IEC 60255-1518 min read

How to Calculate the TMS of an Upstream IDMT Relay for a 0.3 s Grading Margin

Part 1 of the protection series. Step by step from the downstream relay time to the upstream TMS, checked at the maximum fault current — and why the margin only grows at lower currents.

Scenario

Downstream relay R1 (11 kV feeder)IEC Standard Inverse, pickup 400 A, TMS 0.10
Upstream relay R2 (11 kV incomer)IEC Standard Inverse, pickup 1,000 A, TMS to be found
Maximum fault at the feeder13.12 kA (from IEC 60909)
Grading margin0.3 s (numeric relays, vacuum breakers)

Step-by-step calculation

Step 1: R1 operating time at maximum fault

t = 0.14 × TMS / ((I/Is)^0.02 − 1)
M = 13,120/400 = 32.8; t1 = 0.14 × 0.10 / (32.8^0.02 − 1)
t1 = 0.194 s

Step 2: Required R2 time

Step 3: R2 time at TMS = 1

Step 4: Required TMS

TMS = t2 / t(TMS = 1)
TMS = 0.494 / 2.650 = 0.186
Select TMS = 0.19 → t2 = 0.503 s, margin 0.31 s

Step 5: Check the margin across the range

Fault currentR1 (s)R2 (s)Margin (s)
2 kA0.4281.9061.48
4 kA0.2970.9460.65
8 kA0.2270.6260.40
13.12 kA0.1940.5030.31

Result summary

CheckRequirementActualStatus
Margin at maximum fault≥ 0.3 s0.31 s✓ PASS
Margin at 2 kA≥ 0.3 s1.48 s✓ PASS
R2 needs a TMS of 0.19, giving 0.503 s at 13.12 kA and a 0.31 s margin over R1. For two relays with the same curve and a higher upstream pickup, the margin is smallest at the maximum fault current.

Key insight: Always set the margin at the maximum fault current the downstream relay sees. If the curves are of different types (for example SI over EI), check the whole range — they can cross.

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Frequently asked questions

Why 0.3 s and not 0.4 s?

0.4 s is a traditional margin for electromechanical relays and oil breakers. Numeric relays with fast breakers commonly use 0.25–0.3 s; Part 4 shows how the margin is built up.

What sets the pickup?

Above the maximum load current (with margin) and below the minimum fault current the relay must detect, including its back-up zone.

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