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If I connect super capacitors in series, I'm able to raise the voltage by the sum of the capacitors' voltages, but I lose 1/2 of each of the Farad capacity for each of them in series. Correct?

If I connect these super capacitors in parallel, I have the sum of the Farad capacity without loss of charge. Correct?

So if I instead use an up-converter to increase the voltage instead of connecting the super capacitors in series, I would have a huge amount more storage available. Is my logic correct here?

Even if the up-converter is only 92% efficient, It beats the 50% loss per capacitor in series. Right?

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    \$\begingroup\$ No. Calculate the energy instead. E=CU^2/2 \$\endgroup\$ – winny Jul 30 '20 at 8:07
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Take two one Farad capacitors and charge them to 1V. Energy stored in each is 0.5 Joules.

As capacitance is defined as the ratio of charge over voltage, the charge in each capacitor is one Coulomb.

Connect them in in series, and you have two Volts across them. Discharge them through an integrating current meter, and one Coulomb charge is measured, as the same current path flows through both. The total capacitance of the combination is 1 Coulomb/2 Volts = 0.5 Farad. The total energy is one Joule, either because you have two capacitors at 0.5J or from the formula ½C∙V².

Connect them in parallel, you have one Volt across them. Discharge them, and you measure two Coulombs as the current from each are added together. So the total capacitance of the combination is 2 Coulomb/1 Volts = 2 Farad. The total energy is one Joule, either because you have two capacitors at 0.5J or from the formula ½C∙V².

But in both, you can swap the capacitors over from one topology to the other at a given state of charge and the capacitors themselves don't change.

If both capacitors are fully charged, so the energy stored is at the maximum, and switching topology would mean "a huge amount more storage available", then you could charge them in one topology then use a relay to switch them to the other, and use this extra energy to charge another bank, and so create a perpetual motion machine. In reality, the amount of energy storage capacity is the same.

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Regardless how you connect the caps, the energy is the same (1/2CV^2). So you just need to convert the voltage up or down depending upon your application. There might be secondary issues like ESR or something like that, but those are secondary to the energy problem. So stack and buck down, or parallel and boost up. Nice thing about stacking is the current that charges one cap also charges the other. One caution, you don't want to overcharge either cap. So you need to have some method to control the charging if one cap is smaller than the other (i.e. V(Cbot) rises faster than V(Ctop). Bottom line is both caps need to be managed so they aren't overstessed.

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