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Overcurrent relay coordination
Put up to six relays, reclosers and fuses in series on the same time-current plot, referred to a common voltage, with the transformer damage curve, the inrush and the faults at each bus, and check the margin between each pair over the whole fault range.
IEC 60255-151dependent-time curvesIEEE C37.112inverse curvesIEEE C57.109through-fault damage curve
Results
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Charts
- Damage, inrush and faults F1…F6
- Criterion just met (headroom 0)
Calculation and details
Operating times at each fault level
Times in s. Recloser: fast / slow. Fuse: melting / clearing. «—»: does not operate at that current. Currents referred to the reference voltage.
Pair-by-pair check
What it means for coordination
What it is for
In a radial system, each fault must be cleared by the closest device, and the upstream one only acts if that device fails. This tool puts the relays, reclosers and fuses of one path on a single plot, from the lateral up to the substation main, with the transformer damage curve and its energization current. Use it to review existing settings before changing them, to prepare a settings proposal or to verify by hand what the coordination software gives.
How it is calculated
IEC 60255-151dependent-time characteristicIEEE C37.112inverse curves
IEC: t = TMS · k / (Mα − 1) · IEEE: t = TD · (A / (Mp − 1) + B) · M = I / IsThe same constants as the inverse-time curve calculator. The curve starts at 1.1 times Is; the instantaneous 50 operates at its time from its pickup. For definite time, the dial is the time. If the pickup is entered in secondary amperes, it is converted to primary with the CT ratio: Iprim = Isec · CTprim / CTsec.
Referral to a common voltage
Iref = Idevice · kVdevice / kVrefAll currents, pickups, faults and fuse points are referred to the reference voltage with the voltage ratio, as for a three-phase fault through an ideal transformer.
Coordination criterion per pair
Δt(I) = tup(I) − tdown(I) ≥ CTI, for Imin ≤ I ≤ Imax of the downstream zoneIEEE 242 tabulates minimum intervals by relay technology, because the CTI must cover the breaker time, relay overtravel and a safety margin. The tool defaults to 0.3 s between relays and 0.2 s when a fuse is involved (Tiza's convention, labelled as such). A downstream fuse enters with its total clearing curve; an upstream fuse, with its minimum melting curve.
- Between fuses: total clearing of the downstream fuse ≤ 75 % of the minimum melting of the upstream fuse, the usual rule of thumb for expulsion fuses, which covers preloading, tolerances and predamage.
- Recloser and fuse (fuse saving): k · tfast ≤ tmelt, and the slow curve above the fuse total clearing with the fuse CTI. When fuse saving is lost only at high currents, the status is a warning and the tool shows up to which current it holds.
Each pair is green if it passes over the whole range, yellow if the upstream device operates later but without the margin, and red if it operates first (or if the downstream fuse clearing exceeds the upstream fuse melting).
IEEE C57.109through-fault damage curve
Thermal: 2 pu → 1800 s · 3 pu → 300 s · 4.75 pu → 60 s · 6.3 pu → 30 s · 11.3 pu → 10 s · 25 pu → 2 s · Mechanical: I²t = K, K = 2 · (1/Z)²Categories of three-phase liquid-immersed transformers: I up to 500 kVA, II up to 5000 kVA, III up to 30,000 kVA and IV above. Category I follows the thermal curve, with I²t = 1250 up to 35 times (up to 300 kVA) or 40 times (up to 75 kVA). In categories II and III with frequent faults, and always in IV, the curve drops from 0.7/Z (category II, 4.08 s) or 0.5/Z (III and IV, 8 s) following I²t = K down to 1/Z at 2 s. For III and IV, Z = ZT + ZS. The points come from the tables of the SKM application guide on C57.109. With a Dyn transformer and a phase-to-ground fault on the secondary, the primary relay sees 0.58 times the per-unit current and the curve is shifted by that factor.
The transformer protection must stay below the damage curve where it operates, and above the inrush point: common practice places it at 8 to 12 times In for 0.1 s; use your transformer's value if you know it.
Limits of the calculation
- It checks one radial path with three-phase faults. It does not calculate the short circuit: you enter the maximum and minimum fault of each zone. It does not cover ground faults (51N), directional relays or contributions from distributed generation.
- The voltage referral assumes ideal transformers. For unbalanced faults through a Dyn the current seen on the primary changes; here that effect is only applied to the damage curve with the shift factor.
- Fuse points are joined with straight lines on log-log scales; below the first point the fuse is taken as not operating and above the last one the curve is extended with constant I²t. Use points from the manufacturer's curve, with its tolerance and preloading.
- The recloser k factor comes from its manufacturer according to the reclosing interval and the number of fast operations; the preloaded value of 1 compares the curves as they are.
- It does not include the minimum response time of the relays, CT saturation or curve limits above 20 times pickup. Confirm each setting with the device manual.
- The preloaded CTIs are Tiza's convention, not a value from a standard; the coordination study sets the value for each case.
Frequently asked questions
How is coordination between two overcurrent relays checked?
Plot both time-current curves referred to the same voltage and compare the backup relay time with the time of the relay closest to the fault, for every fault current in the zone of the latter, from minimum to maximum. The difference must be equal to or greater than the chosen coordination time interval (CTI) over that whole range, not only at the maximum fault.
What is the IEEE C57.109 transformer damage curve?
It is the through-fault withstand curve of the transformer: the time it can carry each multiple of its rated current without thermal or mechanical damage. It has a thermal part (for example, 2 times rated for 1800 s and 25 times for 2 s) and, for category II to IV transformers with frequent faults, a mechanical part I²t = K calculated from the impedance. The transformer protection must stay below that curve and above the energization (inrush) point.
How does fuse saving with a recloser work?
The recloser opens first on its fast curve, before the lateral fuse can melt, and recloses: if the fault was temporary, service returns without replacing the fuse. If the fault persists, the recloser slow curve waits for the fuse to clear the lateral. It works as long as the fast curve, multiplied by the manufacturer's k factor, stays below the fuse minimum melting curve; at high currents that margin is lost.
We do your protection coordination study
We model the system, calculate the short circuit at each bus and set relays, reclosers and fuses so the one closest to the fault clears first. Then we load the settings into the relays, remotely as well.