1.5 Equivalent <strong>system</strong> for a network with <strong>unearthed</strong> or impedance-earthed neutralA few def<strong>in</strong>itions and assumptions are givenbelow <strong>in</strong> order to def<strong>in</strong>e the equivalent <strong>system</strong>for this network (see fig. 4 ):c the neutral po<strong>in</strong>t is <strong>unearthed</strong> or earthed by animpedance (Z N ) of high value (normally 1 kΩ to2kΩ) whose earth connection is equivalent to aresistance (R B );c the load frames are <strong>in</strong>terconnected either fullyor by group. For EMC reasons (see“Cahier Technique” no. 187), it is advisable to<strong>in</strong>terconnect all the application frames of thesame <strong>in</strong>stallation and to connect them to thesame earth connection (resistance R A );c the earth connections (R A and R B ) are<strong>in</strong>terconnected (<strong>in</strong> most cases), or separate.NB: Two earth connections are considered to beseparate if they are more than 8 m apart;c each live conductor has, with respect to earth,an impedance made up of a resistance and acapacity.N321NPEZ NRN R1 R2 R3CN C1 C2 C3R BR AFig. 4 : equivalent <strong>system</strong> of a network with <strong>unearthed</strong> or impedance-earthed neutral.Cahier Technique Schneider Electric no. 178 / p.8
2 <strong>The</strong> 1 st <strong>in</strong>sulation fault with the <strong>IT</strong> <strong>earth<strong>in</strong>g</strong> <strong>system</strong>In normal operat<strong>in</strong>g conditions, safety of personsis guaranteed when contact voltage is less than50 V as per standard IEC 60364 (NF C 15-100).When this contact voltage is exceeded, thesestandards require automatic open<strong>in</strong>g of thecircuit. <strong>The</strong> follow<strong>in</strong>g section shows how use ofthe <strong>IT</strong> <strong>earth<strong>in</strong>g</strong> <strong>system</strong> for network operationprevents tripp<strong>in</strong>g on the first <strong>in</strong>sulation fault.2.1 Calculat<strong>in</strong>g fault currents and contact voltage on the first faultGeneral case (resistive fault)Should a fault with a resistive value Rd occurbetween phase 3 and the earth, a fault current I dflows <strong>in</strong> the neutral impedance and <strong>in</strong> thecapacities C1, C2 and C3 (see fig. 3a).Assum<strong>in</strong>g that the phase-to-earth capacities arebalanced (C1 = C2 = C3 = C), the fault currenthas the follow<strong>in</strong>g value:1+ 3j CωZI d =UN0Rd+ ZN + 3jCωZNRd<strong>The</strong> capacitive current is written as:3j CωZI c =UN0Rd+ ZN + 3jCωZNRdand the current <strong>in</strong> the impedance Z N :UI N =0Rd+ ZN + 3jCωZNRd<strong>The</strong> contact voltage U C (contact voltage betweenthe frame of a faulty device and another frame orthe earth) is calculated from the fault current I dflow<strong>in</strong>g <strong>in</strong> the earth connection resistance R A ofthe application frames if they are not<strong>in</strong>terconnected, else R B (only network earthconnection):U C = R A I d .Case of the full faultThis paragraph calculates the configurationgenerat<strong>in</strong>g the highest contact voltage (U C ):thus for a fault occur<strong>in</strong>g on a frame with earthconnection separate from that of Z N .By application of the above formulae, whereR d = 0, we obta<strong>in</strong>:UI d = 0ZN+ 3j CωUUc = R 0AZ N + 3j Cω<strong>The</strong> capacitive current is equal to:I C = +3j Cω U 0and the current <strong>in</strong> impedance Z N :UI N = 0ZNIn the various examples below, studied forZ N = ∞ (<strong>unearthed</strong> neutral) and Z N = 1 kΩ(impedance-earthed neutral), the calculationsare made for a network <strong>in</strong> the <strong>IT</strong> <strong>system</strong>,400 VAC (U 0 = 230 V), where:R A , earth connection resistance = 10 ΩR d , <strong>in</strong>sulation fault value = 0 to 10 kΩ.c Case 1:Low capacity network (e.g. limited to anoperat<strong>in</strong>g theatre)C1 = C2 = C3 = C = 0.3 µF per phase.c Case 2:Power network, whereC1 = C2 = C3 = C = 1.6 µF per phase.c Case 3:Very long power network, whereC1 = C2 = C3 = C = 10 µF per phase, i.e.roughly 40 km of cables!<strong>The</strong> results of all these calculations, grouped <strong>in</strong>the table <strong>in</strong> figure 5 , confirm the low faultvoltage (≈ 20 V <strong>in</strong> the most unfavourable cases),ensur<strong>in</strong>g cont<strong>in</strong>uity of operation, without risk forpersons, of a network designed us<strong>in</strong>g the <strong>IT</strong><strong>system</strong>. <strong>The</strong>y prove that addition of animpedance between the neutral and the earthhas very little effect on contact voltage.R d (kΩ) 0 0.5 1 10Case 1 Z N = ∞ U C (V) 0.72 0.71 0.69 0.22C R =1µF I d (A) 0.07 0.07 0.07 0.02Z N = 1 kΩ U C (V) 2.41 1.6 1.19 0.21I d (A) 0.24 0.16 0.12 0.02Case 2 Z N = ∞ U C (V) 3.61 2.84 1.94 0.23C R =5µF I d (A) 0.36 0.28 0.19 0.02Z N = 1 kΩ U C (V) 4.28 2.53 1.68 0.22I d (A) 0.43 0.25 0.17 0.02Case 3 Z N = ∞ U C (V) 21.7 4.5 2.29 0.23C R =30µF I d (A) 2.17 0.45 0.23 0.02Z N = 1 kΩ U C (V) 21.8 4.41 2.26 0.23I d (A) 2.18 0.44 0.23 0.02Fig. 5 : comparison of fault currents and contactvoltages on a first fault.Cahier Technique Schneider Electric no. 178 / p.9