I'm designing a circuit, which includes a voltage multiplier. The schematics of voltage multiplier is well known, but it usually appears without any specific values for capacitors and diodes:
My quesitions are:
- How do I calulate necessary capacitance of C1-C4 with respect to output current and switching frequency?
I assume, the minimum capacitance should be high enough to store the energy which I plan to draw from the output over 1 period of time (1/f). But I'm not sure this is also the optimal value of capacitance.
- What voltage should the capacitors and diodes be rated for?
My assumption here is, that the parts "see" a smaller voltage than the output, so they could be rated to this smaller voltage, plus some safety margin.
- What type of capacitors and diodes are best suitable for this application?
So far, I considered using ceramic capacitors and fast diodes (e.g. UF4007).
As a guideline, you can assume following parameters of the multiplier:
Vi = 250 Vp, Vo <= 1000 Vdc, Iout = 20 mA, f = 50 kHz
Best Answer
The parts "see" just the voltage of one stage. So, the peak to peak voltage of the AC and add a bit to be on the safe side.
As for actually figuring the values, that's... somewhat complicated.
You have a good start on the parameters: input voltage, output voltage, output current, frequency of the input AC.
Add in the allowable ripple voltage (or rather, voltage drop,) and the forward voltage of your diodes I think you have all the needed parameters.
This equation is at the heart of the whole thing:
$$ E_{out} = 2nE_{pk} - \frac {I_{load}}{6fC} (4n^3 + 3n^2 - n) $$
I got it from this site. I recommend you read it through. I learned a lot of stuff from it that I had been trying to figure out myself.
You need to extend it to include the forward voltage of the diodes:
$$ E_{out} = 2nE_{pk} - \frac {I_{load}}{6fC} (4n^3 + 3n^2 - n) - 2nV_{f}$$
What that does is to compute your output voltage from several parameters:
Plug in your capacitance and other info, and see if it meets your needs. If the output is too low, first try to use larger capacitors. Then try more stages.
Notice how \$n\$ plays a very large part in the impedance of the multiplier. Each stage drastically increases the resistance of the multiplier.
If I assume 100nF capacitors, four stages, and diodes with \$V_{f}\$ of 1V, and take your \$250V_{pk}\$ input and goal of 1000VDC output, then I come up with the following:
$$ E_{out} = 2\times4\times250V - \frac {0.02A}{6\times50000Hz\times(1.0\times10^{-7}F)} (4\times4^3 + 3\times4^2 - 4) - 8\times1V$$
If I haven't fumbled it somewhere, that works out to 1792VDC with the 20mA load on it. That's a pretty good bit too high, but you see the idea.
You'll want to use diodes rated for at least 500VDC. The UF4007 you mentioned is rated for 1000V, so that's good. From the recovery time and the mention of the junction capacitance being measured at 1MHz, I'd say it looks fast enough for use at 50kHz.
You worked an example yourself, and came up with 2.8 microfarads. I find 2.2 microfarad capacitors rated for 1000V at $190 each.
I find 100 nanofarad capacitors rated for 1000V starting at $0.50. If you can live with a higher voltage with no load, it will probably be cheaper to go with more stages with the 100nF parts.