Transmission Line Basics
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Transmission Line Basics Transmission Line Basics Presentation Transcript

  • Transmission Line Basics II - Class 6Prerequisite Reading assignment: CH2TARA SAIKUMAR
  • 2Real Computer Issues data Dev a Dev b Signal Measured Clk here Switch ThresholdAn engineer tells you the measured clock is non-monotonicand because of this the flip flop internally may double clockthe data. The goal for this class is to by inspectiondetermine the cause and suggest whether this is a problemor not. Transmission Lines Class 6
  • 3Agenda The Transmission Line Concept Transmission line equivalent circuits and relevant equations Reflection diagram & equation Loading Termination methods and comparison Propagation delay Simple return path ( circuit theory, network theory come later) Transmission Lines Class 6
  • 4Two Transmission Line Viewpoints Steady state ( most historical view) Frequency domain Transient Time domain Not circuit element Why? We mix metaphors all the time Why convenience and history Transmission Lines Class 6
  • 5 Transmission Line ConceptPower Frequency (f) is @ 60 Hz Power Wavelength (λ ) is 5× 10 m 6 Plant ( Over 3,100 Miles) Consumer Home Transmission Lines Class 6
  • PC Transmission Lines 6Signal Frequency (f) is approaching 10 GHz Integrated Circuit Stripline Wavelength (λ ) is 1.5 cm ( 0.6 inches) Microstrip T PCB substrate Cross section view taken here Stripline W Via Micro- FR4 Dielectric Copper Trace Cross Section of Above PCB Strip Signal (microstrip) Ground/Power T Copper Plane Signal (stripline) Signal (stripline) Ground/Power Signal (microstrip) W Transmission Lines Class 6
  • 7Key point about transmission line operation Voltage and current on a transmission line is a function of both time and position. I2 V = f ( z, t ) I1 I = f ( z, t ) V1 V2 dz The major deviation from circuit theory with transmission line, distributed networks is this positional dependence of voltage and current! Must think in terms of position and time to understand transmission line behavior This positional dependence is added when the assumption of the size of the circuit being small compared to the signaling wavelength Transmission Lines Class 6
  • 8Examples of Transmission LineStructures- I Cables and wires (a) Coax cable (b) Wire over ground (c) Tri-lead wire (d) Twisted pair (two-wire line) Long distance interconnects + + - (a) (b) - - + - + - (c) (d) Transmission Lines Class 6
  • 9Segment 2: Transmission line equivalentcircuits and relevant equations Physics of transmission line structures Basic transmission line equivalent circuit ?Equations for transmission line propagation Transmission Lines Class 6
  • 10E & H Fields – Microstrip Case How does the signal move Signal path from source to load? Y Z (into the page) X Electric field Remember fields are setup given field Magnetic an applied forcing function. (Source) Ground return path The signal is really the wave propagating between the conductors Transmission Lines Class 6
  • Transmission Line “Definition” 11 General transmission line: a closed system in which power is transmitted from a source to a destination Our class: only TEM mode transmission lines A two conductor wire system with the wires in close proximity, providing relative impedance, velocity and closed current return path to the source. Characteristic impedance is the ratio of the voltage and current waves at any one position on the transmission line V Z0 = I Propagation velocity is the speed with which signals are transmitted through the transmission line in its surrounding medium. c v= εr Transmission Lines Class 6
  • 12Presence of Electric and Magnetic Fields H I I + ∆I I I + ∆I + + + + E V V + ∆V V V + ∆V H I I + ∆I I I + ∆I - - - -  Both Electric and Magnetic fields are present in the transmission lines These fields are perpendicular to each other and to the direction of wave propagation for TEM mode waves, which is the simplest mode, and assumed for most simulators(except for microstrip lines which assume “quasi-TEM”, which is an approximated equivalent for transient response calculations).  Electric field is established by a potential difference between two conductors. Implies equivalent circuit model must contain capacitor.  Magnetic field induced by current flowing on the line Implies equivalent circuit model must contain inductor. Transmission Lines Class 6
  • 13T-Line Equivalent Circuit  General Characteristics of Transmission Line Propagation delay per unit length (T0) { time/distance} [ps/in] Or Velocity (v0) {distance/ time} [in/ps] Characteristic Impedance (Z0) Per-unit-length Capacitance (C0) [pf/in] Per-unit-length Inductance (L0) [nf/in] Per-unit-length (Series) Resistance (R0) [Ω/in] Per-unit-length (Parallel) Conductance (G0) [S/in] lR0 lL0 lG0 lC0 Transmission Lines Class 6
  • 14Ideal T Line Ideal (lossless) Characteristics of Transmission Line lL0 Ideal TL assumes: Uniform line lC0 Perfect (lossless) conductor (R0→ 0) Perfect (lossless) dielectric (G0→ 0) We only consider T0, Z0 , C0, and L0. A transmission line can be represented by a cascaded network (subsections) of these equivalent models. The smaller the subsection the more accurate the model The delay for each subsection should be no larger than 1/10th the signal rise time. Transmission Lines Class 6
  • Signal Frequency and Edge Rate 15 vs. Lumped or Tline ModelsIn theory, all circuits that deliver transient power fromone point to another are transmission lines, but if thesignal frequency(s) is low compared to the size of thecircuit (small), a reasonable approximation can beused to simplify the circuit for calculation of the circuittransient (time vs. voltage or time vs. current)response. Transmission Lines Class 6
  • 16T Line Rules of Thumb So, what are the rules of thumb to use? May treat as lumped Capacitance Use this 10:1 ratio for accurate modeling of transmission lines Td < .1 Tx May treat as RC on-chip, and treat as LC for PC board interconnect Td < .4 Tx Transmission Lines Class 6
  • Other “Rules of Thumb” 17 Frequency knee (Fknee) = 0.35/Tr (so if Tr is 1nS, Fknee is 350MHz) This is the frequency at which most energy is below Tr is the 10-90% edge rate of the signal Assignment: At what frequency can your thumb be used to determine which elements are lumped? Assume 150 ps/in Transmission Lines Class 6
  • When does a T-line become a T-Line? 18  Whether it is a bump or a mountain depends on the ratio of its When do we need to size (tline) to the use transmission line size of the vehicle analysis techniques vs. (signal lumped circuit wavelength) analysis?  Similarly, whether or not a line is to be considered as a transmission line depends on the ratio of length of the line (delay) to the wavelength of Wavelength/edge rate Tline the applied frequency or the rise/fall edge of the Transmission Lines Class 6 signal
  • Equations & Formulas How to model & explain transmission line behavior
  • 20Relevant Transmission Line Equations Propagation equation γ = ( R + jωL)(G + jωC ) = α + jβ α is the attenuation (loss) factor β is the phase (velocity) factor Characteristic Impedance equation ( R + j ωL ) Z0 = (G + jωC ) In class problem: Derive the high frequency, lossless approximation for Z0 Transmission Lines Class 6
  • 21Ideal Transmission Line Parameters  Knowing any two out of Z0, Td, C0, and L0, the other two L0 can be calculated. Z0 = ; T d = L0 C0 ;  C0 and L0 are reciprocal C0 functions of the line cross- T0 sectional dimensions and C0 = ; L0 = Z 0 T 0 ; Z0 are related by constant me. 1  ε is electric permittivity v0 = ; C0 L0 = µε; ε 0= 8.85 X 10-12 F/m (free space) µε ε ri s relative dielectric constant µ = µr µ0 ; ε = εr ε0 .  µ is magnetic permeability µ 0= 4p X 10-7 H/m (free space) µ r is relative permeability Don’t forget these relationships and what they mean! Transmission Lines Class 6
  • Parallel Plate Approximation 22 Assumptions TC TEM conditions ε TD Uniform dielectric (ε ) between conductors WC TC<< TD; WC>> TD ε * PlateArea Base T-line characteristics are C= function of: d equation Material electric and WC  F  WC  pF  magnetic properties C0 ε⋅ ⋅  8.85 ⋅ε r ⋅ ⋅  TD  m  TD  m  Dielectric Thickness (TD) TD  F  T D  µH  Width of conductor (WC) L0 µ⋅ ⋅  0.4 ⋅π ⋅µ r ⋅ ⋅  Trade-off WC  m  WC  m  TD ; C0 , L0 , Z0  TD µr Z0 377 ⋅ ⋅ ⋅Ω WC ; C0 , L0 , Z0  WC εrTo a first order, t-line capacitance and inductance canbe approximated using the parallel plate approximation. Transmission Lines Class 6
  • 23Improved Microstrip Formula Parallel Plate Assumptions + WC Large ground plane with TC zero thickness ε TD To accurately predict microstrip impedance, you must calculate the effective dielectric constant. From Hall, Hall & McCall: 87  5.98TD  Z0 ≈ ln  Valid when: εr + 1.41  0.8WC + TC  0.1 < WC/TD < 2.0 and 1 < r < 15 εr + 1 εr − 1 TC εe = + + F − 0.217( εr − 1) 2 12TD WCTD 2 1+ WC 2 You can’t beat  WC  0.02(εr −1)1 − a field solver WC  for TD <1 F=  TD  0 for WC >1 TD Transmission Lines Class 6
  • 24Improved Stripline Formulas Same assumptions as WC TD1 used for microstrip ε TC apply here TD2 From Hall, Hall & McCall:Symmetric (balanced) Stripline Case TD1 = TD2 60 4(TD1 + TD1)  Z 0 sym ≈ ln   0.67π (0.8WC + TC )  εr   Valid when WC/(TD1+TD2) < 0.35 and TC/(TD1+TD2) < 0.25 You can’t beat aOffset (unbalanced) Stripline Case TD1 > TD2 field solver Z 0 sym(2 A, WC , TC , εr ) ⋅ Z 0 sym(2 B, WC , TC , εr ) Z 0offset ≈ 2 Z 0 sym(2 A,WC , TC , εr ) + Z 0 sym(2 B,WC , TC , εr ) Transmission Lines Class 6
  • 25Refection coefficient Signal on a transmission line can be analyzed by keeping track of and adding reflections and transmissions from the “bumps” (discontinuities) Refection coefficient Amount of signal reflected from the “bump” Frequency domain ρ=sign(S11)*|S11| If at load or source the reflection may be called gamma (ΓL or Γs) Time domain ρ is only defined a location The “bump” Time domain analysis is causal. Frequency domain is for all time. We use similar terms – be careful Reflection diagrams – more later Transmission Lines Class 6
  • Reflection and Transmission 26 Incident 1+ρ Transmitted ρ ReflectedReflection Coeficient Transmission Coeffiecent Zt − Z0 ρ Zt − Z0 τ (1 + ρ) "" → "" τ 1+ Zt + Z0 Zt + Z0 2⋅ Zt τ Zt + Z0 Transmission Lines Class 6
  • 27Special Cases to Remember A: Terminated in Zo Zs − Zo Zo ρ = Zo Zo = 0 Vs Zo + Zo B: Short Circuit Zs − Zo ρ = 0 Zo = −1 Vs 0 + Zo C: Open Circuit Zs ∞ − Zo Zo ρ= =1 Vs ∞ + Zo Transmission Lines Class 6
  • 28Assignment – Building the SI Tool Box Compare the parallel plate approximation to the improved microstrip and stripline formulas for the following cases: Microstrip: WC = 6 mils, TD = 4 mils, TC = 1 mil, εr = 4 Symmetric Stripline: WC = 6 mils, TD1 = TD2 = 4 mils, TC = 1 mil, εr = 4 Write Math Cad Program to calculate Z0, Td, L & C for each case. What factors cause the errors with the parallel plate approximation? Transmission Lines Class 6
  • 29Transmission line equivalent circuits andrelevant equations  Basic pulse launching onto transmission lines  Calculation of near and far end waveforms for classic load conditions Transmission Lines Class 6
  • Review: Voltage Divider Circuit 30 Consider the simple circuit that RS contains source voltage VS, source RL VS VL resistance RS, and resistive load RL. The output voltage, VL is RL easily calculated VL = VS from the source RL + R S amplitude and the values of the two series resistors. Why do we care for? Next page…. Transmission Lines Class 6
  • 31Solving Transmission Line ProblemsThe next slides will establish a procedure that will allow you to solve transmission line problems without the aid of a simulator. Here are the steps that will be presented: Determination of launch voltage & final “DC” or “t =0” voltage Calculation of load reflection coefficient and voltage delivered to the load Calculation of source reflection coefficient and resultant source voltage These are the steps for solving all t-line problems. Transmission Lines Class 6
  • Determining Launch Voltage 32 TD Rs A B Vs Zo 0 Vs Rt (initial voltage) t=0, V=Vi Z0 Rt Vi = VS Vf = VS Z 0 + RS Rt + RS Step 1 in calculating transmission line waveforms is to determine the launch voltage in the circuit.  The behavior of transmission lines makes it easy to calculate the launch & final voltages – it is simply a voltage divider! Transmission Lines Class 6
  • 33Voltage Delivered to the Load TD Vs Rs A B Vs Zo Rt 0 (initial voltage) t=0, V=Vi (signal is reflected) t=2TD, ρΒ = Rt − Zo V=Vi + ρB(Vi) +ρA(ρB)(Vi ) t=TD, V=Vi +ρB(Vi ) Vreflected = ρ Β (Vincident) Rt + Zo VB = Vincident + Vreflected Step 2: Determine VB in the circuit at time t = TD  The transient behavior of transmission line delays the arrival of launched voltage until time t = TD.  VB at time 0 < t < TD is at quiescent voltage (0 in this case)  Voltage wavefront will be reflected at the end of the t-line  VB = Vincident + Vreflected at time t = TD Transmission Lines Class 6
  • 34Voltage Reflected Back to the Source Rs A B Vs Zo 0 Vs ρA ρB Rt TD (initial voltage) t=0, V=Vi (signal is reflected) t=2TD, V=Vi + ρB (Vi) + ρA B )(Vi ) (ρ t=TD, V=Vi + ρB (Vi ) Transmission Lines Class 6
  • Voltage Reflected Back to the Source 35 − Zo Vreflected = ρ Α (Vincident) ρ Α = Rs Rs + Zo VA = Vlaunch + Vincident + VreflectedStep 3: Determine VA in the circuit at time t = 2TD The transient behavior of transmission line delays the arrival of voltage reflected from the load until time t = 2TD.  VA at time 0 < t < 2TD is at launch voltage Voltage wavefront will be reflected at the source  VA = Vlaunch + Vincident + Vreflected at time t = 2TDIn the steady state, the solution converges to VB = VS[Rt / (Rt + Rs)] Transmission Lines Class 6
  • 36Problems Solved Homework Consider the circuit shown to the right with a resistive load, assume propagation RS I1 I2 Z0 ,Τ 0 delay = T, RS= Z0 . VS V1 l V2 RL Calculate and show the wave forms of V1(t),I1(t),V2(t), and I2(t) for (a) RL= ∞ and (b) RL= 3Z0 Transmission Lines Class 6
  • 37Step-Function into T-Line: Relationships Source matched case: RS= Z0 V1(0) = 0.5VA, I1(0) = 0.5IA Γ S = 0, V(x,∞ ) = 0.5VA(1+ Γ L) Uncharged line V2(0) = 0, I2(0) = 0 Open circuit means RL= ∞ Γ L = ∞ /∞ = 1 V1(∞) = V2(∞) = 0.5VA(1+1) = VA I1(∞) = I2 (∞) = 0.5IA(1-1) = 0 Solution Transmission Lines Class 6
  • 38Step-Function into T-Line with Open Ckt At t = T, the voltage wave reaches load end and doubled wave travels back to source end V1(T) = 0.5VA, I1(T) = 0.5VA/Z0 V2(T) = VA, I2 (T) = 0 At t = 2T, the doubled wave reaches the source end and is not reflected V1(2T) = VA, I1(2T) = 0 V2(2T) = VA, I2(2T) = 0 Solution Transmission Lines Class 6
  • 39Waveshape:Step-Function into T-Line with Open Ckt I1 IA I2 RS I1 I2Current (A) 0.75IA Z0 ,Τ 0 l 0.5I A VS V1 V2 Open 0.25IA 0 Τ 2Τ 3Τ 4Τ Time (ns) VA V1 This is called V2 “reflected waveVoltage (V) 0.75VA switching” 0.5VA 0.25VA Solution 0 Τ 2Τ 3Τ 4Τ Time (ns) Transmission Lines Class 6
  • 40Problem 1b: Relationships Source matched case: RS= Z0 V1(0) = 0.5VA, I1(0) = 0.5IA Γ S = 0, V(x,∞ ) = 0.5VA(1+ Γ L) Uncharged line V2(0) = 0, I2(0) = 0 RL= 3Z0 Γ L = (3Z0 -Z0) / (3Z0 +Z0) = 0.5 V1(∞) = V2(∞) = 0.5VA(1+0.5) = 0.75VA I1(∞) = I2(∞) = 0.5IA(1-0.5) = 0.25IA Solution Transmission Lines Class 6
  • 41Problem 1b: Solution At t = T, the voltage wave reaches load end and positive wave travels back to the source V1(T) = 0.5VA, I1(T) = 0.5IA V2(T) = 0.75VA , I2(T) = 0.25IA At t = 2T, the reflected wave reaches the source end and absorbed V1(2T) = 0.75VA , I1(2T) = 0.25IA V2(2T) = 0.75VA , I2(2T) = 0.25IA Solution Transmission Lines Class 6
  • Waveshapes for Problem 1b 42 I1 IA I2 I1 I2 RS 0.75IA Z0 ,Τ 0 Current (A) l 0.5IA VS V1 V2 RL 0.25IA 0 Τ 2Τ 3Τ 4Τ Time (ns) I1 VA I2 Note that a 0.75VA properly terminated Voltage (V) 0.5VA wave settle out at 0.5 V 0.25VA Solution 0 Τ 2Τ 3Τ 4Τ Time (ns) Solution Transmission Lines Class 6
  • Transmission line step response 43  Introduction to lattice diagram analysis  Calculation of near and far end waveforms for classic load impedances  Solving multiple reflection problems Complex signal reflections at different types of transmission line “discontinuities” will be analyzed in this chapter. Lattice diagrams will be introduced as a solution tool. Transmission Lines Class 6
  • 44Lattice Diagram Analysis – Key Concepts V(source) Zo V(load) Vs Rs The lattice diagram is a 0 TD = N ps Vs Rt tool/technique to simplify the accounting of ρsource ρload reflections and waveforms V(load) Time V(source)  Diagram shows the boundaries 0 a (x =0 and x=l) and the reflection A’ coefficients (GL and GL ) N ps A  Time (in T) axis shown b vertically  Slope of the line should 2N ps c B’ indicate flight time of signal Particularly important for multiple reflection problems using both 3N ps B microstrip and stripline mediums. d  Calculate voltage amplitude C’ for each successive reflected 4N ps e wave  Total voltage at any point is the 5N ps sum of all the waves that have reached that point Transmission Lines Class 6
  • Lattice Diagram Analysis – Detail 45 ρ ρ source load V(source) V(load) 0 Vlaunch 0 Time Vlaunch N ps Vlaunch ρload Vlaunch(1+ρload) Time 2N ps Vlaunch ρloadρsource Vlaunch(1+ρload +ρload ρsource) 3N ps Vlaunch ρ2loadρsource Vlaunch(1+ρload+ρ2loadρsource+ ρ2loadρ2source) 4N ps Vlaunch ρ2loadρ2source V(source) Zo V(load) Vs Rs 5N ps0 Vs TD = N ps Rt Transmission Lines Class 6
  • Transient Analysis – Over Damped 46 V(source) Zo V(load) Assume Zs=75 ohms 2v Zs Zo=50ohms0 TD = 250 ps Vs=0-2 volts Vs Zo  50  Vinitial = Vs = (2)  = 0.8 ρ source = 0.2 ρ load = 1 Zs + Zo  75 + 50  Time V(source) V(load) Zs − Zo 75 − 50 0 ρ source = = = 0.2 0.8v Zs + Zo 75 + 50 0v Zl − Zo ∞ − 50 500 ps 0.8v ρ load = = =1 Zl + Zo ∞ + 50 0.8v 1000 ps 1.6v Response fr om lattice diagram 0.16v 2.5 1500 ps 1.76v 2 0.16v 1.5 V olt s 1.92v 1 Sour ce 2000 ps 0.032v 0.5 Load 0 2500 ps 0 2 50 500 750 1000 1250 Tim e , ps Transmission Lines Class 6
  • Transient Analysis – Under Damped 47 V(source) Zo V(load) Assume Zs=25 ohms 2v Zs Zo =50ohms0 Vs TD = 250 ps Vs=0-2 volts Zo  50  Vinitial =Vs = (2)   =1.3333 ρsource = −0 . 3333 ρload = 1 Zs + Zo  25 +50  Time V(source) V(load) Zs − Zo 25 −50 0 ρsource = = = −0.33333 1.33v Zs + Zo 25 + 50 0v Zl − Zo ∞ −50 500 ps 1.33v ρ = = =1 1.33v load Zl + Zo ∞ +50 1000 ps 2.66v Response from lattice diagram -0.443v 3 1500 ps 2.22v -0.443v 2.5 2 Volts 1.77v 1.5 2000 ps 0.148v 1 Source 0.5 Load 2500 ps 1.92 0 0.148v 0 250 500 750 1000 1250 1500 1750 2000 2250 Time, ps 2.07 Transmission Lines Class 6
  • Two Segment Transmission Line Structures 48 X X Rs Zo1 Zo2 Rt Vs TD TD A= a A = b + e T3 T2 B = a+c+d B = b + e + g + i ρ1 ρ 2 ρ3 ρ4 C = A+ c+ d + f + h C = b + e + g + i + k + l a a = vi Z o1 vi = Vs TD A c b Rs + Z o1 b = aT2 2TD Rs − Z o1 c = aρ 2 d e ρ1 = Rs + Z o1 3TD B g A’ d = cρ 1 f Z o 2 − Z o1 4TD ρ2 = e = bρ 4 h i Z o 2 + Z o1 B’ f = dρ 2 + eT3 5TD C j k Z − Zo2 ρ 3 = o1 g = eρ 3 + dT2 Z o1 + Z o 2 l Rt − Z o 2 h = fρ 1 C’ ρ4 = Rt + Z o 2 i = gρ 4 T2 = 1 + ρ 2 j = hρ 2 + iT3 T3 = 1 + ρ 3 k = iρ 3 + hT2 Transmission Lines Class 6
  • 49Assignment Previous examples are the preparation Consider the two segment transmission line shown to the right. Assume RS= 3Z01 and Z02= 3Z01 . Use Lattice R I1 I2 I3 S diagram and calculate Z01 ,Τ 01 Z02 ,Τ 02 l1 l2 reflection coefficients atV S V1 V2 V3 Short the interfaces and show the wave forms of V1(t), V2(t), and V3(t). Check results with PSPICE Transmission Lines Class 6