FAQ- Why a cap is useless!

well, i dont wanna confuse anybody with complex terminology but if your light are dimming u're using more power than you are making, how to fix it? first check your battery or alternator, (please check with factory specs, dont just go to the near autoparts) and if somebody ask me, go for an extra battery is cheaper and you can find better results (an optima battery in the trunk always make my day) but what kind of result you will got? forget about best quality, higher spl (a.k.a bass), just your system playing in the right way.
You can find another things, but if you really need it is beacuse you're really in the car audio industry

And remember if somebody tell you, you need this and this before you listen your system he's a salesman not an installer

 

I read most of the posts in this thread and the only thing posted was "i'm right, your wrong" but NO solutions for the arguement that caps are worthless. If you think a cap is worthless then what is your solution for preventing your lights from dimming when the bass hits? The last post offers a solution but wouldn't the alternator be overwhelmed with having to charge two batteries? Sounds to me like you'd need to either buy a bigger alernator or 2 alternators to power the two batteries. But if you bought a bigger alternator you wouldn't have a need for the 2nd battery. I don't know that much about electricity but i do know that neither side is offering a solid solution on the arguement nor any proof that there arguement is the correct version. So instead of creating a huge thread where its only disagreement, why doesn't someone find out the right solution to the problem?

 

Better batteries, bigger alternator, upgraded wiring. All you need.

This isn't a thread about how to improve electrical problems; it's a thread about why a capacitor is pretty useless.

 

yeah im not sure that you guys know what your talking about i bought one just to complete my set i had 2 audiobahn chrome intake amps and figured why not get it just for looks. I always listened to ppl like you that said a cap was junk but to tell you the truth it made the perfomance of my stereo jump dramaticaly. Its like night and day. And who wants to spend all that money on alternators and batterys and all taht stupid crap. 90 bucks and you have a great looking chrome cap. with a blue display. seems win win to me.

 

ohh yeah and i forgot to mention any one heard of clipping its because your amps want to pull to much wattage or amperage and it cant get ahold of it quick enough hense the purpose of the cap. It holds voltage right there whenever the amps need it. Thats the reason a cap has to be no more then a foot or so away from your amp. If you guys like distortion and floppy bass go ahead and save a hundred bucks but for those who put hundreds in a system and want the benefits spend the extra hundred. Ive got 1100 watts rms of audiobahn.

 

p-p-p-placebo! lol

My friend (the topic starter) bought a stinger capacitor (far better "brand" than audiobahn) and he said the same thing until... we looked at the actual numbers.

Voltage meter was about the same during heavy bass, surprised surprise. (by "about the same" i mean the voltage was a tad lower with the capacitor).

It is completely inferior for solving power problems, but great for voltage stabilization. That's about it.



Bottom line:

Power problem- get H/O alt and/or better battery

SQ freak- cap could help slightly


"..The audio industry is the only place i know of where you can publish specs that show your product is useless and still be able to sell them------and whats worse is that technically ignorant people will argue against the math!!!!!!!..............RC.."

 

I know a guy that put in a 10F cap. not a little *****-ass .5F or 1F...a f'n 10F cap. Guess what...it didn't do ****. His lights still dimmed just as badly and nothing sounded any better. You bring me your car and I will trick you ears into hearing whatever you WANT to hear.

Now, on to clipping. Clipping is hardly caused by the amplifier not receiving enough voltage. If an amp doesn't receive enough voltage, it simply won't reach it's maximum power under optimum conditions and under extremely low voltages, will simply shut off. Clipping has to do with the signal that the amplifier sends to the speaker. A normal signal that a speaker should receive should resemble a sine wave (yellow in my rough ass drawing ).

When you start adjusting the gain, you're actually adjusting the amplitude of the sine wave. Once the peaks of the sine wave reach the peak power the amplifier can put out, the peaks begin to flatten out. This is when clipping occurs. If you push the gains too far, the amplitude of the wave becomes so large, that the sine wave turns into a square wave (black in my rough ass drawing).

 

/begin longpost


Not to try to say Im an end-all expert, but i have 2+ years of military, college level training backed by years of military experience with electronics.

A bit more than the guy who took the DeVry class and made a power supply.

Im going to lay out a few FACTS here. Capacitors may do a lot of things, but realistically they only have a few effects. The rest are sort of, like side-effects, in a circuit. Hang with me here, I wont get overly technical.



Here is the major one, I'll bold it for you. Capacitors, OPPOSE a change in voltage, up or down.

How do they do this?

By relasing that stored energy everyone talks about, and charging and drawing power. Because of that everyone seems to think of them as little batteries. This is why that guy's DC power supply steadied out, the cap resisted the voltage jumps.

Charge and Discharge times in a cap are inversely porportional to each other, and are dependent on type/material, and size. (Farads). Charging one of those caps takes a good 30 secs to a minute, right off the battery. So if it discharges just as fast, how is that slow trickle of miniscule power going to aid the amp? You cannot change the discharge time of a cap, and it doesnt under any circumstances like hard bass notes. Caps used in real electronics are on the micro and pico farad scale .000001 and .000000000001 farads, respectively. They charge, and discharge extremely fast.

These gigantic caps, just like any other cap, can add to line voltage, then immediately start to charge exactly as fast as it discharged. Hmm, this is ideal when you have a load(anything that draws power) that stays constant, but you have an unreliable supply.

However, we have the opposite, an unreliable load (power demands constantly varying), with a reliable (12-14.4V) supply. The cap is between the amp and the battery, not the between the battery and the lights. More likely than not, people curing their headlight dimming problem are probably choking their amp somewhat.(Saving headlights from the amp, rather than the other way around) It might not be noticable if you're not already near saturation (full gain or so) on your amp, but these caps are an expensive way to keep your headlights from dimming. The likelyhood also exists that these caps can actually detract from power quality in certain situations, due to their slow response times.

Moral of the story?

There is no magic bullet to increase power output when it simply doesnt exist in the first place.

Higher quality and greater number of batteries will increase the amount power available. Higher power alternator will provide the current to keep said batteries charged in any condition. A flashlight uses 2 AA batteries, not 1 and a capacitor right? Not an exact example, but the cap would discharge so fast when the battery started to die, then draw on it, and make it worse.

But anyway, thats is how you cure dimming lights, and increase the power to your amp.


Oh, there is a component called an inductor, which opposes a change in CURRENT, which might be helpful. However, they are basically a coil of wire and inherently noisy, and like to introduce interference into nearby electronics, you'll never see them on an external component level in a vehicle.

/endlongpost

 

Figured that'd shut you guys up.

 

Don't worry, it's only a matter of time before somebody comes back in here with their misconceptions and flat out wrong "facts".

Excellent post by the way. It's nice to see someone that really knows the truth about what's going on.

 

Thanks.

Either way, there will always be people who will insist on spending $100's on caps to make themselves feel better.

 

wow...VERY informative thread! i was thinking of getting a cap -- not because of dimming issues, but because my rpm's in neutral went from a solid 1000 to 500 after installing my 2 10's and a 600w mono amp....i'm figuring a more powerful alternator would be a better purchase.

 

Thanks for the post, but i was a little confused about the 30 second charge statement.

Unless your cap has a trickle charge, a car will charge a 1 farad cap in under a second. I have a 1F cap (used for a capacitor car for physics class) and i used my car to charge it. It would suck all of the power out of the alternator and even make the RPM's drop on the engine during the immense power transfer.

 

Hmm, interesting. You could be correct. My experience with car size capacitors consists of working with one of my friends on his system, he insists on having a cap. We had to charge his, but we didnt have his truck on. We connected the little board with the lights on it on the top, hooked it up to the battery and charged it. Maybe it wasnt making good contact and didnt charge right away.

Even in that case, the second the cap gets done adding to the power from its own reserve, it immediately begins to draw on the source again, effectively negating any possible gain.

 

Quick note: I have my Bachelor of Science in Electrical Engineering from the University of Texas at Austin. But, most of my study, along with my daily job is in computer chip design. I'm pretty good with component level stuff, and I have built a few amps, but I haven't really sat down and crunched numbers in a long time. So if there is a flaw in the following, or my math is incorrect, then kindly point it out. These are all rough estimates based on assumptions, but it should get the point across.

The following assumptions are made:
You have a 130 amp alternator that will output 90 amps at 14.4 volts when everything electrical in the car (except the audio system) is on.
You have a $60 battery from NAPA that has 540 CCA (which means nothing). The important number is the 270 test load amps at 12 volts.
You are running one amp on a 4 ohm load.
Your wire from the battery to the cap has a total resistance of 1 ohm.
You have a 1 Farad cap.

The following formulas and definitions are needed:

Capacitance (symbol C) is measured in Farads (symbol F).
Energy (symbol E) is measured in Joules (symbol J).
Power (symbol P) is measured in watts (symbol W).
Time (t) is measured in seconds (symbol s).
Voltage (V) is measured in volts (V).
Current (I) is measured in Amperes (I).
Resistance (R) is measured in ohms.
Electrical charge (q) is measured in coulombs (c)

V=IR, P=VI, P=E/t, C=q/(V2 - V1), E = (1/2) * q * (V2 - V1)^2

To charge a cap: Vf = Vi * (1 - e^(-t/(R*C)))
To discharge a cap: Vf = Vi * e ^ (-t/R*C))
time constant of a cap = R * C

Your amps is giving you 90 A of current at 14.4 V which equals 1296 watts. Now if you complete exhaust this must power, your amps will pull from the battery, which is 270 A at 12 V giving you 3240 watts extra. So, unless you run more that 4536 watts, you will never drop below 12 volts. Im going to assume that no one will pull that much power and therefore never go below 12 volts.

When we charge the cap, the R is 1 ohm (for the wire). When we discharge, the R is 4 ohm (for the speaker load).
You can approximate the time need to charge a cap from 0 to full voltage (in our case 12 volts) by multiplying 5 times the time constant. Or 5*R*C = 5 seconds. But once it gets to at least 12 volts, it will never drop below 12 volts. So a discharge from 14.4 V to 12 V would take 0.7 seconds and a charge from 12 V to 14.4 V would take 1.8 s. So if you draw more than 1296 watts for more than 0.7 seconds, your cap will no longer do you any good until it gets the 1.8 seconds it needs at 14.4 volts to get back to fully charged.

Now, from C=q/(V2 - V1), and E = (1/2) * q * (V2 - V1)^2 we can derive that W = (1/2) * C * (V2 - V1)^2 where V2 is 14.4 and V1 is 12. Or, simply, the amount of energy released in dropping a 1 Farad cap voltage from 14.4 V to 12 V is 2.88 J. And since that discharge happens in 0.7 s, the power delivered is a whopping 4.11 watts of power. But only for those 0.7 seconds, then it is gone until it recharges back to 14.4 V. So, after you use all of your 1296 W from the altenator (which is continuous power), you will have an extra 4.11 W for 0.7 seconds.

It has been suggested that it will help make the power supply line a little smoother. This is really a ridculous arguement, but this is a negligible effect because the small caps in the amp's power supply will filter any small amount of wiggle in the line.

It was also mentioned that clipping has nothing to do with the amp, but this isn't true. Clipping is caused when you try to amplify the signal past it's limit (usually within 87% of the rail voltage). So if the voltage drops on the main power wire, and the amp doesn't shut off, and the gains remain the same, the signal will clip. But like I said, you need to pull more than 4536 watts to drop under 12 volts so that sould never be an issue.

So what's the answer? Use that $100 or so to get a better (or another) battery. Or get yourself a bigger altenator. Or you can give your money to me and I'll tell you that your system sounds better.

 

I'm not sure if that was directed at me, but I'm pretty sure that I was the only one that tackled the clipping argument.

I know i didn't say that clipping had nothing to do with the amp. I said that clipping had very little to do with the amp not receiving enough voltage. An amp will shut down long before the low voltage causes clipping. Other than that, our explanations are the same.

Great explanation on the crapacitor though. I think we might have finally turned the tide.

 

This is what I said, without getting into ohm's law and capacitance and RC time constants
However, this is important as well, since you have both the basic electronic theory, that I posted, and the technical data that urainium posted. Seems like a tight case to me.

Again, we all offer different explanations that come to the same conclusion:

Dont waste your hard-earned cash.

 

NavyStanger, excellent post.

The only thing i'm gong to comment on is Urainium's claim that more than 4000 watts required to drop below 12V.

My friend and i have the same exact equipment. The subwoofer amplifer puts out a ~1500x1 (1 ohm) which i have verified (approx) with multimeter. We also have a Directed 500 four channel amp which is wired at its recommended 4 ohm load per speaker... giving 75x4 rms a piece.

My friend bought a HO 200 amp alternator (still 14.4V output) and the best battery he could buy, which i'm sorry i can't remember the exact specs (lets just assume better than stock). Oh, and to ensure that wiring isn't the problem, he got zero guage that branched off to four guages for the amplifiers.


Dropping down below 12V was entirely possible even with a better power supply than you mentioned. So maybe your calculations weren't "real world" enough, i don't know.

We have also been to countless audio competitions and ran in classes with far less than 4000 watts. Everyone we talked to could take their voltage below 12.


Either way, I have no doubt in my mind that caps are completely worthless. I have tested it in the real world, and thanks to Urainium, we now have a scientific explanation for why they do not help.


Thank you for all who contributed!

Locked

 

Why a Capacitor is NOT useless (corrections need to be made to the FAQ-Why a CAP is useless!!!! thread)

I am writing these because of all the incorrect post that have been posted to the Why a CAP is useless!!! Thread. The thread appears to have been brought to and end by Urainium with some people agreeing with what he wrote. While the post by Urainium is very insightful with lots of equations and calculations it is also incorrect.

The problem with Urainium’s example is that he has made a very bad error based on an extremely poor assumption that any engineer with his education background should have realized. While the formulas and equations themselves are correct he poorly assumed the resistance of the wire between the cap and the battery being to high. He as also incorrectly assumed the amplifier’s resistance. These misassumptions threw off his RC calculations therefore ruining any supporting evidence.

“Your wire from the battery to the cap has a total resistance of 1 ohm” – Urainium

If the resistance is 1 ohm your amplifier would not even work because there would be too much voltage drop across the wire. An amplifier is capable of drawing anywhere from say 1 amp to 100’s of amps. If the wire were 1 ohm then by I=V/R you could only have a maximum total current draw of 12 amps. 12 amps X 12 volts means we could only have at max a 144 watt amplifier. This is just not going to happen. The actually resistance in the wire between your battery and amplifier is many times smaller than 1 ohm. This is why your capacitor example is flawed and I will post corrected charge and recharge times for you later.

According to Introductory Circuit Analysis Seventh Edition by Boylestad, if we were using a 10 AWG# wire the resistance per 1000 ft. would be .9989 ohms. Let’s say we are using 15ft. of wire from battery to Cap. That comes to .015 Ohms for 15 ft. of wire. Now I did not calculate the wire from the negative of the battery to the cap because as we all know it is grounded as soon as possible directly to the car. So for this example we will use 15ft. of wire.

Let’s not get the capacitor charge/discharge calculations just yet. We have one more major error to fix.

“When we discharge, the R is 4 ohm (for the speaker load).” – Urainium

If we assume that we have 1 amplifier with 1 speaker attached then by I=V/R the most we could apply to this speaker (assuming a 100% efficient amplifier) is 3 amps. Well we all know that our amplifiers put out well more than three amps to our speakers so I will give us an estimated resistance of the amplifier to the battery by using known values and ohms law. Let’s say that we have an amplifier that outputs 500 Watts to the speakers and the amplifier is 70% efficient.

Output Watt / % efficiency = Total Watts consumed by amps and speakers
500 Watts / .70 = 714.28 Watts (lets keep it simple and say 714)

Now lets find the resistance of the amplifier

Amps
Watts / Voltage = Amps
714 Watts / 12 Volts = 59.5 Amps

Ohms
V / I = R
12 Volts / 59.5 Amps = .20 Ohms

Now we can do the RC circuits.

Charging the Cap

Let’s say we are using a 1 Farad cap. 5(time constants) * R * C = Time to charge in seconds. This comes to .075 seconds to fully charge a cap. Now I can tell you there is a problem here in that the calculated current draw at first is well more than what the battery could supply. So this .075s is not the actually amount of time but it is MUCH faster than the time you would get using Urainium’s incorrect model.

I’m not going to go into the rest of the charging and discharging of the capacitor at this point since I have already busted the model that for some reason this board has accepted to be correct. It will take a little longer for the cap to charge from 12 to 14 (about .045 sec) than to discharge from 14 to 12 (about .03 sec if the cap is the sole source, will probably last about .06 seconds when working together with the charging system). But at this point you need to take into account that music is transient and will be changing its current draw continuously so it would be interesting to see a chart of capacitor charging vs its discharging taking into account the ever changing current demand on the system

Why a capacitor is not always useless

I can speak from experience that some lower quality amplifiers will greatly benefit from the addition of a properly installed capacitor. I have also found that higher quality amps really don’t gain any improved sound quality from a capacitor.

By no means is a capacitor worthless and it does not increase the load on the charging system anymore than if it wasn’t there. A capacitor is not going to act as a band-aid for a weak charging system in anyway but it will act to stabilize voltage in any system. Its sole purpose is to oppose changing voltage.

If you think that a capacitor is totally useless then I invite you to take out the large stiffing caps that are inside your amps. Removing these caps will reduce performance and sound quality of your amplifiers. Adding more capacitance may help in some situations but never hurt. Before deciding if a cap will not help you should try one on a trail basis from a friend and see if it helps any. If it doesn’t then you know you don’t need one.

Bachelor of Science in Electrical Engineering from Purdue