CT saturation
The key idea
A current transformer must push its secondary current through the relay burden, and that takes volt-seconds — flux in the core. A fault with a DC offset drives the flux one way with every cycle instead of letting it swing evenly about zero. When the core is full, the CT stops transforming, and the relay stops seeing the very fault it exists to catch.
The idea
A CT is a transformer, and a transformer works by flux. To drive current through the relay and the wiring, the secondary winding must produce a voltage across that burden, and a voltage across a winding builds flux in the core. The flux at any instant is the running integral of that voltage: the total volt-seconds so far.
A clean symmetrical wave keeps that integral small. The current goes equally positive and negative, so the integral swings equally about zero and returns every half cycle. The core never holds more than one half cycle's worth of volt-seconds.
But a fault is not a clean symmetrical wave. As the X/R ratio page shows, it arrives with a DC offset underneath. The offset has one sign only, so its contribution to the integral never comes back. The flux climbs in one direction, cycle after cycle, for as long as the offset lasts. A fully offset fault needs roughly (1 + X/R) times the flux of a symmetrical one, and at X/R = 20, no core of a sensible size can hold that much.
When the core reaches its saturation flux, it simply cannot accept more. The magnetizing branch swallows the primary current instead of passing it to the secondary, and the current at the relay collapses toward zero. The CT recovers only when the wave swings back far enough to remove some flux. What the relay receives is a chopped remnant of the real current — no peaks, with long flat gaps between the pieces that survive.
Three things push a CT toward that limit: more burden (more voltage for the same current, so the volt-seconds pile up faster), more offset (a higher X/R, so the one-way climb lasts longer), and a smaller core (less room for flux in the first place).
Try it
Start at the defaults and look for the point where the solid trace peels away from the dashed one. Then reduce the burden, or grow the core, until the solid trace follows the dashed one for the full five cycles.
saturated at 8 ms · RMS error 82%
- ideal secondary: the current a perfect CT would copy
- what the relay receives
- saturated: the core is on its limit and the CT does not transform
The core reached its flux limit. This model uses an ideal core. The CT copies perfectly until the running integral of the burden voltage reaches the saturation flux. The secondary current then falls to zero, until the wave reverses far enough to reduce the flux. A real CT has a rounded knee, and it can start with remanent flux from the last fault. It therefore saturates earlier and less sharply than this model.
Why it matters
- The relay underreaches. A saturated CT delivers less RMS current than the fault really carries, so an overcurrent element measures a smaller fault: it slides further out on its time-current curve and trips late. Distort the wave badly enough and the element never picks up at all.
- Differential schemes can misoperate. If one CT in a differential zone saturates and its partner does not, the two currents stop matching, and that mismatch looks exactly like an internal fault. This is why differential relays carry harmonic restraint and saturation-detection logic.
- Burden is a design number. Lead length, relay input impedance and the CT's own winding resistance all add up, and doubling the burden doubles the volt-seconds the core must supply for the same fault current.
- The sizing rule uses X/R directly. The CT's knee-point voltage must exceed the symmetrical requirement multiplied by (1 + X/R), a factor that comes straight out of the same short-circuit study that produced the fault current.
The math, if you want itOptional — the page reads completely without it
A fully offset fault contains the one-sided term the X/R page derives. In per unit of the symmetrical peak, which is what the widget plots:
the offset secondary current
i(t) = e−t/τ − cos(ωt) · τ = X/Rω
The core flux is the running integral of the voltage the CT must produce across the burden. For a resistive burden Rb that voltage is simply Rb·i, so:
flux is volt-seconds
λ(t) = ∫0t Rb · i dt · saturation when |λ| ≥ λsat
Integrate the two terms separately. The cosine integrates to a sine of amplitude 1/ω, the symmetrical requirement. The offset integrates to τ = X/(ωR), which is X/R times bigger. Their sum is the standard sizing factor:
the flux-doubling rule, approximately
λoffsetλsym ≈ 1 + X/R
The factor is an approximation: it assumes a fully offset fault, ignores the flux the negative half cycles remove, and takes no account of remanence. Relay engineers therefore specify a CT accuracy class and a maximum burden together: the class fixes λsat, the burden fixes how fast the CT spends it, and the sizing factor says how much it needs.
See it in Phasor
Phasor holds the burden of each CT circuit and the X/R at each fault location, so it can identify a CT that saturates before the relay downstream of it gets to decide. The same short-circuit results that set the relay curves also set the sizing factor the CT must meet.