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This is the given circuit with output voltage across RL resistor with input voltage amplitude 5V

Output across A and B.

Simulated circuit with output measured between A and B nodes

{ The first image is original circuit.} { The second image is the output across RL resistor with ac source amplitude voltage 5V. } {Simulated circuit with output measured between A and B nodes.}

To me it seems like the wave is rectified and smoothened using capacitors. Also capacitors are storing energy and adding it to the negative wave part.

Can someone please tell me what this circuit and how this output is achieved?

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  • \$\begingroup\$ Show the precise circuit you simulated and indicate what nodes you are measuring. Don't forget to show component values, component names and the operating frequency (along with any source impedances). \$\endgroup\$ – Andy aka Oct 12 '20 at 12:52
  • \$\begingroup\$ To me it seems like the wave is rectified and smoothened using capacitors. Also capacitors are storing energy and adding it to the negative wave part. And the capacitors are.... in series. So the voltages add up. It is a combination of two halve wave rectifiers. I'm not sure what you mean by "what this circuit is representing". Many circuits have a name like "halve wave rectifier" and all circuits have a function like "convert AC into DC". That's it. \$\endgroup\$ – Bimpelrekkie Oct 12 '20 at 12:52
  • \$\begingroup\$ Half-wave rectification \$\endgroup\$ – Finbarr Oct 12 '20 at 12:54
  • \$\begingroup\$ I have added the circuit which I simulated. \$\endgroup\$ – Prajit Kumar Oct 12 '20 at 12:59
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It's a voltage doubler circuit. The output voltage (ignoring diode drops) is double the peak voltage of the AC input.

Since the capacitors charge on alternate half-cycles, the total voltage goes up in a somewhat lumpy fashion.

The output capacitor C3 causes the output to rise more slowly and in lumps than if you had two larger caps in series. At the first power-on it will tend to reverse-bias the capacitor which is not being charged, which is not good if the capacitors are polarized types.

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  • \$\begingroup\$ You can take a look at the circuit which I simulated. \$\endgroup\$ – Prajit Kumar Oct 12 '20 at 13:01

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