Cuarto año Protection · tool
Transformer fault current (infinite-bus estimate)
Enter the transformer kVA, secondary voltage and nameplate %Z. The tool gives the full-load current and the most current a bolted fault at the secondary terminals could draw if the supply behind it were infinitely stiff.
- Secundario: corriente a plena carga—
- Multiplicador (100 ÷ %Z)—
- Corriente de falla en las terminales del secundario (barra infinita)—
Mostrar el cálculo
Estimación del peor caso: la corriente de falla real es menor una vez que se cuentan la impedancia de la red de suministro y la de los conductores. Las capacidades nominales de los equipos y el estudio mismo salen de su libro de código y del ingeniero responsable del proyecto.
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How it works
- FLA = kVA × 1000 ÷ (√3 × V)three-phase (÷ V alone for single-phase)
- Multiplier = 100 ÷ %Ztimes full-load current
- Isc = FLA × 100 ÷ %Zbolted fault, infinite bus
Worked: 500 kVA, 480 V, 5.75 %Z
- FLA = 500,000 ÷ (1.732 × 480) = 601 A.
- Multiplier = 100 ÷ 5.75 = 17.39.
- Isc = 601 × 17.39 = about 10,460 A at the terminals, before any conductor or utility impedance.
Preguntas frecuentes
What does "infinite bus" mean?
It assumes the utility behind the transformer is infinitely strong — it can supply any current without its voltage sagging — so the only thing limiting a fault is the transformer’s own impedance. Real supplies are not infinite, so the real fault current is lower. That makes this the worst case at the transformer terminals, which is the useful number for a first look.
Why does a lower %Z give a bigger fault current?
%Z is the percentage of rated voltage it takes to push full-load current through the transformer with its secondary shorted. A 5% transformer reaches full-load current with only 5% voltage applied, so at 100% voltage it would pass 100 ÷ 5 = 20 times full-load current into a bolted fault. Halve the %Z and you double the fault current.
Why is there a tolerance field?
The %Z on the plate is a measured or design value with a manufacturing tolerance — the actual unit may be a little lower. Some engineers allow for that by using a slightly lower %Z, which gives a slightly higher (more conservative) fault current. Enter the tolerance you have been told to use, or leave it at 0.
Is this the number for the panel down the hall?
No. Every metre or foot of conductor adds impedance and lowers the fault current further downstream. Point-to-point methods and study software add those impedances in. The equipment interrupting and withstand ratings needed at each point are an engineering and code-book question.
Ver también
- Available fault current — the lesson behind this tool
- Transformador: corriente a plena carga y relación
- Overcurrent protection and coordination
- Protective relaying concepts
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