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Is this a bad idea? Supercapacitor bank to crank engine, fed by the main battery

Thread starter #1
I'm thinking of putting together a source of backup power that could crank the engine and start the car, even if the battery is discharged and/or cold. A bank of supercapacitors would provide the power to turn the engine. 6 x 2.7V/500F in series with balancing circuitry, to total 16.2V/83F max. The supercapacitors would be charged from the main car battery via a voltage boost circuit, so even when main battery voltage drops below 12V, the supercapacitors can still be charged to about 14V. A diode would prevent power returning to the battery, and the starter would be the only thing connected to the supercapacitors.

Here is the general order of the circuit I have in mind:

car battery > diode > voltage booster > supercapacitors > starter

Could this damage the car's battery or electronics? Am I overlooking anything here? Thanks for any input.



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What is the effective series resistance of your 83F capacitor? The peak in-rush starter motor current (depending on engine displacement) to get the engine turning is like 500A, and then the cranking current until the engine starts is > 100A.


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I would say yes, it is a bad idea.

Keeping MikeMI's post in mind, note the graph below:
The voltage, across ALL types of caps (an "Ultra" type in this case) when under a load, drops very quickly, vastly reducing the power (I times E) that can be delivered.

They simply aren't suitable for your application at a reasonable price.

Not that it can't be done, For instance: but the cost:
is remarkable. Note also that it's only rated at 100A.
Thread starter #5
Thanks for the replies. I hadn't considered the internal resistance of the capacitors. The ones I was looking at (on eBay) don't even list it, so not going to look further into that particular item.

I see Maxwell units, as suggested by cowboybob, have more complete specs. 1900A is overkill for a car starter, but a smaller model such as this one might be usable?

At 15V and 19 mOhm, I calculate a peak of 789A. Seems this should be sufficient to get the starter going, as the starter current drops dramatically after the initial rush of current.


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At 15V and 19 mOhm, I calculate a peak of 789A.
That would be the current into a dead short, not into a starter motor, if 19 mOhm is the cap resistance. Or is 19 mOhm the starter resistance?


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BTW, Welcome to ETO, Sam5253!

I think the major downside of using caps (of whatever flavor) as a "battery", in particular as you envision, is that their full voltage level immediately starts to drop from its peak as soon as a load is applied. This has the effect of, in a like manner, of quickly reducing the power transferred to the load, as graphically represented in the graph in post #3.

This effect can be lessened, to a degree, by massively over building the circuit by increasing the total capacitance to, in my opinion, ridiculous and very expensive high values for automotive use.

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