Four Special Differential Protections And Their Respective CT Requirements

Special cases of differential protection

Although the manufacturers of relays dedicated to differential protection impose the necessary CT secondary characteristics required for proper operation, it is useful, for understanding and avoiding errors, to possess minimum knowledge of this type of protection.

A differential protection monitors an area limited by CTs which measure incoming and outgoing currents. If the outgoing currents are not consistent with the incoming currents, this is normally because a fault has occurred in the protected area.

Now, let’s examine following protections with their respective CT requirements:

1. 1. High impedance protection,
   2. Pilot wire protection,
   3. Percentage differential protection and
   4. Low impedance differential protection.

According to the type of protection and its use, relay manufacturers have had to apply a variety of principles varying in complexity in order to guarantee the stability of their relays against transient phenomena likely to cause nuisance tripping of this protection.

1. High impedance differential protection

This type of protection is normally used for protection of motors, generators, busbars as well as for “restricted earth fault” (REF) protection of transformers.

General

This type of protection is used to protect an area with the same voltage level. In healthy conditions, the input current i’_e is identical to the output current i’_s and thus the differential current i’_d = 0 (see figure 1)

A high fault current can flow through the monitored area and cause CT saturation, hence the risk of nuisance tripping of the protection (non stability). The “stability” of the relay is obtained by connecting it in series with a “stabilising” resistance R_st.

This results in (1): (R_st + R_p) ≥ (R_ct + 2 R_L) × I_ssc / Ir

where:

• I_ssc = maximum through current observed at the CT secondary,
• I_r = relay secondary setting current.
• R_st may vary between a few ohms and a few hundred ohms (exceptionally it may be greater than 1000 ohms).

For the relay to operate properly at I_r if a fault occurs in the area, the knee point voltage V_k must be greater than:
2 × (R_st + R_p + R_ct + 2 R_L) × Ir .

As a rule R_ct + 2 × R_L are negligible compared with R_st + R_p thus (2):
V_k ≥ 2 × Ir × (R_st + R_p)

By combining the equations (1) and (2) we find (3):
V_k ≥ 2 × I_ssc × (R_ct + 2 × R_L)

The result of these observations is that the CTs are optimized if R_ct and V_k are as low as possible and if the through current (seen from the CT secondary, i.e. I_scc) is defined without excess.

Whatever the application where the high impedance differential is used, all the CTs must have:

1. The same ratio,
2. The same magnetising curve (same minimum V_k),
3. The same maximum R_ct,
4. And comply with expression (3). For V_k, since the relevant CTs are not at the same distance from the relay, take the maximum R_L.

For this protection, a maximum value of the magnetizing current I_o must also be defined at V_k / 2 according to the required sensitivity.

For the relay to detect a current I_r, a voltage V_s = V_k / 2 must be developed at the terminals of the parallel-connected CTs. For this purpose, the minimum primary current I_rms really detected by the relay will be:

I_rms = n × (I_r + ρ × I_o)

where:

• n = CT ratio and
• ρ = number of parallel-connected CTs (there may be many of them on a busbar protection!)

Application to the “motor” differential protection

The maximum through current for which the motor must remain insensitive is in this case the motor starting current:
I_ssc = I_st (seen at the secondary).

If you do not know this current Ist, you know that:
I_st < 7 × I_n motor

Application to the “generator” differential protection

The maximum through current is in this case the short-circuit current supplied by this generator only. If you know the generator subtransient reactance X’’ %, the following will be taken:

I_ssc = I_n × 100 / X”

If this value is not known, X’’ % = 15 will be taken.

Note – The peak voltage at the CT secondary must be calculated using:
I_ssc maxi = I”_generator + I_sc network

Application to the “busbar” differential protection

In this case, the through current is equal to switchboard I_sc:
I_ssc = switchboard I_sc seen from the CT secondary

Application to the “restricted earth fault” differential protection of transformers (REF)

In the case of figure 3a , this protection detects the insulation faults at the transformer secondary windings and up to the CTs located downstream.

In this case also, we shall calculate R_st and V_k based on the maximum through current in the CTs for a fault outside the protected area. As a first approximation, we can say that this current is less than the current limited by the transformer
impedance, i.e.

I_through = P_sct / (U_n × √3)

where P_sct = P_n × 100 / Z_sc (transformer short-circuit power)

If we know the upstream short-circuit power (P_u), a more accurate value can be calculated replacing:
P_sct with (P_sct × P_u) / (P_sct + P_u)

The through fault current must then be converted into I_ssc seen at the CT secondary.

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2. Line or cable differential protection with pilot wires

A relay of this type is installed at each end of the cable or the line. On the pilot wires, each relay reproduces a voltage which is the image of the sum:

a×I₁ + b×I₂ + c×I₃ + d×I_h

If the two voltages are different, both relays trip.

In this case also, class X CTs are required, and each manufacturer gives an empirical formula for the minimum knee point voltage V_k.

An example of the minimum knee point voltage required:

V_k mini = 0,5 ×N × k_t × I_n × (R_ct + X × R_L)

where N, k_t and X are constants associated with the relay response time, its sensitivity and its type of connection.

Another example:

V_k mini = 50/I_n + I_f × (R_ct + 2 × R_L)

where

• I_n is the CT nominal secondary current (1 or 5 A),
• I_f is the through short-circuit current, seen at the CT secondary.

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3. Percentage biased differential protection for transformers

The term “percentage differential” stems from the fact that the operating threshold increases with the through current.
Simple comparison of the currents in each upstream phase with the currents in the same downstream phases is not suitable for transformer differential protection.

This is because:

1. The upstream and downstream currents of a power transformer do not have the same amplitude or the same phase angle,
2. When the transformer is energised, its magnetising current is only seen upstream,
3. The presence of an earthing generator in the protected area (e.g. earthing the transformer neutral) can trip the protection, while the fault is, for example, located on a downstream feeder.

Precautions to be taken to solve these problems:

The aim is to ensure that the relay sees upstream and downstream currents of the same amplitude and in phase during normal operating conditions. This can be achieved by intelligent choice of CT ratio and connections.

Furthermore, it must be noted that all the “transformer differential” relays are immunized to the 2^nd order harmonic blocking their operation when the transformer is energised.

CT voltage V_k

In 99 % of cases, a class X is requested. The minimum knee point voltage is imposed and depends on the resistance of the secondary winding “Rct” of the CT and of its real load R_r. More complex specifications are sometimes mentioned, which include the X/R ratio of the network or the magnetising current of the power transformer.

However, faced with the problems that users have in obtaining all these parameters, relay suppliers sometimes provide simplified empirical formulas which lead to a slight oversizing.

Examples of minimum knee point voltage imposed for the Schneider Electric’s Sepam protection relay:
V_k mini = A × I_b (R_ct + 2 × R_L)

where:

• 2 × R_L = total resistance of the secondary wiring,
• Rct = CT secondary winding resistance,
• I_b = power transformer nominal current seen at the CT secondary,
• A = constant depending on transformer power

Some suppliers take into account the through current, for example:

on the star side of the power transformer,

and Vk ≥ 4 × I_f × (R_ct + 2×R_L + R_p) on the delta side of the power transformer. The through current will be defined in the same way as for the restricted earth fault protection.

NOTE: Use of matching CTs leads to different expressions of the knee point voltage for the main CTs which must take into account the extra load that they represent.

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4. Low impedance differential protection

This protection is used for busbar differential protection. It is very costly and space-consuming, as it requires a large number of modules and matching CTs which need one or more cubicles according to switchboard size (see figure 5).

In the case of a double busbar switchboard, the protection must be continually “informed” on the position of the transfer switches in order to direct the currents of each feeder and incomer to the relay associated with monitoring of the busbar on which this feeder or incomer is connected.

However, as saturation can be tolerated, knee point voltage requirements are less severe than for high impedance differential protection.

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Reference // Current transformers: how to specify them by P. Fonti (Schneider Electric)