Transformer sizing and prospective short-circuit current

Two questions, two calculators

Transformer selection asks how large the unit must be to carry the load, and how large a fault current it can deliver. The transformer size calculator answers the first from load, power factor and growth margin. The transformer short-circuit current calculator answers the second from kVA, voltage and impedance. Both are screening estimates; selection and protection need the applicable standard, supplier data and qualified review.

Neither result is a protection setting. The sizing result gives full-load current for cable and device selection by others, and the fault result gives an upper bound that must be compared with equipment ratings in a proper study.

Sizing formulas

The calculator uses:

1. Load kVA = kW ÷ PF.
2. Required kVA = load kVA × (1 + margin ÷ 100).
3. Select the first standard rating at or above the requirement. The list used is 25, 50, 63, 100, 125, 160, 200, 250, 315, 400, 500, 630, 800, 1000, 1250, 1600, 2000, 2500 and 3150 kVA; availability varies by region and maker.
4. Rated secondary current = kVA × 1000 ÷ (√3 × V) for three-phase, or kVA × 1000 ÷ V for single-phase.

Enter maximum demand rather than the sum of nameplates unless every load can run together. The same current conversion is available in the kVA to current calculator.

Loading is the load kVA divided by the selected rating. A result near 100 % means no remaining capacity once the margin is used up; a very low result may mean a smaller rating would suit, or that the margin is generous. The calculator reports it so the choice can be judged. Impedance, cooling class, ambient temperature and the loading policy that the maker allows are not part of the calculation and can change which rating is appropriate.

Worked example 1: 320 kW three-phase demand

Maximum demand is 320 kW at PF 0.85, with 20 % spare capacity and a 415 V secondary.

1. Load = 320 ÷ 0.85 = 376.47 kVA.
2. Required = 376.47 × 1.20 = 451.76 kVA.
3. Next standard rating = 500 kVA, so the stated load is 376.47 ÷ 500 = 75.29 % of rating.
4. Rated secondary current = 500 × 1000 ÷ (√3 × 415) = 695.60 A.

These match the calculator. The margin is a project choice, not a code value.

Short-circuit formulas

The fault calculator assumes an infinitely strong upstream supply, so only the transformer impedance limits the current:

• Rated current I_n = kVA × 1000 ÷ (√3 × V).
• Prospective fault current I_sc = I_n × 100 ÷ Z%.
• Fault level (MVA) = kVA ÷ (Z% × 10).

Z% is the percentage impedance from the nameplate or data sheet. The result is the three-phase value at the secondary terminals. A lower impedance gives a higher fault current.

The figure is an upper-bound estimate at the transformer terminals. Cable and busbar impedance lower the current at downstream points, while motor contribution can raise it. Arc resistance is ignored. Equipment breaking and withstand ratings, discrimination and protection settings must come from a complete short-circuit study, not this single number.

Worked example 2: fault level of the selected unit and a comparison

Take the 500 kVA, 415 V unit above with an assumed impedance of 5 %. I_n = 695.60 A, so I_sc = 695.60 × 100 ÷ 5 = 13,912 A (13.91 kA), the fault level is 500 ÷ 50 = 10 MVA, and the fault current is 20 times rated.

A second case: a 150 kW load at PF 0.90 with 25 % margin at 400 V gives 166.67 kVA, 208.33 kVA required and a 250 kVA standard rating (66.67 % loaded, 360.84 A). At 4 % impedance its terminal fault current is 9,021 A (9.02 kA, 6.25 MVA, 25 × rated). The table shows how impedance changes a 630 kVA, 400 V unit (rated current 909.33 A).

The comparison table shows why impedance is a design parameter: moving from 6 % to 4 % raises the fault level by half (100 ÷ 4 against 100 ÷ 6). A low-impedance unit gives better voltage regulation but a higher fault duty on downstream equipment, so the two calculators should be read together.

Common mistakes

• Summing nameplate ratings of loads that never run together, or conversely using an average instead of maximum demand.
• Entering kVA as kW, or forgetting power factor.
• Using line-to-neutral voltage for the three-phase conversion.
• Using the primary voltage for current at the secondary.
• Applying the terminal fault level to a remote board without cable impedance.
• Ignoring motor starting, harmonics, ambient temperature or cooling class, which the sizing calculator does not model.

Checks before you trust the result

• Confirm the demand figure from measurements or a documented load schedule.
• If power factor will be corrected, size on the corrected value and see power factor correction sizing.
• Take impedance from the actual nameplate or data sheet.
• Check the rated current against the secondary cabling and the loading policy.

A useful cross-check on the sizing result is the load current: for the first example, 376.47 kVA at 415 V gives 523.7 A, which is 75.29 % of the 695.6 A rated current, consistent with the loading figure. Do the same sanity check on every result before it goes into a record.

Frequently asked questions

Why is the fault current a multiple of rated current?

Only the transformer impedance limits it, so the multiple is 100 ÷ Z%.

Does a larger margin always help?

It adds capacity but also raises fault current and cost, and a lightly loaded unit may run less efficiently.

Can I use this for protection settings?

No. Use a full study.

Record the data

Record the demand source and date, power factor, margin, selected rating, nameplate impedance, voltage and the calculated rated and fault currents. Update the record whenever load is added or the upstream supply changes.

Also note the assumed upstream condition: the fault figure presumes an infinitely strong supply, so recording that assumption beside the result prevents it being mistaken for a measured or utility-supplied value.