Transformer Calculator: Full-Load Amps and Fault Current
Enter the transformer rating in kVA and its primary and secondary voltages to get the full-load current on both sides. If you also enter the impedance from the nameplate, the calculator gives the maximum fault current the transformer can deliver on its secondary terminals. Works for single-phase and three-phase units.
The formulas
Full-load current comes straight from the apparent power. For a single-phase transformer:
I = S × 1000 / V
For a three-phase transformer, with V as the line-to-line voltage:
I = S × 1000 / (√3 × V)
where S is the rating in kVA. Use the same formula on each side with that side’s voltage. The kVA is the same on both sides, ignoring losses, so the current changes in inverse proportion to the voltage.
The maximum secondary fault current, assuming the primary is fed from a source of unlimited strength, is:
Isc = IFL, secondary × 100 / %Z
A transformer with 5 % impedance therefore delivers at most twenty times its full-load current into a bolted fault on its terminals.
Example: 75 kVA, 480 V to 208Y/120 V
- Primary: 75 000 ÷ (1.732 × 480) = 90.2 A.
- Secondary: 75 000 ÷ (1.732 × 208) = 208.2 A.
- With Z = 5 %: 208.2 × 100 ÷ 5 = 4164 A, about 4.2 kA at the secondary terminals.
Using the results
Overcurrent protection. The full-load currents are the starting point for sizing the primary and secondary protective devices. Under NEC 450.3(B), for transformers up to 1000 V, primary-only protection is set at no more than 125 % of the rated primary current, with other combinations allowed when the secondary is protected as well. In the 75 kVA example, 125 % of 90.2 A is 112.8 A, and the next standard size up may be used where the rule permits it. IEC installations follow the manufacturer’s guidance and the national wiring rules instead, but the full-load current is the same figure.
Inrush. When a transformer is switched on it can draw a magnetising inrush of 8 to 12 times its full-load current for a few cycles, more for small toroidal units. The primary device has to ride through that, which is why transformer primaries are commonly protected with time-delay fuses or with breakers having a D or K characteristic. See MCB trip curves B, C and D.
Fault current and breaking capacity. The fault current figure is a conservative upper limit, because it ignores the impedance of the supply network and of the cables. It is the number to compare with the breaking capacity of the devices on the secondary side. If the panel is rated 10 kA and the calculator shows 4.2 kA, you have margin; if it shows 25 kA, the board needs a higher rating or the fault current needs a closer look with the network impedance included. Nameplate impedance also has a manufacturing tolerance, typically ±7.5 % under IEEE and ±10 % under IEC 60076, so do not cut the margin fine.
Voltage of the secondary. Remember that a 208Y/120 V secondary gives 120 V line to neutral. Single-phase loads connected line to neutral draw their current from one phase only, so on an unbalanced board one phase can be near its full-load current while the others are lightly loaded. Balance the circuits across the phases when you allocate them.
Related: kVA to amps, kW to kVA, power factor calculator, transformer symbols.