All about Electrical Engineering: Restricted Earth Fault (REF) protection

REF protection is Based on Kirchhoff's current law :

Current flowing into a node (or a junction) must be equal to current flowing out of it. Or in other words vector sum of currents in and out from a node is zero

REF protection is a basic protection used in many equipment in Power system like transformers, reactors, generators etc.

For REF protection is applied to windings having electrical couplings. Magnetic coupling is ignored for this purpose. For example we may consider following single phase two winding transformer:

Here both windings are electrically isolated and magnetically coupled. current entering from primary side I11 shall be same as current leaving primary side I12. Similarly, current entering from primary side I21 shall be same as current leaving primary side I22. Therefore

I11 + I12 = 0
I21 + I22 = 0

In this example we have used two REF relays, one for primary and other for secondary side. In some cases, in transformers having small transformation ratios, single REF can be used for cost saving. In that case

I11 + I21 + I12 + I22 = 0

Setting calculation: The simplified circuit is shown as below:

In normal condition I11 and I12 will be equal and opposite to each other, these will cancel each other and there will be no current flow through relay coil.

The typical setting for transformers is ~15% of full load current. For example transformer has full load current of 400A and CT ratio is 500/1A. CT ratio should be same for all CTs connected to REF relay. The current setting for REF relay shall be 400 x 15% = 60A primary (0.12A secondary).

In some relays the setting is entered in Volts. Voltage is calculated by multiplying setting current with stabilizing resistor value. Say stabilizing resistor is 1000 ohm, the voltage setting will be 120V.

Value of stabilizing resistor: CT saturation sometimes occur during high fault currents, we need stable relay operation during CT saturation. Stabilization resistor is used for avoiding relay operation during CT saturation during through faults. Its value is calculated on maximum through fault current of the protected equipment.

Let us assume CT2 (Neutral side CT) saturates during through fault. It will not generate any output and will act as resistor as per its secondary winding resistance.

In this case let us assume percentage impedance of transformer to be 10%. Maximum through fault current will be 400A/10% = 4000A Primary (8A Secondary). If CT secondary resistance is 6 ohm and lead resistance is 4 ohm, Voltage developed in case of maximum fault current will be 8A x (4 + 6) = 80V. Now our requirement is that for this voltage relay should not operate. In other words stabilizing resistor should restrict the relay current below set value.

Now keeping 150% safety margin the current produced by 120V throgh relay coil shall be 0.12A. For this requirement stabalizing resistor shold be 120V / 0.12A = 1000 ohm. In the above case of CT saturation, Voltage developed was 80V, the current through relay coil will be 80V / 1000 ohm = 0.08A. This is much below the set value of 0.12A. Therefore relay will not operate during this CT saturation, which meets our requirement.

Varistor: Stablizing resistor is in series to relay circuit, it may have values upto 2 K ohm. During fault condition, there may be high voltage across the CT terminals due to higher value of relay circuit. To protect equipment (lCT, cables relays etc) and persons working in relay panels varistor is used to limit the voltage below ~300V.

Three Phase transformers: For three phase transformers the example is as below: