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Current transformer saturation

Compare the voltage a protection CT would have to develop during an asymmetrical fault with what its class allows, estimate the time to saturation and see how the secondary current and the core flux distort.

IEEE C37.1101 + X/R criterion and time to saturationIEC 61869-2§ 5.6.202protection CTs, classes P, PR and PXIEEE C57.13class C relaying accuracy

Example cases
Current transformer saturation

Only the basic data; the rest is taken with typical values.

E1Current transformer
A
VA

20 for a 5P20.

V

Terminal voltage at 20 × In with the standard burden.

A

Only when you use a tap of a multi-ratio CT: the class drops in proportion. Empty = same ratio.

Ω
E2Fault
%
E3Connected burden
Ω

If you do not know the lead resistance, calculate it with the CT burden tool. For a ground fault on wye-connected CTs, count out and back.

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

Required voltage

— V

Available voltage

— V

Time to saturation

— ms

Factor Ks(1 − λ)

—

Secondary fault current

— A

Maximum fault without saturation

— kA

Secondary current
  • With saturation
  • Ideal, Iprim/CTR
  • 1st simulated saturation
  • ts from the formula
Full asymmetry, resistive burden and an ideal saturating core. Vertical divisions are cycles. The grey line marks the IEEE C37.110 ts, which is conservative; the yellow one, the first saturation in the simulation.
Core flux
  • Actual flux
  • Flux the fault would demand
  • ±ψsat
In per unit of ψsat = √2 · Vavail / ω. It starts at the remanent flux; where the dashed line goes above 1, the core can no longer follow.

Step-by-step calculation

What it means for the protection

    What it is for

    A saturated CT delivers less current than flows in the primary, and with a distorted shape. That can delay a 51, shorten the reach of a 21 or make a differential element operate on an external fault. Use this tool to check in minutes whether the chosen CT withstands the maximum fault with its actual burden, to compare cores or classes before buying, and to understand an event report where the secondary current «flattens» after a few milliseconds.

    How it is calculated

    Secondary current and burden

    If = If,prim / CTR · Zs = RCT + Rleads + Zrelay

    With Zrelay = VA / In² if the relay burden is given in VA. A resistive burden is assumed.

    IEEE C37.110criterion to avoid saturation

    Vs = If · Zs · (1 + X/R) / (1 − λ)

    It is the voltage the CT needs to avoid saturation with full asymmetry. With remanent flux λ in the same polarity as the DC component, the core only has (1 − λ) of its flux available.

    IEC 61869-2§ 5.6.202classes P and PXIEEE C57.13class C

    Available voltage by class

    • Class P: accuracy limiting EMF EALF = ALF · In · |RCT + Zb|, with Zb = rated burden / In² at cos φ = 0.8 (cos φ = 1 below 5 VA). The sum is a phasor sum: adding arithmetically overestimates the voltage.
    • Class PX: the rated knee-point voltage Ek. It is a conservative value, because full saturation is somewhat higher.
    • Class C (IEEE C57.13): VC is the terminal voltage at 20 · In with the standard burden, power factor 0.5. The EMF is Vx = 20 · In · |RCT + Zstd∠60°|, with Zstd = VC / (20 · In). On a tap, VC is multiplied by the ratio used over the full ratio.

    IEEE C37.110time to saturation, resistive burden

    Ks = Vavail / (If · Zs) · ts = −(X/R)/ω · ln(1 − (Ks(1 − λ) − 1) / (X/R))

    If Ks(1 − λ) < 1, the CT saturates even with symmetrical current. If it is equal to or greater than 1 + X/R, it does not saturate. The formula takes the peak AC flux from the first instant, so it is conservative: with a small Ks the simulation gives a longer time, and the tool shows both.

    Waveform and flux

    i(t) = √2 · If · (e−t/T − cos ωt) · ψ = ∫ Zs · i dt

    Here T = (X/R)/ω. The flux starts at λ · ψsat, with ψsat = √2 · Vavail / ω. While the flux sits at the limit and the current pushes it outward, the output is taken as zero. It is the simplified ideal saturating core model and only serves to illustrate the waveform.

    Limits of the calculation

    • It is an estimate, not a study. It assumes a resistive burden, full asymmetry, remanence in the unfavourable polarity and a core that switches abruptly from unsaturated to saturated.
    • A real CT saturates gradually along its excitation curve. The IEEE PSRC model, which uses that curve, gives smoother waveforms.
    • It takes the knee-point voltage (PX), the accuracy limiting EMF (P) or the 10 % error voltage (C) as full saturation, which keeps the result on the safe side.
    • A saturating CT does not by itself mean the relay will misoperate. Many numerical relays tolerate a given time to saturation, and each manufacturer publishes the CT requirement of its relay, such as a minimum time to saturation or a dimensioning factor.
    • Confirm the data with the CT test report (winding resistance and knee point) and with the relay manual.

    Frequently asked questions

    How do you know whether a current transformer saturates during a fault?

    Compare the voltage the CT would have to develop with the voltage it can deliver. The voltage required to avoid saturation with full asymmetry is Vs = If · (Rct + Rb) · (1 + X/R), with If in secondary amperes (IEEE C37.110 criterion). If it is higher than the CT saturation or knee-point voltage, or than the accuracy limiting EMF of a class P CT (IEC 61869-2), the CT will saturate at some point during the fault.

    How long does a CT take to saturate?

    It depends on how far the CT is from meeting the criterion. With the factor Ks = Vsat / (If · Zs), the IEEE C37.110 expression is ts = −(X/R)/ω · ln(1 − (Ks − 1)/(X/R)). It is conservative: it assumes the peak AC flux from the first instant, so the real CT usually lasts somewhat longer. If Ks is below 1 the CT saturates even with symmetrical current; if Ks is equal to or greater than 1 + X/R it does not saturate without remanence.

    What does remanent flux do to CT saturation?

    Remanent flux takes up part of the core margin. If it has the same polarity as the DC component of the fault, the CT only has (1 − remanence) of its saturation flux available, so it saturates sooner. Closed-core CTs can retain high remanence, around 80 %; gapped CTs, such as IEC 61869-2 class PR, limit it.

    We test your CTs on site

    We measure the excitation curve, knee point, winding resistance and actual secondary burden, and check against the protection study whether the CT withstands the maximum fault. Field work in Colombia, or elsewhere on request.

    Request a quoteSee design and studies

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