I'v heard that a delta/delta or Wye/Wye transformers do not make any phase shift. In delta/wye or wye/delta transformers, There is a 30 degrees phase shift between the primary and secondary coils. So, Why is there a phase shift ? and Why is it 30 degrees ?.. Why not 60 or 120 ?

I googled and I found calculations with phasor diagrams that proves the 30 degrees phase but I'm confused because of too many calculations and diagrams, I didn't get it.

Would you give me a simple answer, please ? and I prefer the physical meanings and concepts rather than equations and mathematics.

Thank you very much,


Let's call the 3 phases A, B and C and let's say we notionally have a neutral wire. Neutral is basically 0V in the system.

The "A" phase voltage (to neutral) is my chosen reference that all other voltage phase angles are measured from hence, V\$_B\$ is 120 lagging V\$_A\$ and V\$_C\$ is 120 degrees leading V\$_A\$.

OK so far?

What about the voltage between line A and line B (aka V\$_{AB}\$) - this is called line voltage (not to be mistaken with voltages between phase and neutral). Line voltages are \$\sqrt3\$ times bigger than phase voltages.

OK so far?

If you are not just examine what happens here: -

enter image description here

If you use trigonometry and resolve all the triangles you can find the length of V\$_{AB}\$ - it is \$\sqrt3\$ times bigger than either A or B to neutral.

It's also 30 degrees leading A and this is where the 30 degrees comes from.

So, a delta primary will receive primary line voltages of V\$_{AB}\$. V\$_{BC}\$ and V\$_{CA}\$.

Given that a transformer doesn't inherently phase shift anything (other than the trivial cases of 0 degrees and 180 degrees), any secondary winding voltage must be in phase with their respective primary voltage no-matter whether the secondary is connected delta or wye.

OK so far?

Then you have it because, a delta primary works with line voltages and these are 30 degrees shifted to their nearest phase voltage. The secondary outputs are also 30 degrees shifted and hence a wye secondary will produce a phase voltage that is 30 degrees shifted from the equivalent (but not directly connected to) phase voltage on the primary.

It's trivial to do the wye-delta transformer so I'll leave it to someone else.

  • 1
    \$\begingroup\$ Heh. Looks like one of my illustrations. :) \$\endgroup\$ – Li-aung Yip Sep 20 '15 at 5:05
  • 1
    \$\begingroup\$ @Li-aungYip well thank you dear chap. I found it on the internet (images) kind of randomly so I never bothered to track down the source but thank you. \$\endgroup\$ – Andy aka Sep 20 '15 at 8:55
  • 1
    \$\begingroup\$ It was actually on EE.SE in the first place, so fair game. \$\endgroup\$ – Li-aung Yip Sep 20 '15 at 9:54
  • 1
    \$\begingroup\$ electronics.stackexchange.com/questions/92678/… \$\endgroup\$ – Li-aung Yip Sep 20 '15 at 12:30
  • \$\begingroup\$ Good answer Andy. Also, note that the 30 degree shift only occurs under balanced conditions. Further, the phase shift to negative sequence will be -30 when positive sequence experiences +30 and vice versa. \$\endgroup\$ – relayman357 Sep 17 at 18:23

What you are looking for my friend is an intuitive understanding of why we have the phase shift between the HV winding and the LV winding of the Delta Wye transformer connection. You won't really understand this unless you sit down and starting drawing the phasor diagrams as illustrated by Andy.

I would recommend watching this video tutorial that goes through the entire phasor diagram process step by step: http://gpac.link/1OfN591

From what I understand --- lets suppose you have a Transformer Turns Ratio of 1:1 --- if you look at the line-to-ground voltage on the Delta side and compare it to the line-to-ground voltage on the Wye side --- you will see a 30 degree phase shift and square root 3 magnitude difference.

Why does this occur? It's easier if you just watch the videos :) http://generalpac.com

I hope this helps :)


You might find it more intuitive to see it as waveforms. See image below:

Wye-Delta transformer Waveform

  • \$\begingroup\$ Welcome to EE.SE. Your graphic representation is a little piece of artwork, however, it does not explain the root cause. I recommend you add some more explanation in text form. \$\endgroup\$ – Ariser Sep 16 at 20:11
  • \$\begingroup\$ Hello Ariser. The graphic was provided as a simple aid to understanding the vector diagram's 30° displacement. It represents something like what would be seen with a two channel oscilloscope, triggering on Channel A (for Voltage A to neutral, that is, Va0) on one set of the transformer's windings with Channel B being connected across Voltage A and Voltage B (that is, Vab) on the other set of windings. Voltage B to neutral (Vb0) on the first set of windings would not be seen on the oscilloscope, but is shown in the graphic for reference and clarity. Hope this helps. \$\endgroup\$ – Richard Oct 17 at 1:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.