Help me understand capacitors

[jmanpc 5,000+ posts]

So I've read over and over why capacitors don't work, etc., but no one really gets into the nitty gritty math. I've looked up some stuff, and seen that a Farad is 1 Ampere x 1 Second / Voltage.

I've come to conclude that in the setting of car audio, that 1 Farad = 12 Amperes x 1 second / 12 volts.

So, a capacitor can provide 12 amps for 1 second. However, this assumes that the capacitor remains at 12v throughout the process, which it obviously does not.

Can someone explain to me how a 50A current draw would affect a 1 farad capacitor? Like, can we mathematically demonstrate how long it would take the voltage to drop to say from 14.4v to 11v, under a 50A current draw on a 1F cap? How much amperage would it actually provide to an amplifier?

Help me further my understanding, people.

Click to read more...

 

[jmanpc 5,000+ posts]

ttt

 

[ultimate157 5,000+ posts]

50A current draw from a 14v source is equal to ~0.28 ohms of resistance.

The RC time constant is a fixed time interval equal to resistance times capacitance. It takes ~5 time intervals to fully discharge a capacitor.

280x10^-3 ohms x 1 F = .28 seconds. Five times this is ~1.4 seconds. This is from fully charged to fully discharged.

Since the discharge curve of a capacitor resembles exponential decay, it takes a very short time for the fully charged capacitor (sitting @ 14v) to reach 11v. This would happen within 1 time constant, or less than .28 seconds. (Edit: By looking at a discharge curve, 80% charge is remaining at just over 1/5th the RC time constant... in this example that is approximately 0.056 seconds)

It happens VERY quickly when using a capacitor as a voltage source.

You have to remember this is using the capacitor as the only source. Complimenting a 14vdc source will decrease the discharge rate.

 

[ngsm13 5,000+ posts]

nG

 

[ultimate157 5,000+ posts]

What is the time interval of that graph?

 

[ngsm13 5,000+ posts]

Like less than a second...

Reinforces the fact that a capacitor is a band-aid for a weak electrical system...

nG

 

[IDSkoT 5,000+ posts]

Caps are basically a small battery. They charge up, and once they're charged they stay charged until your system needs the extra boost.

In theory they're good... but in reality they're bad. In essence they're just a weak second battery. So why not spend another $100-50 dollars and buy another battery?! //content.invisioncic.com/y282845/emoticons/wow.gif.23d729408e9177caa2a0ed6a2ba6588e.gif

From my understanding, they add more of a strain on your electrical system and most caps are only 1 farad [i believe an average battery is about ~10,000 farad... probably more]

 

[jmanpc 5,000+ posts]

50A current draw from a 14v source is equal to ~0.28 ohms of resistance.
The RC time constant is a fixed time interval equal to resistance times capacitance. It takes ~5 time intervals to fully discharge a capacitor.

280x10^-3 ohms x 1 F = .28 seconds. Five times this is ~1.4 seconds. This is from fully charged to fully discharged.

Since the discharge curve of a capacitor resembles exponential decay, it takes a very short time for the fully charged capacitor (sitting @ 14v) to reach 11v. This would happen within 1 time constant, or less than .28 seconds. (Edit: By looking at a discharge curve, 80% charge is remaining at just over 1/5th the RC time constant... in this example that is approximately 0.056 seconds)

It happens VERY quickly when using a capacitor as a voltage source.

You have to remember this is using the capacitor as the only source. Complimenting a 14vdc source will decrease the discharge rate.

Excellent. Thank you.

 

[Nate_100 10+ year member]

Wow!

 

[helotaxi 5,000+ posts]

I did the math a while back and the usable current reserve on a cap comes out to around 0.5 A*sec/F. Basically nothing.

 

[mrogowski 5,000+ posts]

50A current draw from a 14v source is equal to ~0.28 ohms of resistance.
The RC time constant is a fixed time interval equal to resistance times capacitance. It takes ~5 time intervals to fully discharge a capacitor.

280x10^-3 ohms x 1 F = .28 seconds. Five times this is ~1.4 seconds. This is from fully charged to fully discharged.

Since the discharge curve of a capacitor resembles exponential decay, it takes a very short time for the fully charged capacitor (sitting @ 14v) to reach 11v. This would happen within 1 time constant, or less than .28 seconds. (Edit: By looking at a discharge curve, 80% charge is remaining at just over 1/5th the RC time constant... in this example that is approximately 0.056 seconds)

It happens VERY quickly when using a capacitor as a voltage source.

You have to remember this is using the capacitor as the only source. Complimenting a 14vdc source will decrease the discharge rate.

Wow. That was sure a blast form the past. You dusted off a few neurons for me.. //content.invisioncic.com/y282845/emoticons/cool.gif.3bcaf8f141236c00f8044d07150e34f7.gif

 

[ultimate157 5,000+ posts]

Wow. That was sure a blast form the past. You dusted off a few neurons for me.. //content.invisioncic.com/y282845/emoticons/cool.gif.3bcaf8f141236c00f8044d07150e34f7.gif

//content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif Currently going to school for electrical engineering.

 

[audioarsonal 10+ year member]

Caps are basically a small battery. They charge up, and once they're charged they stay charged until your system needs the extra boost.
In theory they're good... but in reality they're bad. In essence they're just a weak second battery. So why not spend another $100-50 dollars and buy another battery?! //content.invisioncic.com/y282845/emoticons/wow.gif.23d729408e9177caa2a0ed6a2ba6588e.gif

From my understanding, they add more of a strain on your electrical system and most caps are only 1 farad [i believe an average battery is about ~10,000 farad... probably more]

Sorry but they are not like a small battery.They DO NOT produce any current of there own.They only STORE a charge to stabilize voltage.But as th graph displays it is for such a short time that it is ineffective.

You have to have a realitvly stable voltage to begin with in order for a cap to have any chance of keeping up.

 

[helotaxi 5,000+ posts]

Sorry but they are not like a small battery.They DO NOT produce any current of there own.They only STORE a charge to stabilize voltage.

Hate to break it to you, but a battery does not produce any current of its own either. It just stores the charge it is given just like a cap. The difference is that a battery stores it as chemical energy and the cap stores it in an electrical field. The battery has a much higher energy density and thus a much larger capacity.

 

[Lakota 5,000+ posts]

Sorry but they are not like a small battery.They DO NOT produce any current of there own

That's what a generator does.