# A little help understanding 3 phase waveforms and summation

• InquisitiveOne
InquisitiveOne
Hello all,

I'm in the process of learning about 3 phase power and how to wire loads to a generator. I've searched high and low with as many sentence structures as I could think of in the google search bar and I can't seem to find the answers I'm looking for. I'm really hoping you guys can help.

I'll be referencing 208V single phase loads connected to 208V 3 phase service/generator.

I understand the individual phases measure 120V to ground, are separated by 120º, and connecting a load across two phases results in single phase 208V. I also understand the formula used to calculate the resultant combined voltage is one legs voltage to ground times the square root of 3. There are a ton of pictures on line and videos on YouTube that show the 2 (or 3) 120V phases as viewed on an oscilloscope, separated by 120º. But, there doesn't seem to be a single source that shows the combined 208V waveform. Looking at the separate waveforms makes me think the resultant waveform might have a couple bumps positive and a couple bumps negative, but I also recall learning somewhere that the two out of phase sinewaves sum to become a single sinewave.

Question 1) Do the two 120V phases sum to a single 208V pure sinewave?

I'm feeling 95% positive the answer is yes.

To help better understand the relationship and how the two phases combine, I set up and Excel sheet to help visualize everything. Except, I'm running into a confusing issue. Two sinewaves of identical frequency and 120V amplitude, separated by 120º are summing to a sinewave centered between the to being summed, but the combined amplitude is only 120V. If I change the separation angle to 60º, the summed sinewave peaks at 207.841, the same exact figure I get when completing the equation I put up above.

I've double and triple checked all my Excel formulas and have watched several YouTube videos that also verify two identical sinewaves separated by 120º sum to the originals amplitude.

Question 2) Do you have any idea why the spreadsheet indicates the right 208V voltage when separated by 60º, no increase in summed amplitude when separated by 120º, yet measuring across two legs of 3 phase service, separated by 120º does in fact sum to an increased voltage?

Is it my charting in Excel? Or, is there something about the way 208V single phase is created that makes it add up as though the phases were actually 60º apart?

I'm a little confused about that.

I have one other question that I'll save for later. The answers to these two may help me get there on my own.

Thank you guys, very much, for any help or guidance in better wrapping my mind around this concept.

InquisitiveOne said:
Question 1) Do the two 120V phases sum to a single 208V pure sinewave?
Yes, but the phases are a differential voltage, so they subtract and are plotted here referenced to "neutral", plotted as white. Frequency = 50 Hz.
Here on the left are the "lines" or "legs", R, B and Y; Red, Blue and Yellow.
The "phases" are R-B, B-Y and Y-R; magenta, orange and gray.
Notice the amplitude of 120 Vrms is 170 Vpeak.
The amplitude of the "phases" is then 208 Vrms, or 295 Vpeak.

Klystron, DaveE and berkeman
InquisitiveOne said:
Question 1) Do the two 120V phases sum to a single 208V pure sinewave?
Have you come across Phasor Notation in all your reading? It makes questions like this one very straightforward. By 'freezing' the constant phase rotation (e.g. 50 cycles per second), you get a simple bit of trigonometry to find how the Phasors (vectors) combine.

DaveE
##sin(\omega t) - sin(\omega t +\frac{2}{3} \pi) = ##
##sin(\omega t) - sin(\omega t)cos(\frac{2}{3} \pi) - cos(\omega t)sin(\frac{2}{3} \pi) = ##
##sin(\omega t) - sin(\omega t)(\frac{-1}{2}) - cos(\omega t)(\frac{\sqrt{3}}{2}) = ##
##sin(\omega t)(\frac{3}{2})-cos(\omega t)(\frac{\sqrt{3}}{2}) =##
##\sqrt{3}[sin(\omega t)(\frac{\sqrt{3}}{2})-cos(\omega t)(\frac{1}{2})] =##
##\sqrt{3}[sin(\omega t)cos(\frac{1}{6} \pi) - cos(\omega t)sin(\frac{1}{6} \pi)]=##
##\sqrt{3}[sin(\omega t - \frac{1}{6} \pi)]##

Note that you don't sum the phase voltages, you subtract them. Think of holding a voltmeter with + and - probes.

i.e. the Potential Difference.

I really appreciate all of you. You have all been a big help in both, wrapping my mind around how and why it is the way it is, as well as making the appropriate changes to my Excel graphs that indicate the relationship between phases correctly... which really was as simple as subtracting one phase from the other instead of adding them.

Thanks again. You guys (and gals) are awesome!

sophiecentaur, DaveE and Baluncore

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