GE SR745 Differential Slope: Calculating Test Currents

Daniel Price9 min read
GE FanucTechnical ReferenceWiring & Electrical
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Problem: Predicting the Trip Point Instead of Trusting It

Secondary injection testing of a percentage differential element on a transformer relay is only a valid test if you can calculate the expected trip current independently of the relay. Without that number you are recording whatever the relay does and calling it a pass. The worked example below uses a real setting and application set:

Item Value
Transformer 16 MVA, 13.8 kV / 4.16 kV, Dyn1
Winding 1 CT 1200:5 (ratio 240)
Winding 2 CT 3000:5 (ratio 600)
Differential pickup 0.2 x CT
Slope 1 25 %
Breakpoint 1 x CT
Slope 2 98 %
Scope note: the settings above are the only relay data available. Every formula in this article is presented with its assumption labeled. Before you sign a test sheet, confirm the restraint definition, the per-unit base for "x CT", and the above-breakpoint characteristic shape against the SR745 instruction manual for your firmware revision.

Step 1 - Base Currents and CT Secondary Values

Compute rated line current on each winding, then the corresponding CT secondary current. Three-phase: I = MVA x 1000 / (sqrt(3) x kV_LL).

Quantity Winding 1 (13.8 kV) Winding 2 (4.16 kV)
Rated primary current 16000 / (1.732 x 13.8) = 669.4 A 16000 / (1.732 x 4.16) = 2220.5 A
CT ratio 240:1 600:1
CT secondary at rated load 2.789 A 3.701 A
Fraction of 5 A CT rating 0.558 x CT 0.740 x CT

The two secondary currents are not equal, so the relay must apply an internal magnitude (tap) correction derived from the MVA and voltage settings you entered. Balanced through-current therefore corresponds to a secondary injection ratio of:

k_mag = I2_sec / I1_sec = 3.701 / 2.789 = 1.327
I_W2_inject = 1.327 x I_W1_inject   (magnitude only, before phase compensation)

Step 2 - Resolve the "x CT" Base Before You Do Anything Else

A setting of 0.2 x CT can be referenced to two different bases, and the numbers differ by nearly a factor of two on winding 1. Verify which one the manual specifies; do not assume.

Interpretation Pickup on W1 Pickup on W2 Breakpoint (1 x CT)
Base = CT rated secondary (5 A) 1.000 A 1.000 A 5.000 A on each CT
Base = transformer rated current 0.558 A 0.740 A 2.789 A / 3.701 A

If the relay normalizes both windings to a common per-unit base (the usual reason a relay asks for MVA and kV), the second row is the one that produces an internally consistent characteristic and the one your spreadsheet should carry. Prove it during commissioning with the pickup test in Step 5.

Step 3 - Phase Compensation for Dyn1

A Dyn1 transformer introduces a 30 degree displacement between winding 1 and winding 2 currents plus zero-sequence isolation. A numerical differential relay removes this internally by applying a delta-type combination to one winding's currents. Two consequences for testing:

  1. Three-phase balanced injection: phase-shift the winding 2 source set by the compensation angle so the differential currents read near zero. Confirm this by reading the relay's metered differential per phase - target is a few percent of tap, not exactly zero, because of CT ratio mismatch and source error.
  2. Single-phase injection into the delta-compensated winding: the compensation algebra (for example (Ia - Ib) / sqrt(3)) splits one injected phase current into two differential elements at 1 / sqrt(3) = 0.577 of the injected magnitude. Your calculated trip current on that winding must be divided by 0.577 (multiplied by 1.732) to get the amps the test set must output.
Which winding carries the compensation, and whether the divisor is sqrt(3) or 3, is firmware- and setting-dependent. Determine it empirically: inject 1.00 A into phase A of one winding with the other winding open, then read the relay's metered per-phase differential currents. If A and B both show ~0.577, that winding is delta-compensated. If A alone shows ~1.00, it is not.

Step 4 - Dual-Slope Characteristic Equations

Define Id = operate (differential) current and Ir = restraint current, both in the relay's per-unit base. Below the breakpoint the element operates when:

Id > max( Pickup , S1 x Ir )        for Ir <= Breakpoint
with Pickup = 0.2, S1 = 0.25, Breakpoint = 1.0

Note that 0.2 / 0.25 = 0.8 pu: below 0.8 pu restraint the pickup setting governs, above it the 25 % slope governs.

Above the breakpoint there are two common implementations. They diverge badly at 98 % slope, so identify yours:

Ir (pu) Form A - continuous
Id = S2 x Ir - (S2 - S1) x BP
Form B - discontinuous
Id = S2 x Ir
0.5 0.20 (pickup governs) 0.20 (pickup governs)
1.0 (BP) 0.25 0.25
1.5 0.74 1.47
2.0 1.23 1.96
3.0 2.21 2.94

Form B implies a vertical step of 0.73 pu at the breakpoint - a region where increasing fault current can move the operating point from trip to no-trip. If the relay's own characteristic plot shows no step, you have Form A. Test one point at Ir = 2.0 pu: 1.23 pu versus 1.96 pu operate current resolves it in a single shot.

Step 5 - Restraint Definition and the Spreadsheet

The restraint quantity is the third variable manufacturers do not standardize. For a two-winding relay with compensated per-unit currents I1 and I2:

Definition Ir Ir with I1 = 1.0, I2 = 0 (one-end injection)
Maximum max(|I1|, |I2|) 1.000
Average (|I1| + |I2|) / 2 0.500
Sum |I1| + |I2| 1.000

Operate current is the phasor sum of the compensated winding currents: Id = |I1 + I2|, where the polarity convention is such that through-current cancels. Build the sheet in this order:

  1. Inputs: W1 test amps, W1 angle, W2 test amps, W2 angle, MVA, kV1, kV2, both CT ratios, pickup, S1, BP, S2.
  2. Normalize: I1_pu = I_W1 / 2.789, I2_pu = I_W2 / 3.701 (transformer-rated base) or divide by 5.0 (CT-rated base). Keep the base selectable with a flag cell.
  3. Compensate: apply 0.577 and the phase rotation to whichever winding the relay treats as delta.
  4. Compute: Id, Ir, then the threshold from the Step 4 equation with the correct branch.
  5. Verdict: IF(Id > Threshold, "TRIP", "NO TRIP"), plus a margin cell (Id / Threshold - 1) x 100 in percent.

Step 6 - Test Procedure and Verification

  1. Metering check first. Inject a known single-phase current into each winding separately and compare the relay's metered secondary amps to the test set. Any error above roughly 2 % means the CT ratio or base settings are wrong, and every later result is invalid.
  2. Pickup test. Winding 2 open, ramp winding 1. With one end open, Id = Ir = I1_pu, so the slope line (Id = 0.25 Ir) can never be reached and the element trips at the pickup setting. Predicted W1 secondary amps = pickup_pu x 2.789 A, divided by 0.577 if W1 is the compensated winding. Record the ratio of measured to predicted - that ratio tells you conclusively which per-unit base of Step 2 is in use.
  3. Slope 1 test. Inject opposing currents at a restraint point of about 0.5 x BP and again just below BP. Hold I2 fixed, ramp I1 until trip, and compare to the sheet. Expect the measured slope to land near 25 % with a few percent tolerance.
  4. Breakpoint / slope 2 test. Run the Ir = 2.0 pu point to select Form A or Form B, then add a third point at Ir = 3.0 pu to confirm the 98 % gradient.
  5. Harmonic inhibit. Second-harmonic inrush restraint and fifth-harmonic overexcitation restraint will block an otherwise correct trip. Confirm the test set output is a clean sinusoid, or disable the inhibit functions for the slope test and restore them afterward. A "failed" relay that trips only with a pure sine source is a test-set harmonic problem, not a relay problem.
  6. Timing. Log the operate time at 2x the calculated threshold on each slope segment and trend it across the fleet.

Interpreting Repeat Failures

Before condemning a population of relays, eliminate these in order, since each one produces symptoms identical to a defective differential element:

Symptom Likely cause Check
Trip point high by 1.73x Delta compensation factor omitted from the calculation Compare metered per-phase Id to injected amps
Trip point off by 1.33x Wrong CT ratio or wrong MVA/kV entered, so tap correction is wrong Recompute 2.789 A / 3.701 A and compare to metering
Trip point off by 2x on one-end injection Restraint uses average, sheet assumes maximum Repeat pickup test at two restraint levels
No trip well above the characteristic Harmonic inhibit active, or element output not mapped to the trip relay Read harmonic content; check output relay assignment
Trip near the breakpoint but not above it Form B step discontinuity at 98 % slope Plot the relay's own characteristic curve

Document the resolved answers to the three ambiguities - per-unit base, restraint definition, above-breakpoint form - once per relay model and firmware revision, and reuse them across the fleet. That single page turns the spreadsheet from an estimate into an acceptance criterion.

What secondary current corresponds to rated load on a 16 MVA 13.8 kV / 4.16 kV transformer with 1200:5 and 3000:5 CTs?

Winding 1: 16000 / (1.732 x 13.8) = 669.4 A primary, divided by a 240:1 ratio = 2.789 A secondary. Winding 2: 16000 / (1.732 x 4.16) = 2220.5 A primary, divided by 600:1 = 3.701 A secondary. Balanced through-current therefore corresponds to an injection ratio of 1.327.

Why does single-phase injection produce differential current in two phases?

The relay applies an internal delta-type combination to one winding to correct the Dyn1 30 degree displacement. Injecting a single phase into that winding splits into two elements at 1 / sqrt(3) = 0.577 of the injected magnitude, so the required test current is 1.732 times the calculated per-phase operate current.

Where does the 25 % slope actually start controlling the trip point?

Below 0.8 pu restraint the 0.2 pu pickup setting is the higher of the two thresholds and governs. Above 0.8 pu restraint, 0.25 x Ir exceeds 0.2 pu and the slope characteristic governs, up to the 1 x CT breakpoint.

How do I determine whether slope 2 is continuous or stepped at the breakpoint?

Test a single point at 2.0 pu restraint. A continuous characteristic (Id = 0.98 x Ir - 0.73 x BP) trips at 1.23 pu; a discontinuous one (Id = 0.98 x Ir) trips at 1.96 pu. The measured value identifies the implementation unambiguously.

The relay would not trip above the calculated characteristic - is it defective?

Check harmonic restraint first. Second-harmonic inrush and fifth-harmonic overexcitation blocking will inhibit a valid differential trip if the test set output contains distortion. Disable those inhibits for the slope test and restore them before returning the relay to service.

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