CT sizing calculator: knee point and 5P10 / 5P20 check (IEC 61869-2)

Tool · Protection

CT sizing calculator

Check if a protection CT is big enough for your fault current. Enter the nameplate (like 400/1 A 5P20 15 VA), the relay and the cable. The tool finds the real accuracy limit factor and the knee point voltage, per IEC 61869-2.

CT nameplate
Relay and cable
Fault

Real ALF with your burden
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CT limit voltage Eal
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Voltage needed
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Meet the CT: from the substation to the core

Follow the four steps to see where a CT sits and what is inside it.

Incoming line1Busbar2Disconnector3Circuit breaker4CTring around the conductor5Outgoing line6Relay panel7CT wires in cable trench8Conductorthis is the CT primary1CT ringno contact with the conductor2Secondary wiresgo to the relay panel3400/1 A5P20 · 15 VAcut along theconductorS1S2Primary conductorline current flows through1Resin casing2Insulation3Secondary windingcopper turns, seen cut4Iron corethin steel strip, wound5Magnetic fluxout on top, in below6Terminals S1, S2to the relay7

    Step-by-step working

    How to use this calculator

    1. Copy the CT nameplate: ratio, class (5P or 10P), accuracy limit factor and rated burden. Take the winding resistance Rct from the CT test report.
    2. Enter the relay burden from the relay manual, and the cable from the CT to the relay.
    3. Enter the highest fault current that flows through the CT. You can find it with our fault current calculator.

    The tool tells you if the CT stays accurate up to that fault current. If it does not, it shows by how much it falls short.

    What 5P20 and 15 VA mean

    A protection CT nameplate like 400/1 A 5P20 15 VA has four parts:

    • 400/1 A is the ratio. 400 A in the primary gives 1 A in the secondary.
    • 5P is the class. P means protection. The composite error stays below 5% up to the accuracy limit.
    • 20 is the accuracy limit factor (ALF). The error limit holds up to 20 times rated current, so up to 8000 A here.
    • 15 VA is the rated burden. The promise above is only true when the connected burden is 15 VA or less.
    ClassRatio error at rated currentPhase error at rated currentComposite error at ALF × rated current
    5P±1%±60 minutes5%
    10P±3%not specified10%

    Source: IEC 61869-2. Class PX CTs are specified differently, by knee point voltage and winding resistance, and are used for differential protection.

    The formulas

    Rb = rated VA ÷ Isn²
    Ractual = Rrelay + ρ × L × loop ÷ A
    Eal = ALF × Isn × (Rct + Rb)
    ALF′ = ALF × (Rct + Rb) ÷ (Rct + Ractual)

    Eal is the secondary limiting EMF: the voltage the CT can produce before its error gets too big. The knee point voltage of a class P CT is roughly 0.8 × Eal. ALF′ is the real accuracy limit factor with your actual burden. A light burden makes ALF′ much bigger than the nameplate ALF.

    Worked example

    A 400/1 A 5P20 15 VA CT has Rct = 2 Ω. It feeds a numerical relay (0.1 VA) through 50 m of 2.5 mm² copper. The maximum fault is 10,000 A.

    • Rated burden: Rb = 15 ÷ 1² = 15 Ω.
    • Cable at 75 °C, out and back: 0.02097 × 100 ÷ 2.5 = 0.839 Ω. Relay: 0.1 Ω. Actual burden = 0.939 Ω.
    • Eal = 20 × 1 × (2 + 15) = 340 V.
    • Needed: (10,000 ÷ 400) × 1 × (2 + 0.939) = 25 × 2.939 = 73.5 V.
    • 340 V is more than 73.5 V, so the CT is fine. Its real ALF is 20 × 17 ÷ 2.939 = 115.7.

    Now take a 400/5 A 5P10 15 VA CT with Rct = 0.3 Ω and 100 m of 2.5 mm² cable. The cable alone is 1.68 Ω, almost three times the rated burden of 0.6 Ω. The real ALF drops to 4.5, but the fault needs 25. This CT saturates. This is why 1 A secondaries are preferred when the cable is long.

    Safety

    Never open the secondary of a CT while primary current flows. With no secondary current to oppose it, the core saturates and very high voltage appears at the terminals. Always short the secondary at the test block first.

    Questions people ask

    What is the knee point voltage of a CT?

    It is the point on the excitation curve where a 10% rise in voltage needs a 50% rise in magnetising current. Above it, the core saturates and the CT stops copying the primary current well.

    Should I choose a 1 A or 5 A secondary?

    For long cable runs, choose 1 A. Cable burden grows with current squared, so a 5 A CT sees 25 times the cable burden in VA for the same cable.

    What can I do if the CT fails the check?

    Use a bigger cable, shorten the cable, choose a 1 A secondary, choose a higher ALF or rated burden, or choose a higher CT ratio. A relay with a lower burden also helps.

    What is the extra factor K?

    Some relays need the CT to stay accurate during the DC offset at the start of a fault. Differential and distance relays often ask for a factor of 2 or more. The relay manual gives the value. For simple overcurrent relays, K = 1 is common.

    Is the relay burden in VA or ohms?

    Relay manuals usually give VA at rated current. The tool turns it into ohms with R = VA ÷ Isn².

    Related

    What 5P10 and 5P20 meanThe CT class in plain words Fault current calculatorFind the fault current to enter here IDMT relay calculatorCheck relay trip times

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