Ugh... The more I look around the more lost I get

 

Then I just found this. Ugh I'm about ready to give up. I'll pay someone to build me something because this is over my head and I'm only getting myself frustrated

 

1st observation: Are you using aluminum for the shorting bars/straps?

Copper would be the metal of choice here. Some copper pipe hammered flat would be my suggestion.

2nd observation: If the straps are just being attached 'willy-nilly' to provide us with a picture then OK. Otherwise I STRONGLY suggest you think about your series parallel connections and re-visit the pictured strap configuration. Appear to be lots of problems in those connections the way they are arranged vs. what you describe as your desired arrangement.

 

1. Yes, aluminum. I know copper might be a little bit more conductive, but it wasn't worth it for the price lol. They are 1/4" by 2" bars. Should be more than adequate given that the strapping bars you can purchase are only 1/8" x 1." Let me put it this way: Each strap has 6x the cross sectional area of 0 gauge wire. Is there something to be gained by switching to copper? Maybe. But I'm not sure it's worth the cost

2. I guess I'm not seeing the issue? 7 paralleled pairs wired in series. Arranged that way to fit in my space constraints. It looks a bit funky but it functions as intended

 

2. I guess I'm not seeing the issue? 7 paralleled pairs wired in series. Arranged that way to fit in my space constraints. It looks a bit funky but it functions as intended

I'm confused. This arrangement has the equivalent capacitance of 857 F. If you distribute the voltage equally, your supply would be 17.5 V (7 * 2.5 V).

How do you calculate the amount of power in the capacitors?

 

I'm confused. This arrangement has the equivalent capacitance of 857 F. If you distribute the voltage equally, your supply would be 17.5 V (7 * 2.5 V).

How do you calculate the amount of power in the capacitors?

Short them with a screwdriver and see how much shorter it is after the shower of sparks has settled.

 

Short them with a screwdriver and see how much shorter it is after the shower of sparks has settled.

Are you taking up welding? It works better with a car battery.

 

I'm confused. This arrangement has the equivalent capacitance of 857 F. If you distribute the voltage equally, your supply would be 17.5 V (7 * 2.5 V).

How do you calculate the amount of power in the capacitors?

And if I were to limit to 2.4v it would be 16.8v. If I were to go to the absolute maximum they would be at 18.9v.

I guess I'm really not sure what you are asking?


Ideally I'd like to have the power distributed as evenly as possible regardless of voltage

 

And if I were to limit to 2.4v it would be 16.8v. If I were to go to the absolute maximum they would be at 18.9v.

I guess I'm really not sure what you are asking?


Ideally I'd like to have the power distributed as evenly as possible regardless of voltage

I guess what I want to know is how much power can you get from this configuration. If you have a load that requires 100 mA, how long will it be powered?

i.e. Amount = voltage * amperage * seconds = ? W·s

 

It doesn't work quite that way, since the voltage and the current will both decay as the charge depletes.

A coulomb of charge is 1 amp-second. Capacitance is coulombs per volt, for instance 1 Farad is 1Coulomb/V

At 2.7V and 3000F, that's 2.7V x 3000F = 8,100 coulombs = 8,100 amp-secs or 2.25Ah.

But again, you cannot then simply multiply by the starting voltage to get power. You need to integrate the area under the curve. Or you can cheat and use the formula. It's almost 11kW•s

 

Okay I see they are connected properly. My apologies.

Now. Is this some kind of new capacitor that utilizes a chemical reaction?>

Won't the charge self distribute among the capacitors? why the need for 'balance'? Again, these are not chemical powered are they? They simply store energy in the form of an excess of electrons gathered on the surface of a conductive material.

 

I don't see any practical solution other than careful selection and matching of individual capacitors, just as you would do with battery cells as you build a pack.

If you need to protect each cell by constraining it's voltage between 0V and 2.7V (or whatever range), in principle you want a window comparator and a bypass switch. A MOSFET conducts in either direction, so that is a candidate for the switch. But that switch would have to handle the entire bypass current, which I think is 500A if you expect 1000A from two parallel strings. Allowing for safe operation and a reasonable life, you'd need transistors rated to 1000A or so, at least for seconds-long time periods if not continuous. Such a transistor is not cheap.

Now that I think about it, I think you'd need more than one switch for each cell. One to switch the bypass loop and another to switch the "main" path. You cannot just add the bypass since this would short the capacitor being bypassed. Not cool. A relay to switch from one path to the other for each cell is the logical equivalent but I'm not sure you can get a SSR 1000A. Never looked.

 

At 2.7V and 3000F, that's 2.7V x 3000F = 8,100 coulombs = 8,100 amp-secs or 2.25Ah.

But again, you cannot then simply multiply by the starting voltage to get power. You need to integrate the area under the curve. Or you can cheat and use the formula. It's almost 11kW•s

Is this for 1 capacitor, or for the bank of capacitors as he has them arranged?

 

Each one individually

 

I'm seriously behind the tech curve on these new fangled caps. The info at the link helped me understand. My thanks to the helpful answers all here have provided.

http://www.tecategroup.com/ultracapacitors-supercapacitors/ultracapacitor-FAQ.php

 

I see there's a circuit shown there for active balancing. I'm quite sure I don't understand it all.

 

Instead of balancing, don't you "just" need a way to stop charging when any individual cell reaches the target voltage?

Yeah, for this kind of current I think you're exactly right.

Between picking cells that are well-matched and cutting off charge when one cell hits 2.7V, you would get something like 80% of the possible energy of the stack - probably cheaper to spend an extra $200 on making the stack 20% larger than putting a ton of huge MOSFETs in there.

Kind of a brute force solution, but then this is kind of a brute force project it seems...

 

Hah, I've seen that Tecate Group active balance circuit before, too - not even going to look at it, out of my league.

Instructable on Supercap Balancing

That one makes sense, although it's set up for a very low-power input.

 

Okay I see they are connected properly. My apologies.

Now. Is this some kind of new capacitor that utilizes a chemical reaction?>

Won't the charge self distribute among the capacitors? why the need for 'balance'? Again, these are not chemical powered are they? They simply store energy in the form of an excess of electrons gathered on the surface of a conductive material.

Apparently supercapacitors are pretty critical on working voltage, add to that tolerances on capacitors in general is pretty wide - its easy for a capacitor in the chain to end up charging to a higher voltage than designed for.

 

Reason I picked 2.4v is suppose I were to limit each cell to 2.7v, would it not be possible, in theory, to have the bank charged to 16.8v, 6 capacitors charged to 2.7v, 1 charged to .6v. then if the bank were to drop below 12.6v, it would cause the final capacitor to reverse charge, damaging it I've been able to do that when playing around with smaller capacitors, but haven't had the balls to try it with these

That does not sound possible (to me). The thing that seems most relevant is that voltage in a capacitor bank is a dependent variable - charge is the determining factor.

The charge on each negative plate equals the charge on each adjacent positive plate. For the charge on one to be reversed, they would all have to be reversed...right?


When you say that you have done this with smaller capacitors, do you mean you have set up a series bank, charged it up, then rapidly discharged it - and some of the capacitors have ended up with negative charge while others are positive?

I think this could only happen if you put in some wacky extra circuitry to make it happen.