Capacitor Ratings

What are some of the diferences between capacitors used between your battery and amps, and the capacitors used within an amplifier of or before ac electric motors?  How are they rated?  They have a microfarad spec, voltage, amp, and or watt rating don't they.

Is a 1 Farad capacitor actually 1,000,000 microfarads?  My instructor in my fundamentals of electricity class had a capacitor he used on an AC motor to stabalize the power consumption which was 50 micofarads in capacity (It was actually 2x25uf wired in parallel).  The two togheter were about the size of a 1farad capacitor you buy for car audio amps.  What are the differences between these two?

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Poly Dollies

correction farads are a measurement of the storage of energy.  Caps store the farads.  sorry.

While you are not completely wrong, boxmaker, you are FAR from accurate... Capacitors ARE rated in farads, (named after Michael Faraday, the discoverer of electro-magnetic induction, electro-magnetic rotations, the magneto, etc.) it's true - this is the capacity of charge stored. They also have voltage ratings. The voltage times the capacitance equals the ENERGY STORED, in joules, or watt-seconds. One farad at one volt equals one joule, which is also one volt at one amp for one second... at this drain, this capacitor charge would last one second, and then be (for all intents and purposes) dead, until it is recharged. By doing the math, you can see that a 1 farad capacitor, charged to 14.4 volts, would be a total charge capacity of 14.4 joules. If you amp is demanding 300 amps from the power supply, it won't get it from the battery! - but the cap CAN deliver! 14.4J divided by 300A equals .048 seconds. This means the cap can supply 300 amps at 14.4 volts for 48 milliseconds - if you have two farads, that time frame goes to .096 seconds - almost one tenth of a second! PLENTY of time for the battery to come up with the huevos to supply 300 amps...

(side note: I have some capacitors at my house that are actually more storage capacity than that, even though they are only very small capacitance. They just happen to be rated at (one of them) 20,000 volts and 14 microfarads, but the other is rated 50,000 volts and 7 microfarads. I use them in parallel (21 microfarads at around 18,000 volts) to shrink quarters... pretty cool, really! These came from Lawrence Livermore Labs, and were used in the original Shiva Fusion Reactor Project, as energy discharge caps in the laser arms.)

A capacitor is built of plates - conductive plates and insulating plates - called dielectric, of which there are THOUSANDS - anything that is an insulator, from air to the skin on your hand - can be a dielectric, all with various characteristics that make them suitable for various flavors of capacitors. Capacitance is determined by how large the conductive plates are, (the surface area) and how close they are together - the larger they are, the more capacitance, and the closer they are, the more capacitance. If you want voltage capacity, you will trade off capacitance, because your dielectric must be thicker, to stand off the higher voltage, BUT if you make the plates bigger, you can end up with a capacitive value close to where you were before you desired a high voltage... The reverse also holds true - if you want higher capacitance, you can get the plates closer together, at the expense of voltage rating, but if you make the plates bigger, you will regain the lost capacity - are you following me?

The capacitors used for car stereos are low voltage, (EXTREMELY thin dielectric - so the plates are close together) but high capacitance - close together plus EXTREMELY large surface area on the plates; on the order of an acre or so - and I mean that literally - they get this huge surface area by acid etching the aluminum foil of the plates (and they use some VERY exotic manufacturing processes in the production of the foil as well) The motor start/run capacitors you refer to that your instructor showed you are (relatively) high voltage devices, which means they do not have much plate area, but the plates are separated by a fairly thick dielectric.

There is the long answer to your original question.

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It all reminds me of something that Molière once said to Guy de Maupassant at a café in Vienna: "That's nice. You should write it down."

I just wanted to say, because i read what Boxmaker wrote that i have two 22 uf, 100v capacitors connected in parallel in each of my front car speakers and it solved the problem that was too much bass. Instead of doing a lot of nothing, it completely solved the problem, so it helps.

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Nuno Gomes

i've got to give it up to haemphyst. that was a great explanation of capacitors.

for you newbies out there this is exactly why it is better to have a cap than a second battery, capacitors react faster than bateries!!!!

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";P

you sound like you know a thing or two.

what other projects have you done aside from automotive?

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";P

wow, the knoledge here never seases to amaze me.   That is the most in depth explanation that i have read since class in 99   haha

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2009 0-1000 Trunk WR 154.0DB 2009 1001+ Trunk WR
2007 USACI World Champion
2007 World Record
2006 USACI Finals 2nd Place

Can you divide 14.4J by 300amps?  You said joules are watt-seconds so a watt-second divided by amps is equal to seconds? I was a bit concused with the units.

House voltage is at 120v and car voltage is at 14.4 so if you were to have a one farad capacitor in each case the one for the house would be huge because of the voltage capacity right (greater dielectric fluid needed). Our instructor was saying something like that.

Capacitance adds up in the opposite way a resistor does right.  In parallel the capacitance increases (like having a bigger/longer capacitor) and in series it's 1/Cap(total) = 1/Cap(1) +1/Cap(2) + 1/Cap(3)...   What are some advantages of connecting capacitors in series?  Higher voltage rating?

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Poly Dollies

One amp at one volt for one second equals one joule. If you divide joules by seconds, you get volt-amps, so yes... you can divide time by current, leaving voltage and time. Since voltage,in this case, is fixed at 14.4, this leaves time as the unknown, which is what you are solving for with 14.4J/300A.

(Here is an interesting page I found on the construction of capacitors, and the types of dielectrics, life-spans, and various other goodies in reference to them.)

A 1F cap, rated for 120 volts AC, using the SAME materials, would be about 500 times the volume of a 1F, 20v cap, because you need a cap rated for 340v DC, which is twice the peak voltage that would be coming from the wall - 120 X 1.414 = 170. So you need a dielectric that is 16 times thicker than the existing dielectric (to stand off the voltage), and to get the capacitance back up to the 1F, you need to increase the plate size by 16

Capacitors in series behave the same as resistors in parallel. The math is a little weird, and I don't know how to type it, but I think it goes a little like this... (1/((c1/1)+(c2/1)+(c3/1)...(cN/1))). The voltage capacity does increase as well, but only up to the voltage rating of the lowest rated one, times the number of capacitors in the string. What this means is if you have 3 caps in series, and two of them are 50 microF 100 volt, but the third one is a 50 volt piece, them the string would be rated for 150 volts.

Parallel is MUCH easier - add the values of the capacitors together, and the voltage is the same as the lowest rated one in the array.

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It all reminds me of something that Molière once said to Guy de Maupassant at a café in Vienna: "That's nice. You should write it down."