I have a very simple capacitive power supply schematic that I'm using to teach myself some of the underlying math and concepts. Let me be clear up front - I am not planning on building this - so I'm not concerned about its safety or cost or anything. I'm just trying to get the math right so I can understand how it works.


simulate this circuit – Schematic created using CircuitLab

In the above schematic, R1 is a load that I want to apply 3.3v across and that I expect to draw 220mA. I sized C2 for a 1% ripple at 120hz (since it's a full-wave rectifier) using the formula \$V_{pp}=\frac{I}{2 \pi fC}\$ and got \$\frac{220mA}{2 \pi \cdot 120hz \cdot .033V} = 8.842mF\$.

I still need to size C1, and that's where I'm running into trouble. I know that C1 and the R1/C2 circuit must drop a total of 120V, and I don't yet know the total current or impedance of the whole 120V circuit. But! I can calculate the total impedance of R1/C2.. and thus I can calculate the current that will flow through the bridge.. which must be the total current drawn from the mains.

C2's reactance at 120hz by \$X=\frac{1}{2 \pi fC}\$, is \$\frac{1}{2 \pi \cdot 120hz \cdot 8.842mF} = 0.15 \Omega\$. (Sniff test #1 - this seems super low.)

The total R1/C2 impedance would then be \$Z=\frac{1}{\frac{1}{15} + \frac{1}{0 - .15j}}\$ - or, as I worked it out, \$Z = .0667 - .149985j\$. The effective impedance of that is \$|Z| = \sqrt{.0667^2 + .149985^2}\$, or \$.164135\Omega\$. 3.3v applied to that will flow a bit over 20.1A. (Sniff test #2 - crazy high.)

Ok, I guess.. now that we know the total current draw and the combined impedance of the rectified circuit, let's solve for C1.. $$ 120v = 20.1A \cdot \sqrt{(.0667 + 0)^2 + (.149985 + X_{C1})^2} \\ X_{C1} = 5.81979\Omega\\ C1 = \frac{1}{2 \pi \cdot 120hz \cdot 5.81979\Omega} = 227.893 \mu F $$

However if I put in 227.893\$\mu F\$ for C1, and then run a simulation, I get 53v across R1:

enter image description here

Where am I going wrong?

  • 1
    \$\begingroup\$ What's the purpose of C1? Usually you'd have a fuse with maybe 1-3 Ohms DC resistance in place of C1 , this will help limit the current draw. \$\endgroup\$
    – User7251
    Nov 9 '17 at 19:16
  • 2
    \$\begingroup\$ As far as I understand it, C1's reactance causes a voltage drop without actually dissipating any power? Otherwise I'm also not certain what the advantage of a capacitor is, over a resistor, in this type of power supply. en.m.wikipedia.org/wiki/Capacitive_power_supply \$\endgroup\$
    – tophyr
    Nov 9 '17 at 19:25
  • 2
    \$\begingroup\$ I just want to point out that the equation for \$C_1\$ shouldn't use \$120Hz\$, it should be \$60Hz\$. Though this doesn't solve your problem, but whoever that will attempt to solve it will now at least use the correct frequency. \$\endgroup\$ Nov 9 '17 at 20:43
  • 1
    \$\begingroup\$ Ooh nice catch @HarrySvensson \$\endgroup\$
    – tophyr
    Nov 9 '17 at 20:44

I believe your \$C_2\$ is correct so I won't touch that one.

Regarding \$C_1\$, we want it, on average, to push 220 mA through R1. I will do an approximation, assuming that the diodes are ideal. So you should be within 10% of the actual answer.

The RMS value for a sine wave is \$\frac{A}{\sqrt2}\$ where \$A\$ is the amplitude of the sine wave.

\$220\$ mA DC \$\rightarrow 220\$ mA \$×\sqrt2≃311\$ mA AC

The maximum voltage across \$C_1\$, once we're in steady state mode
will be \$120\$ V \$-2V_f-\frac{3.3}{2}\$ where \$V_f\$ is the forward voltage of the diodes. I'll assume 0.75 V.

So we have an RMS current, a voltage and a frequency.

We also know this: \$Q=I×S\$ and \$C=\frac{Q}{V}\$

Where \$Q\$ = charge, \$S\$ = time, \$I\$ = current, \$V\$ = voltage

In our case \$S = \frac{1}{120}\$ s, \$I = 311\$ mA, \$V = 120-2V_f-\frac{3.3}{2}=116.85\$ V

\$C=\frac{0.331×\frac{1}{120}}{116.85}=23.778\$ µF

*puts \$23.778\$ µF into the simulator*

Hmm, I've messed up somewhere, but I am on the right track at least. The current through \$C_1\$ is \$1×\sin(2\pi60t)\$ A (according to simulation). I'm no rocket scientist.. so let's just scale that 1 A down to 331 mA.

\$C=\frac{0.331×\frac{1}{120}}{116.85}×0.331=7.37595\$ µF

*puts \$7.37595\$ µF into the simulator*

3.1 V across our 15 Ω load. Ehh, it was an approximation and an inverse rocket science. The error was \$\frac{3.3-3.1}{3.3}=6\%\$, less than 10% as I stated.

The reason for why it's not 100% correct is because there are dead time when the diodes are not active, and my approximation implied that there was no dead time. Which is why my approximation gave an answer that was less than 3.3 V.

I do not encourage you to mark this as the correct answer since this is just an approximation. But hey, it beats 53 volts.

Edit, try #2

Converting from radians per second to Hz is a factor \$2\pi\$, the "Hz" I used above was actually in radians per second, so the semi correct equation should be:

$$ \begin{align} C &= \frac{(220\sqrt{2}\text{ mA})((120-2\text{V$_{diode}$}-3.3)\text{V})}{2\pi60\text{ Hz}}\\\\ C &= \frac{(220\sqrt{2}\text{ mA})((120-1.5-3.3)\text{V})}{2\pi60\text{ Hz}}=7.16 \text{ µF} \end{align} $$

This makes much more sense, I tested other frequencies and other voltages and the equation seems to always be off by 10% = I get 200 mA through the 15 ohm resistor instead of 220 mA when I try it with your parameters. I believe the following picture which is the current through \$C_1\$ explains why:
Off by 10% cause we're not getting a full sinewave

Here's a link to the simulator

So if you want an equation that is "good enough" for your particular setup, then just multiply the equation above by 1.1 so you're no longer off by 10%. But this 10% is dependent on the type of diodes, the frequency and the load, but hey, at least it's in the ballpark area of the correct answer which is good enough I guess.

  • \$\begingroup\$ I'm not able to follow why we're scaling C down by .331 again - it certainly seems to fit, but I can't figure out why. \$\endgroup\$
    – tophyr
    Nov 11 '17 at 2:05
  • \$\begingroup\$ A couple other things I've thought of and investigated so far (but that don't seem directly related to the math in this answer) - in my question text, I had been applying a full 3.3V AC across C2, but it really only actually sees .033V AC - the vast majority of the voltage on that side is DC. Adjusting for that, C1 needs to pass a peak 440mA (during ripple valley), and average 330mA (??... I think). Also, the current through C1 when sized approximately-correct, seems to be a nearly perfect square wave. Not sure how significant that is. \$\endgroup\$
    – tophyr
    Nov 11 '17 at 2:07
  • \$\begingroup\$ \$C = \frac{1}{2 \pi \cdot 60hz \cdot \sqrt{(\frac{120}{.33A})^2 - 7.5^2}} = 7.296 \mu F\$, which is really close to what you got! Still low however. I was really hoping to get this exact, but your mention of the diode switching stuff has me unsure of whether or not I need to account for that with the simulator. Without the ripple capacitor, I can size C1 to provide exactly 3.3V to R1.. so I want to say the simulator's not putting any non-ideal behavior in.. but I'm not sure. \$\endgroup\$
    – tophyr
    Nov 11 '17 at 2:20
  • \$\begingroup\$ Edit to comment above: voltage after properly-sized C1 is a square wave (not current) \$\endgroup\$
    – tophyr
    Nov 11 '17 at 2:26
  • \$\begingroup\$ @tophyr if you want to make it exact then you definitely have to take the real world problems into account. Like the dead time of the diodes. I assume you've seen the current of \$C_1\$, during the zero transition no current will flow because the diodes are not active. But if you just want to get it done, put 7.95 µF which appears to be good enough for you and call it a day. Either way, I'm out. \$\endgroup\$ Nov 11 '17 at 6:15

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