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Circlotron 19th May 2003 02:31 PM

50 Farad 18 volt capacitor. Yoicks!
Look under Products / The C.A.P.

jan.didden 19th May 2003 02:37 PM

lots of power
So, you can charge your amp at night, play it disconnected from mains during the day???;)
Any idea about the price?

Jan Didden

KevinLee 19th May 2003 02:39 PM


Rockford Fosgate has a 100 farad cap for car audio. When you have a system drawing 2 or 3 thousand amps of current, I guess you might need one of these.


Circlotron 19th May 2003 02:42 PM

I think I saw a either a 5 or a 15 Farad one for USD$899 (!) A bit like asking abou the fuel consumption of a Rolls Royce - if you need to know then you can't afford it. Probably costs even more if you buy it charged up.

Circlotron 19th May 2003 02:46 PM

Mine's bigger than your's.
I have seen pics of EPCOS ultracaps at 2700 Farads but only 2.7 volts each. Used for starting big diesel engines. :cool: Only have a limited (but reasonably large) amount of charge / discharge cycles though.

jan.didden 19th May 2003 02:58 PM

50 farad
What was again the definition of a farad - wasn't it Amps x Secs, so 50 farad would be 50 amps for a sec or 10 amps for 5 secs? Come to think of it, not really THAT impressive compared with a good battery, isn't it? $899!

Jan Didden

Circlotron 20th May 2003 03:38 AM

Re: 50 farad

Originally posted by janneman
What was again the definition of a farad - wasn't it Amps x Secs, so 50 farad would be 50 amps for a sec or 10 amps for 5 secs? Come to think of it, not really THAT impressive compared with a good battery, isn't it? $899!

Jan Didden

A Farad will supply 1 amp for 1 second and decrease by 1 volt. So 50 Farads will supply 10 amps for 5 sec and go down from say 14 to 13 volts.

With batteries vs caps, a car battery IIRC has an ESR of about 10 milliohms depending on temperature, state of charge etc. Caps can go way lower. Whether it really makes a worthwile difference is another thing... Open to opinions here. My car audio thing is barely suitable for listening to the news. :cannotbe:

zx3chris 20th May 2003 04:22 AM

the manufacturers of car audio caps claim that they will keep your system voltage stable at 14volts during peak--this of course doesnt happen bc the usable joules of a capacitor are so small it is negligable. btw, these super caps, the alumapro for example, (which sells for 500), is just a bank of aerogel capacitors.. they are about $1 a peice retail, and if u put a bunch in series to bring up the operating voltage, then a bunch in parallel to bring the capacitance back up, you can make one yourself for under 100$ for 50-100 farad... in any event, tests have been done showing that these caps which try to keep the system voltage at ~14 during peaks actually do not work--- i have a very long tutorial written by richard clark which ill post after this--its very elementary but its written to people who know nothing---a good read however


zx3chris 20th May 2003 04:23 AM

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. Lets 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? Lets 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.
E=I x R so...
E=100 x .017 = 1.7
E=300 x .017 = 5.1

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 it's output. An amplifier has ESR. A powersupply has ESR. A battery has ESR, and yes a cap has ESR. Components have ESR because we do not have perfect conductors to make things from. 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 lets say we have built the largest cap in the universe. Billions and Billions of Farads. It's 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 charge the cap 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?
The voltage on the cap side would be 14.2 volts, because regardless of load that huge cap will maintain 14.2.
On the load side, at 100amps draw the voltage would be 12.5, and at a 300amp draw the voltage would be 8.9.

Alex first let me say that you almost got your answers right. The 8.9 should have been 9.1 but since you were so close it is obvious that you understand what's happening. Just a simple math error. I'm not so good at it either. But don't get ahead of the class. You may think you see where we are headed but you are really going to be surprised to see where we are actually going to end up. As for my comments to cade. Yes I was upset at his comments. I could not believe the attitude of his post. I took it not as a sincere inquiry but a matter of fact statement with a "dare to disprove" presentation. Yes his math was right but it had little to do with the issue he raised. He provided enough facts in his posts to prove himself wrong. I thought everyone could see that. And when people give testimonials to support a failed assumption I really get confused. Audio just seems to be filled with that. Everybody seems to make such a big deal that lots of competitors use these things. That still doesn't mean anything if the facts don't add up. It just means they are mistaken. Let me tell you a story that is true. I can let you call the man involved if you would like to check out it's authenticity. A friend of mine was on a pit crew at Indy several years back. The team he was on was tearing up the track in qualifying. They had the pole position. Everyone was watching them intently to see what their secret was. Early in the day someone had accidently dropped a hammer and put a small dent in one of the tailpipes on the car. Somebody must have spotted it. The next day almost every body that was running that kind of car had a dent in exactly the same place! Now you would think that by the time someone got to Indy ------- I studied Psychology in college and had to give it up. I have no trouble with physics but I just don't understand people. Already I can see that from your comments on the last questions you can see he is wrong if we go no farther. But you can bet that even when it is as plain as the sun in the noon sky there will still be a few that will say "i don't care about the facts I know what i hear"

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 cap 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.
Watts = volts x amps
battery: 12.8 volts/65 amps
watts = 12.8 x 65 = 832 watts/second
3600 seconds = one hour
832 x 3600 = 2,995,200 watts/hour
one joule = one watt/second
2,995,200 joules

zx3chris 20th May 2003 04:24 AM

Lesson 4
In the actual real world the voltage of the battery would drop a little from it's open circuit voltage of 12.8 v with a 65 amp load. In the case of the yellow top it's 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. 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 tonights 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?
20 F cap.
10*(14V^2)=1960 J
10*(10V^2)=1000 J
1960 J @ 14V - 1000 J @ 10V= 960 J
If I understood the fuel pump analogy you are saying that you can only use 960 J of the 1960 J (49%) before the battery is nearly depleted, leaving 1000 J unused ?
Buzz-- you are exactly correct. The simple facts are that only about half the joules in the cap can ever be used by the system. David I seemed to have lost you a little. The point was that the battery energy is almost 95%+ available to the system and only half the caps energy is ever available. Even when drained to 10 volts one half of it's charge is at a potential too low to ever be use. This means that of the wopping 1960 joules availale only 960 of them are even usable! Please check Buzzes post and make sure you get this and anyone else who is following. The problem I have tonight is that I wanted to continue with this post with lesson 5 but after reading the post by powertrip I am completely surprised. Instead of the lesson I was going to do as part of my continuing explanation I feel it is absolutely necessary to cover his comments. Powertrip it is obvious why you may have posted your first post. At this point I will say I am sorry for what I said and I really feel sorry for you. To be in a technical job and make those comments about the series parallel thing. It is clear that you don't have even an elementary understanding of this subject. In fact if no one following can explain to powertrip what is wrong with his logic tonight then I am so far ahead of everyone that I might as well stop right now. Please somebody prove to me that somebody in car audio knows enough to tell him that his post doesn't make a bit of sense. If nobody can I will point out what is wrong with his post and then stop wasting my time on this subject.

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