Calculating Differential Impedance of Parallel PCB Traces Without Ground Plane Jim Hagerman, IBiS Networks

Transcription

1 Calculating Diffeential Impedance f Paallel PCB Taces Withut Gund Plane Jim Hageman, IBiS Netwks The Pblem While facing a ecent design challenge, I came upn the need t calculate the diffeential impedance f a pai f taces n a cicuit bad when n gund plane was pesent. A few days seaching the Intenet tuned up vey little elevant infmatin. I did lcate ne idiculusly cmplicated fmula that used elliptic integals f the fist kind, which t be hnest, was way ve my head. I needed smething simple. Paallel Wies Ftunately thee was a gd place t stat. A lng time ag smebdy wked ut a vey gd appximatin f the impedance f a pai f paallel ( twisted pai) wies in a dielectic. It is given as: 120 2s ln d We can mdify this a bit clse t u desied esult by enclsing half f the css sectin in a dielectic f. The inductance f the taces will nt change, but capacitance will incease, theeby lweing impedance. Half f the electic fields ae in ai, the the half in dielectic. Theefe, the ective pemittivity will be: ( 1) 1 + = + = Hweve, tw pblems emain. One is that the PCB is nt infinitely thick. The the is that PCB taces ae nt und. Simulats The bvius slutin t this pblem is t emply ne f the many elect-magnetic simulats available. But we dn t all have access t ne! And they can be athe

2 expensive. Ftunately, I was lucky t get a 15-day evaluatin license f the Si8000 simulat fm Pla Instuments. With this handy tl I was able t easily calculate esults f many cnditins. It was based n these data that I evese engineeed an appximate fmula. PCB Thickness The fist challenge was hw t deal with the ective pemittivity due t finite thickness f the PCB. We knw it shuld ange fm 1.00 t ( + 1)/2 as the thickness ges fm ze t vey thick. But what functin t use? While bwsing thugh my 35 yea ld CRC Handbk, I fund the actangent() t have just the ight ppeties. F psitive values the actangent ges fm ze t π/2. It was a simple matte f scaling the amplitude and adding an ffset f 1.00 t get a wking esult. It simplified t: 1 η h = 1 + actan π w Flat Cnducts Given the physical similaities t the und wie cnfiguatin, the new fmula takes the same fm: χ 2s ln w With a fcus n lw cst, this analysis ptimizes esults based n FR4 ( = 4.3) mateial with ealistic cnduct sizes and spacing.

3 Using a cnduct width f 20 mils and a spacing f 50 mils, the simulat gave an impedance f 256 hms in ai. With a PCB thickness f 63 mils, the esult was 163 hms. Using a little algeba and (2s/w) = 5, the value f χ can be calculated as ughly 160. We nw have the esult: 160 2s ln w Nw we just need t cuve fit the dielectic. The endpints ae aleady cect, with f eithe 1.00 ( + 1)/2. Using the 163 hms esult fm the abve cnditins, we btain that must be Reaanging u fmula, η can be calculated as: η = w h ( 1) π tan = Runding ff t simplify things, we end up with the appximatin f: 1 2h = 1+ actan π w This was pltted f an FR4 cicuit bad as a functin f the thickness t tace width.

4 Results The fllwing plt shws the calculated vesus simulated esults f tw diffeent bad examples. The lwe tace is u 63 mil thick FR4 bad; the uppe is f a Rges 4003 ( = 3.55) 30 mil thick bad. The fmula gives supisingly accuate esults with an e f nly a few pecent ve this useful ange. Oh, I fgt t mentin, this analysis is based n 1 unce cppe, a tace thickness f 1.2 mils. Simulatins wee un t test the sensitivity f this paamete.

5 The ect f slde mask was als investigated, which had a simila ect, deceasing the impedance. The lwe tace is with slde mask. Cnclusin This newly deived empiical fmula pvides vey gd estimates f impedance ve a wide ange f cnditins and mateials. I fund it vey useful f ptimizing the feed f my cicuit bad antenna pject.