Per-Unit System Calculator (Base Impedance and Base Change)

Power system engineers express impedances, voltages and currents as fractions of chosen base values. This per-unit method removes transformer ratios from the maths and makes fault studies much quicker, as shown in this introduction to the per-unit system in power systems.

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Type in any field to change it. Press × on a value to calculate that field instead.

How to use the per-unit System calculator

Choose a base power and base voltage, and enter an actual impedance in ohms. You get the base impedance, base current and per-unit impedance. Enter a new base to convert the per-unit value — for example from a transformer’s own rating to a 100 MVA system base.

For more background, read How to Calculate Short Circuit Fault Current.

Per-unit System formulas

Quantity Formula
Base impedance Zb = kVb² ÷ MVAb
Base current Ib = MVAb ÷ (√3 × kVb)
Per-unit Zpu = ZΩ ÷ Zb
Change of base Znew = Zold × (Vold/Vnew)² × (Snew/Sold)

Base impedances

Voltage 10 MVA base 100 MVA base
0.415 kV 0.01722 Ω 0.001722 Ω
3.3 kV 1.089 Ω 0.1089 Ω
6.6 kV 4.356 Ω 0.4356 Ω
11 kV 12.1 Ω 1.21 Ω
22 kV 48.4 Ω 4.84 Ω
33 kV 108.9 Ω 10.89 Ω
66 kV 435.6 Ω 43.56 Ω
132 kV 1742 Ω 174.2 Ω
220 kV 4840 Ω 484 Ω

Example: transformer on a system base

A 2.5 MVA transformer with 6.25% impedance on its own base, converted to 100 MVA:

Z = 0.0625 × (100 ÷ 2.5) = 2.5 pu on 100 MVA.

Short-circuit with per-unit

Fault current in pu = 1 ÷ Ztotal,pu; multiply by base current for amps. Adding source, transformer and cable impedances in per unit is how quick fault-level studies are done by hand.

Frequently asked questions

What is the per-unit system?

Expressing voltages, currents and impedances as fractions of chosen base values. It removes transformer ratios from calculations and makes equipment of different sizes easy to compare.

How do I calculate base impedance?

Z_base = (kV_base)² ÷ MVA_base. For 11 kV and 10 MVA, Z_base = 12.1 Ω.

How do I change a per-unit value to a new base?

Z_new = Z_old × (V_old ÷ V_new)² × (S_new ÷ S_old).

What is the link between % impedance and per unit?

Percentage impedance is per-unit × 100. A transformer with 6% impedance has 0.06 pu on its own rating.

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