CT Saturation Calculations for Protection Relay Applications

CT Saturation Calculations for Protection Relay Applications

Current transformer (CT) saturation is the single most common cause of misoperation in numerical overcurrent, differential, and distance protection schemes. When a CT saturates, its secondary current waveform becomes clipped or distorted, leading to under-reach, over-reach, false differential currents, and failed breaker tripping. This technical reference walks through the analytical method — based on knee point voltage, internal resistance (R_ct), and connected burden — used by protection engineers to verify that a CT remains in its linear region during the maximum credible through-fault condition. The worked example uses a 1200/1 10P10 CT exposed to an 18 kA primary fault, terminated at a Siemens SIPROTEC 7SJ numerical overcurrent relay.

CT Protection Class Designations: 10P10, 5P20, and C-Class Explained

A protection CT is described by two standardized characteristics: an accuracy class letter (P or C, occasionally X or K) and a numerical accuracy limit factor or secondary voltage rating. Misreading these ratings is the leading source of CT misapplication.

A 1200/1 10P10 CT is therefore warranted by the manufacturer to deliver a 1 A secondary current with less than 10 % composite error up to 12 × 1200 = 14,400 A primary, provided the connected burden does not exceed the rated burden. An 18 kA fault exceeds the 10P10 design point by 4,800 A. Whether the CT survives this without unacceptable saturation is not answered by the class rating alone — the knee point voltage and full excitation curve must be examined.

Applicable Standards Governing CT Performance

Three documents govern the design, test, and reporting of CT saturation behavior:

• GE Vernova — CT Saturation Tolerance for 87L Applications: covers CT fundamental theory, AC and DC saturation, simulation models, and the factors driving a CT into saturation for line differential protection.
• Schweitzer Engineering Laboratories (SEL) — Beyond the Knee Point: A Practical Guide to CT Saturation: an in-depth treatment of the KPV method, including worked examples and excitation-curve interpretation.
• IEEE C57.13:2016 — Standard Requirements for Instrument Transformers: defines accuracy classes, reporting of excitation curves, and the requirement that manufacturers supply the 1-second short-time withstand rating of a CT with secondary shorted, expressed in primary amperes.

The Knee Point Voltage Method

The knee point voltage (KPV or V_knee) is defined by IEC 60255 as the secondary voltage at which a 10 % increase in applied voltage produces a 50 % increase in excitation current. Graphically, it is the "elbow" of the secondary V–I excitation curve plotted on log–log paper.

The fundamental current-transformer equivalent circuit is:

V_secondary = I_excitation × jX_m  // magnetizing branch
V_secondary = (I_secondary + I_excitation) × (R_ct + Z_external)

For a fault at the relay terminals, with primary current I_pri, the maximum expected secondary EMF before saturation begins is:

E_sec.max  =  V_knee − I_excitation × R_ct   ≈  V_knee   (ignoring the small magnetizing drop)
I_sec.max  =  V_knee / (R_ct + R_burden + R_lead)
I_pri.max  =  CTR × I_sec.max

Where:

• V_knee — knee point voltage from the manufacturer excitation curve (volts, RMS)
• R_ct — CT secondary winding DC resistance, measured or from the test report (ohms)
• R_burden — impedance of relay current input at operating frequency (ohms)
• R_lead — one-way loop resistance of the secondary cabling (ohms); use R_lead = 2 × ρ × L / A, taking the round-trip
• CTR — current transformer ratio (e.g. 1200:1 = 1200)

Once I_pri.max exceeds the available fault current, the CT begins to produce significant composite error; above approximately 1.5 × I_pri.max, the secondary waveform is heavily clipped and the relay may fail to operate.

Worked Example: 1200/1 10P10 CT at 18 kA Fault

Use the calculation method stated in the discussion between Kevin Bosch and the responding engineer: assume typical tap data, solve for the required CT primary current limit, then compare to the available 18 kA.

Given:

Step 1 — Required CT excitation data (must come from the manufacturer):

V_knee   ≈  18  V   (typical 10P10 1200/1 wound-primary CT, 1 sec rating basis)
R_ct     ≈  0.20 Ω  (1 A secondary — measured cold DC, not under excitation)

Step 2 — Combined secondary impedance:

Z_sec_total  =  R_ct + R_burden(7SJ) + R_lead
             =  0.20 Ω + 0.05 Ω   + 0.40 Ω
             =  0.65 Ω            (all values illustrative)

The Siemens 7SJ80x input impedance at 50 Hz is approximately 0.02 to 0.05 Ω per phase at rated current (per the SIPROTEC 5 manual), negligibly small compared with the cabling in a 1 A secondary circuit. The cabling therefore dominates.

Step 3 — Maximum linear secondary current:

I_sec.max  =  V_knee / Z_sec_total
           =  18 V / 0.65 Ω
           =  27.7 A_secondary

Step 4 — Equivalent linear primary current limit:

I_pri.max  =  CTR × I_sec.max
           =  1200 × 27.7 A
           ≈  33 250 A primary

Step 5 — Compare to available fault:

Saturation_margin  =  I_pri.max / I_fault_available
                   =  33 250 / 18 000
                   ≈  1.85 ×   —  CT remains largely linear at the first-cycle peak

A saturation margin above 1.3 typically indicates acceptable steady-state accuracy. Below 1.0 the CT will produce composite error in excess of 10 % and protective relay operating times may be delayed. The 1200/1 10P10 in this example is generously over-rated for an 18 kA primary fault, provided the cabling is kept short and the manufacturer's published V_knee is at least 18 V.

What if the burden increases? Re-running the same arithmetic with R_burden+R_lead=1.5 Ω (longer cable run, undersized conductors, or a junction box):

I_sec.max      =  18 V / (0.20 + 1.5) Ω = 10.6 A
I_pri.max      =  1200 × 10.6 = 12 720 A_primary
Sat_margin     =  12 720 / 18 000 = 0.71   —  CT now saturates severely

This is the practical reality of protection engineering: a CT suitable on paper can be rendered useless by the length or gauge of the secondary cabling.

Calculating Internal Resistance (Rct) and Reading Excitation Curves

Two CT parameters are critical and routinely missing in field installations: the DC secondary winding resistance R_ct, and the points on the V–I excitation curve that bracket the operating point. They are always provided on the manufacturer's approval drawing package.

Measuring Rct in the field

1. Isolate the CT from all secondary wiring and from the relay.
2. Short the secondary with a calibrated low-resistance bridge lead at the CT terminals.
3. Measure across the short with a 4-wire Kelvin bridge or a micro-ohmmeter at DC.
4. Record ambient temperature; correct to 75 °C using R₇₅ = R_amb × (234.5 + 75)/(234.5 + T_amb) (copper constant 234.5).

For 1 A secondaries of 1200 A ratios, expected R_ct is generally 0.1 to 0.5 Ω for measuring-class types and up to 1.5 Ω for protection-class iron-core designs. A 200:5 (40:1) CT would typically have R_ct an order of magnitude lower.

Reading a published excitation curve

Excitation curves are plotted on log–log axes. Locate the elbow using a 45 ° tangent: V_knee is at the intersection. Read the excitation current I_e at V_knee — this value times 0.1 approximates the magnetizing drop contribution to error. Use the curve to find E at the desired secondary EMF, then evaluate percent error:

%Error  ≈  (I_e / I_total) × 100,   where  I_total  =  I_sec + I_e

A composite error of 10 % at the operating point corresponds to V_operating ≈ 0.78 × V_knee; at V_knee, error is roughly 30 to 50 %. The 10P10 point is therefore meaningfully below V_knee.

Burden Analysis for Siemens 7SJ Relays

The SIPROTEC 7SJ80 series (7SJ801, 7SJ802, 7SJ803, 7SJ804) presents a very small burden to the CT secondary, typically 0.02 to 0.05 VA at rated current per phase for the 1 A variant. Converted to impedance:

Round-trip cable resistance

For copper conductors at 20 °C:

R_loop  =  2 × ρ × L / A   =   0.0353 Ω/(m·mm⁻²) × L/A   for round trip

Sample values, 1 A secondary CT:

AC vs DC (Transient) Saturation

The calculation above evaluates symmetrical steady-state behavior. Two further phenomena must be considered for completeness:

AC saturation

Repeated peak flux density is determined by V_sec alone. If V_sec exceeds 0.78 × V_knee at any peak, the secondary waveform becomes clipped on alternate half-cycles. Numerical relays sample at 16 to 64 samples/cycle; even small clipping will produce measurable RMS error.

DC (transient) saturation

During the first 30 to 100 ms after a bolted fault, primary current is offset by an asymmetric component:

i_pri(t)  =  I_peak_sym  ×  ( e^(−t/τ_pri)  ×  cos(ωt + φ)  −  cos(φ)  ×  e^(−t/τ_pri) )

The X/R ratio of the upstream network determines the DC time constant τ_pri. A fully-offset fault can drive the CT flux to 2 × peak flux (per IEEE C37.91). For 7SJ relays, transient saturation is generally tolerated because the relay's definite-time and IDMT elements have intentional coordination time delays of 100–300 ms; DC offset has largely decayed by then.

Antuokzio's rule of thumb for DC offset

V_transient  ≈  V_sec_sym  ×  (1 + X/R)

If V_transient > V_knee, apply an "S" (or "K" per IEC 60044-6) rated CT, or switch to a non-cores-balance scheme. The S-class CT has a controlled remenance flux below 10 % of saturation flux, achieved through annealing and a gap in the core.

Commissioning Verification Procedure

1. Pre-energization. Verify CT polarity markings (P1 towards source, P2 towards load) match the relay wiring diagram; verify the star-point ground at the relay panel only, never at the CT.
2. Secondary loop check. With the CT secondary shorted to ground at the CT terminal block, measure loop resistance from the relay panel. Confirm against the calculation. Open-circuit is forbidden.
3. Polarity and ratio test. Inject from a primary injection set (or use the burden's secondary injection). Confirm ratio at 20 %, 50 %, 100 % rated current — record deviation in %.
4. Excitation curve verification. With CT secondary disconnected from the relay, apply 50 Hz AC to the secondary through a variac. Capture the V vs I curve at 0.25 A, 0.5 A, 1 A, 2 A, 5 A, and 10 A excitation. Compare to the manufacturer's factory curve. Should match within 10 %.
5. Primary injection at relay setting. Inject the 50 % and 100 % of fault current at the relay input terminals. Verify pickup, operating time, and direction discrimination.
6. Three-phase fault simulation. If a primary injection set is available, perform a true three-phase primary fault and capture relay COMTRADE files; verify the CT and relay report the correct ratio and angles.
7. Documentation. Issue a Protection Settings Report, signed-off test sheets, and an "as-left" excitation curve.

Acceptance criteria

Troubleshooting CT Saturation in Service

If the 7SJ relay records misoperation events that point to the CT (slow trip, no trip on close-in fault, unexpected restraint in differential), check the items below in priority order:

Symptom matrix

Diagnostic and Calculation Tools

• ETAP Current Transformer Sizing & Saturation module — automated KPV-based check, multi-fleet library, side-by-side CT comparison.
• GE Vernova CT Saturation Tolerance for 87L Applications (PDF) — canonical reference covering AC/DC saturation, relay models, and 87L-specific reporting.
• SEL — Beyond the Knee Point: A Practical Guide to CT Saturation (PDF) — definitive treatment of the KPV method, worked relay-grade examples.

Quick-Reference: Sizing the CT for a New Installation

1. Determine I_{fault,max,pri} at the relay location (including DC offset contribution V_transient).
2. Calculate Z_sec,total = R_ct + R_burden,relay + R_{burden,lead,loop}.
3. Compute V_sec,max = I_{fault,max,sec} × Z_sec,total.
4. Require V_{knee,manufacturer} ≥ 1.5 × V_sec,max. (1.5 is the typical factor including transient DC overshoot.)
5. If requirement fails: choose (a) a higher-ratio CT, (b) a larger cable cross-section, (c) a higher-V_knee CT, or (d) an S-class CT for X/R > 17 networks.
6. Document V_knee, R_ct, full excitation curve, rated burden, short-time current rating, and S-class certification in the approval package.

Frequently Asked Questions