High impedance protection system is a simple technique that requires that all CTs used in the protection scheme have relatively

o Similar magnetizing characteristic.

o The same ratio and class “CT ratio error”.

o Low secondary winding impedance.

o Polarity shall be considered for parallel connection.

These CTs shall be installed in all ends of the protected object. In order to make the scheme, all CTs belonging to one phase shall be connected in parallel.

In the below single line diagram, the protected zone is defined from the CT location.

The CTs are connected in parallel for each phase and connected to the relay as shown below

As shown below, The Fault is outside the Busbar Differential Protection Zone.

The Faulted Feeder current = Summation of all Current Sources [Incomer - Bus Section – any other Source Feeder]

The BB CT located on the Faulted feeder would be saturated due to high current flow, hence the differential current would flow through the BB protection Relay = Mal-Operation

A. Series Resistor

For external faults, even with sever CT saturation of some of the current transformers, the voltage across the CT paralleling point shall not rise above the relay V_Setting. To achieve that a suitable value for setting resistor R_Series shall be selected in such away that the saturated CT provides a much lower impedance path as compared with the measuring branch.

Thus, in case of an external of external fault causing CT saturation, the non-saturated CTs will drive most of the spill current through the secondary winding of the saturated CT and not through the measuring branch.

By adding High Series Resistance, the current would pass through the saturated CT which has less resistance compared to the added series resistance.

Some High Impedance protection relays offering built in resistance, and others external Series Resistor shall be added. In both cases the high impedance relay operation would be through a setting voltage.

The Series Resistance must be a suitable value so that,

o Even with external faults and severe saturation of current transformers the voltage across the parallel branch shall not arise above the relay V_Settings

o The saturated CT provides a much lower impedance path as compared with the measuring branch.

B. Metrosil

To limit the peak voltage developed by current transformers under internal fault condition to a value below the insulation level of the CT.

C. CT Wire Supervision

CT wire supervision is a relay connected in parallel with the BB protection relay after the CT parallel combination and at the start of the BB protection system.

Consider ‘R’ phase wire loose in CT-2, voltage would increase across the loop that would lead to BB protection relay operation. This is not a fault current, but it causes unnecessary trip. To avoid this trip CT supervision relay is used.

The voltage setting of this relay is very low with time delay So this relay will be operated for leakage current, and Its contact will short the CT circuit to avoid false tripping. Main relay setting is greater than this value and instantaneous. So, for heavy internal bus bar fault Main protection will operate immediately.

D. Shunt Resistor

Shunt resistor is playing an important role that,

o Allow the calculations for Busbar protection system minimum Fault current setting

o Provide system stability during a problem in one of the CTs [CT Broken Leads]

Let’s have an example of 132kV Busbar Protection, below is the given data

A. CT Stability Criterion:

To prevent maloperation of the relay during saturation of the CT on external fault, the relay setting Vs must be greater than the voltage produced by the maximum secondary through fault current flowing through the cable resistance.

Since the CT Knee point is greater than twice the calculated voltage for stability then the CT is adequately sized.

The voltage is proportional to the CT secondary resistance and the CT secondary leads. Typical practices employed to keep this voltage as low as possible include:

• Using the highest and optimum possible CT ratios

• Using the lowest resistance leads possible

• Keeping CT lead lengths as short as possible

• Paralleling all the CT secondaries to a junction point that is equidistant from and as close as possible to the CTs.

B. Protection Relay Setting Current:

As per the standard practice for most cases I_Setting shall be 0.1A

C. Series Resistor [Stabilizing Resistor]:

As per the standard practice for most cases I_Setting shall be 0.1A

Resistor Continuous Power Rating à P = I_Setting² x R_Series = 0.1² x 1750 = 17.5 Watt

D. Calculation for Primary Fault Sensitivity:

To calculate the minimum operating current

As shown in the below graph, CT Magnetization Curve is linear up to the knee point so we can calculate approximately the magnetizing current at any voltage less than the V_Knee.

E. Calculation for Shunt Resistor:

To avoid the Busbar Protection Relay mal-operation during CT abnormal conditions (CT Broken Leads) of any feeder in the busbar zone, the primary operating current shall equal to 1.2 of the rated current of the feeder.

I_{Busbar_System} = 1.2 x 1200 = 1440 A

Which equals in secondary with respect to CT ratio of Busbar = 1440 / 2000 = 0.72 A

While I_Operating < I_{Busbar_System} then Shunt resistor must be calculated to draw the difference

F. Calculation for Metrosil:

Under in-zone fault conditions, a high impedance relay constitutes an excessive burden to the current transformers, leading to the development of a high voltage; the voltage waveform will be highly distorted, but the peak value may be many times the nominal saturation voltage.

Metrosil is required if the V_P [Peak Voltage across relay under maximum internal fault condition] > 2000V

I_{Busbar_System} = 1.2 x 1200 = 1440 A

Which equals in secondary with respect to CT ratio of Busbar = 1440 / 2000 = 0.72 A

While I_Operating < I_{Busbar_System} then Shunt resistor must be calculated to draw the difference

Factors C & 𝞫 from the Metrosil Manual / Data sheet ((Below the tables))

For calculating C factor, take the average where, C= (405+495)/2 = 450