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.
Quote:
Originally Posted by jakobnev
That should really be: Q=dU*C
4V*500F/3600s=0.56Ah
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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!
Quote:
Originally Posted by Old Tele man
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?