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Author Topic: Capacitors internal resistance and charging (solved - data error)  (Read 8129 times)

Offline capthook

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Why isn't a capacitor collecting all the power sent to it?

I have a pulse motor with separate generator coils.  I'm trying to collect and store the power generated in a capacitor(s).

The coils output: 6V @ 20mA = 120mW ( 120 milliJoules/second )

When connected to a 36V 4700uF electrolytic capacitor, it collects 3V in 1 second:

1/2(.0047x3x3) = (0.02115 x 1,000) = 21.15 milliJoules/second

So - 120 mJ out : 21 mJ on capacitor!

I could only figure the resistance of the capacitor was the problem - it VERY quickly rises to 1,000's of ohms!

So how am I supposed to efficiently collect/store my output on a capacitor when it appears the HUGE,RAPID resistance prevents my very small amps from getting through?

A battery has an internal resistance measured in milliohms!

So - forget the capacitor idea?  Battery the only way to go?  Or is there a circuit/capacitor/method where it would work?  What about large Farad, small voltage supercapacitors? (like the ones from Nesscap?)

Yaaaarrrrghh!!!     >:(

CH
« Last Edit: April 22, 2008, 08:41:47 AM by capthook »

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Offline Jokker

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Re: Capacitors internal resistance and charging
« Reply #1 on: April 21, 2008, 03:22:04 PM »
I'm quite beginner in this area, i got some theory, but i got no experience.

But so sure that there is a huge difference between  material of capacitor

Theres the pic...  what describe frequency ...

(http://www.imagehosting.com/out.php/i1709719_capacitor.bmp)


So u need learn more about how to charge tease things, I'm quite sure that it takes some time.
And theres something up with inductive voltage ...

Offline gyulasun

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Re: Capacitors internal resistance and charging
« Reply #2 on: April 21, 2008, 04:21:15 PM »
Hi capthook,

Question is how you mean your generating coils:  you wish to collect the voltage pulse created in your coil after the input pulse is switched off  or you rotate magnets inside or near your coil to induce voltage across it?
In the first case when you have the induced voltage pulse due to the collapsing magnetic field (when your input pulse goes to zero voltage again), it is best to use high speed / ultra fast type diodes to rectify the induced pulse (called also as flyback pulse) because the this pulse appears immediately after switchoff and you do not wish to be late to utilize it. This needs diode types like MUR120 (1A, 200V) or see here some other types out of many: http://www.vishay.com/diodes/rectifiers/ultrafast-recovery-plastic-axial/  and choose for the smallest reverse recovery types (trr in nanosecond).  For MUR120, trr=25ns
If you cannot easily obtain those types, you could use some cheap and more readily available 1N4148 or 1N914 (trr=5ns!) diodes connected in parallel to increase forward current rating,  for this select some diodes for very similar forward voltage drop by a digital multimeter set in its Diode test range because this way current sharing will be more universal among them if paralleled.
Besides this diode switch time problem, another thing to consider is to find the best capacitor value for your coil induced voltage /current capability, i.e. to match the cap to the coil in this respect. You can test capacitors of different values, starting from 10uF/350V or 22uF/250V and going up in uF value.

If your output power is not the given by the coil's flyback pulse but comes about  by normal induction due moving magnets, then really fast diodes are not neccessary, you may wish to find the smalles forward drop diodes for loosing the least possible output power across the diodes and also you need to test different capacitor values for the best match.
If you may be willing to say some more details on your setup, it is possible to give some more pieces of advice. (how you generate the 6V at 20mA)

rgds,  Gyula

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Re: Capacitors internal resistance and charging
« Reply #2 on: April 21, 2008, 04:21:15 PM »
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Offline capthook

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Re: Capacitors internal resistance and charging
« Reply #3 on: April 22, 2008, 12:16:45 AM »
Thank you for the replies!

I'm sorry for being so incomplete with the details of my question...... :-[

I have 12 coils placed below a N/S/N/S 8 magnet rotor @ 120 RPM's.

The coils are in a 3-phase, star connection so that the voltage adds.

The output is then rectified through a 6 diode (1N5818 30V 1A) schottky full-wave bridge.

When metered through a 10ohm resistor - 6V @ 20 mA = 120 mW (120 milliJoules/second )

When connected directly to a 35V 4700uF electrolytic capacitor:  3 volts in 1 second ( 5.8V in 2s)
1/2(.0047x3x3) = (0.02115 x 1,000) = 21.15 milliJoules/second

So;  120mJ out : 21mJ collected at capacitor

When measuring current through capacitor ? it very quickly drops to 5 mA in 2 seconds and continues to drop.  It seems the current drop is due to additional resistance in the capacitor as it charges.

However, it?s a 35 volt capacitor that is only charged to 6 volts in 2 seconds ? 17% of full charge ? but the resistance of the capacitor has already gone to over 1,000 ohms.

Is it that the small current I?m producing (20 mA) is too small to be practical in charging a capacitor?
Or do I need a bigger capacitor?  What about a supercapacitor of 10F @ 2.7V and changing to a "delta" wiring so that amps, rather than volts, add?

CH

Offline fritznien

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Re: Capacitors internal resistance and charging
« Reply #4 on: April 22, 2008, 05:41:32 AM »
at 5.8volts your cap has near full charge. how much current do you expect from .2 volts?
the resistance of a cap is measured in one of 2 ways, the dielectric leakage which should be near infinite
resistance and the resistance to current flow in and out off the plates.
  4.7 milliamperes will charge your cap by one volt per second ( constant current).
35 seconds to reach 35 volts, you only have 6 volts.
 by the way 6 volts into 10 ohms is 600 ma how do you get 20?

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Re: Capacitors internal resistance and charging
« Reply #4 on: April 22, 2008, 05:41:32 AM »
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Offline capthook

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Re: Capacitors internal resistance and charging
« Reply #5 on: April 22, 2008, 08:41:18 AM »
by the way 6 volts into 10 ohms is 600 ma how do you get 20?

ahhh - the error is revealed!!

Thanks for the sharp eye fritznien!

I'm working on a number of devices... turns out I mixed up my data sets.

Using the correct data resulted in accurate charge being stored by capacitor.

Thanks for finding my error!!!

CH

 

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