Scott connection for three-phase to two-phase

The Scott (T) connection uses two single-phase transformers to convert a balanced three-phase supply into a balanced two-phase supply (two voltages equal in magnitude and 90° apart), or the reverse. It is used for two-phase loads such as electric furnaces and for traction supply.

Connections

• Main transformer: primary of $N_1$ turns connected across two lines B and C, with a centre tap D.
• Teaser transformer: primary of $\frac{\sqrt3}{2}N_1$ turns (86.6%), connected between line A and the centre tap D.
• Both secondaries have the same number of turns $N_2$ and feed the two phases separately.

            A
            |
            |  Teaser primary
            |  (0.866 N1 turns)
            |
   B -------D------- C
     N1/2       N1/2
   Main primary (N1 turns,
   centre tapped at D)

  Secondaries (N2 turns each):
   Teaser sec -> V_T (phase 1)
   Main sec   -> V_M (phase 2)

Voltage relations (phasor diagram)

Let the line voltage be $V_L$.

              A
             /|\
            / | \
           /  |  \   VAD = 0.866 VL
          /   |   \  (perpendicular
         /    |    \   to BC)
        B-----D-----C
          VL/2  VL/2

• Voltage across the main primary: $V_{BC} = V_L$.
• D is the midpoint of BC, so $V_{AD}$ is the height of the equilateral triangle:

$$V_{AD} = \sqrt{V_L^2 - \left(\frac{V_L}{2}\right)^2} = \frac{\sqrt3}{2}V_L = 0.866\,V_L$$

and $V_{AD}$ is at 90° to $V_{BC}$.

• Volts per turn on the teaser: $\frac{0.866V_L}{0.866N_1} = \frac{V_L}{N_1}$, the same as on the main. Therefore the secondary voltages are

$$V_{M} = \frac{N_2}{N_1}V_L, \qquad V_{T} = \frac{N_2}{0.866N_1}(0.866V_L) = \frac{N_2}{N_1}V_L$$

They are equal in magnitude and 90° apart — a balanced two-phase supply.

Current relations (balanced two-phase load)

• Teaser primary current $I_A = \frac{2}{\sqrt3}\frac{N_2}{N_1}I_2 = 1.155\,\frac{N_2}{N_1} I_2$, flowing in line A.
• It divides equally at D into the two halves of the main primary, where it combines with the main load current; the resulting line currents $I_B$ and $I_C$ are equal in size to $I_A$ and 120° apart, so the three-phase side sees a balanced load.

Points to note

• If the neutral of the three-phase side is wanted, it is at the point on the teaser one-third of its length from D (i.e. at $\frac{N_1}{2\sqrt3}$ turns from D).
• The two transformers can be identical units if each has 50% and 86.6% tappings, so they are interchangeable.