I want to generate a negative voltage from a possitive rail. To do so, I'm reading this article. Here is the schematics it is depicting, assument the input step is +10V:
I understand that when the capacitor is charged, there will be a potential difference of 9.7V, since the first plate is at 10V and the second one needs to be at 0.7 because of
However, the text also says:
When the positive plate drops rapidly from 10V to 0V, this 9.3V potential difference must be maintained due to coupling. Therefore, the negative plate drops to -9.3V.
I don't get that. My understanding is that if the voltage in the first plate drops sharply to 0V, C1 won't discharge through D1, because D1 won't be forward biased when its anode goes below 0.7V. Therefore, those 0.7V should discharge through the former C1 positive plate, which is now at 0V (so GND).
How is that the negative plate drops to -9.3V? how is that possible if D1 is supposed to force its anode to be at +0.7V with respect to ground?