CT saturation
The key idea
A current transformer (CT) must push its secondary current through the relay burden. That action takes volt-seconds, which is flux in the core. A fault with a DC offset makes the flux increase in one direction with every cycle, so the flux no longer swings equally about zero. When the core is full, the CT no longer transforms the current. The relay then does not see the fault that it protects against.
The idea
A CT is a transformer, and a transformer works by flux. The secondary winding must produce a voltage across that burden to drive current through the relay and the wiring. A voltage across a winding builds flux in the core. The flux at any instant equals the running integral of that voltage. That integral is the total volt-seconds up to that instant.
A clean symmetrical wave keeps the flux small. The current goes equally positive and negative. The integral therefore moves equally on each side of zero, and it returns every half cycle. The core never holds more than the volt-seconds of one half cycle.
A fault is not a clean symmetrical wave. The X/R ratio page shows that the current arrives with a DC offset under it. That offset has one sign only, so its contribution to the integral never returns to zero. The flux therefore increases in one direction with every cycle, for as long as the offset lasts. A fully offset fault needs approximately (1 + X/R) times the flux that a symmetrical wave needs. At X/R = 20, no core of a sensible size can supply that much flux.
When the core reaches its saturation flux, it cannot accept more flux. The magnetizing branch then takes the primary current instead of the secondary winding. The current at the relay falls toward zero. The CT recovers only when the wave reverses far enough to remove some flux. The relay therefore receives a chopped remnant of the real current. The remnant has no peaks, and long flat intervals separate the parts that remain.
Three conditions bring a CT closer to that limit. More burden needs more voltage for the same current, so the volt-seconds accumulate faster. More offset means a higher X/R, so the flux increases in one direction for longer. A smaller core holds less flux.
Try it
Start at the default values. Look for the point where the solid trace moves away from the dashed one. Then decrease the burden or increase the core size 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. An overcurrent element therefore measures a smaller fault. It moves further out on its time-current curve and trips late. If the distortion is large enough, the element does not pick up at all.
- Differential schemes can misoperate. If one CT in a differential zone saturates and the other does not, the two currents no longer match. The difference is the same as the difference that an internal fault produces. Differential relays therefore carry harmonic restraint and saturation-detection restraint.
- Burden is a design number. Lead length, relay input impedance, and the winding resistance of the CT all add together. If you double the burden, you double the volt-seconds that the core must supply for the same fault current.
- The sizing rule uses X/R directly. The knee-point voltage of the CT must exceed the symmetrical requirement multiplied by (1 + X/R). That factor comes from the same short-circuit study that produces the fault current.
The math, if you want itOptional — the page reads completely without it
A fully offset fault contains the one-sided term that the X/R page derives. The equation below uses 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 that 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. That amplitude is the symmetrical requirement. The offset integrates to τ = X/(ωR), which is X/R times bigger. The sum of the two gives the standard sizing factor:
the flux-doubling rule, approximately
λoffsetλsym ≈ 1 + X/R
The factor is an approximation. It assumes a fully offset fault. It ignores the flux that the negative half cycles remove, and it 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 uses that flux, and the sizing factor gives the amount that the CT needs.
See it in Phasor
Phasor holds the burden of each CT circuit and the X/R at each fault location. It can therefore identify a CT that saturates before the relay downstream of it makes a decision. The same short-circuit results that set the relay curves also set the sizing factor that the CT must meet.