how long does an 8 farad cap lasts?

How long does an 8 farad cap lasts on charging a battery on 2500 rms of power ?...The setup is in my room so thats why i'm using battery and cap...

Are you asking how much extra playing time the cap is going to give you?

well that too, but just wanna know the life time of the current in the cap when under a frequent useage

I still do not understand your question, but the cap, without the battery connected will run your amp for less than a couple seconds.

daniel gt1 wrote:

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How long does an 8 farad cap last....

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Over 1,000 times less than a $35 NP7-12 battery (AGM 12V 7AH).

oldspark wrote:

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daniel gt1 wrote:

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How long does an 8 farad cap last....

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Over 1,000 times less than a $35 NP7-12 battery (AGM 12V 7AH).

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That may be true, but the instantaneous current that the cap can VERY safely supply would freakin' melt to a pool of plastic and lead the 12V7AH battery... You've hawked this battery now at least twice as the savior of voltage drops and sags, and it seems as though you believe that it will do much more thna a cap will...

You're no more correct in your assumptions than anybody else believing that a cap is their solution. Here's why:

The internal resistance for the AGM 12V 7AH battery you are COMPLETELY and (as far as I am concerned) inaccurately in love with is around .025 ohms, allowing for a MAXIMUM discharge current for that battery of 40A for 5 seconds. Ohm's law tells us that equal voltage sources wired in parallel will provide equal current to the load. If your battery can only provide 40A, and the primary battery will only be able to provide the same thing, then your secondary battery is as worthless as is ANY cap. An 80A drain? That's about an 800 watt Class D amplifier. Beyond that, you will be pulling more than you AGM 12V 7AH battery can safely supply, deeply discharging it, and providing an additional load for the alternator to charge - even more than a cap! Pulling that much from that battery and having an alternator under the hood (a current source of in most cases a MINIMUM of 65 to 80A) you will not be able to limit the charging current effectively to 2.1A (INTITAL, not a continuous current by ANY means)

A decent 8F cap will have an internal resistance of somewhere around .002 ohms, about 10 times lower, for an order of magnitude better peak current capacity in the same time frame, i.e. four HUNDRED amps for the same 5 seconds!

Now, caps simply don't work in time frames that long. They don't. They want to dump ALL of the current as fast as possible. The internal resiatance is the limiting factor!

8F times 14.4V = 115.2J of energy.
115.2J /.002 ohms internal resistance = 57,600A! Peak instantaneous current

Let's say it dumps in .5 seconds. This is four THOUSAND amps! .050 seconds? Fourty thousand amps! Now, granted, the 40kA math is far fetched, but it demonstrates to you the dramatic currrent superiority of caps over AGM batteries.

NEITHER will do what anybody thinks they do, in the way they are pitched!

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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 know, and I apologize.
I'm having a fundamental go at a very basic level.

Do you happen to know the input (power) impedance of the amplifier involved?


PS - Granted, for a 200A as load described, I'd use a larger battery. My point was merely how long the battery would last compared to a capacitor. How long would the 8F last? [Noting the energy stored at 14.4V is ~830J not 115.2J (even though only ~570J of that can be used if discharged to 8V), but what is the energy in the NP7-12 from 12.7 to 30% discharge. That is my point.

I wasn't considering transient response in this thread.
Since you brought the issue up, I'd love to see a transient analysis - hence why I asked above for the amp's PSU impedence. (S-domain?)

I have never seen an amplifier list the spec for "Power input resistance"

And hence the plot thickens....
I wonder why not?

Can someone calculate or measure it?

FYI Mr I - I updated my reply during/after your reply....

ok guys..here's the question in a clearer format


" how long...does the current in the 8 farad cap lasts , when charging the battery\or being used for powering the amp ?"

I do not know how to calculate such a number.  But I have seen 200 amps mentioned here, so do the math at 200 amps @ 13.8 volts and I come up with (.)069 Ohms of DC resistance.  But we have no way of knowing exactly how much current the amp is drawing.

i am an idiot wrote:

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I do not know how to calculate such a number.  But I have seen 200 amps mentioned here, so do the math at 200 amps @ 13.8 volts and I come up with (.)069 Ohms of DC resistance.  But we have no way of knowing exactly how much current the amp is drawing.

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Exactly.  Ohms Law calculate the power supoply internal resistance given the voltage and current.  The capacior's time delay is T= C (in uf) X R (in megohms.)

T = 8,000,000 uf X 0.000000069 Mohms = 0.552 seconds

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Aha!
So the amp is 69mR (R = Ohms).
My undersized battery is 25mR.
A real AGM battery to supply 200A would be say 2 - 7mR.
And an 8F cap ~2mR.

What is the relevance of ESRs of 2-10mR when the load has a DC-Resistance far greater than that (69mR; and higher if not at full output)?   

After that, rather than getting into time (domain) analysis, or trying to estimate the effective capacitance of a batteries surface charge (since it is obvious its sub~12.7V charge well exceeds a cap), does anyone have voltage displays of a cap versus substituted battery? (I'm sure the Capacitor suppliers do.)

Or we could stick with DC - how a discharged capacitor gets around the need for current limiting when connected to a battery, or how big a capacitor would be required to melt a 12V battery (let's start with a Canon camera's 12V batt - do you think an 8F capacitor can?)   

That's if we continue down this hijacked path.

Personally I like Daniel's reply - that is the original question.

So far we have the $35 7AH battery lasting 10 times longer.
However that is using the Cap's time constant (say 2/3rds of 14.4V - ie to ~4.5V).
[ I thought I was being pretty generous with the Cap's ~570J if discharged to 8V. ]

As for current - let's stick with a nice round 200A. That's the worst case for the battery (since higher discharge rates mean lower battery capacity).

Fisrt off, I'd like to apologize for the tone of this comment:

haemphyst wrote:

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The internal resistance for the AGM 12V 7AH battery you are COMPLETELY and (as far as I am concerned) inaccurately in love with is around .025 ohms

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It was 2AM, and I had just had a REALLY bad World of Warcraft beatdown! (Spirestone server, look for Splashlog!) (LOL) My apologies, still, as the tone was unneeded and unprofessional.

haemphyst wrote:

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8F times 14.4V = 115.2J of energy.
115.2J /.002 ohms internal resistance = 57,600A! Peak instantaneous current

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I think I'm standing by my calculations...
1F @ 1V is 1J.
1F @ 14.4V is 14.4J
8F @ 14.4V is 115.2J

I am only asking, (because if I am wrong, I want to know why I am wrong) but from where is your number derived?

I might also mention that the 56kA figure I mentioned was for an infinitely short period of time... Real world numbers will be significantly less than that.

oldspark wrote:

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Since you brought the issue up, I'd love to see a transient analysis - hence why I asked above for the amp's PSU impedence. (S-domain?)

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That should actually be fairly easy to figure. The instantaneous voltage divided by peak current demand, in amps, will provide us with the internal resistance at that moment. The internal resistance of a power supply is dynamic, though, it isn't a constant. It changes with the output demands. This is likely why you haven't ever seen it as a "published" number. (That, and the fact that John Q. Public would generally have no idea what it means, nor would he even care, I think.)


:::::EDIT:::::
Doh... answered already!

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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."

You are understating capacitive energy.

Charge = CV.

Energy = 0.5CV**2, hence 0.5 x 8 x 14.4 x 14.4 = 830J.
E @ 12.7 = .5 x 8 c 12.7 x 12.7 = 645J, hence 185J discharged.
Etc, etc.


I still think solve the original situation and compare to my 7AH - assuming it can supply 5x its stated max current draw (whther that be an "i-squared T" rating, or lifecycle, or?)
Say the discharge time for 2500W from 14.4V to 12.7V, and then maybe 12.7 to 10.7 for 7AH & 8F etc.


Then if you wish, we can discuss transient response.
IE - why worry about 0.025R into a load that is a MINIMUM of 0.06R? The minimum load resistance is at least twice that of the 7AH.


The reason I used 7AH was from an example where I calculated somewhere between 1,000 - 10,000 the "capacity" for the given load etc (they were not too specific) & a 1F cap.

I only chose the 7AH then for being the best bang for buck.
Incidentally, they were citing a cap that claimed to have an ESR well in excess of 0.1R (0.6 from memory) whereas I offered a lower ESR solution.   
These days I'd probably use a LiPo battery.

And when dealing transients etc, I'm trying to reflect real life, hence "The instantaneous voltage difference divided by peak current demand" etc.
Otherwise with instantaneous current etc, it's as practical as the >2000HP from my 70HP engine...

PS - Daniel, I'll let others calculate times in answer to your question - though haemphyst has given an indication (at full load - assume 100 times longer at 1/100th power etc).

But in short, if you are looking to use a cap to power your system - forget it! An iPod maybe, but the energy content of a capacitor is far less than a battery.
On a volume comparison basis, ratios of 100's if not 1,000's apply .
If in doubt, try cranking your car using the 8F cap.
Then look at the batteries in those emergency starter packs (sometimes as small as 12V-7AH though now usually 15AH etc).

And you probably have an idea of how long your two fully charged batteries will keep your system running.
As I said i the other thread - the cap is negligible.

Remember - a battery is often modelled as a HUGE capacitor - ie, in system analysis - a voltage source in parallel with a capacitor (with ESR etc).

Alas this thread is sounding a lot like a recent thread I mentioned - namely Capacitors, Taken From The Writings Of Richard Clark...
At least in here I have indicated Pulse-R's misunderstanding in his "this is just rubbish...AHHAHAHAHAHAHAHA" to Richard's "Only half the electrons have a potential over 10 volts" (Pulse-Rs link at the end of the above link.
(There are other errors, and Pulse-R also complicates things but introducing a much broader transient/frequency consideration. I here we may still be having problems determining simple DC response - ie, hold-up/discharge time.)

And remember that a capacitor does not magically recharge itself or "increase voltage".
It is merely a charge storage vessel. It gets the charge from a battery or voltage source (with higher voltage than itself).
A battery not only exhibits capacitive behaviours, but is also a generator - it converts chemical energy to electrical.