Autotransformer (autoformer) useful for voltage changes ≤ 25 % - dimensioning

Question:
Which values are important when dimensioning an autotransformer for voltage regulation?

Answer:
An autotransformer consists of a tapped coil. Part G is used both by the primary and by the secondary side. The auxiliary winding Z is used for voltage changes. The rated autotransformer power Srat is calculated as follows:
Only the kVA rating St is important for the dimensioning of the autotransformer. The kVA rating and, accordingly, the required iron core only depend on the additional voltage Uz.

If U1> U2:

• 1-phase: St = (U1-U2) · I2 = Srat · (1-ü) = Sn · (1-U2/U1)
• 3-phase: St = (U1-U2) · I2 · root(3) = Srat · (1-ü) · root(3) = Srat · (1-U2/U1) · root(3)

If U1< U2: (The ratio ü must be < 1)

• 1-phase: St = (U2-U1) · I2 = Srat · (1-ü) = Srat · (1-U1/U2)
• 3-phase: St = (U2-U1) · I2 · root(3) = Srat · (1-ü) · root(3) = Srat · (1-U1/U2) · root(3)

Advantages of the autotransformer:

1. less costly
2. small design / light weight
3. little iron and copper losses
4. high efficiency
A disadvantage is the conductive connection of the two electric circuits (no electrical isolation). For this reason, autotransformers cannot be used for converting high voltage into low voltage and low voltage into extra-low voltage. Autotransformers are useful for voltage variations limited to ≤ 25 %.

When using autotransformers at the power input of frequency inverters, it is absolutely necessary to use mains chokes or mains filters because e. g. a 20 % voltage boost through the autotransformer will also reduce the short-circuit voltage Uk of the transformer by approx. 20 %. If you use an isolation transformer, mains chokes and mains filters are, however, not required.

Example:

The motor voltage of a 30 kW motor (Irat =56 A) ^*) with inverter supply shall still be 400 V at rated load. The total voltage losses in the entire drive axis e. g. amount to 10 % = 40 V. The rated mains voltage is 400 V (U1).

         U2 = U1 + 10 % = 400 V + 40 V = 440 V
St  = (U2 - U1) · I2 · root(3) = (440 V - 400 V) · 56 A · 1.73 = 3.88 kVA

       --> an autotransformer with a kVA rating of 5 kVA is selected

*) For simplification, the motor current is equated with the mains current.