Calculation of minimum levels of short-circuit current

If a protective device in a circuit is intended only to protect against short-circuit faults, it is essential that it will operate with certainty at the lowest possible level of short-circuit current that can occur on the circuit

In general, on LV circuits, a single protective device protects against all levels of current, from the overload threshold through the maximum rated short-circuit current breaking capability of the device. The protection device should be able to operate in a maximum time to ensure people and circuit safety, for all short-circuit current or fault current that may occur. To check that behavior, calculation of minimal short-circuit current or fault current is mandatory.

In addition, in certain cases overload protective devices and separate short-circuit protective devices are used.

Examples of such arrangements

Figure G43 to Figure G45 show some common arrangements where overload and short-circuit protections are achieved by separate devices.

Fig. G40 – Circuit protected by aM fuses

Fig. G41 – Circuit protected by circuit breaker without thermal overload relay

Fig. G42a – Circuit breaker D provides protection against short-circuit faults as far as and including the load

As shown in Figure G43 and Figure G44, the most common circuits using separate devices control and protect motors.

Figure G42a constitutes a derogation to the basic protection rules, and is generally used on circuits of busways (busbar trunking systems), lighting rails, etc.

Variable speed drive

Figure G42b shows the functions provided by the variable speed drive, and if necessary some additional functions provided by devices such as circuit-breaker, thermal relay, RCD.

Protection to be provided	Protection generally provided by the variable speed drive	Additional protection if not provided by the variable speed drive
Cable overload	Yes	CB / Thermal relay
Motor overload	Yes	CB / Thermal relay
Downstream short-circuit	Yes
Variable speed drive overload	Yes
Overvoltage	Yes
Undervoltage	Yes
Loss of phase	Yes
Upstream short-circuit		Circuit-breaker (short-circuit tripping)
Internal fault		Circuit-breaker (short-circuit and overload tripping)
Downstream earth fault (indirect contact)	(self protection)	RCD ≥ 300 mA or CB in TN earthing system
Direct contact fault		RCD ≤ 30 mA

Fig. G42b – Protection to be provided for variable speeed drive applications

Conditions to be fulfilled

The protective device must fulfill:

• instantaneous trip setting Im < Isc_min for a circuit-breaker
• fusion current Ia < Isc_min for a fuse

The protective device must therefore satisfy the two following conditions:

• Its breaking capacity must be greater than Isc, the 3-phase short-circuit current at its point of installation
• Elimination of the minimum short-circuit current possible in the circuit, in a time tc compatible with the thermal constraints of the circuit conductors:

[math]\displaystyle{ tc \le \frac{k^2S^2}{Isc_{min}\, ^2} }[/math] (valid for tc < 5 seconds)

where S is the cross section area of the cable, k is a factor depending of the cable conductor material, the insulation material and initial temperature.

Example: for copper XLPE, initial temperature 90 °C, k = 143 (see IEC60364-4-43 §434.3.2 table 43A and Figure G52).

Comparison of the tripping or fusing performance curve of protective devices, with the limit curves of thermal constraint for a conductor shows that this condition is satisfied if:

• Isc (min) > Im (instantaneous or short time delay circuit-breaker trip setting current level), (see Fig. G43 )
• Isc (min) > Ia for protection by fuses. The value of the current Ia corresponds to the crossing point of the fuse curve and the cable thermal withstand curve (see Fig. G44 and Fig. G45)

Fig. G43 – Protection by circuit breaker

Fig. G44 – Protection by aM-type fuses

Fig. G45 – Protection by gG-type fuses

Practical method of calculating Lmax

In practice this means that the length of circuit downstream of the protective device must not exceed a calculated maximum length: [math]\displaystyle{ \definecolor{bggrey}{RGB}{234,234,234}\pagecolor{bggrey}L_{max}=\frac{0.8\ U\ S_{ph} }{2 \rho I_{m} } }[/math]

The limiting effect of the impedance of long circuit conductors on the value of short-circuit currents must be checked and the length of a circuit must be restricted accordingly.

For protection of people (fault protection or indirect contacts), the method to calculate the maximum circuit length for TN system and IT system (second fault) is presented in Earth-fault current calculation and Implementation of protections.

Two other cases are considered below, for phase-to-phase and phase-to-neutral shortcircuits.

1 - Calculation of L_max for a 3-phase 3-wire circuit

The minimum short-circuit current will occur when two phase wires are short-circuited at the remote end of the circuit (see Fig. G46).

Fig. G46 – Definition of L for a 3-phase 3-wire circuit

Using the “conventional method”, the voltage at the point of protection P is assumed to be 80% of the nominal voltage during a short-circuit fault, so that 0.8 U = Isc Zd, where:

Zd = impedance of the fault loop
Isc = short-circuit current (ph/ph)
U = phase-to-phase nominal voltage

For cables ≤ 120 mm², reactance may be neglected, so that [math]\displaystyle{ Zd=\rho \frac{2L}{Sph} }[/math]

where:

ρ = resistivity of conductor material at the average temperature during a short-circuit,
Sph = c.s.a. of a phase conductor in mm²
L = length in metres

The condition for the cable protection is Im ≤ Isc with Im = trip current which guarantees instantaneous operation of the circuit breaker..

This leads to [math]\displaystyle{ Im \le \frac{0.8 U}{Zd} }[/math]  which gives [math]\displaystyle{ L \le \frac{0.8\ U\ S_{ph}}{2 \rho I_{m}} }[/math]

For conductors of similar nature, U and ρ are constants (U = 400 V for phase-to phase fault, ρ = 0.023 Ω.mm²/m for copper conductors), so the upper formula can be written as:

[math]\displaystyle{ L_{max}=\frac {k\ S_{ph}}{I_{m}} }[/math]

with Lmax = maximum circuit length in metres

For industrial circuit breakers (IEC 60947-2), the value of Im is given with ±20% tolerance, so Lmax should be calculated for Im+20% (worst case).

k factor values are provided in the following table, for copper cables, taking into account these 20%, and as a function of cross-section for Sph > 120 mm²

Cross-section (mm2)	≤ 120	150	185	240	300
k (for 400 V)	5800	5040	4830	4640	4460

2 - Calculation of L_max for a 3-phase 4-wire 230/400 V circuit

The minimum Isc will occur when the short-circuit is between a phase conductor and the neutral at the end of the circuit.

A calculation similar to that of example 1 above is required, but for a single-phase fault (230V).

• If Sn (neutral cross-section) = Sph

Lmax = k Sph / Im with k calculated for 230V, as shown in the table below

Cross-section (mm2)	≤ 120	150	185	240	300
k (for 230 V)	3333	2898	2777	2668	2565

• If Sn (neutral cross-section) < Sph, then (for cable cross-section ≤ 120mm²)

[math]\displaystyle{ L_{max}= 6666 \frac{Sph}{Im}\frac{1}{1+m} }[/math]

[math]\displaystyle{ m=\frac{Sph}{Sn} }[/math]

Tabulated values for Lmax

Based on the practical calculation method detailed in previous paragraph, pre-calculated tables can be prepared.

In practice, the tables Fig. F25 to Fig. F28 already used in chapter Protection against electric shocks and electrical fires for earth-fault calculation can also be used here, but applying the correction factors in Fig. G51 below, to obtain Lmax value related to phase-to-phase or phase-to-neutral shortcircuits.

Note: for aluminium conductors, the lengths obtained must be multiplied again by 0.62.

Circuit detail
3-phase 3-wire 400 V circuit or 1-phase 2-wire 400 V circuit (no neutral)		1.73
1-phase 2-wire (phase and neutral) 230 V circuit		1
3-phase 4-wire 230/400 V circuit or 2-phase 3-wire 230/400 V circuit (i.e with neutral)	Sph / S neutral = 1	1
	Sph / S neutral = 2	0.67

Fig. G51 – Correction factor to apply to lengths obtained from Fig. F25 to Fig. F28 , to obtain Lmax considering phase-to-phase or phase-to-neutral short-circuits

Examples

Example 1

In a 3-phase 3-wire 400 V installation the short-circuit protection of a 22 kW (50 A) motor is provided by a magnetic circuit breaker type GV4L, the instantaneous short-circuit current trip is set at 700 A (accuracy of ±20 %), i.e. in the worst case would require 700 x 1.2 = 840 A to trip.

The cable c.s.a. = 10 mm² and the conductor material is copper.

In Fig. F25 the column Im = 700 A crosses the row c.s.a. = 10 mm² at the value for Lmax of 48 m. Fig. G51 gives a factor 1.73 to apply to this value for 3-phase 3-wire circuit (no neutral). The circuit breaker protects the cable against short-circuit faults, therefore, provided that its length does not exceed 48 x 1.73 = 83 metres.

Example 2

In a 3L+N 400 V circuit, the protection is provided by a 220 A circuit breaker type NSX250N with micrologic 2 trip unit having instantaneous short-circuit protection set at 3000A (±20 %), i.e. a worst case of 3600 A to be certain of tripping.

The cable c.s.a. = 120 mm² and the conductor material is aluminium.

In Fig. F25 the column Im = 3200 A (first value > 3000A, as the table already includes the +20% on Im in its calculation) crosses the row c.s.a. = 120 mm² at the value for Lmax of 125 m. Being a 3-phase 4-wire 400 V circuit (with neutral), the correction factor from Fig. G51 to apply is 1. In complement, as the conductor is aluminium, a factor of 0.62 has to be applied.

The circuit breaker will therefore protect the cable against short-circuit current, provided that its length does not exceed 125 x 0.62 = 77 metres.