OK, I know how a voltage doubler works ... or do I? Rereading the explanation of its operation, there's something I just cannot wrap my brain around.
So this, as we all know, is the classic voltage doubler circuit:
**broken link removed**
All the explanations invariably start with this: on the first negative cycle on the input, the first (leftmost) diode conducts, charging the leftmost capacitor with the polarity shown:
**broken link removed**
But here's where it goes off the tracks for me. Now the first positive-going cycle comes. But isn't it going to be in opposition to that charged capacitor?
**broken link removed**
Look at the way the capacitor charges on the negative cycle: the capacitor plate closest to the positive terminal, the rightmost plate, gets a positive charge. Right?
So now when the positive cycle comes, it's going to want to charge the plate closest to the positive terminal--the leftmost plate--positively. Am I right? But that plate already has a negative charge, not a positive one. So instead of doubling the charge, shouldn't it just cancel it out to close to zero?
I must be missing something really obvious here, since it's well known that this circuit actually works. So what am I missing here?
So this, as we all know, is the classic voltage doubler circuit:
**broken link removed**
All the explanations invariably start with this: on the first negative cycle on the input, the first (leftmost) diode conducts, charging the leftmost capacitor with the polarity shown:
**broken link removed**
But here's where it goes off the tracks for me. Now the first positive-going cycle comes. But isn't it going to be in opposition to that charged capacitor?
**broken link removed**
Look at the way the capacitor charges on the negative cycle: the capacitor plate closest to the positive terminal, the rightmost plate, gets a positive charge. Right?
So now when the positive cycle comes, it's going to want to charge the plate closest to the positive terminal--the leftmost plate--positively. Am I right? But that plate already has a negative charge, not a positive one. So instead of doubling the charge, shouldn't it just cancel it out to close to zero?
I must be missing something really obvious here, since it's well known that this circuit actually works. So what am I missing here?
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