Using one super capacitor to charge another

[raylito]

I have a pair of 50F 2.7V capacitors in series. I charge them up to 5V.

I then have another pair of 50F 2.7V capacitors at zero to 0.5volts. i.e. almost fully discharged.

I wish to use the first pair of caps to charge up the second pair. I do so by connecting them in parallel. There is an initial high current draw, and both pairs reach about 2.6V as expected. (V1+V2)/2

Is there any way (such as using a Joule thief or similar) to almost fully empty the first pair of capacitors so that the charge on the second pair is higher than 2.6V or the equilibrium voltage? i.e say over 4V-4.5V?

I want to transfer as much of the charge from the first pair to the second pair.

Is it possible? :-D

 

[WimRFP]

Theoretically it can be done with an inductor and a switch that interrupts the current when the current wants to flow back (mosfet). Practically that inductor will become impractically large to keep current well below a point where ohmic losses will eat your energy.

A feasible solution is using a step-down/step-up converter (switching regulator). First the converter has to work as a current controlled step down converter (like a buck type), when output voltage equals input voltage, the converter has to change operation to a boost-type circuit (like in a PFC circuit). Because of the voltages in your example, you need synchronous rectification to get high efficiency.