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I replied to this thread regarding a question asked by @BjornFanden , and I was asking if someone could verify my calculations so we do not cause any confusion. I think it is better to start a new thread for me, if that is a bad move please remove this thread again.

It was asked if full cancellation of the 5th and 7th harmonic were possible when operating two identical non-linear loads in parallel were one load is connected in D-d and the other D-y, introducing a 30 deg phase shift.

30 degree by itself is not enough for shifting either 5th or 7th harmonic 180 degrees, but because of the delta-wye configuration it seem to be possible and this is also stated in litterateur and internet pages (I do not have any references available at the moment), but none of them shows why complete cancellation is happening. Therefore I hope someone can verify my calculations below.

So here are the calculations, demonstrating full cancellation:

In the delta-delta the currents are the same at primary and secondary.

[itex]

a1 = x1 = sin(5 \omega t)\\b1 = y1 = sin(5 \omega t - 120^{\circ})\\c1 = z1 = sin(5 \omega t + 120^{\circ})

[/itex]

for the delta-wye we have on the secondary side:

[itex]

x2 = sin(5 \omega t+30^{\circ}\times 5) = sin(5 \omega t+150^{\circ})\\

y2 = sin(5 \omega t - 120^{\circ}\times 5 +30^{\circ}\times 5) = sin(5 \omega t -90^{\circ})\\

z2= sin(5 \omega t + 120^{\circ}\times 5 +30^{\circ}\times 5) = sin(5 \omega t +30^{\circ})

[/itex]

Going from secondary to primary we have to include the [itex]\sqrt3:1[/itex] ratio and thus:

[itex]

A2= (x2-z2)/\sqrt3 = (sin(5\omega t+150^{\circ})-sin(5\omega t +30^{\circ}))/\sqrt3 = -sin(5 \omega t)

[/itex]

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# 5th and 7th harmonic cancellation with phase shift

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