diy solar

diy solar

Building the sickest ® VAWT ever. Brilliant minds unite please!!

Your alternator? There is no neutral. You just have 3 windings.
It would apply to 3-phase 4 wire (plus ground) from a generator or the grid.

Well, are your windings delta or wye? If wye, there is a neutral. Don't connect load to that or reference that. Only use rectifier (+) and (-). Whether delta or wye.
 
nah not delta . star. it is not the voltage levels I am worried about being able to get at. It is the amps that should be wicked(TM). Thus with my limited knowledge I arrive at a star configuration.

But left or right. It is easy enough to try both and then see in practice how car lights actually explode!!
I will tell you this my friends.

That I do not care if I did it wrong as I am still open to learn how to do it better.

Yet when in a star configuration things just are smoother when eventually hooked up to a battery cell.

Yes, In my experience, which is crap yes I admit, Delta does not make sense to me yet.
 
The circuit diagram shown for the rectifier implies a Y configuration since it only shows one wire per phase. Tie the other ends of the coils together, but do not connect that to the - pin or anything else.
 
The circuit diagram shown for the rectifier implies a Y configuration since it only shows one wire per phase. Tie the other ends of the coils together, but do not connect that to the - pin or anything else.
yes thank you Gary. But it is already configured like that.

I will do some research where I am derailing, if at all. perhaps everything is how it is supposed to be and I have to radically rethink the coils.
 
ouch!!

Everything is as it is supposed to be.

But it seems I am aiming wayyyy to low to reach a vDC of 52 after rectification.

I looks I should be aiming vor 109 vAC(pp) per phase. Which will require a drastic rethinking of how I intend to increase the RPM of the PMA.


Here I am trusting my AI buddy again with some relevant talks;
Me:
if I want to reach a vDC of 52. What should the vAC(pp) per phase before rectification be?

-----

To calculate the required peak-to-peak AC voltage per phase (before rectification) needed to achieve a DC output voltage of 52 V, let's work backward using the relationships we've discussed.

Calculation Steps​

  1. Target DC Voltage:
    • Let VDC=52 VVDC=52 V.
  2. Formula for DC Voltage after Rectification:
    • For a 3-phase rectifier, the approximate DC voltage is given by:VDC≈VL(RMS)×1.35VDC≈VL(RMS)×1.35.
    • Rearrange to solve for VL(RMS)VL(RMS):VL(RMS)=VDC1.35≈521.35≈38.52 VVL(RMS)=1.35VDC≈1.3552≈38.52 V.
  3. Relating RMS to Peak-to-Peak Voltage:
    • RMS voltage of a sinusoidal wave is related to the peak voltage as:VL(RMS)=Vpeak2VL(RMS)=2
  • Vpeak.
  • Solve for the peak voltage:Vpeak=VL(RMS)×2≈38.52×2≈54.47 VVpeak=VL(RMS)×2
≈38.52×2
    • ≈54.47 V.
    • Peak-to-peak voltage is twice the peak voltage:Vpp=2×Vpeak≈2×54.47≈108.94 VVpp=2×Vpeak≈2×54.47≈108.94 V.

Conclusion​

To achieve a DC output voltage of 52 V after rectification using a 3-phase bridge rectifier, the required peak-to-peak AC voltage per phase (before rectification) should be approximately 109 V.
 
I think the answer is closer to 30V.

If this was 3 coils wired Delta, then 52V per coil would put out 52V peak from full-wave rectifier.
With 3 coils wired Wye, then 30V per coil would put out 52V peak.

Depending on how much current drawn, and how big a filter capacitor, need higher voltage from coil so DC voltage doesn't sag lower.
Perhaps RMS vs. peak is about right.

I think your AI assistant was thinking of single phase. Not split phase and not 3-phase. Well, at least it does mention 3-phase. But probably 3-phase Delta not Wye.
 
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