Three-Phase Transformer Connections (Wiring Diagrams Included)

Contents

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Key learnings:

Three-phase Transformer Connections Definition: A three-phase transformer connects its primary and secondary windings in star or delta configurations for various electrical applications.
Star Connection: In star connection, three windings connect at one end to form a neutral point, used for creating a neutral terminal.
Delta Connection: In delta connection, windings form a closed loop, creating a triangle-like shape, providing a path for the supply to the junction points.
Types of Connections: Three-phase transformers can connect in Delta-Delta, Star-Star, Delta-Star, and Star-Delta configurations, each with specific voltage and current characteristics.
Open Delta Connection: This connection operates with two transformers, maintaining three-phase power with a lower load capacity when one transformer is out of service.

What is a Transformer Connection?

A three-phase transformer connects each three-phase winding in star (Y) or delta (Δ). The primary and secondary choices together set the voltage ratio, the phase shift and whether a neutral is available.

The bank can be three single-phase units or one three-limb (or five-limb) core. One three-phase core uses less iron and copper than three separate units of the same rating.

Generation, transmission and distribution are three-phase. Three-phase transformers step that voltage up or down.

Three-phase Connections

Each winding is connected in one of two ways:

Star connection
Delta connection

Star Connection

In a star (Y, wye) connection, one end of each of the three windings joins at a common point. That point is the star or neutral point. A neutral conductor can be taken from it.

The free ends are the three line terminals. The diagram looks like a Y, so it is also called a Y-connection or wye connection.

The star connection is shown in the figure below.

Delta Connection

In a delta (Δ) connection the three windings form a closed loop. The three line conductors join the three corners. There is no star point.

The loop looks like the Greek letter delta, so it is called a delta or mesh connection.

The delta connection is shown in the figure below.

Factors Affecting Choice of Connection

Choice of star or delta depends on:

a neutral for grounding and single-phase loads
a path for zero-sequence current and third harmonic current
voltage to earth and insulation class
phase shift if units must run in parallel
copper, iron and bushing cost
behaviour on earth faults and open-phase faults

Three Phase Voltages and Currents in Star and Delta Connection

Line voltage is the voltage between two lines. Phase (winding) voltage is the voltage across one winding. In star, that winding voltage is also the line-to-neutral voltage.

In star, line current equals winding current. Each line feeds only one winding.

$I_{LL} = I_{LN}$

On a four-wire star as in the figure, Kirchhoff’s law around two phase voltages 120° apart gives a line voltage of √3 times the line-to-neutral voltage.

$V_{LL} = \sqrt{3} V_{LN}$

In delta there is no neutral. Line voltage equals winding (phase) voltage. The stored latex still writes V_LN for that winding voltage.

$V_{LL} = V_{LN}$

Three Phase Transformer Connections

Each three-phase transformer has a primary winding set and a secondary winding set.

Each set can be star or delta. The usual four combinations are:

Delta-Delta (Δ-Δ)
Star-Star (Y-Y)
Delta-Star (Δ-Y)
Star-Delta (Y-Δ)

The first letter is the primary. The second letter is the secondary.

Delta-Delta (Δ-Δ) Connection

In a delta-delta bank both windings are delta. The connection is shown below.

Primary ends are A1A2, B1B2 and C1C2. Secondary ends are a1a2, b1b2 and c1c2.

A1 and a1 have the same polarity. Winding A1A2 pairs with a1a2.

The phasor diagram below is for lagging power factor cos ф. Magnetizing current and leakage-impedance drops are omitted.

phasor diagram of delta delta Δ Δ connection — Phasor Diagram of Delta-Delta (Δ-Δ) Connection

On each delta, line voltage equals winding voltage. Primary line voltages V_AB, V_BC and V_CA are in phase with Vab, Vbc and Vca (Dd0). The voltage ratio equals the turns ratio.

$\frac{V_{AB}}{V_{ab}} = \frac{V_{BC}}{V_{bc}} = \frac{V_{CA}}{V_{ca}} = N$

With balanced load, line current is √3 times winding current. Neglecting magnetizing current, the current ratios are:

$\frac{I_{AB}}{I_{ab}} = \frac{I_{BC}}{I_{bc}} = \frac{I_{CA}}{I_{ca}} = \frac{I_{A}}{I_{a}} = \frac{I_{B}}{I_{b}} = \frac{I_{C}}{I_{c}} = N$

Primary and secondary voltages are in phase, so this is a 0° (Dd0) connection.

Reverse the secondary delta (swap the sense of each winding) and the voltages are 180° apart: a Dd6 connection.

The 180° wiring and phasors are shown below.

delta delta Δ Δ connection 180 degree connection — Delta-Delta (Δ-Δ) Connection 180-degree connection

Here secondary corners b1c2, c2a2 and a1b2 are joined. The phasors show secondary voltage opposite to primary voltage.

Delta-delta is used when:

It can supply a balanced or unbalanced three-phase load (no single-phase-to-neutral load).
Triplen harmonics circulate inside the delta and do not appear in the line-to-line voltage.
If one unit of a three-unit bank is lost, the other two can run as an open-delta (V-V) bank at reduced rating.

There is no neutral, so this bank is not used where a line-to-neutral load or a grounded-wye source is required on that winding.

Star-Star (Y-Y) Connection

In a star-star (Y-Y) bank both windings are star. The connection is shown below.

Line current equals winding current. Line voltage is √3 times phase voltage. With magnetizing current ignored, the voltage phasors look like the 0° delta-delta set.

For an ideal transformer the voltage ratio is

$\frac{V_{AN}}{V_{an}} = \frac{V_{BN}}{V_{bn}} = \frac{V_{CN}}{V_{cn}} = N$

The current ratio is

$\frac{I_A}{I_a} = \frac{I_B}{I_b} =\frac{I_C}{I_c} = N$

With no neutral conductor, an unbalanced load shifts the star point and the phase voltages go unequal. Ungrounded Y-Y is a poor choice for unbalanced load.

Y-Y is also sensitive to the third-harmonic part of magnetizing current.

Magnetizing current is not a sine wave: it includes a triplen (third-harmonic) component. A closed path for that current is needed if the flux is to stay close to a sine wave.

The three triplen magnetizing currents are in phase with each other. They add at the star point instead of cancelling.

If the star points are isolated, those currents have no return path. The flux picks up a third-harmonic component and both windings show a third-harmonic voltage to earth.

That extra voltage sits on the fundamental. In textbook sketches the peak can approach about twice the usual value. The exact peak depends on core saturation and whether the star point is grounded.

Two common fixes are:

Solid Grounding of Neutral

Solidly earth the primary star point to the source neutral. The triplen magnetizing current can then flow in the neutral and the large third-harmonic voltage is avoided.

The same conductor also returns unbalanced load current.

The neutral then carries 150 Hz (or 180 Hz) current, which can couple into nearby communication circuits.

Provide Tertiary Winding

If you cannot run a neutral, add a third winding on the same core: a tertiary.

That tertiary is connected in delta. The bank is then Y-Δ-Y: the closed delta gives a path for triplen current without a neutral conductor.

Delta-Star (Δ-Y) Connection

In a delta-star (Δ-Y) bank the primary is delta and the secondary is star.

The connection is shown below.

delta star Δ y connection — Delta-Star (Δ-Y) Connection

On the delta primary, line voltage equals winding voltage.

On the star secondary, line voltage is √3 times phase voltage.

The line-to-line voltage ratio is then:

$\frac{V_{LP}}{V_{LS}} = \frac{V_{PP}}{\sqrt{3}V_{PS}}$

$\frac{V_{PP}}{V_{PS}} = N$

$\frac{V_{LP}}{V_{LS}} = \frac{a}{\sqrt{3}}$

The phasors below are for lagging power factor and balanced load.

phasor diagram delta star Δ y connection — Phasor Diagram Delta-Star (Δ-Y) Connection

Secondary phase voltage V_aN leads primary V_AN by 30°. V_bN and V_cN lead V_BN and V_CN, by the same angle. IEC vector group Dy11 is this +30° case.

Reverse one winding set and the secondary lags the primary by 30° (Dy1). That wiring is shown below.

delta star Δ y 30 degree connection — Delta-Star (Δ-Y) -30-degree Connection

Star-Delta (Y-Δ) Connection

In a star-delta (Y-Δ) bank the primary is star and the secondary is delta. The connection is shown below.

Star primary: line voltage is √3 times phase voltage. Delta secondary: line voltage equals winding voltage.

The line-to-line voltage ratio is:

$\frac{V_{LP}}{V_{LS}} = \frac{\sqrt{3} V_{PP}}{V_{PS}}$

$\frac{V_{PP}}{V_{PS}} = N$

The phasors are the Δ-Y set with primary and secondary swapped. Line voltages are shifted 30°. Yd11 is the +30° case.

Reverse one winding set for a 30° lag (Yd1). That wiring is shown below.

star delta y Δ 30 degree connection — Star-Delta (Y-Δ) 30-degree Connection

Y-Δ is Δ-Y with the windings swapped. Either bank can feed an unbalanced three-phase load. The delta gives a path for triplen current.

The delta holds the Y-side voltages in balance for triplen magnetizing current and lets those harmonics circulate without a neutral conductor.

Open Delta Connection (V-V Connection)

If one unit of a delta-delta three-unit bank is lost, the other two can still feed a three-phase load at reduced rating. That emergency bank is an Open Delta connection (V-V).

Only two transformers are in circuit.

The open-delta connection is shown below.

open delta connection v v connection — Open Delta Connection (V-V Connection)

Primary line voltages are V_AB, V_BC and V_CA. The two remaining secondaries give V_ab and V_bc. There is no winding from a to c, yet a voltage still appears between a and c.

$V_{ab} + V_{bc} + V_{ca} = 0$

(1) $\begin{equation*} V_{ca} = - V_{ab} - V_{bc} \end{equation*}$

Write the primary line voltage as V_P and the secondary as V_S.

$V_{AB} = V_P \angle 0^\circ$

$V_{BC} = V_P \angle -120^\circ$

$V_{CA} = V_P \angle +120^\circ$

Take leakage impedance as zero so the secondary voltages follow the primary angles.

$V_{ab} = V_S \angle 0^\circ$

$V_{bc} = V_S \angle -120^\circ$

Put those phasors into the stored equation:

$V_{ca} = - V_S \angle 0^\circ - V_S \angle -120^\circ$

$V_{ca} = -V_S - \left[ -cos(120) V_S - j sin(120) V_S \right]$

$V_{ca} = -V_S - \left[ -0.5 V_S - j 0.866 V_S \right]$

$V_{ca} = -0.5V_S + j 0.866 V_S$

$V_{ca} = V_S \angle +120^\circ$

The voltage across a and c then has the same magnitude as the other secondary line voltages and sits 120° from them. With zero leakage the secondary set is still a balanced three-phase set.

A closed delta-delta bank uses three transformers at the bank rating. Open delta uses two of those units.

Two of three units look like 66.7% of the bank. The three-phase rating is lower than that, because the two windings do not share current the same way as a closed delta.

Let V_2B and I_2B be rated secondary winding voltage and current. In delta, winding voltage equals line voltage. Closed-delta line current is √3 times winding current.

Closed delta-delta bank rating:

$S_{\Delta-\Delta} = \sqrt{3} \times line \, voltage \times line \, current$

$S_{\Delta-\Delta} = \sqrt{3} V_{2B} (\sqrt{3} I_{2B})$

$S_{\Delta-\Delta} = 3 V_{2B} I_{2B}$

With one unit removed, line current I_2B equals that unit’s rated winding current (not √3 times it).

Open-delta rating:

$S_{V-V} = \sqrt{3} V_{2B} I_{2B}$

The ratio of those ratings is

$\frac{S_{V-V}}{S_{\Delta-\Delta}} = \frac{\sqrt{3} V_{2B} I_{2B}}{3 V_{2B} I_{2B}}$

$\frac{S_{V-V}}{S_{\Delta-\Delta}} = \frac{1}{\sqrt{3}}$

$\frac{S_{V-V}}{S_{\Delta-\Delta}} = 0.577$

Open delta can supply 57.7% (1/√3) of the original three-unit bank rating if each remaining unit stays at rated current.

The stored latex below is algebraically off. The 73.2% figure equals √3 minus 1: the extra current in each remaining unit if you keep the original 100% bank load (line current still √3 times rated winding current).

$\frac{ S_{V-V} - S_{\Delta-\Delta}}{ S_{\Delta-\Delta}} \times 100 = \frac{\sqrt{3} V_{2B} I_{2B} - 3 V_{2B} I_{2B} }{ V_{2B} I_{2B}} = 73.2%$

Do not keep the original load. Drop the load to the 57.7% rating or the two windings will overheat.

Scott Connection

A Scott (T) connection uses two single-phase transformers to convert three-phase to two-phase, or the reverse.

The two units are wired together. They sit on separate cores, so they are not magnetically coupled to each other.

One unit is the main transformer. The other is the teaser (auxiliary) transformer.

The Scott connection is shown below.

Lines are A, B and C. The main transformer is centre-tapped at D and is connected across B and C.

Main primary is BC. Main secondary is a1a2.

The teaser primary is connected from line A to the mid-tap D, not from A to C. Teaser secondary is b1b2.

Balanced line voltages V_AB, V_BC and V_CA give the phasors below.

phasor diagram scott connection — Phasor Diagram Scott Connection

Take V_BC as the reference phasor.

$V_{BC} = V_L \angle 0^\circ$

$V_{CA} = V_L \angle -120^\circ$

$V_{AB} = V_L \angle +120^\circ$

D splits winding BC in half, so BD and DC have equal turns.

If the main primary has T_P turns, BD and DC each have T_P/2.

Voltage on BD equals voltage on DC and is half of the voltage across BC.

$V_{BD} = V_{DC} = \frac{1}{2} V_{BC} = \frac{1}{2} V_L \angle 0^\circ$

Voltage from A to D is

$V_{AD} = V_{AB} + V_{BD}$

$V_{AD} = V_L \angle 120^\circ + \frac{1}{2} V_L \angle 0^\circ$

$V_{AD} = V_L \left[ \frac{-1}{2} + j \frac{\sqrt{3}}{2} \right] + \frac{1}{2} V_L$

$V_{AD} = j \frac{\sqrt{3}}{2} V_L$

$V_{AD} = 0.866 V_L \angle 90^\circ$

Teaser primary voltage is 0.866 (√3/2) times the main primary voltage and is 90° from it.

Teaser primary voltage is V_AD, teaser secondary is V_2a.

Secondary teaser voltage V_2a leads main secondary V_2m by 90°, as in the figure below.

Equal volts per turn (same core flux density) means the teaser primary must have 0.866 T_P turns.

The two secondaries then have the same voltage rating. V_2a and V_2m are equal in size and 90° apart: a balanced two-phase pair.