Lets keep this simple, Basic TX Line theory shows us that we can represent any tx line with a = combination of lumped elements R, L, C and G. So you can get a feeling for this just think about R and the resistance = of the copper of the line, L as its inductance, C as the capacitance = between the conductors and G and the conductance between the conductors. = You can estimate these quantities in a number of ways such as:- 1. Look them up in a book for the particular structure (normally given = per unit length) 2. Calculate them using a handbook method such as those given in the = classic book by grover or others 3. Use an E-M simulator to determine them for you. The E-M simulator works out the values of R L C and G from the field = distributions it found on its simulation runs. However, the way it = arrives at the values depends on what quantity the simulator found to = start with and how it found it. Most simulators will give you pretty = good values for C and L but you need to take care with the R and G. The = values of R and G are the losses in the system. Some simulators use = formulae to account for skin effect etc so don't be surprised to find = that high frequency losses are not that accurate! If the simulator is a = true 3D beast and you have a good fine mesh into the metal/Substrate you = should get a good idea of R and G. Heres the basics of how you get at R L C and G For Capacitance C =3D Q/V (By definition) then we use Maxwell's First Equation [Guass's Law for Electric = Fields]which is the surface integral of the electric flux density to get = Q. We get V from Maxwell's Third Equation [Faraday's Law] which is = simply the integral of the electric field over the distance of interest. = The solver can do all this sad maths for you. For Inductance L =3D Flux linkages/Current The linkages or flux you get from the flux = density multipled by area. Flux densisty comes from Amperes law.=20 For resistance At DC and low frequencies the resistance of any conducting structure is = easily ascertained from the conductor cross sectional area 'A', length ' = l' and material conductivity 'r':- Rdc =3D r.l /A However, at higher frequencies the resistance of a conducting structure = becomes a function of frequency, conductor geometry and the material = properties of its local environment. Not surprisingly, it is very = difficult to determine precise values of resistance at high frequencies. = =20 Your probably asleep now so that will be enough steve =20 Steve Rogers B.Eng (Hons) C.Eng IEE RF Design Engineer Micromill Electronics Limited Leydene House Waterberry Drive=20 Waterlooville Hampshire PO7 7XX Tel: +44 (0) 23 9236 6600 Fax: +44 (0) 23 9236 6673 Registered No. 1456922 (England). =20 Registered Office Brook Road Wimborne, Dorset BH21 2BJ *********************************************************************** "This email and any attached files are confidential and may be legally = privileged.If you are not the addressee, any disclosure, = reproduction,copying, distribution,, or other dissemination or use of = this communication is strictly prohibited. If you have received this = transmission in error please notify the sender immediately and then = delete this email. It is the policy of Micromill Electronics Limited that no legally = binding statements, representations or commitments (collectively = 'statements') may be made by email. Any such statements must be = confirmed either by facsimilie or by post before they will have legal = effect. The sender of this email is not authorised to commit the company = in any way and the addressee is hereby formally notified of that fact." *********************************************************************** -----Original Message----- From: karan bagga [mailto:kbagga31@xxxxxxxxx] Sent: Monday, August 11, 2003 9:11 AM To: si-list@xxxxxxxxxxxxx Subject: [SI-LIST] need help on RLGC Hi=20 I am new to this and need to know the basics on how to represent a PCB = Track using RLGC matrix ? How does 3d extractor works and extracts the RLGC matrix for a given PCB = track or a via. Any pointers or links will be of great help Thanks in Advance, Best Regards Karan --------------------------------- Do you Yahoo!? Yahoo! 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