I've simulated the following schematic on Cadence, and another less professional simulator called 'everycircuit'.

enter image description here

I initially had a C1 = C2 = 10uF which simulated correctly on both simulators. Then I realized that I did not have the correct voltage rating 10uF in my kit, but did have 470pF rated for 500V.

So I go into Cadence to test out the 470pF value, and it did not like that at all. Gave me a much lower voltage than what was expected. Everycircuit didn't seem to care at all, gave me the correct output.

I've since purchased some 10uF capacitors, but they are HUGE and would cause problems fitting them in between the rest of the components. I've already soldered the 470pF ones onto the PCB and am wondering if it will be necessary to spend the time desoldering the tiny pFs.

Does the capacitance values matter? Is Cadence just being finicky again?

enter image description here

  • 1
    \$\begingroup\$ If frequency is 50 or 60HZ, C1 and C2 may need to be 10uF to 100uF. Use the equation for capacitive reactance to find maximum current being supplied at a given frequency and voltage. 10uF @ 60HZ and 120 VAC will give you only 500 mA of current max. \$\endgroup\$
    – user105652
    Nov 29 '18 at 23:51
  • \$\begingroup\$ 470pF can be ok at 1 MHz signal detector circuit, but it's not especially high capacity energy storage at 50...60Hz mains powered DC power supplies. You should add something about the frequency , voltage and the load. \$\endgroup\$
    – user287001
    Nov 29 '18 at 23:58
  • \$\begingroup\$ Alright. I'm going in boys. Big Bertha 1 and 2 are gonna get soldered. \$\endgroup\$
    – snowg
    Nov 30 '18 at 0:04
  • \$\begingroup\$ (Continued) at 1 mA load current the voltage drops in 470pF cap about 2000V in one millisecond. \$\endgroup\$
    – user287001
    Nov 30 '18 at 0:06

In order to properly simulate the mean output voltage, you'll need to make an assumption about the load, \$R_L\$. Also presumably you know the driving frequency. It's also possible that the real operating behavior of your diodes would make a difference.

Broadly speaking, I would expect the capacitors in a 500V supply to take up some space on the board.

  • \$\begingroup\$ 500V has nothing to do with the problem. 470pf stores no energy in a 50/60Hz application such as this. \$\endgroup\$ Nov 30 '18 at 0:08
  • \$\begingroup\$ Should I be worried about the way these are mounted to my PCB? I'm using through-hole. I guess I can just always lift it by the ginormous blue things? \$\endgroup\$
    – snowg
    Nov 30 '18 at 0:21
  • \$\begingroup\$ Regarding restraints, you may want something on those caps, if the supply hits the floor the wrong way the connections could be sheared off, depending on how robust they are. \$\endgroup\$
    – isdi
    Nov 30 '18 at 0:56
  • \$\begingroup\$ @JackCreasey I've been researching capacitors a lot lately, but nobody ever talks about THE ENERGY!!! It's always "current decays exponentially, voltage time constant" etc... How much energy does a capacitor store??? I was trying to make sense of it by saying that the voltage rating of the capacitors is what determined how much energy they could store. Obviously a 500V pF is not gonna be able to store as much as a 500V uF. Is charge/energy the same? \$\endgroup\$
    – snowg
    Nov 30 '18 at 1:32
  • \$\begingroup\$ @snowg stored energy in a cap is = 0.5 * C * U^2 where U = the voltage charged to the capacitor. That can be proved mathematically with elementary input power integration and the formula is virtually in every basic electricity textbook. \$\endgroup\$
    – user287001
    Nov 30 '18 at 3:18

The doubler is essentially two half-wave rectifiers. You can estimate the peak-to-peak ripple of the doubler from:

\$V_R = \frac{V_{IN(peak)}}{R_L\cdot f \cdot C}\$

Where f is the frequency (valid for Vr << Vin)

So if you are willing to have a 5% peak-to-peak ripple, then

\$C \ge \frac{20}{R_L\cdot f}\$

This ignores the capacitance of the diodes, which may not be accurate for very low value load output capacitors, as well as making the conservative assumption that the current is constant at peak.

Incidentally, the photo you have shows a film capacitor on the right. Nothing wrong with that, they have a very long life, but they are relatively large for values exceeding a microfarad or so. In most applications, an electrolytic capacitor would be chosen. A typical 10uF/450V electrolytic cap is 10mm diameter and 22mm tall. A 47uF is more like 16mm diameter and 32mm tall.

  • \$\begingroup\$ I didn't choose Big Bertha. She chose me! \$\endgroup\$
    – snowg
    Nov 30 '18 at 1:08

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