Star-delta transformer cannot parallel with star-star

Parallel operation requires that the secondary line voltages of the two banks be exactly equal in magnitude and in phase. A star–delta transformer does not meet this condition with a star–star or a delta–delta bank.

Reason: phase displacement

• In a star–star or delta–delta transformer, the secondary line voltage is in phase with the primary line voltage (0° shift).
• In a star–delta (or delta–star) transformer, the secondary line voltage is displaced by 30° from the primary (lagging for Yd1, leading for Dy11).

When the primaries are on the same supply, the secondary line voltages of the two banks differ by 30°.

  V (Y-Y secondary)
   ^
   |    30 deg
   |  /  V (Y-D secondary)
   | /
   +----------------->

The voltage difference across the paralleled terminals is

$$\Delta V = 2V\sin\frac{30^\circ}{2} = 0.518\,V$$

This is about 52% of the rated secondary voltage. It appears across the very low internal impedances of the two transformers, so a huge circulating current flows, even at no load. The windings overheat and may be damaged, and the load cannot be shared properly.

Also: the voltage ratios differ in form. Star–delta has $V_{L2}/V_{L1} = 1/(\sqrt3\,a)$ compared with $1/a$ for star–star, so the turns would have to be redesigned. Even if the magnitudes matched, the 30° displacement cannot be removed by any change of turns ratio.

Hence star–delta can be paralleled only with another star–delta, or with a transformer that has the same phase displacement (the same vector group), such as delta–star of the right group.