Has anyone here ever chatted about a diode cascade? I asked ChatGPT a bit about it yesterday and got some interesting answers.
The voltage at the output increases while maintaining the same input-output current. Since P=U*I, theoretically, a higher power output is possible at the output compared to the input. With a stepped cascade of 12V 12mA input, one could achieve several watts of gain.
https://en.wikipedia.org/wiki/Voltage_multiplier
Have I overlooked something?
ChatGPT Part:
The voltage at the output increases while maintaining the same input current. Since P=U*I, theoretically, a higher power output would be possible at the output compared to the input. With a stepped cascade of 12V 12mA input, one could achieve several watts of gain.
If energy from the surroundings is introduced into the system, similar to how a heat pump works, this could explain the increased power output.
But how to "introduce " energy from surroundings to it?
Don Smith also has schematics of several voltage multipliers in his papers.
(11-11-2024, 11:51 AM)Andy Wrote: [ -> ]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?
The same current at the beginning and at the end is internesting thing. I assumed that when voltage goes up, current goes down. If not in this case (in this multiplier), it sounds interesting. I know that ChatGPT sometimes talks nonsense. I am open to this new idea. I will also try to search in this direction.
Andy
Edit: I asked ChatGPT about it and here is the answer:
"In a voltage multiplier circuit, like the Cockcroft-Walton multiplier or others, the voltage indeed increases at each stage while the available current decreases."
Thanks for checking.
As you said, chat GPT is sometimes not entirely trustworthy
(11-11-2024, 04:00 PM)Andy Wrote: [ -> ]The same current at the beginning and at the end is internesting thing. I assumed that when voltage goes up, current goes down. If not in this case (in this multiplier), it sounds interesting. I know that ChatGPT sometimes talks nonsense. I am open to this new idea. I will also try to search in this direction.
Andy
Edit: I asked ChatGPT about it and here is the answer:
"In a voltage multiplier circuit, like the Cockcroft-Walton multiplier or others, the voltage indeed increases at each stage while the available current decreases."
In a voltage multiplier circuit, each capacitor stage needs time to charge up to its target voltage. When you have a multiplier with multiple stages (e.g., 4x or more), the input current gets effectively divided across these stages. This means the charging process is spread out, and each capacitor charges more slowly as a result.
While GPT's interpretation is partially correct, there’s a key point to understand: As you increase the number of stages, it will indeed take longer for the entire multiplier circuit to reach higher voltage levels. You can still discharge this stored energy at nearly the same current as your input, but it will be delivered as a pulse (a brief surge of current, also known as a Joule pulse).
The main downside here is a trade-off in output frequency. The output frequency becomes lower than the driving frequency due to the longer charging times and inherent losses (from diodes and other components). These losses introduce a significant efficiency drop, meaning the input power isn’t fully converted to useful output power, resulting in a noticeable input-output discrepancy.
Hi Joel, when I saw your video about Nonlinear Dual System, I asked myself if it is possible to "inject current" into the voltage multiplier also. But I think that it is the same principle as you told in video, the voltage multiplier is just source of HV DC. I also found some image but YT doesn't allow me to put the link in comment. So I upload it here.
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