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Transformer vector groups and fault current

Short answer

Transformer vector groups determine the zero-sequence path. A Yg-Δ provides a low-impedance earth path on the grounded side but blocks zero-sequence on the delta side. Unearthed windings or Δ-Δ arrangements block zero-sequence current, drastically reducing L-G fault current. Positive/negative impedance magnitudes are unchanged, so L-L fault currents remain unaffected. Neutral earthing impedance adds 3Zn and further limits earth faults.

How do the different vector groups of transformer affect the fault current in the power system network?

The vector group of a transformer (Yy, Yd, Dy, Dd, with the neutral earthed or not) decides how the zero-sequence current can flow through it and what phase shift it gives. This changes the zero-sequence network and therefore the earth-fault current. The positive and negative sequence impedances are the same for all groups.

Effect on sequence networks

  • Positive and negative sequence: every group behaves as a series leakage reactance XTX_T. Only a phase shift of ±30∘\pm 30^\circ appears in Dy/Yd units (positive sequence shifted one way, negative the other). The magnitudes of balanced and L-L fault currents are not changed.
  • Zero sequence: zero-sequence currents are in phase in all three lines, so they need a return path through an earthed neutral or a circulating path inside a delta.
Vector groupZero-sequence equivalentEffect on earth fault current
Yg–Yg (both neutrals earthed)Series Z0Z_0 between both sidesZero-seq current passes through; fault fed from both sides
Yg–ΔYg side connected to reference through Z0Z_0; Δ side openActs as an earthing source on the Yg side; blocks zero seq to the Δ side
Y–Δ (Y not earthed)Open on both sidesNo zero-seq path; no earth fault current through it
Yg–Y (one neutral isolated)Open circuitBlocks zero-seq current
Δ–ΔOpen on both sidesNo earth fault current through it
Neutral earthed through ZnZ_n3Zn3Z_n in series in zero-seq pathEarth fault current reduced

How it affects fault current

  1. L-G fault current If=3EZ1+Z2+Z0+3ZfI_f = \dfrac{3E}{Z_1+Z_2+Z_0+3Z_f} depends strongly on Z0Z_0. A Δ-winding or an unearthed star makes Z0Z_0 seen from that side very large (open), so the L-G fault current falls sharply; for an isolated system it becomes almost zero.
  2. A Δ/Yg transformer feeding a fault on its star side provides a low-impedance zero-sequence path, so the L-G current can even exceed the three-phase fault current when Z0<Z1Z_0 < Z_1.
  3. Yg–Yg units let the zero-sequence current pass to the other side, so earth faults are fed from both sides and are seen by relays on both sides.
  4. Neutral impedance ZnZ_n appears as 3Zn3Z_n and is used deliberately to limit earth fault current.
  5. Phase shift of Dy/Yd units shifts the positive and negative sequence currents by ±30∘\pm30^\circ. The current magnitudes on the other side are the same, but the phase currents are distributed differently, e.g. an L-G fault on the star side of a Dy transformer appears as current in two lines on the delta side.
  6. L-L faults have no zero-sequence component, so the vector group does not change their magnitude.

Example

For a generator connected through a Δ/Yg step-up transformer, an L-G fault on the HV side is fed only through the transformer's own Z0Z_0 (the generator's zero sequence is isolated by the delta). An L-G fault on the delta side gives almost no fault current if neither the generator nor the delta side has an earthed neutral.

✓ Answer checked by Claude Sonnet 5.5

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