Copper saving in an autotransformer

An autotransformer has a single winding, part of which is common to the primary and the secondary. Because the common part carries only the difference current, less copper is needed than in a two-winding transformer.

Circuit and notation (step-down, N1 turns total, N2 turns across the output)

 I1 -->  A o---------+---------o  (input)
                     |
             N1-N2   |
             turns   | I1
                 B o-+-----+---o  (output tap)
             N2      |     | I2
             turns   |     v
                 C o-+---------o  (common)
          V1 across AC, V2 across BC

• Transformation ratio K = N2/N1 = V2/V1 = I1/I2 (K < 1 for step-down).
• Section AB (N1 − N2 turns) carries current I1.
• Section BC (N2 turns) carries (I2 − I1).

Weight of copper ∝ (number of turns × current) for the same current density and mean turn length.

Auto-transformer:
 W_a ∝ (N1 − N2) I1 + N2 (I2 − I1)

Two-winding transformer of the same output:
 W_t ∝ N1 I1 + N2 I2

Using N1 I1 = N2 I2 (ampere-turn balance, so I1 = K I2):

W_a ∝ N1 I1 − N2 I1 + N2 I2 − N2 I1
    = N1 I1 + N2 I2 − 2 N2 I1
    = 2 N1 I1 − 2 N2 I1             (as N1 I1 = N2 I2)
    = 2 N1 I1 (1 − N2/N1) = 2 N1 I1 (1 − K)

W_t ∝ N1 I1 + N2 I2 = 2 N1 I1

W_a / W_t = 1 − K

Saving of copper = W_t − W_a = K × W_t

So the autotransformer needs (1 − K) times the copper of a two-winding unit, i.e. K × 100 % is saved. For K = 0.8 the saving is 80 %. The saving is large only when K is close to 1 (voltage ratios near 1).