CT Transient Saturation (IEEE C37.110)

A current transformer (CT) makes a small replica of a large primary current, for example 5 A for every 1200 A. Every differential, distance and directional element in the collector substation depends on that replica. To drive secondary current through the relay, leads and its own winding, the CT has to develop a voltage. That voltage comes from changing magnetic flux in the core.

A core can only hold so much flux. Past the saturation point the magnetising branch stops behaving like a high impedance and starts absorbing most of the current. The secondary waveform collapses into flat-topped or gapped pulses, while the primary keeps flowing. The relay sees a current that is smaller, distorted and phase-shifted, usually at exactly the moment it most needs an accurate one.

The flux a CT needs is proportional to the time-integral of the secondary voltage, $\lambda = \int R\,i_s\,dt$. A symmetrical sine wave integrates to a flux that swings evenly up and down. A fault with a decaying DC offset integrates to a flux that keeps climbing for several cycles. That one fact drives this whole module.

When a fault starts at an unfavourable point on the voltage wave, the current must start from zero but settle to a sinusoid. The difference is a DC component that decays with the primary time constant:

Transmission-level sources have X/R of roughly 10–20, giving τ of 26–53 ms (1.6–3.2 cycles). Distribution feeders sit nearer X/R 3–6. Near a large main power transformer, at HV bushing CTs and close to generators, X/R of 30–50 or more is normal (τ of 80–130 ms), and those are the worst saturation cases. If the CT is to reproduce the whole offset wave without saturating, its core must support the AC flux plus the integral of the DC term. The bound on that total is $(1+X/R)$ times the symmetrical flux.

The C37.110 screening test for a C-class CT, with every quantity in per-unit, is:

Here $I_f$ is the maximum symmetrical fault current in per-unit of the CT rating (on the tap in use) and $Z_b$ is the total secondary burden in per-unit of the standard burden for the CT's C-class. A C400 CT is rated to deliver 20 × 5 A = 100 A into a 4 Ω standard burden (B-4.0) with at most 10% ratio error, so its standard burden is 4 Ω. This lab counts the CT winding resistance in the loop, which is conservative. Some practitioners include it in the burden and others in the excitation voltage. Be consistent within a study. (The lab's screen counts Rct as burden, while its voltage test credits Rct in $V_{sat}$, so near the boundary the two checks can disagree. Treat either failure as a fail.)

The physical burden is the series loop the secondary current has to drive:

Worked example

Here is a collector substation breaker CT example: maximum external fault 25 kA symmetrical, X/R = 12, 60 Hz, 1200:5 on full ratio, $R_{ct}=0.4\,\Omega$ (representative for a 1200:5 relaying winding; use the manufacturer's value), 50 ft of 10 AWG copper at 75 °C (1.24 Ω per 1000 ft), and a 0.5 VA relay. Find the required saturation voltage and the minimum C-class.

The lab solves the CT equivalent circuit in the time domain. Watch three things together: the secondary current against the ideal ratio current, the core flux climbing toward ±λsat, and the excitation curve against the voltage the burden requires. The preset Saturating CT is a representative feeder CT on its 600:5 tap with 100 ft leads, a shared metering core and 60% remanence. Apply the remedies one at a time and note which ones matter most.

A CT core interrupted near peak flux, or magnetised by DC winding-resistance or polarity testing, can keep a large remanent flux. IEEE C37.110 cites a field survey in which a meaningful fraction of the CTs tested retained high remanence, some in the 61–80% range of saturation flux (check the guide for the exact figures). Treat that range as a worst case, not a typical value. If the remanence sits in the same direction as the next fault's DC offset, the core starts most of the way to saturation.

• Gapless C-class cores keep remanence almost indefinitely in service. Normal load current does not reliably demagnetise them.
• High-speed reclosing into a permanent fault is the classic worst case: remanence from the first trip and fresh DC offset add together.
• Demagnetise CTs after DC tests (winding resistance, polarity checks) as part of commissioning procedure.

34.5 kV collector buses often use high-impedance bus differential. Every CT in the zone is paralleled at a junction point, and a high-impedance voltage element measures across the junction. For an external fault the scheme assumes the worst case outright: the CT on the faulted feeder saturates completely. Its magnetising branch is then a short, so the healthy CTs drive the whole secondary fault current through its winding and leads, and the relay sees

Here $I_F$ is the maximum external fault current in secondary amps, $R_{CT}$ and $R_{lead}$ belong to the saturated CT and its run to the junction, and $K$ is the manufacturer's margin factor (often 1.5–2). Because the calculation assumes total saturation, it needs no X/R or remanence. That is why the scheme is so secure.

Representative example: 25 kA maximum external fault, 2000:5 CTs on full ratio ($I_F = 62.5$ A), R_CT = 0.8 Ω and 300 ft of 10 AWG to the junction ($2R_{lead} = 0.744\ \Omega$ at 75 °C). Then $V_R = 62.5 \times 1.544 = 96.5$ V, and with K = 1.5 the setting is about 145 V.

• Identical ratios and full taps on every CT in the zone. Auxiliary CTs and tapped windings add impedance and break the assumption.
• Dedicated cores with nothing else in the circuit.
• A knee well above the setting. A common rule is a knee voltage of at least about 2 × V_s, so the CTs can drive the relay into operation for internal faults. Use the relay manual's criterion.
• A voltage limiter (MOV) across the relay. For an internal fault the CTs try to drive current into a high impedance, and the peak voltage can reach several kilovolts.
• Check sensitivity. The primary pickup is the CT ratio times the sum of the relay current, every CT's excitation current at V_s, and the MOV leakage. It must sit below the minimum internal fault.

At an inverter-based plant the bigger CT problem is often sensitivity, not saturation. HV CTs are sized for the grid's fault duty, but the plant's own contribution is set by the inverter current limit: typically about 1.0–1.5 pu of rating (from the OEM's model), after a brief sub-cycle transient.

Representative example: a 100 MVA plant (100 MW at unity power factor) at 230 kV has a rated current of about 251 A. At 1.1–1.5 pu it contributes roughly 280–380 A primary. On a 2000:5 CT chosen for 40 kA of grid duty that is only 0.7–0.9 A secondary.

• Relay minimum pickups and the CT's accuracy at low flux now govern. A C-class rating says nothing about either.
• The usual fix for saturation, a higher (full) ratio, pushes the secondary current down further. A higher C-class on the same ratio doesn't change the secondary current, though higher C-classes often come with higher ratios. Sensitive elements may need a different core or ratio, a relay with wide dynamic range, or sequence- and voltage-based elements.
• Ground faults are different: through the YNd main power transformer the plant end is a strong zero-sequence source (Module 4.1), so the shortage is mainly in phase and negative-sequence current.

1. Use full-ratio taps on multi-ratio CTs for relaying wherever the load and pickup settings allow. A lower tap cuts the C-class voltage in proportion and raises secondary current at the same time.
2. Dedicate cores. Give 87T and 87B their own relaying cores and never share them with revenue metering, transducers or panel meters.
3. Specify 1 A secondaries in physically large switchyards and collector sites with long cable runs. Lead burden in per-unit falls by about 25×, but the winding resistance rises by about the same factor (5× the turns of thinner wire), so the net gain depends on how much of the loop is leads. The lab also gives the 1 A design 5× the C-class voltage, which holds for the same core with 5× the turns. How 1 A relaying CTs are rated and specified varies, so confirm the excitation data with the manufacturer.
4. Match CTs within a differential zone where possible (similar C-class and burden), so any saturation is more symmetrical.
5. Screen every protection CT with $(1+X/R)I_fZ_b \le 20$, then check time-to-saturation against the relay's requirement where it fails. Many numerical relay manuals state a minimum saturation-free time; the figure is vendor-specific, so take it from the manual.
6. Demagnetise after DC tests, and consider remanence for reclosing duty.