diode cascade

[Mozart]

But how to "introduce " energy from surroundings to it?

Don Smith also has schematics of several voltage multipliers in his papers.

I assume that some energy is coming from outside due to the principles.
let's think about where that energy might come from.
As a child, I experimented with high-voltage cascades, and I noticed that they became cooler at certain points.
That might be the gain.

Regarding the theory:
The available current (A) remains the same at the input and output of the diode cascade.
ChatGPT confirmed this for me.

An Example.
Let's say we have 12V voltage with a current of 120mA as an example.
Cascade Input: 12V 120mA Formula: P = U * I , 1.44W = 12V * 120mA
Cascade Output: 64V 120mA Formula: P = U * I , 5.52W = 64V * 120mA
Therefore the gain is = 4.08W
Of course, one must also consider the leakage currents of the diodes and the thermal losses in the model.
Am I making a logical error with ChatGPT?

Well, chatgpt might be correct. If you have available 20-100 diodes you can just put them in series  and use your multimeter to check.

I have used such diode chain and despite forward voltage drop the output voltage is increased. I have used 40x byv27 200v 2a diodes and nothing else, go outside and expose them to natural light and you will see 10-30V with very little current, attach them to any battery in series and you get the current from battery and all the voltage from diodes.

I was thinking to make a solar panel out of these diodes as when they are in full sun they generate great voltage and they work even if it is cloudy.
I guess you will pay approx £30 for 2000 diodes and you can make strings of 20 in series and then parallel them.