I´m working on a voltage multiplier with 5 stage, i´m using the same capacitor and i think that my problem is the current load but i actually think that with a resistor in the output of the multiplier but this doesnt work. I actually calculated the output with the ideal equation (5*Vin)-(((n-1)/(c*f))*Ioutput).The output multiplier have to be 5 times the input voltage.

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How I should calculate the value of the load resistor and is the schematic that I made wrong?. enter image description here

Taken of https://www.onsemi.com/pub/Collateral/1N4001-D.PDF

  • \$\begingroup\$ Your post is very unclear and you haven't asked a question. Are you really driving a multiplier with 100 mV power source as shown in your schematic? \$\endgroup\$ – Transistor Mar 6 '19 at 22:08
  • \$\begingroup\$ You refer to "my problem", but you never say what it is. What do you expect to see, and how does what you are seeing differ from what you expect? \$\endgroup\$ – TimWescott Mar 6 '19 at 22:12
  • \$\begingroup\$ @TimWescott I should expect in the output 5 times the value of the AC voltage source \$\endgroup\$ – Juan Mar 6 '19 at 22:15
  • \$\begingroup\$ First, please put that in your question. Second, what does a 1N4001 do when you forward-bias it by 0.1V? \$\endgroup\$ – TimWescott Mar 6 '19 at 22:17

As prompted a couple of times in the comments, the Typical Forward Voltage chart you've added shows that at room temperature you need 0.8 V to get any significant current through those diodes. You are only feeding the circuit with 100 mV hence the diodes don't conduct and you don't get any multiplication.

Your ideal equation

$$ V_{out} = 5V_{in}-\frac {n-1}{Cf}I_{out} $$

(which I haven't verified) does not account for diode voltage drops, as far as I can see. The equation will only hold true when Vin >> 0.8 V.

  • \$\begingroup\$ Thanks for answer my question, is isnt working yet but i will keep trying it \$\endgroup\$ – Juan Mar 6 '19 at 22:55
  • \$\begingroup\$ What are you going to try next? \$\endgroup\$ – Transistor Mar 6 '19 at 23:30
  • \$\begingroup\$ Make the real equation and try but im stuck and i dont know now how to do to this \$\endgroup\$ – Juan Mar 6 '19 at 23:34
  • \$\begingroup\$ Can you not see that your input voltage is too low? You need more like 5 to 10 V instead of 100 mV input. \$\endgroup\$ – Transistor Mar 7 '19 at 7:23

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