Transformer Calculator
Full-load current, turns ratio, short-circuit current, regulation and efficiency.
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- Free preview: each check’s title, a watermarked efficiency curve and the names of each table’s first rows (up to 3), with the currents, the working and every figure hidden.
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Results are estimates from the formulas shown, not a professional design or certification. Have a qualified engineer verify anything safety-critical (structures, electrical installations, gas or pressure systems).
About the Transformer Calculator
One calculator for a transformer’s nameplate. Enter the rating, the two voltages and the impedance, and it gives the full-load currents of both windings for single- and three-phase units, the voltage and turns ratios (they differ by √3 when one winding is delta and the other star), the impedance, resistance and reactance in ohms referred to either winding, and any impedance you enter referred to the other side. From the turns of the windings it works out the secondary voltage they give, and checks it against the nameplate; the line currents go in the inverse of the voltage ratio.
With the impedance it works out the short-circuit current at the secondary terminals, with an infinite source or with the upstream fault level, together with the peak current switchgear must make. With the load losses it splits the impedance into R and X for the voltage drop at your load and power factor, and with the no-load losses it gives the efficiency at every load and the load of maximum efficiency.
Without a pass, the free preview works on your own nameplate and shows each check’s title, the efficiency curve (watermarked, without its figures) and the first rows of each table by name; the currents, the fault level, every other figure, the working, Copy and the CSV and text reports come with a Pro pass.
How to use it
- Choose single- or three-phase and enter the rating (kVA), the primary and the no-load secondary voltage (line to line for three-phase) and the impedance in % from the nameplate.
- For three-phase units, set the winding connections (for example Dyn: delta primary, star secondary) so the turns ratio is right.
- Under Losses, load and power factor, enter the load and no-load losses from the test report and the load you want the voltage drop and efficiency for. Without the load losses, an X/R ratio splits the impedance instead.
- Optionally enter the upstream fault level for a closer short-circuit current, an impedance to refer to the other winding, or the turns of the windings. With a pass, or after unlocking this result, Copy the summary or download it as CSV or a text report.
Examples
Load losses 10.5 kW, no-load losses 1.7 kW, full load at power factor 0.8
I1 = 52.49 A, I2 = 1,333 A; turns ratio 44.00 (voltage ratio 25.40); fault current 26.67 kA with an infinite source; voltage drop 3.83%; efficiency 98.50%, maximum 98.95% at 40% load
Upstream fault level 250 MVA
Fault current 24.70 kA; R/X 0.206 → κ = 1.548, peak 54.1 kA
1,980 : 45 turns, 11 kV delta primary, star secondary
11,000 × 45 ÷ 1,980 = 250 V on each star winding, × √3 = 433 V between the lines
0.01 Ω per phase on the 433 V side of the 11 kV / 433 V transformer
0.01 × (11,000 ÷ 433)² = 6.45 Ω seen from the 11 kV side
Ratios and currents
- Voltage ratio a = V1 ÷ V2, the ratio of the line voltages on the nameplate
- Turns ratio N1 ÷ N2 = ratio of the winding voltages: a delta winding carries the line voltage, a star winding the line voltage ÷ √3. A Dyn 11 kV / 433 V transformer has a = 25.40 but N1 ÷ N2 = 44.0
- Full-load current: I = S ÷ (√3 × V) on each side of a three-phase unit, I = S ÷ V single-phase; a delta winding carries I ÷ √3
- Referring impedances: Z′ = Z × (V_other ÷ V_this)²
Short-circuit current
The impedance on the nameplate is the voltage, in % of rated, that drives rated current through the shorted transformer. So with nothing else limiting it, the secondary fault current is I_k = I2 × 100 ÷ z%. The upstream network adds its own impedance, Z_Q = c × V² ÷ S″_kQ referred to the secondary, split into R and X by its X/R ratio; the two add as complex numbers.
The current has a DC part when the fault starts, so its first peak is larger: i_p = κ × √2 × I_k with κ = 1.02 + 0.98 e^(−3R/X), the formula of IEC 60909-0. Breakers must break I_k and make i_p. The voltage factor c of IEC 60909 can be entered; the classic method uses 1.
Voltage drop and efficiency
The load losses fix the resistance: r% = P_k ÷ S × 100, and x% = √(z%² − r%²). At a load x (fraction of rated current) and power factor cos φ, with a = x(r cos φ + x_T sin φ) and b = x(x_T cos φ − r sin φ), the voltage drop is close to a + b²/2 (in per unit; sin φ is negative for a leading load, which can give a rise). The calculator solves the phasor equation exactly and shows the approximation beside it.
Efficiency: η = x S cos φ ÷ (x S cos φ + P0 + x² Pk). It peaks where the load losses x² Pk equal the no-load losses P0, at x = √(P0 ÷ Pk).
Sources
- IEEE Std 3002.3: recommended practice for short-circuit studies of industrial and commercial power systems
- IEC 60909-0: short-circuit currents in three-phase AC systems (voltage factor c, peak factor κ)
- IEC 60076-1: power transformers, general (ratings, impedance, losses)
Limitations
- The short-circuit current is the three-phase bolted fault at the secondary terminals (or the secondary short of a single-phase unit). Line-to-earth faults, cable impedance, motor contribution and arcing are not included.
- IEC 60909’s transformer correction factor K_T and the minimum-current case are not applied.
- Losses are taken at the temperature they were measured at; the load losses rise with winding temperature.
- Efficiency ignores auxiliary losses (cooling fans and pumps) and stray losses outside the test values.
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Frequently asked questions
What do I get without a pass?
Without a pass, Transformer Calculator shows each check’s title, a watermarked efficiency curve and the names of each table’s first rows (up to 3), with the currents, the working and every figure hidden. Until you unlock it, the result can’t be downloaded or copied. A Pro, Premium or Ultimate pass, a one-time payment that never renews, unlocks the full result. The pricing page lists the passes and their prices.
How do I calculate a transformer’s full-load current?
Divide the rating by the voltage, and by √3 for three-phase: a 1,000 kVA transformer at 433 V gives 1,000,000 ÷ (1.732 × 433) = 1,333 A on the secondary; at 11 kV, 52.5 A on the primary.
How do I find the short-circuit current from the % impedance?
With an infinite source, I_k = full-load current × 100 ÷ z%. A 1,000 kVA, 433 V transformer with 5% impedance gives 1,333 × 20 = 26.7 kA. The supply’s own impedance lowers it; enter the fault level to include it.
How do I find the secondary voltage from the turns?
Divide the primary voltage by the turns ratio N1 ÷ N2 where both windings are connected alike (single-phase, delta–delta or star–star): 11,000 V at 44 : 1 gives 250 V. In a delta–star transformer the star winding’s line voltage is √3 times its winding voltage, so 1,980 : 45 turns on an 11 kV delta primary give 433 V between the secondary lines.
Why is the turns ratio different from the voltage ratio?
The nameplate voltages are line voltages. In a delta–star (Dy) transformer each primary winding sees the full line voltage but each secondary winding only the line voltage ÷ √3, so the turns ratio is √3 times the voltage ratio.
At what load is a transformer most efficient?
Where its load (copper) losses equal its no-load (core) losses: at √(P0 ÷ Pk) of rated load. With 1.7 kW no-load and 10.5 kW load losses that is about 40% of rated.
What is the voltage factor c?
IEC 60909 multiplies the nominal voltage by a factor c to cover the voltage variations of a real network when it calculates maximum and minimum short-circuit currents. Enter the value your study uses; with 1 the calculator gives the classic result.