Calculate the prospective fault current at a low voltage board to IEC 60909, working through the grid source, the transformer and the connecting cable. Returns three-phase, phase-to-phase and phase-to-earth currents, the peak asymmetrical current, induction motor back-feed and the breaking capacity your switchgear needs.
| Symbol | Meaning | Unit |
|---|---|---|
| Ik3 | Symmetrical three-phase fault current | A |
| Ik2 | Phase-to-phase fault current | A |
| Ik1 | Phase-to-earth fault current, approximate | A |
| ip | Peak asymmetrical current, first half cycle | A |
| κ | Peak factor from the X/R ratio | — |
| c | Voltage factor, 1.05 for maximum LV fault | — |
| SkQ | Grid short circuit capacity | MVA |
| %Z, %R | Transformer impedance and resistance | % |
| Imot | Induction motor back-feed contribution | A |
| Sk | Fault level | MVA |
A 1000 kVA distribution transformer at 5 %Z and 1 %R feeds a 415 V main board through 10 m of 240 mm² copper. The utility declares 500 MVA at the HV terminals. The board carries 1500 kVA of running induction motors.
| Input | Unit | Accepted range | Default |
|---|---|---|---|
| Grid fault level | MVA | > 0 | 500 |
| LV bus voltage | V | > 0 | 415 |
| Transformer rating | kVA | > 0 | 1000 |
| Impedance %Z | % | > 0, typically 4–8 | 5 |
| Resistance %R | % | > 0 and ≤ %Z | 1 |
| Cable length | m | ≥ 0 | 10 |
| Cable CSA | mm² | > 0 | 240 |
| Motor load | kVA | ≥ 0 | 1500 |
| Rating (kA) | Typical application | Rating (kA) | Typical application |
|---|---|---|---|
| 6 | Domestic final circuits | 50 | Large LV main board |
| 10 | Small commercial board | 63 | 1000–1600 kVA transformer |
| 16 | Sub-distribution board | 80 | With significant motor load |
| 25 | 500 kVA transformer bus | 100 | Large industrial main board |
| 31.5 | Common MV switchgear rating | 125 | Parallel transformers |
| 40 | 800 kVA transformer bus | 150 / 200 | Generation and heavy industry |
| Rating | Typical %Z | Typical %R | X/R | Isc at LV terminals |
|---|---|---|---|---|
| 100 kVA | 4.0 % | 2.0 % | 1.7 | 3.5 kA |
| 250 kVA | 4.0 % | 1.6 % | 2.3 | 8.8 kA |
| 500 kVA | 4.5 % | 1.3 % | 3.3 | 15.6 kA |
| 1000 kVA | 5.0 % | 1.0 % | 4.9 | 28.1 kA |
| 1600 kVA | 6.0 % | 0.9 % | 6.6 | 37.5 kA |
| 2000 kVA | 6.0 % | 0.8 % | 7.4 | 46.9 kA |
| 2500 kVA | 6.25 % | 0.7 % | 8.9 | 56.2 kA |
Add the impedances of every element between the source and the fault, then divide the driving voltage by the total. IEC 60909 uses a voltage factor c of 1.05 for maximum fault current on low voltage systems, so the three-phase symmetrical current is c times the nominal voltage divided by √3 times the total impedance. The chain is normally the grid source impedance referred to the low voltage side, the transformer impedance from its %Z and %R, and the cable impedance to the point of fault.
Ik3 is the symmetrical three-phase fault, normally the largest and the figure that sets switchgear breaking capacity. Ik2 is the phase-to-phase fault, which for a fault far from the generator is √3/2, about 86.6 % of Ik3. Ik1 is the phase-to-earth fault, and its true value depends on the zero-sequence impedance of the transformer and the earthing arrangement, so it can be either larger or smaller than Ik3. This calculator approximates Ik1 as 85 % of Ik3 and flags that an accurate figure needs zero-sequence data.
Current in an inductive circuit cannot change instantaneously, so a fault starting at a voltage zero produces a decaying DC offset on top of the symmetrical AC component. The first peak can therefore approach 2√2 times the symmetrical current. IEC 60909 quantifies this with the κ factor, which equals 1.02 + 0.98 e−3R/X. A stiff, highly inductive supply with a high X/R gives a κ near 1.8, while a cable-limited fault with a low X/R gives a κ near 1.1.
Yes. During a fault a running induction motor is driven by the inertia of its load and briefly acts as a generator, feeding current back into the fault at roughly six times its full load current for the first few cycles. IEC 60909 requires this contribution to be included where the connected motor rating is significant relative to the transformer. It matters most for the making and peak duty of the switchgear, since the motor contribution decays within a few cycles and has largely disappeared by the time a breaker interrupts.
The breaking capacity must exceed the prospective symmetrical fault current at the point of installation, and the making capacity must exceed the peak asymmetrical current including any motor contribution. Devices are rated in the standard series of 6, 10, 16, 20, 25, 31.5, 40, 50, 63, 80, 100, 125, 150 and 200 kA. Selecting on the symmetrical current alone under-rates the device wherever significant motor load is connected, because the peak duty is what the contacts actually face on closing.
Results are for estimation and preliminary design. A fault study for construction must use manufacturer transformer data, actual cable routes, zero-sequence impedance for earth faults and a minimum-fault case for protection grading. Verify with a qualified engineer.
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