I have this voltage multiplier sextupler (Vin=1V, f=4Hz, l=30µH.)

I would know the output power when the capacitor charged 1.5V.

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  • \$\begingroup\$ It looks like you already have the circuit created in a simulator program...why don't you just run a simulation and find out for yourself? \$\endgroup\$ Jun 16 at 13:02
  • \$\begingroup\$ i want to calculate it theoretically \$\endgroup\$ Jun 16 at 13:29
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    \$\begingroup\$ You should make at least some effort to do this yourself, and show us your work. Then you can ask a specific question if you get stuck. \$\endgroup\$ Jun 16 at 13:41

It's not going to be much.

That thing isn't (quite) a half wave Cockcroft-Walton multiplier, but it is first cousin to one.

Its impedance will be some multiple of the impedance of one of the 47 uF capacitors at 4 Hz. That's some multiple of about 850 ohms. Probably about 10 times, so something like 8.5k.

It will take quite some time to charge that 12 millifarad capacitor.

The low input voltage means that you won't be getting very much voltage at the output. As a sextupler, you would expect it to put out about 6V from a 1V input, but then you have to subtract one diode forward voltage from the output - for each stage. You lose something like 2 or 3 volts in this setup. You might get 3 or 4 volts out of it.

A short circuit of the fully charged C1 would deliver considerable power but only for a short time. Other than that, you are looking at about three volts through about 8.5k. Call it maybe 0.3 milliamperes continuous.

L1 is pretty much useless. It is as good as a short circuit at 4Hz.

  • \$\begingroup\$ thank you very much but why Its impedance will be some multiple of the impedance of one of the 47 uF capacitor i know that in dc the frequency equal 0 that s mean the impedance of C1 towards infinite \$\endgroup\$ Jun 16 at 20:01
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    \$\begingroup\$ The impedances add up. They are partially in parallel and partially in series with each other. The total is difficult to figure, but it will be a multiple of the impedance of one capacitor. \$\endgroup\$
    – JRE
    Jun 16 at 20:03

I want to calculate it theoretically

That's not going to be possible, for most humans: To begin with, even with a Schottky diode like your BAT54, 1 V barely even brings your diode into saturation. In other words, the simplification "diode is either off for V < 0 , or perfectly on for V>0" simply is not good enough. modern BAT54 datasheet.

So this is suddenly not only a system of linear differential equations in time, but a system of strongly nonlinear differential equations. There's not even necessarily analytic solution to these at all! You'd need to numerically approximate them.

This is the stuff the simulation program you've already designed this circuit in was made for. So simulate.


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