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Lect19_20091012_EE429_529

Lect19_20091012_EE429_529 - ECE 529 Introduction to EMC...

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ECE 529 Introduction to EMC University at Buffalo Dr. James J. Whalen Lecture 24 20Nov96 iolci]»> Wed 10 { 131 0 8 Moll
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// Phase Ie 10 // / Phase (R1) r-, lies / // -~ LlSN / \ // . Neutral 10 Product / Neutral (N) «: / . - ) Ie /// -~ ./ Green .// Ie (safely) / , Green (safely) wire wire I I I <:£J Commercial power Spectrum analyzer system 6' FIGURE 9.;' Illustration or the use or a line impedance stabilization network (LISN) in the measurement or conducted emissions of a product. .1 o.( . .1..
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lISN 50 J,H ..•Phase vyy ~. lp 50 ~H -: Product I\- Neutral y ~01 - IN 1 ~11lF - I--- ::-~ ::;:: =;:: 0.1 IlF ( r----. ~F - V- I r-l-- r--- I 1+ I ~ I ~ 500 V:~1kO I 500 Vp ~ 1 kO I . I . Ip LReceiver-.l I - -T- - I (dummy .1 , load) Green wire ...L '·2 FIGURE 9.2 Illustration of the LISN circuit. the input of the test receiver. It is instructive to compute the impedances of these elements at the lower frequency limit, 450 kHz, and the upper frequency limit, 30 MHz, of the FCC regulatory limit. These are D L towel' I~e F I'ef:). 2{;OHz O. ()/A'11L Element "7! LSbklit Z450 kHz 50 I1H •... 'f 71l. 141.3n 9420n 0.1 I1F , .... to .n, 3.54n 1 I1F '" J..n, 0.354n 0.053n 0.0053 n 2(cS~ 2,6J,-K11- 2 a.., ! -L.
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" Phase - "\ - Neutral Green - / wire Phase 1--------1-----1 I Neutral Product I r -.- -I------f I I ~f500 V;~500 I I Green L -;- __ +-_,--_--1 LlSN wire 6..1 fiGURE 9.3 Equivalent circuit of the LISN as seen by the product over its intended frequency range of use. Jl p = 50f p Jl N = 50l N (9.1a) (9.1b)
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. \ + ----------, 10 Ie 1 Ip 1 Phase 1 Ie \~ ,10 10 ~I' IN Product . Neutral ~500 r ~I ~ . +C[ > I ~ (i~. 500 I Ie - I Green -:!::- I 2Ie wire I ISN _________ ...1 r-- I I 1 I I 1 I I I 1 1 I 1 L __ ~ fiGURE fA'i Illustration of the contributions of differential-mode and common-mode current components on the measured conducted emissions. Ip = t; + ID (9.2a) IN = t;- ID (9.2b) Solving these gives ~ 1 ~ ~ (9.3a) ID = 2(IP - IN) ., "" 1 "" "" (9.3b) Ie = 2(JP + IN) The measured voltages are V p = 50(Ie + I D ) (9.4a) V N = 50(Ie - I D ) (9.4b) V p = 50I e , Ie» ID (9.5a) V N = safe, Ie»
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