Help me find/understand voltage equation

[omglol]

I am also interested in this situation with clipping - I understand it, but how did you calculate that 10%? Thanks!

N.B.  the phase voltage inputs go negative of the ground point (blue dot) by one diode Vf drop.  I do hope the existing hardware is designed to allow for that without taking the ADC inputs outside their operating range, otherwise it will clip the negative peaks and if my results are correct, clipping at 0V will cause the calculated power to be about 10% low.

[Ian.M]

The neutral voltage is the average of the phase voltages.

I simmed the ADC input clipping using behavioural voltage sources then plotted the power.   Due to the fact that you cant reuse a calculated trace in other trace expressions, I also added a BV source for the calculated neutral.   See voltages aa, bb, cc and nn in attached sim which are the clipped version of a,b,c, and n.

[IanB]

Everything makes perfect sense, except the equation for neutral voltage. Could you explain it better? Thank you in advance!

The neutral voltage is the average of the phase voltages.

Think of the three phase voltages as points equally spaced around the circumference of a circle, 120 degrees apart. Each phase voltage is then represented by an arrow from the center of the circle to one of those points. The neutral is of course at the center of the circle. Now if you take the average of the coordinates of the three points around the edge of the circle, you will find the center where the neutral is.

[Ian.M]

Yes. I did make the assumption the phases are well balanced.  If you want to handle deriving a neutral for unbalanced three phase systems (either phase angle or voltage imbalance) properly it gets a lot more complex, however the simple average gives you a theoretical neutral that is the closest possible to balanced for that set of phase to phase voltages, and if used as the voltage reference point to calculate the power, will give the same result summed over all the phases as calculating with the actual neutral as the reference point.  Add some imbalance to my sim and try it.   You'll need to add another BV source for the calculated neutral without clipping.

[omglol]

That's simple.  First compute your instantaneous neutral voltage V(n).
   V(n)=(V(a)+(V(b)+V(c))/3
Then for each phase, compute the instantaneous voltage with respect to neutral (example phase A):
   V(a,n)=V(a)-V(n)
Once you have all three phase voltages with respect to neutral, the total instantaneous output power is:
   P=V(a,n)*I(a)+V(b,n)*I(b)+V(c,n)*I(c)
That's the trace in purple on the plot in reply#21.  It then would need a little averaging to smooth it for display - either average over a whole cycle or use something that doesn't require you to store a whole cycle's worth of samples e.g a software low pass filter.

The - in front of the expression in the plot is a SPICE artifact - devices providing power have negative power dissipation.  Ordinary lossy components have positive dissipation.

N.B.  the phase voltage inputs go negative of the ground point (blue dot) by one diode Vf drop.  I do hope the existing hardware is designed to allow for that without taking the ADC inputs outside their operating range, otherwise it will clip the negative peaks and if my results are correct, clipping at 0V will cause the calculated power to be about 10% low.

Additional question - how can I calculate voltage between two phases without calculating instantaneous neutral voltage?

[Ian.M]

Just take the instantaneous difference.
V(a,b)=V(a)-V(b)

The voltage of the reference point used for the measurement drops out as its the same for both.  However, its subject to the same requirements that the ADC (and the maths in the firmware) must not clip V(a) or V(b), and that you must use instantaneous voltages (i.e. sample V(a) and V(b) as close to simultaneously as possible and perform the calculation with each pair of samples).