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Copper saving in auto transformer

Short answer

For same rating and current density, copper weight is proportional to ampere-turns. For an autotransformer W_auto = (1-K)W_two, so copper saving is K times the two-winding copper. Thus saving is largest when K is near unity, where only a small fraction of power is transformed inductively.

Prove with suitable assumption that copper saving in auto transformer is significant when the transformation ratio is nearly equal to unity.

An autotransformer has a single winding, part of which is common to both primary and secondary. Power is transferred partly by conduction and partly by induction, so it needs less copper than a two-winding transformer of the same rating.

Assumptions

  1. Same rating (same V1V_1, V2V_2, I1I_1, I2I_2) for both transformers.
  2. Same current density in all sections, so copper weight ∝ (current × number of turns) of each section (length ∝ turns, cross-section ∝ current).
  3. No-load current and losses are neglected, so N1I1=N2I2N_1I_1 = N_2I_2.
  4. Step-down case with K=N2N1=V2V1=I1I2<1K = \frac{N_2}{N_1} = \frac{V_2}{V_1} = \frac{I_1}{I_2} < 1.
          I1
  o------>----+ A
              S  section AB: (N1-N2) turns,
  V1          S  current I1
              +---------+---> I2  o
              S  B      |
              S section BC: N2 turns,     V2
              S current (I2 - I1)
              S         |
  o-----------+ C ------+---------o

Two-winding transformer

Primary: N1N_1 turns carrying I1I_1; secondary: N2N_2 turns carrying I2I_2.

Wtw∝N1I1+N2I2W_{tw} \propto N_1I_1 + N_2I_2

Autotransformer

  • Section AB has (N1−N2)(N_1 - N_2) turns carrying I1I_1.
  • Section BC has N2N_2 turns carrying (I2−I1)(I_2 - I_1).
Wauto∝(N1−N2)I1+N2(I2−I1)W_{auto} \propto (N_1 - N_2)I_1 + N_2(I_2 - I_1)

Ratio of copper weights

WautoWtw=N1I1−N2I1+N2I2−N2I1N1I1+N2I2\begin{aligned} \frac{W_{auto}}{W_{tw}} &= \frac{N_1I_1 - N_2I_1 + N_2I_2 - N_2I_1}{N_1I_1 + N_2I_2} \end{aligned}

Using N2I2=N1I1N_2I_2 = N_1I_1 and N2=KN1N_2 = KN_1:

WautoWtw=2N1I1−2KN1I12N1I1=1−KWauto=(1−K) Wtw\begin{aligned} \frac{W_{auto}}{W_{tw}} &= \frac{2N_1I_1 - 2KN_1I_1}{2N_1I_1} = 1 - K \\ W_{auto} &= (1 - K)\,W_{tw} \end{aligned}

Therefore

Saving of copper=Wtw−Wauto=K Wtw\text{Saving of copper} = W_{tw} - W_{auto} = K\,W_{tw}

Significance when K is near unity

The saving is KK times the copper of the two-winding transformer, so it grows as K→1K \to 1:

K = V2/V1Copper in auto (× W_tw)Saving
0.10.910%
0.50.550%
0.80.280%
0.950.0595%

Physically, when K≈1K \approx 1 the common section BC carries only the small difference (I2−I1)(I_2 - I_1), and section AB has very few turns, so very little copper is needed. Only the fraction (1−K)(1-K) of the power is transformed inductively; the rest is conducted directly.

Hence autotransformers are economical when the voltage ratio is close to 1 (e.g. 400/440 V, 220/132 kV interconnection, variacs, starting of induction motors), but not for large ratios, where saving is small and the lack of electrical isolation is a safety risk.

✓ Answer checked by Claude Opus 5.5 (high effort)

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