Currently car boosters (a portable unit charged of an outlet and then connected to the car electrical system to start a car when the car battery is dead) typically use batteries - lead-acid, Li-Ion or LiFePO4. Over several years a battery in the booster will wear out.

Would it be practical to use a bank of supercapacitors that are more durable instead of a battery in a booster?

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    \$\begingroup\$ The major difference between supercaps and batteries is not the lifetime but the charge rate. Charging a lead-acid or Li-Ion battery is a chemical process that can take hours, you can charge a supercap in seconds. (for some definitions of "charge" and "a supercap") \$\endgroup\$ Commented Feb 9, 2012 at 9:02
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    \$\begingroup\$ @Kevin Vermeer: Yes, I know that and that is very cool. Yet buying a booster and using it infrequently and then finding its battery naturally worn out is totally uncool. \$\endgroup\$
    – sharptooth
    Commented Feb 9, 2012 at 9:12
  • \$\begingroup\$ Even if functionality per dollar was there, supercaps are huge compared to batteries in energy per volume. Something like an AA battery vs a soda can-sized capacitor. For a stationary application this might not matter much, but in anything portable or movable (phone, car, etc) it pretty much rules out supercaps as the main energy storage. \$\endgroup\$
    – Matt B.
    Commented Feb 9, 2012 at 22:16
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    \$\begingroup\$ @hassan789 -- Do you have some references for that? Supercaps will easily last half-a-million cycles. There are many easy-to-make mistakes in estimating life due to the capacitance recovery effect. Detailed analysis here \$\endgroup\$ Commented Jun 24, 2013 at 3:36
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    \$\begingroup\$ Our customers often use ultracaps instead of batteries specifically because the nearly-infinite lifetime makes them more cost-effective in the long-term. Maxwell ultracaps will keep running long after your first set of lead-acid batteries has died, and probably after the second set has as well. \$\endgroup\$ Commented Aug 6, 2013 at 20:31

4 Answers 4


I started this reply expecting the answer to be "not a chance" but a quick look at specs and prices suggests you could do something which was interesting and possible useful to some extent but that its really impractical and certainly not cost effective so far and is unlikely to be cost effective for a few cycles of Moore's law yet.

Assume starting requires 500 A at 12v for 1 second.
That's far too high in many cases - but lower currents for longer to much longer are common, especially on a very cold morning.
Adjust assumptions to suit.

Energy in a capacitor
\$ = \frac{1}{2}CV^2 \$
\$ = 0.5 \cdot 1 \cdot 144 \$
for 1 Farad at 12 Volt = Say 70 Watt seconds per Farad.

Car starting = \$ 12V \cdot 500A \cdot 1 \$ second as above
= 6000 Watt seconds.

So capacitance required to supply this energy at 12V =
\$ \frac{6000}{70} =~ 100 F\$.

Most supercaps are 2.5V to 3.3V rated for technical reasons.

You can buty modules like this 42V 100F unit that measures 550 x 270 x 110 mm and weighs 13 kg. The note it store 88200 Joule so is 88200/6000 ~= 15 times as large in capacity that the single start solution above.

To build a 12V cap from 3v3 caps requires 4 in series and from 2V5 = 5 in series. Placing capacitors in series reduces capacitance in inverse proportion to quantity so we would need 400F with 3v3 capacitors and 500F with 2V5 ones.

Murphy being active it would be wise to use say 1000F x 12v = 5 x 200F at 2V5.

At this stage it get interesting as we find that eg Digikey WILL sell you supercaps in this range.

Cost is very roughly 10 cents per Farad so a 200F ~~= $20 an 4 cost $80. Say $1000.

A look at the specs shows we are not there yet.
NO max discharge current specified but internal resistance of around 10 milliohm.That's perhaps 200+ Amps at short circuit. Loaded for maximum power transfer at Rload = Rinternal = 10 milliohm say, that's 100+A.
That's not really grunty enough for car starting, and we have not yet looked at voltage droop to extract energy etc.

Note that at \$ \frac{V}{2} \$ a cap has exhausted 75% of its energy.
If a cap is double the energy content needed then draining it to 70% will deliver half the internal energy with the other half stored for next time"

Pretty clearly, a 'battery' that is good for one start is not usually useful. Much bigger caps at bigger charge are needed. And even then it will not be possible to approach the energy capacity of a battery.

So - not practical yet - but slowly heading there. Maybe 10 years (about 7 Moore's law cycles)

470 - 3300 Farad x 2.5 V cells.


Leakage of the above is 0.5C mA - so for a 200 F cap that's 100 mA leakage. A farad will supply that for 10 seconds and drop a volt.

So a 200F cap will take \$ 2 \cdot 200F = 400 \$ seconds to drop a volt. Needs work! Some will be much better than this.

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    \$\begingroup\$ The Ultracap can store 88200 J like you calculated, but only if you supply it with the 42V. In a 12V system it will only store the 7200 J you calculated earlier. \$\endgroup\$
    – stevenvh
    Commented Feb 9, 2012 at 12:54
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    \$\begingroup\$ @stevenvh - Yes. As I've already gone through capacitor energy in the text above that I hope that that would be able to be calculated by "the student" if desired. The whole exercise is a bit moot as nobody is going to do this , so I've not gone into various peripheral points much. eg voltage drop as discharge progresses. I did note that the cap will be at 1/2 energy at 70% of voltage. A 12V car battery under cranking may drop to near 6V [!] - I already covered the supercap internal resistance and maximum power transfer load. But none of that is going to be too relevant if it's not built :-) \$\endgroup\$
    – Russell McMahon
    Commented Feb 9, 2012 at 13:43
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    \$\begingroup\$ Your 10 years estimate does look a bit too high. We are currently at affordable 3kF/2.7V with >200A continuous current at ESRs below 1mOhm and leakage of a few percent per day. \$\endgroup\$
    – PlasmaHH
    Commented Apr 11, 2015 at 22:32
  • \$\begingroup\$ Your initial assumption was 500 amps for 1 second. For an engine which has not caught, you should assume the starter current will be proportional to voltage. So, in your case, at the end of 1 second the starter current will only be 250 amps, which rather violates your original assumptions. The answer is, of course, more farads. \$\endgroup\$ Commented Jul 25, 2021 at 14:46
  • \$\begingroup\$ @WhatRoughBeast Yes, but ... :-). That was a VERY rough scoping estimate, and lower for much longer was also mentioned, and, ... . I have no good feel for motor resistance versus running current when back EMF rises, but I suspect that there may be a very strong surge due to mainly resistive characteristics. BUT they start up the motor and then engage it, so ... ? It would be interesting to measure. || About 250A here \$\endgroup\$
    – Russell McMahon
    Commented Jul 25, 2021 at 22:10

Batteries have a relatively flat curve of voltage over charge, up to a point. Capacitors have a linear curve of voltage over charge.

With batteries, you can just set up the booster battery pack in a way the voltage fits your need over a wide range of charge percentage.

With capacitors, this is not an option, because the voltage would change rapidly with use. To use it at all, you would need some power electronics to supply the correct voltage.

Also, currently the specific energy (stored energy per weight) of supercaps is still lower than what batteries provide. This might change within a few years.

So, right now, supercaps provide less energy, and need additional power electronics, making their use as power source inefficient.

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    \$\begingroup\$ Your power electronics analysis isn't quite correct. Battery voltage drops dramatically too. Remember, in a car, during engine start, the battery voltage will crash more than 50% momentarily as the entire mechanical load of the engine is manifest on the starter motor (it's why your headlights dim during cranking). Remember, a stalled DC motor looks almost like a short circuit. All you would need to keep the capacitor viable is a diode, so that once the engine cranks and the alternator starts producing power, the capacitor bank doesn't load the low-RPM engine and stall it out. \$\endgroup\$ Commented Jun 24, 2013 at 3:41

Very interesting discussion; I appreciate the thorough and detailed calculations. Even though the current technology seems to indicate that this is not a practical application, I found a tinkerer who seems to have had success: http://www.youtube.com/watch?v=GPJao1xLe7w Here is a commercial product designed for installation on 18-wheelers to ensure starting power; in this design one of the four truck batteries are swapped with the ultracapacitor engine start module, but on it's own wiring: http://www.maxwell.com/products/ultracapacitors/products/engine-start-module

Maybe with careful design considerations these ultracapacitor arrays can be useful in some situations.

  • \$\begingroup\$ FYI, same tinkerer has had success with newer 350Farrad 2.5V wired with 6 of them in series to have a max voltage of 15V, because alternators will put out over 14V. See youtube.com/watch?v=z3x_kYq3mHM . Though he's gone back to the 6x 2600F plus a 2.5W solar panel on the dash in youtube.com/watch?v=GUXM1XuLUIs for complete battery replacement. \$\endgroup\$ Commented Nov 28, 2015 at 6:00

As part of my work, I have some tools that compare capacitor banks for a given starting voltage, end voltage, load power, and time. Takes ESR and EOL values into account, too. My databank doesn't have every ultracap in existence, of course, but it's got a number of the most likely suspects.

So let's assume the battery you'd normally use would start at 13.2V unloaded, and drop to 7V when loaded to 500A. We can calculate our power from the low-end voltage, since that's clearly enough to start the car. To pull 3500W for one second from an ultracap and still stay in that voltage range, the best combination I see would be two of these in parallel. So you're talking about >$3k of ultracaps to replace a ~$100 battery. You might could get away with one ultracap module instead of two, especially if you don't use end-of-life values, but you'd have much less overhead, and your ESR losses would go up considerably. Even then, and even with direct-from-the-manufacturer quantity pricing you're still talking about $1500.

So it's doable, and not entirely insane. Whether it's cost-effective or not depends largely on how much your battery costs, and how often you need to replace it in the lifetime of your cap bank.

Regarding how you'd charge the ultracap itself, I don't think that's a problem. The terminal voltage on that capacitance at 3500W load after one second is about 10.2V, so we're talking about 11.5 kJ charge lost in the caps. (So we're delivering 3.5 kJ to the load, and 8 kJ lost in ESR!) That can be charged off a wall socket in just a few seconds. If you want a second shot, and have a wall socket anywhere nearby, you should be fine. And you're nowhere near the voltage limit of the capacitors, which means your charger doesn't need to be particularly smart, like a Li charger would have to be.

Edit: I came across this question again and reran the numbers based on newer tools, pricing, and available parts. The most cost effective solution now appears to be five of these in parallel, with a cost of roughly $600. And that's still assuming EOL values on the caps. For nominal, you'd only need three in parallel. Vast improvement over the last two years! It might actually pay for itself!


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