Differential Protection Slope Formulas
Transformer differential protection formulas: CT secondary currents, the differential and restraint currents, the slope and the operating threshold.
The formulas
| To find | Formula |
|---|---|
| Line current of the transformer | I = S × 1000 / (√3 × V), with S in MVA and V in kV |
| CT secondary current | Isec = I / (CT primary / CT secondary) |
| Differential current | Id = |I1 − I2|, the currents taken into the zone |
| Restraint current | Ir = (|I1| + |I2|) / 2 |
| Slope | slope % = Id / Ir × 100 |
| Operating threshold | Id(op) = Imin + K × Ir |
I1 and I2 are the secondary currents of the two sets of CTs after they are matched in size and phase, so that they are equal for a through load. The relay trips when Id is above the threshold set by the minimum pick-up Imin and the slope K, which gives room for the CT errors, the tap changer range and the magnetising current. Relay makers use a single or a dual slope, different restraint formulas and harmonic blocking for the inrush, so the setting guide of the relay decides. The ratings, the CT ratios and the settings here are examples.
Worked example: CT secondary currents of a 10 MVA, 33 / 11 kV transformer
CT ratios of 200 / 1 on the HV side and 600 / 1 on the LV side.
- HV line current = 10,000 / (√3 × 33) = 175 A, so the CT secondary current is 175 / 200 = 0.8748 A
- LV line current = 10,000 / (√3 × 11) = 524.9 A, so the CT secondary current is 524.9 / 600 = 0.8748 A
- The two currents agree to within 0.01%, so the ratios match without a correction
Worked example: a through fault with a CT error
The secondary currents are 1.00 A and 0.96 A.
- Id = |1.00 − 0.96| = 0.04 A
- Ir = (1.00 + 0.96) / 2 = 0.98 A
- slope = 0.04 / 0.98 × 100 = 4.082%, below a setting of 25%, so the relay restrains
Worked example: an internal fault fed from one side
The secondary currents are 3 A and 0 A.
- Id = |3 − 0| = 3 A
- Ir = (3 + 0) / 2 = 1.5 A
- slope = 3 / 1.5 × 100 = 200%, well above 25%, so the relay trips
Operating threshold for Imin = 0.2 A and K = 25%
| Restraint current Ir (A) | Id(op) (A) |
|---|---|
| 1 | 0.45 |
| 2 | 0.7 |
| 5 | 1.45 |
| 10 | 2.7 |
Frequently asked questions
Why is a restraint needed?
The CTs do not reproduce a large through current exactly, and a tap changer shifts the ratio. Without restraint, these errors would trip the relay on a fault outside the zone.
What does the slope set?
It sets how much of the restraint current may show up as differential current before the relay trips. A high slope gives more stability and less sensitivity to a small internal fault.
Why is the current matched in phase for a delta-star transformer?
The delta winding shifts the phase by 30°. Digital relays correct this inside, and older relays used CT connections of the opposite type.
What blocks a trip on inrush?
A second harmonic restraint. The magnetising inrush of a transformer is rich in the second harmonic, and the relay blocks when that share is high.
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