OT Wye delta transformer 30 degree phase shift - why

leitmotif

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In my wanderings I found a web page referring to 12 pulse rectifiers. 12 pulses are done with 3 phase transformer with a wye and a delta secondary. Standard full wave 3 ph rectifiers are connected to the wye and delta and their DC side is paralleled. This is apparently common in UPS's to improve efficiency by reducing THD. This yields 12 pulse because of a 30 degree phase shift between a wye primary and delta secondary (or vice versa).

If I connect wye wye or delta delta both secondary and primary in phase.

OK now I am trying to find out why there is a 30 degree phase shift between the primary and secondary of wye delta. The other question is why 30 and not 45 or,,,,??

Can someone refer me to good reading material please?

Thanks
Dan Bentler
 
If you have 3 phase AC each phase is 120 degrees apart
convert that to DC you get 6 pulses that are 60 degrees apart.

Now if you have 6 phase AC each phase is 60 degrees apart
convert that to DC you get12 pulses

Now deriving the 6 phase from the origional 3 phases is done with
normally a Delta wound primary and a secondary with a delta winding
and another with a WYE winding (I normally refer to it as Star)

Look at where the mid point of any delta winding would fall in a
360 degree rotation and you will see it coincides with the WYE point
of the WYE winding that shares those same phases with the delta
but you see I hope that that the mid point of Wye winding
is not on the WYE point but 60 degrees away.

I know someone will disagree with this simplistic description
so all power to you when you do better I won't mind.

I think this 6 phase !2 pulse DC is used in high voltage DC power links
that are used in Canada, here in New Zealand ours is i think 700,00 volts pos to neg.
I think the USA is making a backbone DC link at 1 million volts at the moment.
 
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Huh... I never knew that delta-wye or wye-delta caused a 30 degree phase shift in the secondary compared to the primary, but delta-delta or wye-wye doesn't.

So that would mean that you have potential between L1 of the secondary of a Delta-Wye and L1 of the secondary of a Delta-Delta. How would one calculate out how much potential there is there?
 
Monkeyhead,

Why would it ever matter? Normally you can not assume that the secondaries of any two transformers, even though identical and wired to the same primary power supply, are "in-phase". They can be 180 degrees out-of-phase, so the voltage difference could be 2 times the normal voltage.

For the secondary voltages to be in-phase on two identical 3-phase transformers, the primary wires must be connected exactly the same. Even then there may be some small voltage difference due to the different internal impedance of the transformers.
 
To make matters even more complex, 18 pulse drives use a transformer with a delta primary and three secondaries: a delta, a wye, and something called a zig-zag winding. And, if that isn't enough, Robicon used to sell 30 pulse and 36 pulse drives. I can just about imagine what the transformer looked like for these contraptions!
 
Why would it ever matter? Normally you can not assume that the secondaries of any two transformers, even though identical and wired to the same primary power supply, are "in-phase".

I completely understand this. The question is simply a theoretical curiosity as I just don't really understand the math (trig?) that one uses for sine wave calculations.
 
Dan, look at this article. It shows that the 30-degree phase shift is due to the mathematical vector relationship between the Primary and the Secondary transformer wiring.

http://www.rycominstruments.com/images/PhaseIdentifier/PhaseAttributes.pdf

This explained it all using vectors. I generally do not relate well to vectors more than likely because I do not work with them enough.

Gil your explanation came close to those used in Lancie referral.

Thank you all for the help.

Dan
 
I completely understand this. The question is simply a theoretical curiosity as I just don't really understand the math (trig?) that one uses for sine wave calculations.

Now that I can explain.
To remember your trig functions we gotta get down and dirty.
Suzy Can Tell Oscar Has A Hard On Always
SO
Suzy (Sin) Oscar Has Opp / Hyp
Can (Cos) A Hardon Adj / Hyp
Tell (Tan) On Always Opp / Adj

Get yourself a trig table. You will need sin function only.
Select 0 30 60 90. Plot these on X axis of graph paper continue for 120 150 and 180. Plot the corresponding value for each on the Y axis. That is A phase

Now B - it is out of phase (from A) by 120 so at A's starting point it is at 180 -120 thus 60. So plot 60 90 for B

Now C it is out of phase by 240 from A. So it starts at -60 degree when A is at zero so plot 60 and 90 and make them neg.

Now to continue
A for downhill from peak use 60 (for 120) 30 (for 150) 0 for 180. Where it crosses X axis use -30, -60 -90. then -60 -30 and zero
B for downhill from peak use 60 30 0.
C from its max neg at -90 then use -60 -30 then 0.

You now should have a plot of A B and C for one cycle each.

You should be able to use a calculator and dial in a value ie 240 and it will do the "subtraction" for you.

Dan Bentler
 
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Well for whatever reason, the question I posed above drove me crazy all weekend but I think I finally figured it out. I started by trying to understand the relationship between the phases of a 3 phase system where each phase is shifted 120 degrees from the others.

I've forever known that the phase voltage times the square root of 3 will give you your line voltage in a 3 phase system, but I've never known why. After many fruitless google searches, I stumbled across some forum where one of the guys mention that the phase voltages and the line voltage form an isosceles triangle.

This turned out to be the ticket. The attached images show the trigonometric relationship. Use some basic high school trig to determine the length of the phase to phase line (i.e. Line Voltage - shown in yellow).

Basically, the formula is:

LINE_VOLTAGE = 2*PHASE_VOLTAGE * SIN(PHASE_SHIFT / 2)

I guess that some egghead realized that for 120 degrees, this can be further simplified to PHASE_VOLTAGE * SQRT(3), since the sine of 60 is half of the square root of 3.

After applying this to my 30 degree shift, I get 143.386 volts.

And this animated .gif on Wikipedia is awesome.... Cleared up so many misunderstanding I had about trig.

http://en.wikipedia.org/wiki/File:Sin_drawing_process.gif

120.jpg 30.jpg
 
Maybe you mean about three phase rectifier circuit as these
I get this from this https://911electronic.com/rectifier-diode/ about rectifier theory

kvTb4yY4QPOANv-mDbNKkg.png
 

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