Transformer Sizing & Fault Current Calculator — kVA, Z%

What the calculator does

A transformer is sized for the load that actually coincides at the peak, not for the sum of every nameplate on the site. The calculator applies a demand factor to the connected load, converts kilowatts to kVA with the power factor, adds a growth allowance, divides by the loading you are prepared to run the unit at, and rounds up to the next IEC 60076 standard rating. From that rating it works out the full-load current on the primary and secondary sides, and from the nameplate percent impedance the prospective short-circuit current at the low-voltage terminals, first for an infinite source and then for the utility's actual fault level. The fault current is what the LV switchgear has to be rated to interrupt, so it belongs on the same page as the kVA.

Formula

P_demand = P_connected × demand factor S_demand = P_demand ÷ PF
S_required = S_demand × (1 + growth) ÷ planned loading → next standard rating S_n

Three-phase: I_FL = S_n ÷ (√3 × V) Single-phase: I_FL = S_n ÷ V

Infinite bus: I_sc = I_FL ÷ (Z % ÷ 100)
Finite source: Z_s = S_n (MVA) ÷ S_source (MVA) I_sc = I_FL ÷ (Z_t + Z_s)
Fault level = √3 × V × I_sc (three-phase) or V × I_sc (single-phase)

where Z_t is the transformer impedance as a fraction of its own base (4.5 % is 0.045), and Z_s the source impedance converted to the same base: a 250 MVA source looks like 0.315 ÷ 250 = 0.00126 per unit from a 315 kVA transformer. The percent impedance is the fraction of rated voltage that has to be applied to the primary, with the secondary short-circuited, to drive rated current; so with full voltage applied the short-circuit current is the rated current divided by that fraction. The two impedances are added arithmetically; the source term is small enough at distribution level that the difference from a vector sum is negligible.

Worked example

A factory in Faisalabad with 250 kW connected, a demand factor of 70 % from a week of logging, power factor 0.85 at the peak, 20 % growth, a planned loading of 80 %, an 11 kV/400 V three-phase transformer with 4.5 % impedance, and a utility fault level of 250 MVA at 11 kV.

Maximum demand: 250 × 0.70 = 175.0 kW; ÷ 0.85 = 205.9 kVA.
Growth: 205.9 × 1.2 = 247.1 kVA. Planned loading: 247.1 ÷ 0.8 = 308.8 kVA, so the next standard rating is 315 kVA.
Full-load currents: 315 000 ÷ (√3 × 400) = 454.7 A on the LV side; 315 000 ÷ (√3 × 11 000) = 16.5 A on the 11 kV side.
Short-circuit current, infinite bus: 454.7 ÷ 0.045 = 10 104 A, about 10.1 kA.
With the 250 MVA source: Z_s = 0.315 ÷ 250 = 0.00126 (0.126 %), total impedance 4.626 %, I_sc = 454.7 ÷ 0.04626 = 9 828 A, about 9.8 kA; fault level √3 × 400 × 9 828 = 6.81 MVA.
Loading at the peak: 205.9 ÷ 315 = 65.4 % today and 247.1 ÷ 315 = 78.4 % once the growth arrives, within the planned 80 %.

The main LV board therefore needs switchgear with a short-circuit breaking capacity above 10 kA. Since IEC 60909 adds a voltage factor and a transformer correction factor that push the figure higher, and since 10 kA is a common rating limit for miniature circuit breakers, the practical choice is moulded-case switchgear rated at 16 kA or more, confirmed by a proper study.

Standard ratings and their currents at 400 V

Rating (kVA)	Full-load current at 400 V, three-phase	Commonly listed impedance	Short-circuit current at LV, infinite bus
25	36.1 A	4 %	0.9 kA
50	72.2 A	4 %	1.8 kA
100	144.3 A	4 %	3.6 kA
160	230.9 A	4 %	5.8 kA
200	288.7 A	4 %	7.2 kA
250	360.8 A	4 %	9.0 kA
315	454.7 A	4 %	11.4 kA
400	577.4 A	4 %	14.4 kA
500	721.7 A	4 %	18.0 kA
630	909.3 A	4 %	22.7 kA
800	1 154.7 A	6 %	19.2 kA
1 000	1 443.4 A	6 %	24.1 kA
1 250	1 804.2 A	6 %	30.1 kA
1 600	2 309.4 A	6 %	38.5 kA
2 000	2 886.8 A	6 %	48.1 kA
2 500	3 608.4 A	6 %	60.1 kA
3 150	4 546.6 A	6 %	75.8 kA

Ratings from the IEC 60076-1 preferred (R10) series. Full-load current is S ÷ (√3 × 400). The impedance column is the value IEC 60076 lists as commonly used for distribution transformers; the nameplate of the unit you buy governs, and the last column is the rated current divided by that impedance with an infinite source, the conservative figure for LV switchgear.

Assumptions and limitations

One source, no motor contribution. The fault current comes from the utility through the transformer only. Running induction motors on the LV side feed a fault for a few cycles with roughly their locked-rotor current each; IEC 60909-0 includes them, this calculator does not.
Symmetrical rms only. No X/R ratio, DC offset or first-cycle peak. The peak current that switchgear must make and withstand is the rms value times a factor κ that rises with X/R; take it from an IEC 60909 study.
No voltage factor. IEC 60909-0 multiplies the nominal voltage by c (1.05 or 1.10 at low voltage) and corrects the transformer impedance; both raise the result. Use the figure here as a first estimate, not as the switchgear specification.
No harmonics. Variable-speed drives, UPS rectifiers and LED drivers heat the windings more than sinusoidal current of the same rms value. Where such loads are a large share, specify a K-factor or derate the unit.
ONAN rating at the IEC 60076-1 reference ambient: 40 °C maximum, 30 °C monthly average of the hottest month, 20 °C yearly average. Many Pakistani substations exceed all three in summer; derate according to IEC 60076-7 or specify a higher rating.
No tap changer. Off-circuit taps change the ratio, and the currents here are worked out at the voltages you enter, not at a tap position.
The demand factor is an input. The calculator cannot know which of your loads run together; a week of logging on an existing site is worth more than any table.
Not covered: HV protection, earthing, vector group, parallel operation, losses and the cable from the transformer to the board, for which the cable sizing calculator and the power and current calculator continue the arithmetic.

Frequently asked questions

What does percent impedance actually mean?

It is measured in the factory short-circuit test: the secondary is shorted and the primary voltage is raised until rated current flows. That voltage, as a percentage of the rated voltage, is the impedance; the power drawn during the test is the load loss. A 4.5 % transformer needs only 4.5 % of its rated voltage to push rated current through a dead short, which is why, at full voltage, a short gives about 1 ÷ 0.045 = 22 times rated current.

Why does a lower impedance give a higher fault current?

Because the impedance is all that limits the current when the load is a short circuit. Halving the impedance doubles the prospective fault current and everything downstream must interrupt and withstand it. Lower impedance also means better voltage regulation, which is the trade: small distribution units are built at 4 % or so for regulation, larger ones at 6 % or more to keep the fault current within what the switchgear can handle.

Why size at 80 % and not 100 %?

The nameplate rating assumes the IEC 60076-1 reference ambient. In a 45 °C substation the same load runs the windings hotter than the design allows, and insulation ageing roughly doubles for every 6 °C of extra hot-spot temperature. Loading at 80 % cuts the load losses to 64 % of full-load value, holds the winding rise down, and leaves room for the growth allowance to be wrong. The guide goes through the ageing arithmetic.

What happens if the transformer is oversized?

It costs more to buy, and its no-load loss, which is paid for every hour it is energised whatever the load, is larger. A unit that never sees more than 30 % load has its core losses on the bill 8 760 hours a year for very little use. Oversizing also raises the fault current, so the switchgear costs more too. If growth is uncertain, buy for the demand you can justify and plan the space and the HV cell for a second unit.

How do I enter a single-phase transformer?

Choose single-phase and enter the voltages across the windings: 230 V for the secondary of a pole-mounted unit feeding single-phase consumers, and for the primary the voltage the winding is actually connected across, 11 000 V phase-to-phase or 6 350 V phase-to-neutral. The currents are then S ÷ V with no √3.

The nameplate says 11 kV/415 V or 11 kV/433 V, not 400 V. Which do I enter?

The secondary voltage on a nameplate is the open-circuit voltage; under load the terminals drop by the regulation, a few percent, so makers set the no-load ratio above the nominal 400 V. Entering the nameplate no-load voltage gives the strictly correct fault current, and entering 400 V gives one about 4 % (415 V) or 8 % (433 V) higher, which is on the safe side for switchgear. For the full-load current, use the voltage at which the transformer will actually deliver its kVA.

References

IEC 60076-1:2011, Power transformers — Part 1: General — preferred ratings, normal service conditions, impedance tolerance
IEC 60076-7:2018, Power transformers — Part 7: Loading guide for mineral-oil-immersed power transformers — thermal model, ageing rate and loading beyond nameplate
IEC 60909-0:2016, Short-circuit currents in three-phase AC systems — Part 0: Calculation of currents — voltage factor c, impedance correction, motor contribution, peak current
IEEE Std C57.12.00, IEEE Standard for General Requirements for Liquid-Immersed Distribution, Power, and Regulating Transformers

Last reviewed 2026-09-20.

Transformer sizing and fault current calculator

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