Replace your car battery with capacitors!

[redpoint5]

Originally Posted by Old Tele man

So, the LED's are NOT there just to be pretty lights, they are there to "balance/equalize" the charging voltage that each capacitor in the series stack sees while they're being charged.

The property of the LED that balances the capacitor is the Diode, and the Light Emitting feature is not needed, right? So a diode with a forward voltage drop equal to that of the LED could replace the LED?

I think I'll just buy various LEDs and diodes from Norvac next time I'm in Salem and test various combinations. These components are cheap enough and I should have an assortment anyhow.

[redpoint5]

Another thing I don't quite understand is how the capacitor can take a significant load off of the battery when starting the engine. Since the capacitor has a relatively low capacity, the voltage would rapidly deteriorate when starting. Since the capacitor is wired in parallel with the battery, it would maintain the same starting voltage. That means the capacitor would immediately drop below the voltage of the battery and start pulling amps from it.

I'd imagine the battery would supply the majority of the cranking amps, unless voltage sag is significant enough to shift the draw to the capacitor. I would still be afraid of a small LiFePo4 drawing too many amps during cranking. The battery used by Laserhacker is rated for 30C discharge, which is 126A. What is even more troubling is the battery charge rating is 2C, which is only 8.4A. I'd be afraid the alternator would charge the battery too quickly after starting, and I'm not sure the capacitors would significantly buffer the charge.

I suppose this is another area where I will have to test and report my findings. I just ordered a clamp-style DC ammeter, so I hope to get started soon. When I got off work at 1am this morning, the 'ol Cummins with 2 massive batteries decided it didn't want to start... and this reminds me of something; the voltmeter that plugs into the cigarette lighter draws 40mA! I had left this plugged into a lighter socket that retains power with the ignition off, and during my 13hr shift consumed 0.52 Ah.

[kennybobby]

Time for a sanity check...So you could build a 16.2 V, 500F capacitor bank with about 2000 Coulombs, equals 2000 A-seconds, or 0.58 A-hr. With a 0.041 A drain load it would hold up for about 14 hours. Also how much current does the starter motor pull when cranking? The 2000 Coulombs could get used up fairly quickly in cold cranking, e.g. 200 Amps for 10 seconds. Little motorsickle/ATV batteries are rated 12V, 10 A-hr. Is 0.58 A-hr even worth the effort?

[Cobb]

You havent lived til you flash boiled the fluid in a battery.

[freebeard]

It would be great if you could park the car in an open field, put up the lightning rod, take one lightning strike, stow the lightnig rod and be good to go for another 100,000 miles.

[redpoint5]

one point twenty-one gigawatts. Great scott!

http://www.youtube.com/watch?v=mjCRUvX2D0E

I'd be curious to know how many KWh the average lightning bolt delivers, although I think coulombs would be a more appropriate term.

[nackerton]

Originally Posted by Old Tele man

No, it's the COMBINED voltage drops [ 0.7VDC(diode) + 2.0VDC(led) ] that set the VOLTAGE that is maintained across the capacitor that the diode & LED are in parallel with. Both devices are needed; it's the series summation of their voltage-drops that's supposed to ±equal the capacitor's 2.7VDC working voltage.

As stated earlier, all the LEDs+diodes COULD be replaced with precision resistors, but would likely NOT be as precise in voltage balance across each individual capacitor...the LED+diode combination has much tighter variance.

Good info, thanks. I have one problem in my mind: could you easily burn those diodes if capacitor(s) is badly out of balance and charging current is high?

And when the diodes die, capacitor is next...

[foytix]

Read the 1st couple of pages. Sorry if someone has mentioned this. Has anyone thought of a ballistic battery or full lithium ion battery. They are a fraction of the weight of a normal battery. Wish the cost was a fraction though lol.

[Cobb]

Typically its the last cell in a batter towards the negative terminal that is lower than the rest as well as the last battery in a string that has a lower voltage.

Caps go way out of balance more so than anything else and the math you are throwing around is worthless if cap 1 is 100% capacity and the rest are 50% the last 10%.

[redpoint5]

My truck batteries (2x group 27) have finally degraded so much that they cannot start the truck after sitting for a day. I can replace them for about $200, or I can eliminate the batteries entirely and run a supercapacitor.

Someone talk me out of spending $319 on this! The auction is for 6 used 3000 F capacitors with balance circuits. I think I can get the parasitic drain on the truck to 5mA or less, and I have a solar charger in it already.

My only hesitation is that a powerful fuel lift pump runs when the ignition is switched on, and air intake heaters kick on at 180 amps for 8 seconds when it's very cold outside. After all that, the starter can draw up to 400 amps for 3 seconds.

Do you think the capacitors are up to the task of starting when it's cold out? Ideally I'd like to eliminate the battery altogether, but I suppose I could run a smaller battery if necessary.