Circuit Cellar - GIFT

CC-2015-06-Issue 299

Circuit Cellar - GIFT

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CIRCUIT CELLAR • JUNE 2015 #299 48 COLUMNS THE CONSUMMATE ENGINEER T o continue our exploration of shielding principles, we'll examine shielding of cables in this article. Figure 1 models electromagnetic field (EM-field) coupling in a coaxial cable. The cable is exposed to an electromagnetic (EM) wave P. The conductor inside the cable is terminated by load resistors R and the shield is bonded to ground through impedance Z G /2 at each end. Let's consider E-field coupling first. Electric field (E-field) induces current I S in the shield which in turn produces a voltage across resistors R due to the transfer impedance Z T between the shield and the wire. There is also an inherent capacitance (C) between the shield and the inner wire, as you can see in the coaxial cable cross-section in Figure 1. There are essentially two mechanisms causing the electric part of the EM field P coupling to the internal (i.e., shielded) conductor. One is the shield-induced current coupled with the transfer impedance. The other is through shield perforations. There is insufficient space in this article to elucidate all the mathematical origins of constants and variables used in the following equations. I will just focus on the final expressions needed to understand the working principles of the cable shielding. Shield current (I S ) due to e-field irradiation, where S is the length of the cable (b = 2p/l) and bS << 1, is calculated as follows: I = E h + Z Z S S G 0 G × ( ) × × × × Z Z 0 2 β The load voltage across resistor R is caused by the transfer impedance (Z T ): V 1 = I Z S T × Penetration of the e-field caused by the shield perforations also causes current in the inner wire. F S is an empirical value, approximately 0.02 for quality shields with optical coverage of 85%. I = E h + R R S F 0 S W Z Z × ( ) × × × × × 0 2 β I W induces voltage on the load resistor, R. V 2 = I R W × The total voltage induced by the E-field on the load will be: V = V V 2 1 × The shield is bonded to ground through two low impedances, Z G /2. If the shield is grounded at a single point only by impedance Z G , no shield current I S can flow and the interfering voltage (V) on the load resistor will be negligible. Unfortunately, this is true at low frequencies only. Once the length of Shielding 101 (Part 2) In the first part of this series, George introduced the topic of shielding and the physics behind it. This month he tackles the shielding of cables. By George Novacek (Canada) Cable Shielding [2] [3] [4] [1] [5]

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