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Short-circuit current at the secondary of a transformer

Estimate the bolted three-phase fault at the secondary terminals with IEC 60909-0 or with the simple estimate, with or without the upstream source, and see the peak current with full asymmetry. It does not replace a short-circuit study.

IEC 60909-0§ 6.3.3transformer KT correctionIEC 60909-0voltage factor c, table 1IEC 60909-0peak current, factor κ

Example cases
Short-circuit current at the secondary of a transformer
E1Transformer
kVA

From the nameplate.

V

Line to line.

%

From the nameplate, on the rated power base.

Only for the peak current. Use the value from the test report.

E2Upstream source
MVA

Given by the utility at the point of connection.

E3Method

Up to 1 kV; above it, c = 1.10.

Only for the waveform.

Use a decimal point. The calculation runs in your browser and updates as you type.

Three-phase fault I″k

— kA

Peak current ip

— kA

Short-circuit power

— MVA

Rated current

— A

Times rated current

—

IEC 60909 versus simple

— %

Fault current versus source strength
  • IEC 60909-0
  • Simple estimate
  • Infinite bus, IEC
  • Selected source
Source short-circuit power on a logarithmic scale. As the source gets stronger, the current approaches the infinite-bus value, limited only by the transformer.
Current with full asymmetry
  • i(t) with DC component
  • Symmetrical
  • Peak ip
Fault at the voltage zero crossing, with the X/R entered. The vertical divisions are cycles. The peak of the curve matches κ · √2 · I″k within 2 %.

Step-by-step calculation

What it means for your equipment

    What it is for

    The three-phase fault at the secondary terminals is the first figure you need to choose the interrupting rating of a switchboard, check an instantaneous setting or size busbars. This tool gives a quick estimate with the IEC 60909-0 method and compares it with the simple estimate many engineers use from memory, so you can see how far short it falls.

    How it is calculated

    Rated current

    In = Sn / (√3 · Un)

    IEC 60909-0§ 6.3.3KT correctionIEC 60909-0factor c, table 1

    IEC 60909-0 method (maximum initial symmetrical current)

    I″k = cmax · In / (KT · uk + zQ) · KT = 0.95 · cmax / (1 + 0.6 · xT)

    With uk = uk% / 100. The voltage factor cmax is 1.05 or 1.10 at low voltage depending on the system tolerance, and 1.10 above 1 kV. Here xT = uk, which gives the highest current. The upstream source is modelled as ZQ = cQ · UnQ² / S″kQ, which in per unit on the transformer base becomes zQ = cQ · Sn / S″kQ, with cQ = 1.10. With an infinite bus, zQ = 0. Since cmax / KT = (1 + 0.6 · xT) / 0.95, on an infinite bus the result does not depend on cmax.

    Simple estimate

    Ik = In / (uk + Sn / S″kQ)

    Without the c factor or KT, it gives 5 % to 10 % less than IEC 60909, so it is not suitable for choosing the interrupting rating. The short-circuit power is S″k = √3 · Un · I″k in both methods.

    IEC 60909-0peak current

    ip = κ · √2 · I″k · κ = 1.02 + 0.98 · e−3R/X

    The waveform on the screen is i(t) = √2 · I″k · (e−t/T − cos ωt), with T = (X/R) / ω: the fault that starts at the voltage zero crossing, which has the largest asymmetry.

    Limits of the calculation

    • Bolted three-phase fault at the secondary terminals, with the transformer on the nominal tap.
    • Impedances added in magnitude, without separating resistance and reactance; the X/R is used only for the peak current.
    • No contribution from motors, generators or inverters on the low-voltage side, and no cable or busbar impedance up to the fault, so the value is on the high side at the terminals.
    • A study per IEC 60909-0 (or IEEE 551) models the whole network, separates R and X, adds the contributions of rotating machines and converter-based generation, and also calculates the unbalanced faults. That is the result to use for specifying equipment and setting protection.

    Frequently asked questions

    How is the short-circuit current at a transformer secondary calculated?

    The simple estimate divides the rated current by the per-unit impedance: Ik = In / (uk% / 100). IEC 60909-0 adds the voltage factor c and the K_T correction of the transformer impedance, which gives a higher value. For example, 1000 kVA, 440 V and 5.75 % give 22.8 kA with the simple estimate and 24.9 kA with IEC 60909 on an infinite bus.

    What is the IEC 60909 K_T factor?

    It is the impedance correction of a network transformer for the maximum-current calculation: K_T = 0.95 · c_max / (1 + 0.6 · x_T), with x_T the relative reactance. Since K_T is below 1 for usual transformers, it lowers the impedance and raises the calculated fault current.

    What is the peak current ip and what is it used for?

    It is the maximum instantaneous value of the fault current, DC component included: ip = κ · √2 · I″k, with κ = 1.02 + 0.98 · e^(−3R/X) per IEC 60909-0. It is used to check the making capacity of circuit breakers and the dynamic forces on busbars and supports.

    We carry out your short-circuit study

    We model your system with the utility data and your equipment nameplates, calculate the faults per IEC 60909 or IEEE 551, and deliver the interrupting-rating check and the protection coordination.

    See design and studiesSee the coordination study

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