Calculation of electrical cable parameters for pulse propagation model

Abstract

It is necessary to take into consideration nanosecond pulse propagation in bursts immunity tests and in EMC research work. Electrical cable between pulse noise source and equipment influences on malhnction level. The model of pulse propagation in time domain is used for calculafion. Parameters of power cables for the model and effects of propagation are discussed. IntroduciionNanosecond pulses are generated in electrical system in commutation processes. Pulses can lead to malfunction of eiectronic equipment. Amplitude and duration of pulses on the equipment depend on propagation in power cables. The propagation parameters for power cables are not investigated enough. Time constants are not determined. It is important to take into consideration wave effects in cables. Calculation of the time constants is presented for different KNR type cables and for different cable lengths. Calculation of pulse parameters for the line with change of height above steel construction is given in the paper. Propagation effects in cablesIt is possible to distinguish the following cables effects which determines pulse parameters during propagation:-Attenuation of pulse due to losses. -Change value of wave. impedances along a propagation way. r -for losses only in dielectric , l Time parameters tM and r, can be calculated with the following expressions: 2 A2 rM = roMl ; roM = -for Bf = 0, ?r B z, , = rOD1 ; zOD = -for A-& = 0, 3t where zOM is constant time for metal losses (s/ km'), To, is time constant for dielectric losses ( s h ' ) . Parameters70M,ronand length of a cable 1 determine an attenuation in metal and in dielectric. It is not necessary to consider dielectric losses if the following ratio is close to zero [2]: -' D 2 --=-+O,where *OD 1 m=-'OM 2. Attenuation effect TM 'COM.1 m.1 f O DThe attenuation coefficient in frequency domain has three components:The result of calculation (table 1, fig. 1) shows, that the increasing of square sections of cables f?om lmm2 up to 240"' influence on ZoD less, than on T O M . a = A f i + B f +a,,where'the first component characterizes losses in metal, coeficients =4.76.10-8(~~~2), rooo =3.92, the second component -losses in dielectric, and the last one -losses for direct current [ 11.The step response ofthe cable cm be defined by h e ( s h ' ) and m is equal to 250 for the cable KNR 3x1 " ' 9 but =9.8.10-7 (sh2)s rOD =4*29*10-9(slkm2) --following formula: and m=ll for the cable KNR 3x240 m2 ( fig. la,b); It means that it is not necessary to consider dielectric losses in short low current cables.-for losses only in metal 0-7803-9374~0/05/%20.00 6 2005 IEEE