Good day all, I'm toying with the idea of using a small supercapacitor bank to buffer short bursts for acceleration/regen of an electric vehicle (likely an e-bike).

It seems wasteful to put a capacitor bank in parallel with the battery stack (both with independent balancing systems for their respective unit/cell voltages) because the capacitor bank would only provide/accept energy over the small voltage fluctuation when the battery pack provides/accepts current.

What cheap alternatives are there to use the entire capacitor bank capacity over a small voltage fluctuation? An active solution would be a bidirectional DC-DC converter that can fully discharge the capacitor bank when it sees the battery pack voltage sag and then recharge the capacitors when the battery pack voltage jumps up. I imagine this being nonideal though because the DC-DC converter would need to be high power (and therefore heavy/expensive) to handle the relatively infrequent use case of start/stopping.


For the sake of making this fun for us engineers let's throw in some numbers with a more likely situation: we want 20kW for 10 seconds to help accelerate an electric car. 200kJ stored in a supercap bank that has a 5Wh/kg density is about 11kg of capacitors, totally reasonable for an EV! This page cites Maxwell's supercaps as capable of charging/discharging in under 10 seconds: https://batteryuniversity.com/learn/article/whats_the_role_of_the_supercapacitor

To highlight the problem, if we have a sad battery pack with 200mOhm of internal DC resistance cough Leaf, a load of 20kW on the battery would still only sag it's voltage from 360V to about 349V so a supercap bank sized for 200kJ in parallel (3.08F @ 360V) would only provide 12kJ, about 6% of the energy, hardly the capacity we could get if we could extract down to 0V.

Is the only other solution to use a DC-DC converter capable of converting 20kW bursts and tracking a huge voltage swing (following the capacitor charge/discharge voltage profile)?

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    \$\begingroup\$ Welcome to SE EE. I suggest you do some calculations to determine: 1) what amount of energy you would need to store (for that acceleration) 2) determine how much energy actually fits in a "small supercapacitor" 3) determine how many of those "small supercapacitors" you would need. 4) determine if it is even possible to extract that amount of energy in the time that you would need it (hint: not possible). 5) do the same calculation with any capacitor of your choice. \$\endgroup\$ Jul 12, 2019 at 21:17
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    \$\begingroup\$ In my opinion those calculations will show that your idea will be totally impractical. But feel free to do the calculations and prove me wrong! It is much easier to just use batteries which can deliver the required energy during a burst. I was recently asked to look into a project where a capacitor was used to store some energy (from a low-current cell) to perform a cellular connection transmit burst. The idea was totally impractical as a (small) capacitor cannot store enough energy even for that transmit burst and that was only a couple of Joule. \$\endgroup\$ Jul 12, 2019 at 21:18
  • \$\begingroup\$ Does the bike motor have a freewheel? If so then no regen. \$\endgroup\$
    – Transistor
    Jul 12, 2019 at 22:13
  • \$\begingroup\$ @Transistor - Most EV motors don't freewheel as they typically provide regenerative braking as an option. For the sake of this discussion, a motor capable of being forced to regenerate should be assumed. \$\endgroup\$ Jul 12, 2019 at 22:36
  • \$\begingroup\$ @Bimpelrekkie - That can be true if you're using a microfarad range capacitor as an output filter, but supercaps can store kilojoules at reasonable voltages. \$\endgroup\$ Jul 12, 2019 at 22:38

2 Answers 2


You are wasting your time and money using supercaps because each tiny 18650 Li Ion cell has over 10 thousand Farads and you can use all of its Ah capacitance over a small voltage range of 3.7 to 3.0V unlike caps which must be up- converted to use all of its stored energy down to 0V. If you wanted more Jerk for about 100 milliseconds which won’t give you must acceleration boost averaged over 10 seconds.

But imagine baby elephant solution with costly power electronics to satisfy a super wide input Voltage range (>2:1 is wide, 10:1 is super-wide 100:1 is never a good idea, so think again. It’s a great idea to start snowmobiles for << 1s but not drain an e-bike for 10s with a heavy, expensive “white elephant” solution.

But hold this thought for another 10 years and maybe Maxwell will have a super corrosive solution with C60 electrodes that packs more energy/kg.

Also when using higher voltage batteries, you can expect higher ESR from series connections but enjoy less conduction losses for the same power demand since it uses less current.

  • \$\begingroup\$ Using "farad" to describe the energy stored in a battery is just plain wrong. \$\endgroup\$ Jul 15, 2019 at 2:36
  • \$\begingroup\$ @JulieinAustin alas it’s you that are wrong. This is Capacitance which in Amp-hrs can be defined I*dt=CdV for dV = from 3.8 to 3.0 or as per OEM spec. You should know better. Did I say C = E? \$\endgroup\$ Jul 15, 2019 at 3:27
  • \$\begingroup\$ I understand you can get away with what you did, but the comparison is only accurate if you limit the "capacitance" to the voltage range of the battery. We could have an entire discussion about why I disagree with you, but the biggest reason is the OP is asking about an application with a large number of charge and discharge cycles. That poor 18650 isn't capable of the number of charge and discharge cycles what Maxwell, Vishay, and others quote. Unless you limit the depth of charge and discharge, in which cause you've lost a lot of those farads. \$\endgroup\$ Jul 15, 2019 at 3:58
  • \$\begingroup\$ @JulieinAustin My analysis is correct and C declines rapidly with aging as ESR rises at the same time. but a super cap equivalent to a 18650 is a a massive expensive solution that lasts a million cycles instead of 250 to 1500 depending if you know how to extend Charge life. your understanding is incorrect. C is computed from Vi to Vf but supply caps cannot find a booster to utilize all the energy to Vf=0. While batteries do not have that problem. \$\endgroup\$ Jul 15, 2019 at 4:00
  • \$\begingroup\$ Again, I disagree -- lithium chemistry batteries do not CYCLE in the time frame needed for regenerative braking. It's just a fact that supercaps are already being used for the OP's application. What makes the OP's application impractical isn't the impossibility of the application, it's that an e-bike is the wrong application for the technology. maxwell.com/products/ultracapacitors \$\endgroup\$ Jul 15, 2019 at 4:44

There are multiple causes of improbability and impracticality, but they don't all reside in the amount of capacitance at a given voltage.

You can calculate how many volts and how many farads would be needed, and come up with a value which would work. The math is relatively simple as the energy, in ioules, is equal to 1/2 C V^2, where C is in farads and V is in volts. Believe it or not, but for a vehicle in the "e-bike" range, it isn't all that horrible for "e-bike" speeds, weights, and passengers. Many e-bikes are in the 300 to 600 watt motor range, and typically acceleration to full speed will happen within 10-15 seconds. On the high end, you'd need to store about 9,000 watt-seconds, or 9,000 joules.

Believe it or not, this isn't hard. I have a 90F supercap sitting on my desk. It happens to have a maximum working voltage of 5.6 volts, so four of them in series-parallel should make something capable of storing 90 farads at 10 volts. Plugging that into E = 1/2 C V^2 that's 9,000 watt-seconds. I really could get 600 watts out of that for 15 seconds.

[ I forgot to multiply by 0.5, so the answer is actually 4.5 kJ, not 9 kJ, but I hope my point is made ]

Except that you really can't. When that supercap is fully charged, 600 watts at 10 volts is 60 amperes, which is a VERY heavy wire. It is also impossible because a capacitor also has an "equivalent series resistance", and as you know, V = I * R. In this case, V is the voltage drop across the non-ideal-capacitor resistance (ESR), I is the current (60 amps), and R is the ESR value itself. There is also the non-small problem of getting 60 ampere through the small leads which are spot-welded onto the supercap. But more to the point, continuing to extract 600 watts as the voltage approaches zero requires an ever larger amount of current.

So, it isn't just a matter of finding a large enough supercapacitor -- for lower power applications, supercapacitors are more than up to the task. It's just that your specific application requires currents which are beyond the realm of reasonable possibility.

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    \$\begingroup\$ Good answer. Small point: SI units named after a person have their symbols capitalised but are lowercase when spelled out. 'V' for volt, 'A' for ampere, 'K' for kelvin, 'Ω' (capital omega) for ohm, etc. Meanwhile 'k' is for kilo. They also recommend a space between the number and the units (as in '5 apples' rather than '5apples'). \$\endgroup\$
    – Transistor
    Jul 12, 2019 at 22:36
  • \$\begingroup\$ Yeah, but I'm a weirdo. I'm an old weirdo in a country that doesn't use SI units because SI units are evil and cause Socialism. I just don't get the practice with what I should CapItaLiZe. \$\endgroup\$ Jul 12, 2019 at 22:39
  • \$\begingroup\$ Thanks for the response! I hadn't considered ESR as an issue for some reason I assumed supercapacitors were the perfect solution. Without getting into the weeds with power requirements for a specific application, I'd like to focus the answers toward the general implemention of the power management. Truthfully @bimpelrekkie is right that it's impractical for an ebike and I should just buy more capable batteries, so for the greater good of future EV designers, how can we cheaply solve the problem of using the full capacity of an arbitrary, high current, capacitor bank? \$\endgroup\$ Jul 13, 2019 at 2:48
  • \$\begingroup\$ @KentAltobelli - The short answer is to use a higher working voltage. Pretty much whenever the question involves I^2 R losses, increasing V to reduce I is the answer. The cool thing is that increasing V also increases the energy stored. \$\endgroup\$ Jul 15, 2019 at 2:22
  • \$\begingroup\$ If your capacitor is rated to go higher, than yes it'll store exponentially more energy, but I'm assuming that I'm getting the full energy out of however my capacitor bank is configured. You've got me thinking though, the benefit to a lower voltage bank is a lower slew rate for whatever device is going to downconvert or upconvert the battery pack voltage to charge/discharge the capacitors. \$\endgroup\$ Jul 15, 2019 at 3:26

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