will take to discharge (time constant being t=RC, with discharge rate
being proportional to e^-1/t). Considering that at 12V, E=0.5CV^2 yields
very little energy due to being such a low value (72 J/F), the
advantages gained by a cap are minimal at best and can only be
considered a bandaid to a much bigger problem.

Can a capacitor help some systems with a very minimal voltage drop
problem? Yes, but for the price they cost as well as disadvantages cited
in the article, I would not recommend one to anybody since the money can
be easily spent fixing the problem in other ways. While it will very slightly reduce voltage spikes, overall power draw from the battery/alternator will be increased due to the reasons shown in the article. Which is worse? There are so many perspectives on that issue that I don't dare approach it.