waste car battery boosted by a capacitor?

[Christophe]

The maxwell data sheet ( http://www.maxwell.com/images/documents ... module.pdf ou https://www.econologie.info/share/partag ... LvDqdm.pdf ) indicates a capacity of at least 1000F

On page 2, the max current formula is probably a little more realistic than the one above because it takes into account a term in the denominator in addition to the duration (C * ESR in Ohm ... probably an internal resistance)

The delta V given is quite large: 9V and rather incompatible with mounting on a lead battery.

This product stores: E = 1/2 * 1000 * 9² = 40 J there we are in the right.

Anyone have an idea of ​​the price?

[Flytox]

Found this link for orders of magnitude:

http://www.coursstimartinique.fr/tp%20s ... arrage.pdf

Order of magnitude for the starter of an engine from 800 to 1600 cm3:

Nominal current at 10 V: 90 A
Power at 10 V: 900 W
Peak intensity at launch: 500 A
Starter weight: 2.9 Kg

[Christophe]

If the PIC of 500 A under 10 V lasts 0.1 second then an energy of 500 Joules is required!

I did a little comparison with the "big" velleman capacitors: it won't work! Truly not ! You have to do the assembly with ultra capacitors!

The biggest: 6800 µF / 16V with a delta V of 4 gives a useful storable energy of 0.054 Joules! Its recommended retail price: 1.90 €

To make 500 J you would therefore need nearly 10 ... I let you calculate the investment

So, where are ultra capacitors found?

[tigrou_838]

Hello everyone, I don't come very often to participate, but I still read from time to time, and this subject begins to interest me.

the idea of ​​crhistophe and for start-up, so a big demand for power with a car battery, and I think the battery is still mounted on the vehicle.

I will always have a question on the same subject, if I have a car battery on inverter 12v / 220v backup for solar circulators, during an electrical cut demand is less, just the power peak, not example of 300w for a 100w device,
so it would be possible to find condos to improve and especially extend the life of the battery.

thank you all.

[Gaston]

By cons I do not understand your last calculation (the 1st ok):

E = 1/2 C * U² or C = 86400 * 2 / 4² = 21 F no?

There you calculate the energy of a capacitor charged under 4 Volts.

What must be calculated is the difference between the energy contained under 14V and that under 10V, that is:
C = 86400 * 2 / (14²-10²) = 1800 F

It is not linear because there is more energy to recover between 14V and 10V than between 4V and 0V.

[Christophe]

Ok thanks for the clarification

[elephant]

Was the Ford capacitor perhaps for interference suppression or to compensate for irregularities in the continuity of the current?