I'm struggling to understand one part of the article. Could anyone expert in electronics help explain this, please?
> So what if we immediately disconnected the capacitor from the rest of the universe, including its power supply and ground? Well, if we did that after the capacitor was charged, we’d find a differential voltage on the power supply. What if we hooked up ground to the side that had -5V before?
Is this a typo - should this say "to the side that had +5V before?" (ie change minus to plus)?
I think I have a handle on the design, which is that it's charging a capacitor with +5V, changing the charging side to ground, discharging the capacitor through what used to be the input (you can tell I'm not an electrical engineer, I suspect these terms are very faulty) and causing a -5V flow that way. If that's true, I think it's a typo (?); if not, and much more likely, my understanding is flawed. Insight here would be much appreciated.
First you charge the capacitor like so from the 5v supply.
+5v -> +
Cap
0v -> -
Then you disconnect the capacitor and reconnect it this way:
+5v ------> +5v
0v -> + --> 0v
Cap
- --> -5v
Compared to 0v you now have -5v. Consequently you get 10v between the two extremes. The capacitor will deplete fast. So you would repeat the cycle quickly. And add some more circuitry to smooth the output.
> Is this a typo - should this say "to the side that had +5V before?" (ie change minus to plus)?
Yes.
> I'm struggling to understand the design, but I think it's charging a capacitor with +5V, changing the charging side to ground, discharging the capacitor through what used to be the input (you can tell I'm not an electrical engineer, I suspect these terms are very faulty) and causing a -5V flow that way.
That's correct enough. It's all a matter of perspective. I'd probably say you discharge it through what used to be ground, though as the article notes technically ground is the source of electrons - but we usually talk about holes and positive (conventional) current.
Imagine measuring the voltage on a AA battery: 1.5V. Then you flip it around and measure again: -1.5V. Voltage is all relative to where you measure, usually ground. It's a tricky concept.
> So what if we immediately disconnected the capacitor from the rest of the universe, including its power supply and ground? Well, if we did that after the capacitor was charged, we’d find a differential voltage on the power supply. What if we hooked up ground to the side that had -5V before?
Is this a typo - should this say "to the side that had +5V before?" (ie change minus to plus)?
I think I have a handle on the design, which is that it's charging a capacitor with +5V, changing the charging side to ground, discharging the capacitor through what used to be the input (you can tell I'm not an electrical engineer, I suspect these terms are very faulty) and causing a -5V flow that way. If that's true, I think it's a typo (?); if not, and much more likely, my understanding is flawed. Insight here would be much appreciated.