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IEC 60255-1519 min read

Worked Example: IDMT Relay Grading Across a Full Fault-Current Range

Two IEC Standard Inverse relays look correctly graded at one fault current — until the full-range sweep reveals the margin collapses at higher currents.

Scenario

Relay 1 (downstream)Pickup 100 A, CT 1:1, TMS 0.1, IEC Standard Inverse (SI)
Relay 2 (upstream)Pickup 150 A, CT 1:1, TMS 0.25, IEC Standard Inverse (SI)
Test fault current2000 A
Full sweep range105 A – 10,000 A (the calculator's default range)
Required grading margin (CTI)0.4 s

Relay 1 protects a feeder; Relay 2 is the upstream backup device one bus further up. Both must trip in the right order with enough time separation, at every current either relay could realistically see.

Step-by-step calculation

Step 1: Find each relay's multiple of pickup at the test fault current

Eq. 1
M = Ifault / Ipickup
M1 = 2000 / 100 = 20 x M2 = 2000 / 150 = 13.3 x
M1 = 20 x, M2 = 13.3 x

Step 2: Apply the IEC Standard Inverse equation to each relay

IEC 60255-151 defines the Standard Inverse (SI) curve with k = 0.14 and alpha = 0.02.

Eq. 2
t = TMS x (k / (M^alpha - 1))
t1 = 0.1 x (0.14 / (20^0.02 - 1)) = 0.227 s t2 = 0.25 x (0.14 / (13.3^0.02 - 1)) = 0.658 s
t1 = 0.227 s, t2 = 0.658 s

Step 3: Check the margin at this one test point

margin = t2 - t1
margin = 0.658 - 0.227
margin = 0.432 s -> passes the 0.4 s requirement

This is the check most hand calculations stop at. It looks fine.

Step 4: Sweep the full fault-current range instead of one point

Real faults land anywhere in the range a relay can see, not just at one convenient test current. Recomputing the margin at every point across 105 A-10,000 A finds where it is smallest.

Fault currentt1 (Relay 1)t2 (Relay 2)Margin
2000 A0.227 s0.658 s0.432 s
10,000 A (worst case)0.145 s0.399 s0.254 s

IEC inverse curves flatten out at high multiples of pickup, which is why the two relays' trip times converge — and the margin shrinks — as fault current rises.

Result summary

CheckRequirementActualStatus
Margin at the single 2000 A test point≥ 0.4 s0.432 s✓ PASS
Worst-case margin, full 105 A–10,000 A sweep≥ 0.4 s0.254 s at 10,000 A✗ FAIL
As configured, this relay pair passes the single-point check at 2000 A but FAILS the full-range sweep — the settings are not acceptable as-is and need adjustment before commissioning.

Key insight: A comfortable margin at one arbitrarily chosen fault current does not guarantee coordination everywhere. IEC inverse curves compress at high current multiples, so the worst-case point is often at the top of the range, not the bottom — which is exactly why this calculator sweeps the whole range by default instead of checking a single current.

Try it with your own numbers

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Frequently asked questions

What setting change fixes the margin?

Raising Relay 2's time multiplier setting (TMS) from 0.25 to roughly 0.34 restores at least 0.4 s of margin across the entire 105 A–10,000 A range, without affecting Relay 1's own trip times.

Why does the margin shrink at higher fault current for IEC curves?

The IEC inverse-time equation t = TMS x (k / (M^alpha - 1)) approaches TMS x k / M^alpha as M grows, so at high multiples of pickup the trip time is dominated by the curve's shape rather than the TMS offset — two relays on the same curve family converge toward each other's trip times as current rises.

Is 0.4 s always the right grading margin?

0.4 s (400 ms) is a commonly used CTI (coordination time interval) for electromechanical relays, covering breaker interrupting time, relay overtravel and a safety margin. Modern numerical relays with no overtravel are sometimes graded to 0.2-0.3 s — always confirm against your utility's or plant's own protection philosophy document.

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