0
\$\begingroup\$

This is a highly theoretical question.

There are these 100kF, 2.7V capacitors you can buy online. According to my calculations, 5 of them in series would give me the energy capacity close to a 44Ah, 12V car battery (about 0.5kWh), which is convenient, because the voltage rating of the arrangement would be 13.5V.

So, in theory, I could replace my car battery with a capacitor bank that costs 10x more. The problem is that the output voltage is linear to the charge, so the voltage would have to vary from, say 3V (not lower to keep the current manageable) to 13.5V, which is well outside the operating range for car's stuff.

The questions:

  • What kind of circuit/equipment I would need to stabilize [3V-13.5V] to 12V?
  • What would be the efficiency of the process for the high currents needed for a car?
\$\endgroup\$
2
  • \$\begingroup\$ Alibaba.com alert. Alarm bells are ringing etc.. \$\endgroup\$
    – Andy aka
    Jan 4 at 11:37
  • \$\begingroup\$ A boost convertor capable of driving several hundred amps cranking current is likely to be ... sorry about this, but ... a non-starter. Use the capacitors directly for starting, which means they have to be practically fully charged. \$\endgroup\$ Jan 4 at 19:51
1
\$\begingroup\$

Although the charge is linear to the voltage, the energy varies as the voltage squared. This means you only need to swing between V and V/2 to use 75% of the energy capacity. Swinging over a wider voltage range gets you little extra energy, with a wider range needed at your converter, so it's probably not worth doing.

It's not a good idea to stabilise a 3-13 V swing to 12 V if you have alternatives, as that involves both bucking and boosting. Far better to do one or the other, so 5-10.8 V up to 12 V, or 13 to 27 V down to 12 V.

With suitable converter circuitry (multiphase) and with only small voltage ratios (max 3:1) you ought to be able to achieve over 90% efficiency.

Whether it's worth replacing something that's simple, economical and well tried and tested like a car battery with something expensive and complicated is another matter.

\$\endgroup\$
2
  • \$\begingroup\$ Why the gap between 10.8 V and 12 V? \$\endgroup\$
    – lvella
    Jan 4 at 14:49
  • \$\begingroup\$ @lvella supercaps in series come in multiples of 2.7 V, the same reason you used 13.5 V in your OP. To use a boost converter, the input voltage must always be lower than the output voltage. Though, maybe I see what you're getting at. A boost converter has a straight-through mode, and 13.5 V would be perfectly safe for the car electrics. \$\endgroup\$
    – Neil_UK
    Jan 4 at 15:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.