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IEEE C37.91 (guide-consistent)10 min read

Worked Example: Percentage-Bias Differential Protection — Restraint vs. Trip

The same dual-slope 87T characteristic correctly restrains through normal load with CT mismatch, then trips decisively for a genuine internal fault.

Scenario

Minimum pickup (Id0)0.3 p.u.
First slope25%, up to a restraint current of 2.0 p.u.
Second slope60%, above 2.0 p.u. restraint
Case A — normal loadI1 = 1.05 p.u. in, I2 = 0.98 p.u. out (typical CT ratio-error mismatch)
Case B — internal faultI1 = 6.0 p.u. in, I2 = 0.5 p.u. out (winding short collapses the outflow)

Currents are already referred to a common base (CT secondary, ratio and vector-group compensated) as the relay itself would see them.

Step-by-step calculation

Step 1: Compute differential and restraint (bias) current for each case

Id = |I1 - I2| Ir = (I1 + I2) / 2
Case A: Id = |1.05 - 0.98| = 0.07 p.u. Ir = (1.05+0.98)/2 = 1.015 p.u. Case B: Id = |6.0 - 0.5| = 5.5 p.u. Ir = (6.0+0.5)/2 = 3.25 p.u.
Case A: Id = 0.07 p.u., Ir = 1.015 p.u. Case B: Id = 5.5 p.u., Ir = 3.25 p.u.

Step 2: Find the operate threshold at each restraint current

Case A's restraint (1.015 p.u.) is below the 2.0 p.u. knee, so only the first slope applies. Case B's restraint (3.25 p.u.) is above the knee, so the threshold is the value at the knee plus the second, steeper slope beyond it.

Below knee: threshold = Id0 + slope1 x Ir Above knee: threshold = (Id0 + slope1 x knee) + slope2 x (Ir - knee)
Case A: 0.3 + 0.25 x 1.015 = 0.554 p.u. Case B: (0.3 + 0.25 x 2.0) + 0.60 x (3.25 - 2.0) = 0.8 + 0.75 = 1.55 p.u.
Case A threshold = 0.554 p.u. Case B threshold = 1.55 p.u.

Step 3: Compare differential current to the threshold

CaseIdThresholdVerdict
A — normal load, CT mismatch0.07 p.u.0.554 p.u.Restrained — no trip (correct)
B — internal fault5.5 p.u.1.55 p.u.Trips (correct)

Result summary

CheckRequirementActualStatus
Case A — stays restrained through normal loadId < threshold0.07 p.u. < 0.554 p.u.✓ PASS
Case B — trips for a genuine internal faultId ≥ threshold5.5 p.u. ≥ 1.55 p.u.✓ PASS
The dual-slope characteristic does its job in both directions: it stays securely restrained through the everyday CT ratio-error of normal load (Case A), and trips decisively and quickly for a real internal winding fault (Case B).

Key insight: A flat, fixed-pickup differential element would have to be set high enough to ride through worst-case CT mismatch at maximum through-load — which would make it dangerously insensitive to genuine internal faults. The percentage-bias characteristic solves this by letting the pickup threshold rise with restraint current, staying tight (sensitive) at low load and only relaxing (for security against CT saturation) at high through-fault currents.

Try it with your own numbers

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Open Transformer Differential Protection (87T) calculator →

Frequently asked questions

Why does the threshold rise instead of staying at a fixed pickup?

Real CTs have small ratio errors that scale with the current flowing through them, and those errors get worse (relative CT saturation) during high through-fault current. If the pickup stayed fixed at a low value, a large external fault could saturate the CTs enough to look like a differential current and cause a false trip. The second, steeper slope specifically defends against that scenario.

What's a typical range for the two slope settings?

First slopes of 20-30% and second slopes of 50-80% above a knee around 2-8 p.u. restraint current are common starting points, but the right values depend on the specific CTs, their accuracy class and knee-point voltage, and the transformer's tap-changer range — always confirm against manufacturer guidance and a proper CT saturation study for a real installation.

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