A 400/1 A protection CT with a 150 V knee point, checked against the required knee-point voltage for its stated accuracy limit factor β and what happens to its output once the fault current exceeds that limit.
| CT ratio | 400 : 1 A |
| Required accuracy limit factor (ALF) | 20 |
| Knee-point voltage (nameplate) | 150 V |
| CT secondary winding resistance | 2.5 Ξ© |
| Burden | Protection relay, 2.5 VA |
| Lead run | 20 m one-way, 2.5 mmΒ² Cu, go-and-return loop |
The fault-current table sweeps multiples of rated primary current and caps the ideal (linear) secondary output at the rated ALF, showing where the CT's output stops faithfully following the primary current.
| Fault multiple | Primary current | Ideal secondary (linear) | Actual output at ALF cap |
|---|---|---|---|
| 20x (rated ALF) | 8000 A | 20 A | 20 A β still linear |
| 50x | 20,000 A | 50 A | 20 A β capped, CT has saturated |
| Check | Requirement | Actual | Status |
|---|---|---|---|
| Knee-point voltage vs. required for ALF 20 | β₯ 105.6 V | 150 V | β PASS |
| Actual achievable ALF at this burden | n/a (informational) | 28.4 | β PASS |
Key insight: A CT's accuracy limit factor isn't a hard cutoff where the CT stops working β it's the point beyond which the secondary current stops accurately representing the primary current because the core saturates. A protection relay that needs to see the true fault current magnitude at very high multiples (not just detect that a fault occurred) needs a CT with enough real margin above the fault current it will actually see, not just enough to clear the nameplate ALF requirement.
Every input in this example is editable in the live calculator β free, no signup.
Open CT Sizing & Saturation calculator βLead resistance adds directly to the total secondary burden the CT has to drive, and burden is one of the two terms (along with CT winding resistance) that determines the required knee-point voltage β a longer or thinner lead run increases Rb, which increases the required Vk for the same ALF, meaning a CT that comfortably meets its accuracy requirement with short leads can fail to meet it if installed with a long lead run to a distant relay panel.
With this CT's actual achievable ALF of about 28.4, secondary output saturates well before reaching 50x β the relay would see a current signal that plateaus around 20 A regardless of how much larger the actual primary fault current becomes, which could cause a protection scheme relying on accurate magnitude (like some differential or distance protection) to misoperate. This is exactly why the required ALF should be chosen based on the actual maximum fault current the CT needs to represent faithfully, with margin, not just a generic default value.