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Old 12-17-2013, 07:48 PM   #41 (permalink)
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Looks good, but how long a modern car can sit on it's place before clock, alarm system etc. has sucked capacitors too low? 6x 3000F capacitors would be around 2 Ah in capacity I think.

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Old 12-17-2013, 08:10 PM   #42 (permalink)
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another thing to consider is the alternator load, how much current does a low capacitor bank draw from the alternator?
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Old 12-25-2013, 12:54 AM   #43 (permalink)
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I just purchased 6x 350F capacitors ($77 shipped ) to test out in various applications, both with a LiFePo battery and without. I don't have the battery yet, but I do have an ATV with a dead lead acid battery. My first test will be with the ATV, to see how long it can sit and still start with just the capacitors.

The nice thing about having it in the ATV is that if it fails to start, there is a pull cord as a backup.

Next, my truck has 2 very weak batteries that will barely start the Cummins motor after 3 days of sitting. I'm wondering if throwing the capacitors in parallel with the batteries will give it the extra cranking oomph it needs.

Unfortunately, I don't think I can cheaply replace the lead acid batteries with a boost pack because the grid heaters suck down serious amps. Since the speedo on my truck doesn't work, the grid heaters continue cycling even though I exceed the cutoff speed.

Another vehicle I'd like to replace the battery with a boost pack is the TSX. I've already measured the parasitic load at 0.041 Ah, so sizing a LiFePo battery for my needs should be simple.

Last, I have a CBR600 motorcycle that I'm curious if it could loose the lead altogether and just run capacitors. The problem is that it charges using a generator instead of an alternator, and that means it doesn't charge below about 1,500 RPM. In other words, I'm draining the capacitors anytime I'm stuck idling. Even worse, if I idle too much the engine temps rise enough that the E-fans kick on. Fortunately, bump starting a motorcycle is easy enough.

I'll start a new thread when I begin experimenting and report back my findings, but it will be a few weeks to a month before I have anything to share.

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Originally Posted by redneck View Post
I've been very busy lately. I'll try and get this done over the next few weeks.
Any update?

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Originally Posted by nackerton View Post
Looks good, but how long a modern car can sit on it's place before clock, alarm system etc. has sucked capacitors too low? 6x 3000F capacitors would be around 2 Ah in capacity I think.
I just measured my TSX draw at 41mAh today. With 12.4v on the battery, that is just barely more than half a watt drain.

As I've just learned, capacitors don't behave the same way as batteries when placed in series. 6x 3000F capacitors in series has only 500F capacity at ~14V.

The formula (Vmin + Vmax)/2 * F / 3600 = Ah is used to convert F to usable Ah.

I'll assume my car is happy with voltages down to about 10V (correct me if you think it might be different).

(10 + 14) / 2 * 500 / 3600 = 0.83 Ah

If my calculations are right, 6x 3000F capacitors in series and charged to 14v would give me 0.83 Ah and with a drain of 0.041Ah, my car would go for 20hrs before the voltage dropped below 10v and the computer dies.

Someone please analyze my assumptions, formulas, and arithmetic.

Here is a series/parallel capacitor calculator -> Electronics 2000 | Series / Parallel Capacitor Calculator

I wonder why supercapacitors only exist in the 2.5-2.7v range? If someone could just come out with a 15v rated supercapacitor, there would be no need for a series configuration and the resulting loss in capacity. Wiring in parallel to boost capacity would come at zero penalty.
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Old 12-25-2013, 07:17 AM   #44 (permalink)
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That should really be: Q=dU*C

4V*500F/3600s=0.56Ah
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Old 12-25-2013, 10:28 AM   #45 (permalink)
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Old 12-25-2013, 11:41 PM   #46 (permalink)
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I measured the ATV parasitic draw at 0.57 mA! It's so low that it would take an eternity to drain the capacitors. I'm sure the self-drain will be much more substantial. I've got high expectations that the ATV will never need a battery again.

I also bought a 2.5w solar battery maintainer for the truck. Hopefully I can let it sit indefinitely and always have it start. It doesn't get used too frequently since it's a fuel hog.

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Originally Posted by jakobnev View Post
That should really be: Q=dU*C

4V*500F/3600s=0.56Ah
I had pulled the formula from this site:

Ultracapacitors as batteries

Would you please write out the variables in your equation? I'm not familiar with the terms as I've only had a HS intro to electronics course 14 years ago. Thanks for your patience!

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Originally Posted by Old Tele man View Post
This picture shows how all six capacitors are stacked in series because the individual "working voltage" of each capacitor is too low (2.7VDC) for the automotive 12.6-14.7VDC system...which means the total capacitance you end up with is 1/6th of each individual capacitor's 350 Farads...which is quite a reduction.




The LED's are doing the same function as the "voltage-balancing" resistors that I described, but are doing it with a much tighter tolerance. Resistor values are seldom better than 5-10%, but LED and DIODE 'activation' voltages are very precise--much better than 1%.

If you read the sellers instructions, you'll also notice how he mentions that using different colored LEDs changes both their voltage-balance numbers as well as their discharge rates, this is because different color LEDs have different minimum "turn-on" voltages and currents. See the Voltage Drop [ΔV] column in the Color and Materials section/table in this Wiki: Light-emitting diode - Wikipedia, the free encyclopedia
I'm assuming LEDs are used just as a visual indicator, and could be replaced with another diode if it had a similar voltage drop? Does the total voltage drop of the diodes equal the approximate voltage that the individual capacitors would settle at? Couldn't the diode and LED be replaced with a single LED with a 2.7 voltage drop, such as a green or blue?
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Old 12-26-2013, 06:26 AM   #47 (permalink)
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redpoint5 -

Well the formula he had chosen shows the average charge in the capacitors during the discharge, i have no idea why he would be interested in that. What we want to know is the charge Q that we can use. And that's simply the change in voltage dU times the capacitance C.

Now i realize i made a mistake earlier, i divided by an hour (3600s), what i should have done was to divide by the conversion factor between hours and seconds (just 3600 with no unit).
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Old 12-26-2013, 08:56 AM   #48 (permalink)
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I don't think this video of his has been linked to. It shows a balance circuit for the capacitors in his SOLYN1 project.

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Old 12-26-2013, 09:56 PM   #49 (permalink)
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Quote:
Originally Posted by sheepdog 44 View Post
I don't think this video of his has been linked to. It shows a balance circuit for the capacitors in his SOLYN1 project.
I have seen this video, but my questions remain:

1. Can the LEDs be replaced with any diode with a similar voltage drop?
2. Is the goal to have the same voltage drop for the diodes as the capacitors are to be trimmed down to?
3. Can the diode be removed if an LED with sufficient voltage drop is used (possibly green or blue)?
4. Can this balance circuit be used on battery packs such as Li-ion or LiPoFe4, or even NiMh?

EDIT: It seems the answer to #2 is yes, but I'm not sure. Based on this table, the red LED has a 1.7v minimum operating voltage, and a diode has a 0.7v drop. Added, this is 2.4v, or the voltage that each of the 6 capacitors should be trimmed to.

EDIT2: I don't see any reason why this wouldn't work to balance any type of cell in a series circuit, assuming the sum of the forward voltage drop for the diodes equals the max rated cell voltage.

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Old 12-26-2013, 10:37 PM   #50 (permalink)
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No mistake Jakob, you did it right.

Quote:
Originally Posted by jakobnev View Post
... Now i realize i made a mistake earlier, i divided by an hour (3600s), what i should have done was to divide by the conversion factor between hours and seconds (just 3600 with no unit).
The charge is dU(Vmax-Vmin) times Capacitance, with units of volts x farads, which is coulombs. But coulombs = amp-sec, so the conversion of 1hr/3600s is correct in number and in units to get A-hrs.

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