sparkytools

Fourth year 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.

Transformer
  • Secondary full-load current—
  • Multiplier (100 ÷ %Z)—
  • Fault current at the secondary terminals (infinite bus)—
Show the working

    Worst-case estimate: the real fault current is lower once the utility and conductor impedance are counted. Equipment ratings and the study itself come from your code book and the engineer of record.

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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

    1. FLA = 500,000 ÷ (1.732 × 480) = 601 A.
    2. Multiplier = 100 ÷ 5.75 = 17.39.
    3. Isc = 601 × 17.39 = about 10,460 A at the terminals, before any conductor or utility impedance.

    Questions people ask

    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.

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