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I'm not able to actually understand how the circuit behaves as a voltage doubler.From my understanding the circuit would behave as a voltage doubler if during the positive half cycle only D1 would conduct and only C2 capacitor would charge to positive max of input and for negative cycle only D2 would conduct and only C1 would charge to negative max of input and hence the voltage drop across resistor would be 2*max input voltage. However my doubt is during positive half cycle apart from cap C2 wouldn't C1 as well get charged through the resistor similarly during negative half cycle wouldn't C2 also be charged through the resistor. If that is the case would the circuit still act as voltage doubler?

enter image description here

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wouldn't C1 as well get charged through the resistor...[?]

You are correct that current would flow through the resistor and then through C1. C1 would actually be discharged by this action, but by being in the circuit, its voltage (on the - side of the resistor) is added to the voltage of the power supply (D1 to the + side), resulting in double the voltage across the resistor (which is the load). C1 should not discharge all the way; if it does, then there is too much load current and the voltage ripple would be nearing 100%. And so you would want to use a bigger capacitor. The same action applies to C2 during the negative cycle.

I've never seen a full-wave doubler drawn this way, but of course it is correct. I slapped my head when I saw this. And for your benefit, you should search the Web for a full-wave doubler, and look at the images, and see how it's usually drawn. It should make more sense that way.

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  • \$\begingroup\$ How one usually draws a voltage doubler, tripler and quadrupler. Note how the same approach can be used to go even further. (source) \$\endgroup\$ – Mast May 1 '17 at 6:52
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    \$\begingroup\$ Feel free to include any and all of the above comment to improve your answer. Re-directing to "you should search the web" is less helpful than actually showing it. Thought I'd save you some time. \$\endgroup\$ – Mast May 1 '17 at 6:54
  • \$\begingroup\$ Your response is appropriate, but in this case I wanted the OP to do their own search so they could appreciate the breadth of circuits, see the ones that are most popular, and the fact that the one represented here is really quite unusual. \$\endgroup\$ – gbarry Jul 12 '17 at 5:58

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