# IT system - Fault protection

## First fault situation

In IT system, the first fault to earth should not cause any disconnection.

The earth-fault current which flows under a first-fault condition is measured in mA.

Two different examples of earth-fault current calculation are presented below .

### Example 1

In the simplified circuit presented on Fig. F32, the neutral point of the power-supply source is connected to earth through a 1500 Ω resistance. The current through the earthing resistor will be 153 mA in case of a fault (on a 230/400 V 3-phase system). The fault voltage with respect to earth due to this current is the product of this current and the resistance of the PE conductor plus earthing resistor (from the faulted component to the electrode), which is negligible.

Fig. F32﻿Simplified circuit

### Example 2

For a network formed of 1 km of conductors as represented on Fig. F33, the leakage impedance to earth (≈ 1 μF / km) is represented by the capacitors C1 to CN. The capacitive impedance ZC is in the order of 3500 Ω per phase. In normal operation, the capacitive current[1] to earth is therefore:

$I_c= \frac{U_0}{Z_c}= \frac{230}{3500}=66\ mA$ per phase.

Fig. F33﻿Fault current path for a first fault in IT system

During a fault between phase 1 and earth as in Fig. F33, the voltage and current vectors can be represented as shown on Fig. F34.

The fault current passing through the electrode resistance RA is the vector sum of the currents:

• Capacitive currents in the two healthy phases IC2 and IC3,
• Capacitive current in the neutral ICN,
• Current in the neutral impedance, Id1.

Fig. F34﻿Vector representation of voltages and currents in case of fault between phase 1 and earth

The voltages of the healthy phases have (because of the fault) increased to √3 times the normal phase voltage, so that the capacitive currents increase by the same amount: IC2 = IC3 = IC.√3 = 66 x √3 = 114 mA

The neutral voltage is 230V, so the capacitive neutral current is: ICN = IC = 66 mA

These currents are displaced one from the other by 30°, so that the total vector addition Id2 amounts to:

$2(I_c.\sqrt{3}).\frac{\sqrt{3}}{2} + I_c = 3.I_c + I_c = 4. I_c = 4 \times 66 = 264 mA$

e.g. for 3L+N circuits, the capacitive current to earth value increases by a factor of 4 during phase-to-earth fault, compared to its value in normal operation IC

The current Id1 through the neutral resistance is 153 mA (see simplified example above)

The current through the fault to earth is given by the vector sum of the neutral resistor current Id1 (= 153 mA) and the capacitive current Id2 (= 264 mA).

It is then equal to:

$\sqrt {153^2 + 263^2} = 304mA$

Considering for example an earthing resistance RA equal to 50 Ω, the fault voltage Uf is therefore equal to: 50 x 304 x 10-3 = 15.2 V, which is obviously harmless.

### Recommendation

To take full benefit of the continuity of service on first-fault condition provided by the IT system:

• A permanent monitoring of the insulation to earth must be provided, coupled with an alarm signal (audio and/or flashing lights, etc.) operating in the event of a first earth fault (see Fig. F35)
• The location and repair of a first fault is imperative if the full benefits of the IT system are to be realized. Continuity of service is the great advantage afforded by the system. As continuity of service is provided, it is not mandatory to repair the fault immediately, avoiding to operate under stress and urgency.

Fig. F35﻿Example of phase-to-earth Insulation Monitoring Device used in IT system

## Second fault situation

The second fault results in a short-circuit between active conductors (phases or neutral) through the earth and/or through PE bonding conductors (unless occurring on the same conductor as the first fault). Overcurrent protective devices (fuses or circuit breakers) would normally operate an automatic fault clearance.

Particularities and limitations specific to IT system are given in IT system - Implementation of protections.

Fault clearance is carried out differently in each of the following cases (see Fig. F36):

Fig. F36﻿Two different situations to be considered

### 1st case

It concerns an installation in which all exposed conductive parts are bonded to a common PE conductor, as shown in Fig. F37.

Fig. F37﻿Circuit breaker tripping on double fault situation when exposed-conductive-parts are connected to a common protective conductor

In this case, no earth electrodes are included in the fault current path, so that a high level of fault current is assured, and conventional overcurrent protective devices are used, i.e. circuit breakers and fuses.

To adjust the protective devices, the short-circuit current must be calculated, using one of the different methods applicable to TN system, as already presented in TN system - Earth-fault current calculation

• Methods of impedance
• Method of composition
• Conventional method

The first fault could occur at the end of a circuit in a remote part of the installation, while the second fault could feasibly be located at the opposite end of the installation.

For this reason, it is conventional to double the loop impedance of a circuit, when calculating the anticipated fault setting level for its overcurrent protective device(s).

Where the system includes a neutral conductor in addition to the 3 phase conductors, the lowest short-circuit fault currents will occur if one of the (two) faults is from the neutral conductor to earth (all four conductors are insulated from earth in an IT scheme). In four-wire IT installations, therefore, the phase-to-neutral voltage must be used to calculate short-circuit protective levels i.e. :

$0.8\,\frac{U_0}{2\,Z_c}\ge I_a$[2]

where

U0 = phase to neutral voltage
Zc = impedance of the circuit fault-current loop (see TN system - Principle)
Ia = current level for trip setting

If no neutral conductor is distributed, then the voltage to use for the fault-current calculation is the phase-to-phase value, i.e.

$0.8\,\frac{\sqrt{3}\,U_0}{2\,Z_c}\ge I_a$[2]

The settings of overcurrent tripping relays and the ratings of fuses are the basic parameters that decide the maximum practical length of circuit that can be satisfactorily protected, as discussed in TN system - Earth-fault current calculation.

Note: In normal circumstances, the fault current path is through common PE conductors, bonding all exposed conductive parts of an installation, and so the fault loop impedance is sufficiently low to ensure an adequate level of fault current.

### 2nd case

It concerns exposed conductive parts which are earthed either individually (each part having its own earth electrode) or in separate groups (one electrode for each group). If all exposed conductive parts are not bonded to a common electrode system, then it is possible for the second earth fault to occur in a different group or in a separately earthed individual apparatus.

Rules of TT system apply.

Additional protection to that described above for case 1, is required, and consists of a RCD associated to the circuit breaker controlling each group and each individually-earthed apparatus.

The reason for this requirement is that the separate-group electrodes are “bonded” through the earth so that the phase to phase short-circuit current will generally be limited when passing through the earth bond by the electrode contact resistances with the earth, thereby making protection by overcurrent devices unreliable.

The more sensitive RCDs are therefore necessary, but the operating current of the RCDs must evidently exceed that which occurs for a first fault (see Fig. F38).

Leakage capacitance (µF) First fault current (A)
1 0.07
5 0.36
30 2.17

Note: 1 µF is the 1 km typical leakage capacitance for 4-conductor cable.

Fig. F38Correspondence between the earth leakage capacitance and the first fault current