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Three phase transformer connections and applications

Short answer

The four basic connections are star–star (Yy) for small HV units, delta–delta (Dd) for industrial loads, star–delta (Yd) for step-down at substations, and delta–star (Dy) for step-up at generating stations and distribution. Yd and Dy introduce 30° phase shift; Dd and Yy provide 0° shift.

Explain the different three phase transformer connections with neat sketch. Write their application also.

A three-phase transformer bank can be formed from three single-phase transformers or from one three-phase unit. The primary and secondary windings can each be connected in star (Y) or delta (Δ), giving four basic connections.

Let VLV_L be the primary line voltage, and a=N1/N2a = N_1/N_2 the turns ratio per phase.

1. Star–Star (Y–Y)

   Primary (star)            Secondary (star)
  A o--(N1)--+              a o--(N2)--+
  B o--(N1)--+--o N         b o--(N2)--+--o n
  C o--(N1)--+              c o--(N2)--+
  • Line voltage ratio VL1/VL2=aV_{L1}/V_{L2} = a. No phase shift between primary and secondary (0°).
  • Needs a neutral connection, otherwise unbalanced loads and third-harmonic voltages cause trouble.
  • Applications: small high-voltage, low-current transformers, since each phase winding carries only VL/3V_L/\sqrt3 and needs less insulation; small distribution loads with a neutral (4-wire).

2. Delta–Delta (Δ–Δ)

   Primary (delta)           Secondary (delta)
         A                          a
         o                          o
    N1  / \  N1                N2  / \  N2
       /   \                      /   \
    B o-----o C                b o-----o c
        N1                         N2
  • VL1/VL2=aV_{L1}/V_{L2} = a. No phase shift (0°). Line current = 3×\sqrt3\times phase current.
  • Third-harmonic currents circulate in the delta, so the voltages stay sinusoidal. If one transformer fails, the other two can supply 58% of the original load in open delta (V–V).
  • Applications: large low and medium-voltage industrial loads where continuity of supply is needed and the load is balanced.

3. Star–Delta (Y–Δ)

   Primary (star)            Secondary (delta)
  A o--(N1)--+                      a
  B o--(N1)--+--o N                 o
  C o--(N1)--+                 N2  / \  N2
                                  /   \
                               b o-----o c
                                   N2
  • VL1/VL2=3 aV_{L1}/V_{L2} = \sqrt3\,a, with a 30° phase shift between primary and secondary line voltages (for example Yd1, where the secondary lags by 30°).
  • When stepping down, the star primary on the HV side has a phase voltage of only VL/3V_L/\sqrt3, so less insulation is needed, and the delta secondary lets third-harmonic currents circulate.
  • Applications: step-down transformers at the receiving end of a transmission line (substations).

4. Delta–Star (Δ–Y)

   Primary (delta)           Secondary (star)
         A                  a o--(N2)--+
         o                  b o--(N2)--+--o n
    N1  / \  N1             c o--(N2)--+
       /   \
    B o-----o C
        N1
  • VL2/VL1=3/aV_{L2}/V_{L1} = \sqrt3/a (step-up), with a 30° shift. The star secondary provides a neutral.
  • Applications: step-up transformers at the generating station and distribution transformers (11 kV/415 V, Dyn11), giving a 4-wire supply. The delta suppresses third-harmonic currents.

Comparison

ConnectionLine ratioPhase shiftTypical use
Y–Yaa0°Small HV units
Δ–Δaa0°Industrial loads
Y–Δ3a\sqrt3a30°Step-down
Δ–YVL2/VL1=3/aV_{L2}/V_{L1} = \sqrt3/a30°Step-up, distribution

Other connections: open delta (V–V) for 58% capacity, the Scott connection for 3-phase to 2-phase conversion, and zig-zag for earthing.

✓ Answer checked by Claude Opus 5.5 (high effort)

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