What is generic formula for min and max levels of rectified AC voltage?

For example, if i have 1 phase AC voltage with 1 volt peak level and rectify it., i would get some pulsating voltage V = abs(sin(t+0*pi)). Its minimum is 0 volts minimum, and 1 volt maxumum.

Same idea can be applied if i take N separate AC voltages (attention, not N-phase AC system), with syncronized phase offsets (i*(2*pi)/N) for i-th voltage. Each voltage is then rectified and summed up in series as in following examples:

• for 2phases (180deg apart) => series sum yields pulsing voltage with Vmin(2) = 0 volt minimums and Vmax(2) = 2 volt maximums
• 3p (120d) => Vmin(3) = sqrt(3) and Vmax(3) = 2
• 4p (90d) => Vmin(4) = 2 and Vmax(4) = 2*sqrt(2)
• ...

What is a generic analytic formulas for N-phase series system - e.g. what would be Vmin(N) = ? and Vmax(N) = ? voltages?

PS. Illustration: • Are the generators in sync with each other? – winny Sep 1 '18 at 17:36

Fun problem! I think I found the answer. I don't have the time to make a graphic, but the solution seems to be as follows.

The minimum will occur when at least one of the sine waves is equal to zero, aka at phase 0.

The maximum will occur when one of the sine waves is at phase (180/N - (1/2)*(180/N)*mod(N,2)).

The generic solution for min and max V can be generated as a summation of the N sine values with equidistant phases, calculated with the first sine wave at each of the previously mentioned phases respectively.

The equations in radians are below:

$$Vmin(N) =\sum_{i=0}^{N-1} abs(sin(\frac{2\pi i}{N}))$$ $$Vmax(N) = \sum_{i=0}^{N-1} abs(sin(\frac{\pi}{N}-\frac{\pi}{2N}*mod(N,2)+\frac{2\pi i}{N}))$$

• Welcome to EE.SE, Ryan. "The maximum will occur when one of the sine waves is at phase (180/N - 1/2*180/N*mod(N,2))." Can you show how this works out for 4-phase? Figure 4 in my answer shows you should get 45°. Is your formula (a) $\frac {180}{N} - \frac {180}{2N \times (N\ MOD\ 2)}$ or (b) $\frac {180}{N} - \frac {180}{2N}(N\ MOD\ 2)$? – Transistor Sep 25 '18 at 21:21
• Sorry for the ambiguity! The formula is (b), which works out to 45 degrees – Ryan Sep 25 '18 at 21:48
• Got it, thanks. I missed the result of the MOD being zero. – Transistor Sep 25 '18 at 21:55
• So there is no analytic O(1) solution, but O(N), since you have to sum N values? – user197437 Sep 26 '18 at 0:00
• There might be a clever way to decompose those equations, but I can't think of how. – Ryan Sep 26 '18 at 2:08

This answer was made before the OP added the diagram and these words:

"Each voltage is then rectified and summed up in series".

If there are "n" rectified sources of voltage of equal peak amplitude then the maximum peak is n x the individual peak. And, because all "n" generators could align in-phase (from time to time), the minimum voltage is still 0 volts. The rest of my answer is about "n" equidistant and synchronous phases summing in parallel.

The maximum value of the resulting waveform is always going to be 1 volt i.e. it is determined when any of the phases hits a peak. For a 3 phase full wave rectifier, an output peak occurs every 60 degrees and therefore the gap between peaks is 60 degrees: - Halfway into the gap we reach the point that two potentially contributing waveforms have exactly the same voltage and the "baton" for supplying current passes from one waveform to another. Clearly, if the waveform is sinusoidal this point is equal to sin(60) = 0.866 volts.

Hence we can conclude that $V_{MIN} = \sin(90 - 360/4n)$

Where "n" is the number of equidistant spaced phase voltages.

For 3-phase this is sin(60) and for (say) 7-phase it will be sin(77.14) = 0.975 volts.

for 2phases (180deg apart) => series sum yields pulsing voltage with Vmin(2) = 0 volt minimums and Vmax(2) = 2 volt maximums

No that is incorrect, the peak voltage can only be the peak of one sine wave because the others have not reached a peak and are blocked by the rectifiers.

• I am not meaning multiphase system, i mean N separately running generators, and min and max voltages for them, especially Vmax=? – user197437 Sep 1 '18 at 17:08
• for 3 generators, Vmin(N) = sqrt(3), Vmax(N)=2.0V.. need formula for N generators AC phases Vmax(N), Vmin(N)=? – user197437 Sep 1 '18 at 17:09
• In your new example picture, the peaks can be up to n x one generator peak voltage and the troughs can be as low as zero volts because you can't rule out the occasions when all three voltages are perfectly in phase. – Andy aka Sep 1 '18 at 17:32
• @xakepp35 I have prefaced my answer with words that explain. My answer was given before you made it clear that the voltages were wired in series and that the generators might not be synchronized. In that preface I said what would happen with series generator as you lately specified. The original body of my answer remains because it might still be of interest to other readers. I answered your modified question in my added preface. – Andy aka Sep 1 '18 at 18:23
• Then you ought to make clear what your question is trying to uncover. – Andy aka Sep 1 '18 at 18:56

This is a strange problem and I don't have an answer for you but it prompted me to graph the sums for up to 9 phases. The graphs below show the stacked sum of the phase voltages. Figure 1. Single phase with rectification. Figure 2. 2-phases rectified and stacked.

Note that the spreadsheet used has all nine series stacked so the final result shows as 'phase9'. Figure 3. Three phases. This gives the best bang for the Euro in terms of ripple reduction. Figure 4. 4-phases. Figure 5. 5-phases. Figure 6. 6-phases. Figure 7. 7-phases. Figure 8. 8-phases. The flat-spots on the total waveform are probably due calculation in 10° steps. 360°/8 = 45° steps. Figure 9. 9-phases.

I offer this incomplete graphical brain dump from someone who should have been in bed. simulate this circuit – Schematic created using CircuitLab

Figure 1. Since the individual phase voltage sources are all full-wave rectified can we add the vectors as shown? Black represents the maximum DC voltage seen.

My thinking is that the maximum voltage should be the end-to-end DC vector voltage. If this is correct then the maximum voltage becomes a trigonometry problem. I can't offer a reason why this might work, however, and can't see how to find the minimum voltage from the diagram.

• I started like you, but i wrote a pieces of code. So i solved this by numeric integration, i mean using for(...) loops in C++. So, i havent solved this yet, to generate Vmax(N)=? analytic solution.. – user197437 Sep 20 '18 at 13:43