Differential current in relay protection is calculated as the difference between the currents entering and leaving a protected element, adjusted for CT ratios, phase shifts, and transformer tap positions.

Basic Principle

Differential protection relies on comparing currents on both sides of a transformer or protected element. The differential current (Id) is calculated as: Id = |I_primary - I_secondary × (CT_ratio_correction)| Where:

• I_primary = current on the primary side
• I_secondary = current on the secondary side
• CT_ratio_correction = factor to account for differences in current transformer ratios and phase shifts . In normal operation, the differential current is minimal, ideally zero. Any significant non-zero differential current indicates an internal fault, such as winding short-circuits or core damage, triggering the relay to trip .

Steps for Calculation

1. Determine Full Load Currents Calculate the full load current for both high voltage (HV) and low voltage (LV) sides using: I = S / (√3 × V) Where S is transformer rating in VA and V is line voltage .
2. Adjust for Transformer Tap Positions Transformer tap changers affect voltage and current. Calculate currents at extreme taps (±10%) to ensure relay settings avoid unwanted tripping during tap changes .
3. Apply CT Ratios Convert primary and secondary currents to the same reference using CT ratios. If CTs are mismatched, apply a CT matching factor to correct the differential current .
4. Set Differential Pickup (Id_pickup) The relay pickup value should be above normal operating currents but sensitive enough to detect internal faults. It is often expressed in per-unit (pu) of CT secondary current .
5. High-Set Differential (Is-HS1 and Is-HS2)
   • Is-HS1: Minimum differential current for relay operation, ignoring harmonic blocking. Used to prevent tripping during transformer inrush or overflux conditions .
   • Is-HS2: Upper differential current limit. If exceeded, the relay trips instantaneously, even if harmonic blocking is active . Example: For a 70 MVA, 220 kV/6.9 kV transformer with 12% impedance and 400/1 CTs, the maximum fault current on HV side is 1531 A. CT secondary current = 1531 / 400 = 3.82 A. Applying a CT matching factor of 2.17 gives Is-HS1 = 3.82 × 2.17 = 8.3 pu .

Additional Considerations

• Biasing and Slope Characteristics: Modern differential relays use slope characteristics to maintain stability during external faults while remaining sensitive to internal faults .
• Harmonic Blocking: Second and fifth harmonic blocking prevents tripping during transformer energization or inrush currents .
• CT Accuracy: Precise CT performance is critical; errors can lead to false tripping or missed faults .
• Relay Settings Verification: Always verify settings under through-fault conditions and tap variations to ensure reliability . By following these steps, engineers can calculate differential currents accurately and set relay parameters to ensure fast, selective, and reliable protection of transformers and other power system equipment.