Calculators
Calculate short-circuit current at the end of a cable
At the start of a cable the short-circuit current is large, at the far end it is small. That is exactly where it is decided whether the protective device still operates under fault. And that is exactly where nobody recalculates, because the cable "has always run like that".
Result
- Short-circuit current line to protective conductorThe smallest fault current. The disconnection condition depends on it.
- —A
- Disconnection condition
- —
- Three-phase short-circuit current
- —A
- Total loop impedance
- —Ω
- Ratio of I_k1 to I_a
- —
Please fill all fields with valid numbers.
A guide, not a design to standard. We give no warranty for the correctness of the results. The applicable standards and case-by-case verification govern.
The grey values are an example. Simply enter your own.
Matching services
Calculating is the easy part.
A formula gives you a number. Designing a plant also demands installation method, grouping, discrimination, the standards in force and a look at the installed base. That is what we take on.
Method and questions
Short-circuit current at the far end
Fault current at the far end, with a check of the disconnection condition.
Discuss a projectWorked example
The default case the calculator starts with, worked through once. Change the values above and it recalculates immediately.
Inputs
- System voltage (line to line)
- 400 V
- Voltage factor c
- 0.95
- Upstream impedance
- 0.05 Ω
- One-way cable length
- 60 m
- Line conductor cross-section
- 16 mm²
- Protective conductor cross-section
- 16 mm²
- Conductor material
- Copper (κ = 56 m/Ω·mm²)
- Conductor temperature under fault
- 80 °C
- Operating current of the protective device I_a
- 160 A
Result
- Short-circuit current line to protective conductor
- 1,018 A
- Disconnection condition
- Met. The fault current is well above the operating current.
- Three-phase short-circuit current
- 1,653 A
- Total loop impedance
- 0.216 Ω
- Ratio of I_k1 to I_a
- 6.36
Formula and standards
Formula
- Z_loop = Z_upstream + R_L + R_PE
- I_k1 = c · U₀ / Z_loop with U₀ = U / √3
- I_k3 = c · U / (√3 · Z_three-phase)
- Disconnection condition met if I_k1 ≥ I_a
Assumptions and standards
- The upstream impedance is given as a magnitude, its angle is unknown. It is therefore added arithmetically. That slightly overstates the loop and yields the smaller fault current, which keeps the result on the safe side.
- The calculation uses a heated conductor, because the loop impedance is then highest and the fault current smallest.
- The operating current is not the rated current. It comes from the device characteristic and belongs to the required disconnection time.
Why is the rated current of the device not enough?
Because a miniature circuit breaker only trips magnetically at a multiple of its rated current. With a type C characteristic that is around ten times. A 16 A device therefore needs roughly 160 A of fault current to disconnect within the required time, not 16 A.
What does "tight" mean in the assessment?
That the condition is met on paper but without meaningful margin. The inputs are rarely exact. The upstream impedance varies with the state of the network, the conductor temperature with the load. Below 1.25 times the operating current you should measure rather than rely on the calculation.
What if the condition is not met?
You can increase the cross-section, shorten the cable, choose a device with a lower operating current or a different characteristic, or add a residual current device, which disconnects independently of the short-circuit current. Which of the four routes is right depends on the installation.
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