high voltage high capacitance for under $60@100

mctrivia

100F@2.5V

can i put 5 in series to get 20F @ 12.5V

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Philldapill

Yes, you can. It would probably be a good idea NOT to go up all the way to 12.5V, however. Some of the capacitor values will vary, so, some capacitors will experience a higher voltage and could possibly be damaged.

mikediv

MC you have to put capacitors in parallel to add capacitance ...

mctrivia

i only need 10-12V. I knew it worked with traditional caps. was not sure about the super caps though and wanted to ask as they are very expensive.

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mctrivia

i am not adding capacitance. i am dividing capacitance.

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kwinn

Careful mctrivia, the usual -10 +20% tolerance of most capacitors will have an equivalent effect on the voltage when they are in series. Not sure if super caps have the same tolerances, but the effect still applies.

mctrivia

these are +-20%

so if math is correct 9.16V is max safe voltage.

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Post Edited (mctrivia) : 7/27/2009 3:07:54 AM GMT

kwinn

Not really. You have to consider what happens when some are +20% and others - 20%.

To get 12V with 5 capacitors each one would need to be charged to 2.4V which leaves very little room for error. You would need some additional circuitry to balance the voltage on the capacitors unless you want to find out what an overvoltage 20F capacitor sounds like when it explodes.

Of course you could also try matching the capacitors and hoping that they stay matched as they age.

Carl Hayes

When putting capacitors in series for increased voltage, differences in leakage are more significant than differences in capacitance. You're interested in how the DC component (not the AC) will divide itself among the capacitors.

The conventional solution, which I recommend, is to place a resistor in parallel with each individual capacitor. If the resistors are equal, they will equalize the DC across the capacitors.

Choose a resistance value that will pass perhaps 10-50 times the maximum expected leakage current of the capacitors. That's enough to swamp any differences in leakage among individual capacitors.

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mctrivia

So for safety should
*drop voltage to 10v
*place 2.2v zener across each
*place resister accepas each

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Carl Hayes

mctrivia said...
So for safety should
*drop voltage to 10v
*place 2.2v zener across each
*place resister accepas each

With equalizing resistors you can use the full rated voltage; no need to drop to 10v.

Zeners?· Heavens, no.

Place a resistor across each capacitor.

Remember, what you're trying to do is equalize the DC across each capacitor.· To do this, you would like to have capacitors that have the same leakage resistance.· Can't buy them that way.· So, you place a resistor across each one, thus increasing its leakage current by decreasing its leakage resistance.

For example, suppose the worst (leakiest) capacitor has a leakage current of 1 μa, and the best one has only 1/10 as much, 0.1 μa.· The best one has ten times as much leakage resistance and would end up with ten times as much DC across it.· Not good.

So you put a resistor across each capacitor, chosen to pass, say, 50 μa.· That effectively increases the leakage of the worst capacitor to 51 μa, and of the best one to 50.1 μa.· Now they differ by only 2%, and the DC will equalize within 2%.· Much better, eh?

Since I have no clue what levels of leakage current you can actually expect (you'll have to find this out), I have no way to recommend specific values of resistance.

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Post Edited (Carl Hayes) : 7/27/2009 5:27:52 PM GMT

mctrivia

Yes that would work steady state but while applying 2amp DC to charge it they would have little effect until the end of the charging curve. Will have to compute resisters that will keep bellow 2.5 nomatter what.

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Carl Hayes

mctrivia said...
Yes that would work steady state but while applying 2amp DC to charge it they would have little effect until the end of the charging curve. Will have to compute resisters that will keep below 2.5 nomatter what.

Yes; the transient effects will need examining.

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sam_sam_sam

mitrivia

2.5 volt .0005 uA =·5000 ohms

Will have to compute resisters that will keep bellow 2.5 no matter what.

Now as far as going over 2.5 volt why not put a 12 volt regulator

I would use 6 of those cap and the voltage to each one would be 2 volts with a 5000 ohm resister on each cap


Now I could be wrong BUT I am going to try this set up my self and see how this works because I have some 4.5 F @ 2.5 volt one to Play with

Where did you the one that you are talking about· 100·F @ 2.5 volts

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Sam

Post Edited (sam_sam_sam) : 7/27/2009 9:44:38 PM GMT

mctrivia

With 6 I need to use 150f caps. Would add $30 per bank

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sam_sam_sam

mitrivia


Where did you the one that you are talking about· 100·F @ 2.5 volts· could you Post a web Link to them



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Sam

mctrivia

Did in first post

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sam_sam_sam

I send you· a PM

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Sam

Philldapill

I think the (beefy power) zener diodes would be a very good idea. They would totally ensure that no capacitor is getting over 2.5V.

Mind divulging what you are using these for? It SOUNDS like some sort of battery backup, but remember - many of these low voltage super-caps have a fairly high ESR. They are usually used as low voltage, low current, battery backup's in circuits.

mctrivia

the ESR is only 50mOhm which is enought to allow more then enough current.

I need it to supply the 20A 5ms impulse required to blow an ematch under worst case scenario.

my math shows that at 20F even if the power supply was providing nothing to charge the caps the voltage would only drop to about 50%.

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Philldapill

mctrivia, where are you getting this 50% drop? If the caps supply a 20A 5ms impulse, that's a charge of only 0.1C. You don't need to do any calculus to figure this one out. q = CV. Your capacitance is 20F, and your charge is 0.1C, so the change in voltage is only 5mV - unless I'm missing something.

EDIT: Never mind - you're talking about the drop across the ESR. Yes, you're right...

davejames

...I just gotta ask.

Why go to the trouble to put five of these in series instead of just buying a single 20uF @25V? Unless you already have a mess of the 100uF.

DJ

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Instead of:

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I prefer:
"Those who know, do.· Those who understand, teach." (Aristotle)
·

mctrivia

because they do not sell super caps for that high of voltage.

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davejames

...forgive me - but why do you need a "supercap"? Size contraints?

DJ

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Instead of:

"Those who can, do.· Those who can't, teach." (Shaw)
I prefer:
"Those who know, do.· Those who understand, teach." (Aristotle)
·

mctrivia

size a little, cost and weight more so.

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davejames

Unless there is some unique characteristic of a "supercap", I should think any generic 20uF @25V would do...such as these:

http://www.alliedelec.com/Actives-and-Passives/Capacitors/?N=4294821930+4294827693+4294838989+4294821038


DJ

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Instead of:

"Those who can, do.· Those who can't, teach." (Shaw)
I prefer:
"Those who know, do.· Those who understand, teach." (Aristotle)
·

mctrivia

that is 1,000,000 the size i need. not 20uF 20F

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davejames

...sorry - I read "uF" instead of "F".

Why do you need 20 Farads, out of curiosity. Limited supply voltage? Is there any chance of providing a higher voltage to charge the cap bank?

At work we use a 7mF (milli Farad) cap bank charged to 63V to deliver up to 100A. At that current, the bank can only sustain it for 300us (auto shutdown forced by design of the instrument). At lower current settings it can deliver for multiple 10s of ms.


DJ

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Instead of:

"Those who can, do.· Those who can't, teach." (Shaw)
I prefer:
"Those who know, do.· Those who understand, teach." (Aristotle)


Post Edited (davejames) : 7/28/2009 3:34:24 AM GMT

Tubular

Hey McTrivia, I've been playing with some CAP-XX 0.9F modules which have a 30 milliohm ESR and 4.5volt working voltage. I think even these are good for the 20 amps you're after, and they're surface mount. Why the 100F, do you need to do lots of ignitions?

I also have pulled apart my Maxwell boostcap 58F at 15v (6 x 350F "D cell supercaps") to see how the balancing works. They use 6 little active balancing circuits. However for what you want I suspect passive balancing with a resistor across each cell will be fine.

Also regarding the leakage, from what I've observed it seems to be dependent on many factors. Going from 100% to 80% charge happens much much quicker than 80% to 60%...

tubular

mctrivia

yes the small caps could provie the 20A but with 3 in series would give only .3F. It would discharge to 6.9V in only 1ms in 5ms it would be down to .7V

As you can see from the diagram attached with a constant value of R and C the capacitor will discharge much faster at the beginning. The reverse of this is true for charging. the cap will charge rapidly to 60% and then slow down as the voltage approach the same value. I plan to use a constant current source to keep the charging of caps relatively linear but the discharge will be like the diagram.

at 20F and 0.6ohm(20A) in 5ms the cap will discharge to 7.9V I then need to try and recharge back to full before the next shot. of course that is worst case scenario. I will try to keep timing as far as possible apart to minimize strain the customer truly has the final say.

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Tubular

Plugging in R=0.6ohms, C=0.3F, V0=13.5 volts I get Vc=V0e^(-0.005/.18) = 13.13 volts after 5msec??

A second way of calculating would be from an energy standpoint, each "small" supercap would store .5 * .9 * 4.5^2 or about 9 joules, 27J total for the 3. Your load is about 240 watts (20A*20A*0.6 ohms), which for 0.005 seconds is 1.2 J

What are the ematch characteristics like? I assume you need a strike voltage hence the need for multiple caps rather than straight I^2*R current...

The leakage point I was trying to allude to is that it doesn't seem to follow an exponential decay like everything else. Starting from rated 4.5 volts it drops the first volt within a matter of hours, but dropping the second volt takes days. The initial leakage rate is higher than the datasheet would indicate, then perhaps things start to stabilise.

tubular

mctrivia said...
yes the small caps could provie the 20A but with 3 in series would give only .3F. It would discharge to 6.9V in only 1ms in 5ms it would be down to .7V

As you can see from the diagram attached with a constant value of R and C the capacitor will discharge much faster at the beginning. The reverse of this is true for charging. the cap will charge rapidly to 60% and then slow down as the voltage approach the same value. I plan to use a constant current source to keep the charging of caps relatively linear but the discharge will be like the diagram.

at 20F and 0.6ohm(20A) in 5ms the cap will discharge to 7.9V I then need to try and recharge back to full before the next shot. of course that is worst case scenario. I will try to keep timing as far as possible apart to minimize strain the customer truly has the final say.