waste car battery boosted by a capacitor?

[Christophe]

The idea is simple: a lead-acid car battery does not like voltage variations: Voltage peaks and especially voltage drops (starting) ... So why not place a large capacitor (or several small ones in parallel to make a big one, probably less expensive) at the battery terminals?

Voltage drops at start-up would be limited, the capacitor making an energy buffer ...

Question: would it be interesting (batteries, new at least, are already good energy buffers)
Is there a technical contraindication (electrical conflict ??) to do so ?? What minimum capacity for there to be an effect (capacity in relation to the Ah of the battery)?

We may have already talked about it here or there on this forum? Maybe in the discussion of ultra capacitors?

The debate is open...

[izentrop]

It would take a super capacitor and analyze the profitability ???

It was the case with the old engines of strong cubic capacity or the 12 v could go down to 5 v at the time or the starter forced on the maximum of compression.
the first radiocom 2000s sometimes went out. the trick was to put a 10000 µF at the terminals with a non-return diode on the battery side;)
Now the turbos need less power on start-up, I think.

[Christophe]

Ah thanks for the info, so it's doable!

It would take a super capacitor and analyze the profitability ???

This is the purpose of this topic ....

a) calculate the minimum capacity necessary for there to be an effect

b) if a) ok, I don't think we should talk about cost-benefit analysis. The price of some condos is a few euros (20 € at most) and if it can make a battery last much longer it is worth it!

ps: what is a radiocom 2000?

[Christophe]

Concerning a) I think that 5 to 10% of the capacity of the battery could be a correct base to make some tests.

Do you happen to remember the capacity mounted on Ford batteries?

[izentrop]

Not sure about the usefulness of the super buffer condo

ps: what is a radiocom 2000?

the ancestor of the cell phone. The first were assembled only on vehicle. I installed them in the 80s. It was still profitable at the time.
If you forgot to turn it off in the evening, the next day you wouldn't restart

In fact what the lead battery does not like is deep discharge and needs to be stored charged. Especially in winter.
Auto batteries are designed to withstand strong currents of short duration. The value is indicated. http://fr.wikipedia.org/wiki/Batterie_a ... e_batterie

[Gaston]

Maxwell offers a whole set of super capacitors with associated electronics for starting (large) trucks.
These are not directly capacitors in parallel with the battery (the starter is no longer connected to the battery, but only to the capacitor module), but this element is intended to be housed in place of a battery on vehicles that use more than one in parallel.
The idea is that even with the batteries discharged, you can recharge the capacitor over a longer period (a few minutes) and use the stored energy to turn the starter.


Regarding the capacity calculation, the basic equation of the capacitor is I = C dV / dt or C = I dt / dV

So if we want the capacitor to provide an intensity of 100A for one second from 14V to 10V, we need a capacity:
C = 100 * 1 / (14-10) = 25 Farads.

If you want 800A for 5 seconds (enough to power the starter without using the battery), you need C = 800 * 5/4 = 1000 Farads.

The capacities of a few thousand µF do not weigh much compared to the current absorbed by a starter ...

[Remundo]

the usable payload of a condo is Q = C x U, with U = 13V minus the voltage under ampere draw here ...

We realize that the electrical charge coming from the condo will be super limited

already with a 1 Farad condo (which is large), you can only store a few Coulombs.

A current of 1A is 1 C / s

a battery start must draw easy 50A ... the condo will be empty in 1 / 50th second.

Should make more precise calculations by using this reasoning with true credible values ​​of condos (costs, mass and size?). In my opinion there is not much to gain.

[Gaston]

Concerning a) I think that 5 to 10% of the capacity of the battery could be a correct base to make some tests.

It is absolutely huge.
The smallest car battery stores 40Ah in 12V.
In terms of stored energy, it contains around 480Wh or 1728000 Joules.
5% would correspond to 86400 Joules, i.e. the energy stored in an 880 Farad capacitor charged at 14V.

Unfortunately, we cannot use all the energy of the capacitor because the voltage drops as the discharge occurs.
If we admit that we can use the capacitor between 14 and 10V, the 86400 J corresponds to the use of a capacitor of 1800 Farads.

[elephant]

It also seems absolutely useless to me.

[Christophe]

It is absolutely huge.
The smallest car battery stores 40Ah in 12V.
In terms of stored energy, it contains around 480Wh or 1728000 Joules.
5% would correspond to 86400 Joules, i.e. the energy stored in an 880 Farad capacitor charged at 14V.

Unfortunately, we cannot use all the energy of the capacitor because the voltage drops as the discharge occurs.
If we admit that we can use the capacitor between 14 and 10V, the 86400 J corresponds to the use of a capacitor of 1800 Farads.

Indeed, I saw the thing on the "battery" side (increasing the buffer capacity by less than 5% did not seem too efficient to me ...)

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?

It’s huge in any case.

But if Ford rode it for a while, there was an interest.

ps: thanks for the Maxwell link!