Pierre's 170W in 1600W out Looped Very impressive Build continued & moderated

[TinselKoala]

However, I don't think I'll need to go to that high of an input voltage and can easily reduce the quantity which will boost their storage capacity since they're connected in series.

Does it? Consider an example.

Let's use 100 F, 2.7 V caps and put 4 in series. So we now have 25 F capacitance and 10.8 volts maximum. Energy storage capacity at max voltage will be E = (CV²)/2 = (25 x 10.8 x 10. 8) /2 = 1458 J.
Now let's remove one capacitor and calculate energy storage capacity for 3 in series. We now have 33.3 F capacitance and 8.1 volts maximum. Energy storage capacity at max voltage will be E = (33.3 x 8.1 x 8.1)/2 = 1092.4 J.

So while capacitance is increased by removing a capacitor from the series chain, the decrease in maximum voltage more than compensates, and energy storage capacity goes down. Makes sense: fewer capacitors, less energy storage.

However... at the same voltage, say 8.1 volts, the energy storage of the 4-series stack is only (25 x 8.1 x 8.1)/2 = 820.1 J.

So in that sense (equal voltages) the stack with fewer series caps does store more energy.

[seaad]

This is an earth-shattering, paradigm-shifting, Nobel-prize-winning claim

I wish you good luck with the test results Luc!
Arne

[gotoluc]

Does it? Consider an example.

Let's use 100 F, 2.7 V caps and put 4 in series. So we now have 25 F capacitance and 10.8 volts maximum. Energy storage capacity at max voltage will be E = (CV²)/2 = (25 x 10.8 x 10. /2 = 1458 J.
Now let's remove one capacitor and calculate energy storage capacity for 3 in series. We now have 33.3 F capacitance and 8.1 volts maximum. Energy storage capacity at max voltage will be E = (33.3 x 8.1 x 8.1)/2 = 1092.4 J.

So while capacitance is increased by removing a capacitor from the series chain, the decrease in maximum voltage more than compensates, and energy storage capacity goes down. Makes sense: fewer capacitors, less energy storage.

However... at the same voltage, say 8.1 volts, the energy storage of the 4-series stack is only (25 x 8.1 x 8.1)/2 = 820.1 J.

So in that sense (equal voltages) the stack with fewer series caps does store more energy.

Yes, you're right of course as far as total Joule energy value goes if you keep voltage the same. Sorry I didn't write in a way that makes sense to your education.

What I was trying to say is, the less capacitors you have in series the less you divide the capacitors rated Farad value.

Lets say my tests show I only need 10vdc. If I leave the 10 cap in series of 50 Farads each = 5 Farads total storage @ 10V = 250 Joules
However, if I remove 6 caps, I now have 4 caps in series of 50 Farads each = 12.5 Farads total storage @ 10V = 625 Joules
Do you now understand?

Regards
Luc

[listener191]

Those super cap boards are not just PCB's, they have active balancing circuitry to ensure even charge distribution.

This ensures that no single cap is taken over 2.7V during charging.

I have three of those boards.

See https://www.electronicsweekly.com/news/products/analogue-linear-mixed-signal-ics/design-supercapacitor-charger-balancing-2014-08/

Regards

L192

[gotoluc]

First test with magnet as rotor. Without current limitation and magnet rotor the device consumes exactly 1 Amp at 3.50 Volts.

Fr. Premier test avec aimant comme rotor. Sans limitation de courant et rotor magnétique, l'appareil consomme exactement 1 ampère à 3,50 volts.


Video demo: https://youtu.be/w_z7tcq2N9E