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how long does an 8 farad cap lasts?


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i am an idiot 
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Posted: November 22, 2009 at 9:05 AM / IP Logged  
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.
DYohn 
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Posted: November 22, 2009 at 9:16 AM / IP Logged  

i am an idiot wrote:
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.

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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oldspark 
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Posted: November 22, 2009 at 11:17 AM / IP Logged  
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).
haemphyst 
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Posted: November 22, 2009 at 11:19 AM / IP Logged  
Fisrt off, I'd like to apologize for the tone of this comment:
haemphyst wrote:
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
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:
8F times 14.4V = 115.2J of energy.
115.2J /.002 ohms internal resistance = 57,600A! Peak instantaneous current
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:
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?)
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!
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."
oldspark 
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Posted: November 22, 2009 at 12:28 PM / IP Logged  
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...
oldspark 
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Posted: November 22, 2009 at 6:33 PM / IP Logged  
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.
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