Good evening. I am trying to learn about 3-phase "wye" configurations because it interests me. I don't actually have any DIY electrical projects. I understand that in a 208Y/120V system, the voltage between any leg and the neutral is 120V, and the voltage between any two legs is 208V. What I don't understand about the 208V is how the sine wave would be plotted in a graph.

• Would it be +104 to -104 (a span of 208)?
• Or would it be +208 to -208 (a span of 416)?
• Or or a different arrangement that I have not thought of?

Below are some graphs I created to illustrate what I'm asking. Thank you.

+104 to -104


+208 to -208

 

You need to keep in mind you're dealing with three phases which is at least three conductors carrying alternating current voltages that are offset in time by one-third of the period.

 

Maybe easiest to think of 3 phase like this:
3 seperate 120V voltage sources, all with one common wire. (the center Y).

Just like your first graph, they are 120 degrees out of phase. But, the magnitude is 169.7 Volts peak.

To "make" the 208 graph, simply take any two 120V sources and take the first, MINUS the second, using only the non-common ends. Do this graphically for all 3 combinations of the 120V sources.

 

I don't have any graphing program to show you, but basically the RMS voltage of any wave with regard to ground is going to be 120V (peaks are about 170V). The phase to phase (say, A to B) RMS voltage is lower than the sum of those two waves because in a wye configuration the peak of phase A does not "line up" with the peak of phase B at any given time. The peak of phase A is lined up with the slope of phase B, so it is only going to be the 85% or whatever of the actual peak-to-peak differential.

 

RMS = Root Mean Squared
-------------------------

 

Hey guys, I realize that I completely misunderstood what the RMS represents. Thanks for your comments in helping me realize that. I have been doing a bit more reading and I'm going to consult with a co-worker who is an electronics engineer. I may make a post here a few days if I have additional questions. The moderators might consider deleting this thread to avoid confusing people in the future who come across it in a search.

 

There are two topics in this thread that are leading to confusion due to intertwining.
1. The relationship of RMS to peak voltage.
2. Where does 208 volts come from.

If you consider a single phase 120 volt AC source with a square waveform instead of a sine wave, then 120 volt RMS AC drawn on paper to the same scale will have the same area (square inches, square centimeters) between the waveform pulse and the baseline as the 120 volt square wave pulses.

One hundred twenty volts AC square wave measures 120 volts zero (baseline) to pulse height or 240 volts from the pulse depth of the negative pulse to the height of the positive pulse.

One ampere at 120 volts square wave AC represents the same energy as one ampere at 120 volts AC sine wave RMS.

After all is said and done there is a universal fixed multiplier to go from square wave AC at X volts to sine wave AC at X volts peak (and the reciprocal of that multiplier to convert the other way). After we have that multiplier we can finish (terminate; cease) the discussion of RMS vs. peak and resume just discussing 120 volts versus 208 volts versus 240 volts assuming RMS for each of them and disregarding RMS versus peak voltage but rather assuming all of the different voltages we are discussing have the same waveform shape, here, RMS.

 

AllanJ, that is an excellent idea to consider visualizing the signal as a square wave in order to eliminate any distraction caused by the "peak" vs "RMS" concept.

I think what tripped me up was the fact that I was trying to calculate the RMS of each of the three signals independently, but what I should have been considering is the combined RMS of all overlapping signals at any given point in the cycle, averaged over the entire cycle. That's where 208V comes from, but my graphs incorrectly depicted 208V on each signal independently, which makes no sense.

An additional question based on your post: I find that visualizing a single signal using square waves provides clarity, but consider if all three signals of a 3-phase system were square waves. It would appear as a continuous flow of +120V and -120V simultaneously. I think motors wouldn't turn.

 

No need to delete the thread. It will tumble into the archives.