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I constructed a simple voltage doubler on a bread-board. Here's the schematic.

schematic

simulate this circuit – Schematic created using CircuitLab

The output across the 100kΩ resistor was the following. (middle waveform is input, top is output)

enter image description here

Now when I scaled the resistance and capacitance (for the elements on the right) proportionally - resistance went to 1kΩ, capacitance went to 1mF - the output changed as follows.

enter image description here

Why does the output voltage go down so much? I can't seem to analyze the circuit and see why this would happen. Might I have mixed something up?

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    \$\begingroup\$ these voltage doubler circuits cannot output much current, you'd be lucky to get a few milliamps from them and retain the "expected" doubling of voltage. As the load increases, the actual voltage output sags quite a lot. \$\endgroup\$ – KyranF Mar 25 '15 at 1:59
  • \$\begingroup\$ By 'sags', do you mean lowers? If that was true, wouldn't my first output voltage (more resistance - doubled voltage) have to be lower than the second output voltage (less resistance - almost halved voltage)? \$\endgroup\$ – Arturo don Juan Mar 25 '15 at 2:08
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    \$\begingroup\$ load means less resistance. Your circuit operates correctly with high resistance on the "output", but "sags" and is lower when the load is lower resistance \$\endgroup\$ – KyranF Mar 25 '15 at 2:18
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Keeping the left hand cap at 1uF is the problem. Just do the math - it has an impedance of 398 ohms at 400Hz and in your original circuit you had a 100k load. No problem here but, dropping the load to 1kohm means you should increase the left hand capacitance appropriately in order to counter the loading effects. Or increase the operating frequency in order to reduce the 1uF reactance.

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  • \$\begingroup\$ So increasing both capacitance values (left as you said, and right as I said in the question) is necessary? \$\endgroup\$ – Arturo don Juan Mar 25 '15 at 15:59
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    \$\begingroup\$ Left cap is fundamental - you can't get the current needed by the load resistor if this cap is too low in value. Right cap smooths the dc applied to the resistor. \$\endgroup\$ – Andy aka Mar 25 '15 at 16:01
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Minor rant: this would be a LOT easier if you included component designators to your schematic.

I assume that you increased BOTH capacitors to 1000 uF? Are they installed with the correct polarity? The left-hand cap should have (-) on the left & (+) on the right. The right-hand cap should have the (-) lead to ground and the (+) lead to the output.

Measure the AC voltage coming in. Make sure that it hasn't dropped lower than you expected.

Try increasing your load resistance to 10k and see what happens.

So long as your AC source can supply sufficient current, you should see the output voltage doubled less diode voltage drops AND the effect of capacitive reactance.

Equivalent output impedance depends on the capacitive reactance at your input frequency. Larger capacitors mean lower impedance which allows more output current and less sag.

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