How to Calculate Stabilizing Resistor for High Impedance Differential Protection

Calculate stabilizing resistor for Differential Protection:

In three phase alternating power system, the bus bar differential protection needs to be stabilized, because of the differential relay should not trip, if the fault outside of the protective Zone. For example, a through fault occurs outside of the zone means, if the fault is heavy then the current transformer may get saturate, due to this there is an unbalance current in the relay operating coil. Hence relay trips the associated circuit breaker. But actually, the relay should not trip the circuit breaker such a condition (fault outside of the protective zone). In order to avoid or to increase the high impedance differential relay stability we need to add stabilizing resistor while designing high impedance differential relay.

How to Calculate stabilizing Resistor for differential Protection differential Protection
Calculate stabilizing Resistor

How stabilizing resistor Increases the relay stability?

The three identical current transformers CT1, CT2, and CT3 are connected as shown in the figure… the CTs internal resistance such as R1, R2 and R3. Being high impedance differential CT configuration the internal resistance of all CTs are equal R1, = R2 = R3 =RCT The current flow through the Current transformers are I1, I2 and I3. Under Normal condition I1 + I2 + I3 =0. Hence the operating coil do not operate the trip circuit. During the fault condition (fault inside of the protective zone) the I1 + I2 + I3 is not equal to 0. Hence the relay operates the trip mechanism. This is done by normal differential relay. But if the fault is outside of the protective zone i.e through fault, take a worst condition, consider one of the current transformer core got saturated i.e Take CT3 got saturated then the fault current flow through the CT3. Then there is a voltage drop VCT3 devolved across the CT3 is

If the relay configured at panel board side means there, pilot wire (looping wire) resistance will be added. Hence



Here 2 * Rwire means CT has two terminal S1 and S2 hence we need two wire to bring the current transformer’s output to the panel board.

This voltage operates operating over current element (operating coil)….To withstand this voltage we have to add a resistor in series with the over current element (operating coil).

Calculate stabilizing resistor:

In this we are going to use potential divider principle. We are going to divide or balance the voltage across the operating coil by adding the resistor in series.

Let’s us assume the current setting in the over-current element is Is

Then

Here the resistance of the operating coil is very low. Typically

Hence,

If the relay configured at CT terminal side means..



Thus during through fault, for the worst condition of CT saturation, the current through the Relay coil will not be enough to cross setting value of Is and thus the relay do not operate.

Also see:

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