WTB capacitor

[fogelmclovin 10+ year member]

I need an inexpensive capacitor. It's for a 520 RMS amp do it doesn't have to be huge but I do prefer at least 1 farad. I'm looking to spend less than 40 shipped. Obviously I don't care if its used, just not one that's thrashed and 2 years old.

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[fogelmclovin 10+ year member]

quit busting his balls. hes got alot on his plate, and wants to enjoy some bump.
yall were there once, so quit bring pricks and explain what needs to be done. grow up kids.

Thanks man. The kind of smartass replies this CLASSIFIED ad got, are exactly why I don't ask many of the questions I would like to ask on this forum. I have interacted with a few people here. Most are cool, but there is a small percentage of elitests that look down their nose at anyone with less car audio knowledge.

 

[93EXT 10+ year member]

Some of them arent looking down, I know I'd rather see you buy an alt then waste $ on a cap then realise later an alt would of been better idea, BUT it was only 40$ //content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif

I know I paid 200$ about 9 years ago when I bought a 1 Farad, but that kind of stuff has come along way since. and theres only like one shop in town, so no competition= high prices.

 

[fatryan 5,000+ posts]

i love how the vast majority of this forum has jumped on the "cap is just a bandaid and a waste of money" bandwagon when 99% of them dont even know how a cap works. yes when your considering buying like a 1-farad cap new versus a decent HO alt off ebay for maybe...$200, its not very cost effective to buy the cap. but people start talking about caps like theyre going to hurt you. a cap just RESISTS QUICK VOLTAGE CHANGES. it does this with the stored charge inside. thats what its for. quick voltage drops can damage your cars electrical system. and yes, if your running 1500watts off a 105amp stock alt you need a HO alt whether you like it or not. but when you just got a little dimming and you can get a used cap for $25 shipped, thats alot better than spending a couple hundred on a HO alt when you dont have money to be throwing around.

but aside from that, when voltage drops are a problem the first thing you should always ALWAYS do is the big 3. its the most cost effective fix to any electrical dimming issue. may not fix it all but its where you must start.

 

[fogelmclovin 10+ year member]

if your running 1500watts off a 105amp stock alt you need a HO alt whether you like it or not. but when you just got a little dimming and you can get a used cap for $25 shipped, thats alot better than spending a couple hundred on a HO alt when you dont have money to be throwing around.
Thank you.

When I was 16 to 21 I had a few small systems. The biggest amp I ever had was an old Kenwood 921. I don't even remember how many RMS watts it had. Like 125-150 a channel. That little thing made my dash lights dim a little. So with this system doing aroud 720 RMS total, I figured it would be better to have a cap.

 

[fatryan 5,000+ posts]

just make sure you do the big 3 first. it only costs like $20 tops

 

[lilmaniac2 5,000+ posts]

Ok alot of misinformation in this thread, and that link to that "test" alwasy gives me a good laugh. I have a 2 year degree in Digital Electronics and wrote my research paper of capacitors. Here is what I came up with.

1) the usage of a capacitor on the PCB of an amp is completely different than what this kid is wanting to do.

2) The way a capacitor works is it leads in voltage but lags in current ( I think thats right, its been 3 years since I've taken a digi class ) What it does is similar to a water hose. If you have 1 gallon of water and you put it through a large hose its going to empty faster. However if you run it through a reducer it will last longer. the reducer is the capacitor. It starves your amp of the " blood " it needs ie current.

I personally witnessed a SPL test of a small batcap ( 400 I think ) and when we unhooked it the amp did a bit more power and the score went up .3db thats because it didnt have to run through the "reducer"

3) Capacitors do have there place, Voltage conditioners and filters are 2. however they dont work like they are marketed.

The reason why there are so many manufactures of capacitors is b/c they are cheap to make, good mark up, and dont fall under the restrictions that batteries do

 

[93EXT 10+ year member]

Someone with knowledge, thank god //content.invisioncic.com/y282845/emoticons/yourock.gif.93d527d504392657a77fa577298070a1.gif

 

[fatryan 5,000+ posts]

Ok alot of misinformation in this thread, and that link to that "test" alwasy gives me a good laugh. I have a 2 year degree in Digital Electronics and wrote my research paper of capacitors. Here is what I came up with.
1) the usage of a capacitor on the PCB of an amp is completely different than what this kid is wanting to do.

2) The way a capacitor works is it leads in voltage but lags in current ( I think thats right, its been 3 years since I've taken a digi class ) What it does is similar to a water hose. If you have 1 gallon of water and you put it through a large hose its going to empty faster. However if you run it through a reducer it will last longer. the reducer is the capacitor. It starves your amp of the " blood " it needs ie current.

I personally witnessed a SPL test of a small batcap ( 400 I think ) and when we unhooked it the amp did a bit more power and the score went up .3db thats because it didnt have to run through the "reducer"

3) Capacitors do have there place, Voltage conditioners and filters are 2. however they dont work like they are marketed.

The reason why there are so many manufactures of capacitors is b/c they are cheap to make, good mark up, and dont fall under the restrictions that batteries do

what makes you think a cap is a reducer?

 

[hammerhead 10+ year member]

written by Richard Clark which examines capacitors in every detail. Read first...then ask questions.

Lesson 1

Ok “powertrip” how about we have a discussion in basic electrical theory? At the end of this thread you should be the one that can explain to the world that according to ohms law it is impossible for these things to do any good. That is of course if you can admit that they do obey ohms law. We will do this a little at a time so how about you humor me and stick to my questions. We will do them a couple at a time so everyone can follow along. Let’s do a little calculation. Suppose we have a resistor that is .017 ohms (seventeen milliohms). I think that is what you say the ESR of the giant caps is.

The ones I have seen have measured higher but I will give you the benefit of the doubt. According to ohms law how many volts are dropped across .017 ohms if 100 amps of current are flowing? How about if we up the current to 300 amps? Let’s establish the answers to these questions before we go any farther. If we can't agree on the answer to this there is no hope we will ever get to the truth.

Lesson 2

Thanks David you are exactly right. If anyone wants this explained please ask David to clarify it. If everyone is going to follow this and understand fully the final conclusion it is important that no one miss any steps. There will be about ten lessons. Since power trip has left the building we will continue with the rest of the class. ESR stands for equivalent series resistance. This means exactly what it sounds like. It means that if we have a source of voltage it will behave exactly as if it has a resistor of the same value in series with its output. An amplifier has ESR, a power supply has ESR, a battery has ESR, and yes, a cap has ESR. Components have ESR’s because we do not have perfect conductors to make things from.

And now for the homework. Last night we learned that if 100 amps flows through .017 ohms there will be a voltage drop of 1.7 volts. And if the amp flow increases to 300 amps the voltage drop will increase to 5.1 volts.

For the sake of theory only let’s say we have built the largest cap in the universe and it has billions and billions of Farads. Its plates are made of a newly discovered material we'll call unobtanium. This new material has no resistance therefore our super cap has an ESR of ZERO ohms. We then charge the capacitor to 14.2 volts. We then place a resistor with a value of .017 ohms in series with one of the terminals of this cap. The question is: If we place a load that draws 100 amps from this cap what will the resulting voltage be on the load side of the resistor? What will the voltage be on the cap side of the resistor? What about if we increase the load to 300 amps? What will the voltages be on each side of the resistor?

Lesson 3

Ok now that we have studied ESR and understand what it is and it’s effect on the working of a circuit we will move on to another subject. But don’t forget about ESR as it is one of the important final building blocks in our search for truth about caps and we will come back to it. Today we will review the important concepts about total energy storage in a device like a cap. This has been covered in earlier posts (and I will say quite correctly) but I am going to expand on it as well as reiterate it for those who did not get to read it. Besides, I think I can simplify it a little.

In electronics, we measure power in watts. Wattage tells us how much work a device can do. But a wattage rating does not tell us anything about how long we can sustain that work. When we add the element of time to our wattage, we use a value we call Joules. A joule is a watt second. This means that one Joule of energy can provide a watt for a second. Ten joules can provide a watt for ten seconds or ten watts for one second or five watts for two seconds one hundred watts for a tenth of a second, and so on.

The formula for determining the total joules stored in a capacitor is very simple. We take one half the capacitor’s value in farads and multiply it times the squared charge voltage. For example a one farad cap charged to 14 volts would be .5 X (14x14) = 98 or .5 X 196 = 98 Joules. A 20 farad cap charged to 14 volts would be 10 X (14x14) = 1960 Joules.

There is a very important concept to understand about energy storage. A capacitor actually stores electricity.

Batteries don’t. Batteries have the potential to produce electricity by means of a chemical reaction but caps actually store electrons on their plates in the form of an electrostatic charge. In our next two lessons we will learn why this is important to know.

But first, the homework. This is a “think about it question”. We have learned that a Joule is a watt second. A Yellow top battery is rated at 65 amp hours. This means it can provide 65 amps for an hour. The question is how many Joules does this represent? Since this is a thought question, it would really help if whoever answers would show us your math.

Lesson 4

In the actual real world the voltage of the battery would drop a little from its open circuit voltage of 12.8 volts with a 65 amp load. In the case of the yellow top its actual voltage at 65 amps is about 12.2v when fully charged. By the end of the hour it would be down to about 10v. If we use 11 as an average our answer would be........ 2,574,000. Now that's still a lot of joules! Now actually this is not enough to totally kill the battery but at this point there isn't much left in it. This brings us to a very important fact. The energy in a battery will be depleted almost completely by the time it is down to 10 volts.

Lesson 4 (continued)

By the time we have removed those 2.5 million joules from the battery it probably doesn't have more than a hundred thousand joules left. We can almost totally deplete the battery's energy and never drop below 10 volts. This is because the battery doesn't store electricity. It stores chemicals. A chemical reaction produces the electricity. Storing actual electrical charges is very inefficient.

Look at our poor capacitor. Even if we made one as big as a battery it would still only be good for perhaps fifty to one hundred thousand joules---less than that left in a nearly dead battery. But if that were not enough there's more bad news. This exercise will be tonight’s homework.

A capacitor is like a gas tank in a car. The pump can only remove gas down to the pickup point. Any gas below this point can never be removed by the pump. If we charge a 20 farad cap to 14 volts we know from previous lessons that it will contain 1,960 joules. If we use that cap in a system and load it till it drops to 10 volts along with our battery how many joules will we have removed from the cap? How many joules will remain in the cap that we can never benefit from if our system never drops below 10 volts?

Lesson 5

In our last lesson we learned that caps actually store charges on their plates. And of the 1960 joules stored in a 20 Farad cap, 1000 of them sit at a potential below 10 volts. This means there is no way they can ever be used by an operational audio system. Today we will look at another loss factor. It has to do with the loss factor due to the ESR of the cap.

We have already studied voltage drop due to ESR but now let’s view it from an energy/watts standpoint. Let’s clarify things. The power delivered to the stereo by the battery and alternator bypass the cap. They merely flow by its terminals. If the cap charge is lower than the battery/alternator potential current will flow INTO the cap until it reaches equilibrium with the Battery/Alternator. If the B/A potential is lower than the charge potential of the cap current will flow OUT of the cap to the battery and or the amp.

Always remember that voltage always flows from the highest potential to the lowest potential, just like water. Current does not however flow into the alternator even if it is lower than the battery and cap because it has diodes on its output that only let current flow FROM its output. Now whenever any current flows into or out of the cap it must pass thru the ESR of the cap. The resistance is really distributed throughout the cap but it behaves just like it was right on the output terminal as in a series circuit location in the circuit loop does not matter. Now suppose our 20 farad cap is charged to 14.2 volts and we place a load on its output. This load is the same one that we used in lesson 2 to cause 100 amps of current to flow from our unlimited capacity cap. Only now we have our smaller 20 farad cap.

 

[fatryan 5,000+ posts]

no offense but i dont really see how those lessons even explained anything about the issue this forum has with caps.

but let me put it this way: caps charge and discharge with the same time constant (tou=RC). that said, they charge and discharge at the same rate. take a look at graphs of a cap charging and discharging, they are exact inverses of eachother with the time constant at the same point in both graphs. my point behind this is that its untrue to claim that a cap "steals" the current from the amp to recharge itself because at the same time it takes an equal time to discharge. as claimed about everything in a circuit (bat, cap, amp) all have an internal resistance therefore the cap wont "instantly" discharge when the amp needs current. if it did instantly discharge it would HAVE TO instantly charge then too. and on top of that when the cap starts feeding the amp that doesnt mean that the electrical system stops feeding the amp too. you could also simply measure the voltage changes during a time of heavy current draw in your system when a cap is installed and not installed. youll find that the voltage difference between the 2 would be very minimal if even measurable. and since resistance is constant (internal resistance is essentially 0 when compared to other resistors in the circuit), ohm's law then tells us that current is directly proportional to voltage...current in = current out.

im not saying a cap HELPS anything. it doesnt, it just protects your electrical system givin specific circumstances. its use in a car audio DC circuit is very limited, but i still believe it has its place.

 

[hammerhead 10+ year member]

That was a Article written by Richard Clark only part of it I pasted. If you read the article it states Caps are useless in car audio.