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Transmission Line Basics Transmission Line Basics Prof. Tzong-Lin Wu NTUEE 1 Outlines Transmission Lines in Planar structure. Key Parameters for Transmission Lines. Transmission Line Equations. Analysis Approach for Z0 and Td Intuitive concept to determine Z0 and Td Loss of Transmission Lines Example: Rambus and RIMM Module design 2 Transmission Lines in Planar structure Homogeneous Inhomogeneous Microstrip line Coaxial Cable Embedded Microstrip line Stripline 3 Key Parameters for Transmission Lines 1. Relation of V / I : Characteristic Impedance Z0 2. Velocity of Signal: Effective dielectric constant e 3. Attenuation: Conductor loss ac Dielectric loss ad Lossless case 1 L Td Z0 C V p C C 1 c0 1 V p LC e Td 4 Transmission Line Equations Quasi-TEM assumption 5 Transmission Line Equations R0 resistance per unit length(Ohm / cm) G0 conductance per unit length (mOhm / cm) L0 inductance per unit length (H / cm) C0 capacitance per unit length (F / cm) dV KVL : (R0 jwL0 )I dz Solve 2nd order D.E. for dI V and I KCL : (G0 jwC0 )V dz 6 Transmission Line Equations Two wave components with amplitudes V+ and V- traveling in the direction of +z and -z rz rz V V e V e 1 rz rz I (V e V e ) I I Z0 Where propagation constant and characteristic impedance are r (R0 jwL0 )(G0 jwC0 ) a j V V R0 jwL0 Z0 I I G0 jwC0 7 Transmission Line Equations a and can be expressed in terms of (R0 , L0 ,G0 ,C0 ) 2 2 2 a R0G0 L0C0 2a (R0C0 G0 L0 ) The actual voltage and current on transmission line: az jz az jz jwt V (z,t) Re[(V e e V e e )e ] 1 az jz az jz jwt I(z,t) Re[ (V e e V e e )e ] Z0 8 Analysis approach for Z0 and Td (Wires in air) C = ? (by Q=C V) L = ? (by Ψ=L I) 9 Analysis approach for Z0 and Td (Wires in air): Ampere’s Law for H field II H(r)= d 2r c R2 IIR 2) B ds 00 dr ln(1 ) (in Wb) eTS r R 1 2 r 2 R2 10 Analysis approach for Z0 and Td (Wires in air): Ampere’s Law for H field 0 I R2 e ln( ) 2 R1 LI e / 11 The per-unit-length Parameters (E): Gauss’s Law 1) from gauss law D E d s Q S T total q1m E T ds 0 S q 2 0r R1 q 2) V= E d dr C T rR 2 2 0r q R ln 2 2R 01 12 The per-unit-length Parameters (E) qR V ln(2 ) 201R CQV / 13 c. For example Determine the L.C.G.R of the two-wire line. (note：homogeneous medium) S Inductance： r rw2 L= e w1 I e I I where 0I s-r w2 0 I s-r w1 e ln( )+ ln( ) e 2 rw1 2 r w2 (,) I (s-r )(s-r ) 00 =0 ln( w2 w1 ) 2 rw1 r w2 assume s rw1 , r w2 s2 L=0 ln( ) 2 rw1 r w2 14 Capacitance： S 1) e c 00 - - + + r rw2 2 0 w1 - C - 2 + + V s + - - ln ( ) q C/m -q C/m rrw1 w2 qqs-r s-r 2) V= ln(w2 )+ ln( w1 ) 20 r w1 2 0 r w2 q (s-r )(s-r ) = ln(w2 w1 ) 2 0 r w1 r w2 q s2 ln( ) if s rw1 , r w2 2 0 r w1 r w2 q 2 C= 0 the same with 1) approach V s2 ln ( ) rr w1 w2 15 The per-unit-length Parameters Homogeneous structure TEM wave structure is like the DC (static) field structure LG LC So, if you can derive how to get the L, G and C can be obtained by the above two relations. 16 The per-unit-length Parameters (Above GND ) 2C Why? L/2 17 , d. How to determine L,C for microstrip-line. 00 , 1) This is inhomogeneous medium. 11 2) Nunerical method should be used to solve the C of this structure, such as Finite element, Finite Difference... 3) But e can be obtained by 00 eC0 0 0 e C0 where C01 is the capacitance when medium is replaced by 0 medium. 18 Analysis approach for Z0 and Td (Strip line) Approximate electrostatic solution y 1. b x -a/2 a/2 2. The fields in TEM mode must satisfy Laplace equation 2 t (x, y) 0 where is the electric potential The boundary conditions are (x, y) 0 at x a / 2 (x, y) 0 at y 0,b 19 Analysis approach for Z0 and Td 3. Since the center conductor will contain the surface charge, so n x n y An cos sinh for 0 y b / 2 n1 aa odd (,)xy n x n Bn cos sinh ( b y ) for b/ 2 y b n1 aa odd Why? 4. The unknowns An and Bn can be solved by two known conditions: RThe potential at y b / 2 must continuous S| R1 for x W / 2 The surface charge distribution for the strip: s S T| T0 for x W / 2 20 Analysis approach for Z0 and Td 5. R b/2 b/2 V E (x 0, y)dy (x, y) / y(x 0, y)dy | z y z | 0 0 S w/2 |Q (x)dx W(C / m) z s T| w/2 Q W 6. C V 2a sin(nW / 2a) sinh(nb / 2a) 2 n1 (n) 0 r cosh(nb / 2a) odd 1 r Answers!! Z0 v pC cC 7. Td r / c 21 Analysis approach for Z0 and Td (Microstrip Line) y 1. PEC PEC d W x -a/2 a/2 2. The fields in Quasi - TEM mode must satisfy Laplace equation 2 t (x, y) 0 where is the electric potential The boundary conditions are (x, y) 0 at x a / 2 (x, y) 0 at y 0, 22 Analysis approach for Z0 and Td (Microstrip Line) 3. Since the center conductor will contain the surface charge, so nx ny R A cos sinh for 0 y d | n n1 a a |odd (x, y) S nx ny/a | Bn cos e for d y |n1 a Todd 4. The unknowns An and Bn can be solved by two known conditions and the orthogonality of cos function : RThe potential at y d must continuous S| R1 for x W / 2 The surface charge distribution for the strip: s S T| T0 for x W / 2 23 Analysis approach for Z0 and Td (Microstrip Line) 5. R b/2 b/2 nd V E (x 0, y)dy (x, y) / y(x 0, y)dy A sinh | z y z n 0 0 n1 a | odd S w/2 |Q (x)dx W(C / m) | z s T w/2 6. Q W C V 4a sin(nW / 2a) sinh(nd / 2a) 2 n1 (n) W 0[sinh(nd / a) r cosh(nd / a)] odd 24 Analysis approach for Z0 and Td (Microstrip Line) 7. To find the effective dielectric constant e ,we consider two cases of capacitance 1. C = capacitance per unit length of the microstrip line with the dielectric substrate r 1 2. C0 = capacitance per unit length of the microstrip line with the dielectric substrate r 1 C e C0 8. 1 e Z0 v pC cC Td e / c 25 Tables for Z0 and Td (Microstrip Line) 20 28 40 50 75 90 100 Z0 () 3.8 3.68 3.51 3.39 3.21 3.13 3.09 eff L0 0.119 0.183 0.246 0.320 0.468 0.538 0.591 (nH / mm) C0 0.299 0.233 0.154 0.128 0.083 0.067 0.059 (pF / mm) T0 6.54 6.41 6.25 6.17 5.99 5.92 5.88 (ps / mm) Fr4 : dielectric constant = 4.5 Frequency: 1GHz 26 Tables for Z0 and Td (Strip Line) 20 28 40 50 75 90 100 Z0 () 4.5 4.5 4.5 4.5 4.5 4.5 4.5 eff L0 0.141 0.198 0.282 0.353 0.53 0.636 0.707 (nH / mm) C0 0.354 0.252 0.171 0.141 0.094 0.078 0.071 (pF / mm) T0 7.09 7.09 7.09 7.09 7.09 7.09 7.09 (ps / mm) Fr4 : dielectric constant = 4.5 Frequency: 1GHz 27 Analysis approach for Z0 and Td (EDA/Simulation Tool) 1. HP Touch Stone (HP ADS) 2. Microwave Office 3. Software shop on Web: 4. APPCAD (http://softwareshop.edtn.com/netsim/si/termination/term_article.html) (http://www.agilent.com/view/rf or http://www.hp.woodshot.com ) 28 Concept Test for Planar Transmission Lines • Please compare their Z0 and Vp (a) (b) 29 (a) (b) (c) 30 (a) (b) (c) 31 32 Loss of Transmission Lines Typically, dielectric loss is quite small -> G0 = 0. Thus R0 jwL0 L0 1/2 Z0 (1 jx) jwC0 C0 R r (R jwL )( jwC ) a j w 0 0 0 0 L0 R Highly Lossy Near Lossless where x 0 wL0 w • Lossless case : x = 0 • Near Lossless: x << 1 w • Highly Lossy: x >> 1 33 Loss of Transmission Lines •For Lossless case: • For Near Lossless case: R a 0 a 0 2 L0 / C0 L0C0 x 2 L L C L1 O Z 0 0 0 M P 0 N 8 Q C0 L F R I L 1 Time delay T L C Z 0 1 j 0 0 where C 2T / R 0 0 0 0 G J 0 0 C0 H 2wL0 K C0 jwC Time delay T0 L0C0 34 Loss of Transmission Lines • For highly loss case: (RC transmission line) wR C 1 a 0 0 [1 ] 2 2x Nonlinear phase relationship with f introduces signal distortion wR C 1 0 0 [1 ] 2 2x Example of RC transmission line: AWG 24 telephone line in home R0 1 1/2 Z0 [1 ] F R iwLI 2wC 2x Z (w) 648(1 j) 0 0 HG jwC KJ where R 0.0042 / in L 10nH / in C 1pF / in That’s why telephone company terminate w 10,000rad / s(1600Hz) : voice band the lines with 600 ohm 35 Loss of Transmission Lines ( Dielectric Loss) The loss of dielectric loss is described by the loss tangent G tan FR4 PCB tan 0.035 D wC D GZ a 0 (wC tan Z ) / 2 f tan LC D 2 D 0 D 36 Loss of Transmission Lines (Skin Effect) • Skin Effect DC resistance AC resistance 37 Loss of Transmission Lines (Skin Effect) 1 2 s a w 1 length w Rw() area NOTE: In the near lossless region (R / wL 1), the characteristic impedance Z0 is not much affected by the skin effect R(w) w R(w) / wL (1/ w) 1 38 Loss of Transmission Lines (Skin Effect) f (MHz) 100 200 400 800 1200 1600 2000 1 6.6um 4.7um 3.3um 2.4um 1.9um 1.7um 1.5um s f 2.6m 3.7m 5.2m 7.4m 9.0m 10.8m 11.6m R () s ohm ohm ohm ohm ohm ohm ohm Trace 1.56 2.22 3.12 4.44 5.4 6.48 7.0 resistance ohm ohm ohm ohm ohm ohm ohm f Skin depth resistance R = () s 6mil = 4 10-7 H / m Cu 17um (Cu) = 5.8 107 S / m Length of trace = 20cm 39 Loss Example: Gigabit differential transmission lines For comparison: (Set Conditions) 1.
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