Transmission Line Basics II Class 6 Prerequisite Reading

Transmission Line Basics II - Class 6 Prerequisite Reading assignment: CH 2 Acknowledgements: Intel Bus Boot Camp: Michael Leddige

Real Computer Issues Dev a Clk 2 data Dev b Signal Measured here Switch Threshold An engineer tells you the measured clock is non-monotonic and because of this the flip flop internally may double clock the data. The goal for this class is to by inspection determine the cause and suggest whether this is a problem or not. Transmission Lines Class 6

Agenda 3 § 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

Two 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 4

5 Transmission Line Concept Power Frequency (f) is @ 60 Hz Wavelength (l) is 5 106 m ( Over 3, 100 Miles) Consumer Home Transmission Lines Class 6 Power Plant

PC Transmission Lines Signal Frequency (f) is approaching 10 GHz Wavelength (l) is 1. 5 cm ( 0. 6 inches) Microstrip 6 Integrated Circuit Stripline T PCB substrate Cross section view taken here Stripline W Cross Section of Above PCB Copper Trace Via FR 4 Dielectric Micro. Strip Signal (microstrip) T Copper Plane Ground/Power Signal (stripline) Ground/Power Signal (microstrip) W Transmission Lines Class 6

7 Key point about transmission line operation Voltage and current on a transmission line is a function of both time and position. 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

Examples of Transmission Line §Structures. Cables and wires. I (a) (b) (c) (d) Coax cable Wire over ground Tri-lead wire Twisted pair (two-wire line) § Long distance interconnects Transmission Lines Class 6 8

Segment 2: Transmission line equivalent circuits and relevant equations § Physics of transmission line structures § Basic transmission line equivalent circuit § ? Equations for transmission line propagation Transmission Lines Class 6 9

E & H Fields – Microstrip Case How does the signal move from source to load? Remember fields are setup given an applied forcing function. (Source) The signal is really the wave propagating between the conductors Transmission Lines Class 6 10

Transmission Line “Definition” § 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 Propagation velocity is the speed with which signals are transmitted through the transmission line in its surrounding medium. Transmission Lines Class 6 11

Presence of Electric and Magnetic Fields § 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 12

13 T-Line Equivalent Circuit § General Characteristics of Transmission Line Propagation delay per unit length (T 0) { time/distance} [ps/in] Or Velocity (v 0) {distance/ time} [in/ps] Characteristic Impedance (Z 0) Per-unit-length Capacitance (C 0) [pf/in] Per-unit-length Inductance (L 0) [nf/in] Per-unit-length (Series) Resistance (R 0) [W/in] Per-unit-length (Parallel) Conductance (G 0) [S/in] l. R 0 l. L 0 l. G 0 Transmission Lines Class 6 l. C 0

Ideal T Line 14 § Ideal (lossless) Characteristics of Transmission Line Ideal TL assumes: Uniform line Perfect (lossless) conductor (R 0 0) Perfect (lossless) dielectric (G 0 0) We only consider T 0, Z 0 , C 0, and L 0. l. L 0 l. C 0 § 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 1/10 th the signal rise time. Transmission Lines Class 6

Signal Frequency and Edge Rate vs. Lumped or Tline Models In theory, all circuits that deliver transient power from one point to another are transmission lines, but if the signal frequency(s) is low compared to the size of the circuit (small), a reasonable approximation can be used to simplify the circuit for calculation of the circuit transient (time vs. voltage or time vs. current) response. Transmission Lines Class 6 15

T 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 16

Other “Rules of Thumb” § Frequency knee (Fknee) = 0. 35/Tr (so if Tr is § § § 1 n. S, Fknee is 350 MHz) 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 17

When does a T-line become a T-Line? 18 § Whether it is a bump or a mountain depends on the ratio of its size (tline) to the size of the vehicle (signal wavelength) When do we need to use transmission line analysis techniques vs. lumped circuit analysis? § Similarly, whether Wavelength/edge rate Tline Transmission Lines Class 6 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 the applied frequency or the rise/fall edge of the signal

Equations & Formulas How to model & explain transmission line behavior

Relevant Transmission Line Equations Propagation equation is the attenuation (loss) factor is the phase (velocity) factor Characteristic Impedance equation In class problem: Derive the high frequency, lossless approximation for Z 0 Transmission Lines Class 6 20

Ideal Transmission Line Parameters § Knowing any two out of Z 0, § § Td, C 0, and L 0, the other two can be calculated. C 0 and L 0 are reciprocal functions of the line crosssectional dimensions and are related by constant me. e is electric permittivity e 0= 8. 85 X 10 -12 F/m (free space) eri s relative dielectric constant § m is magnetic permeability Z 0 = C 0 = v 0 = L 0 ; C 0 T 0 ; Z 0 1 me m = mr m 0 ; T d = L 0 C 0 ; L 0 = Z 0 T 0 ; ; C 0 L 0 = me; e = er e 0. m 0= 4 p X 10 -7 H/m (free space) mr is relative permeability Don’t forget these relationships and what they mean! Transmission Lines Class 6 21

Parallel Plate Approximation 22 § Assumptions TEM conditions Uniform dielectric (e ) between conductors TC<< TD; WC>> TD § T-line characteristics are function of: Material electric and magnetic properties Dielectric Thickness (TD) Width of conductor (WC) Base equation § Trade-off TD ; C 0 , L 0 , Z 0 WC ; C 0 , L 0 , Z 0 To a first order, t-line capacitance and inductance can be approximated using the parallel plate approximation. Transmission Lines Class 6

Improved Microstrip Formula § Parallel Plate Assumptions + § Large ground plane with zero thickness To accurately predict microstrip impedance, you must calculate the effective dielectric constant. From Hall, Hall & Mc. Call: Valid when: 0. 1 < WC/TD < 2. 0 and 1 < er < 15 You can’t beat a field solver Transmission Lines Class 6 23

24 Improved Stripline Formulas § Same assumptions as used for microstrip apply here From Hall, Hall & Mc. Call: Symmetric (balanced) Stripline Case TD 1 = TD 2 Valid when WC/(TD 1+TD 2) < 0. 35 and TC/(TD 1+TD 2) < 0. 25 Offset (unbalanced) Stripline Case TD 1 > TD 2 Transmission Lines Class 6 You can’t beat a field solver

Refection 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 r=sign(S 11)*|S 11| If at load or source the reflection may be called gamma (GL or Gs) Time domain r 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 25

Reflection and Transmission Incident r 1+r Reflected Transmission Lines Class 6 Transmitted 26

Special Cases to Remember 27 A: Terminated in Zo Zs Vs Zo Zo r = Zo Zo = 0 Zo + Zo B: Short Circuit Zs Vs Zo r = 0 Zo = -1 0 + Zo C: Open Circuit Zs Vs Zo Transmission Lines Class 6 r= ¥ - Zo =1 ¥ + Zo

Assignment – 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, er = 4 Symmetric Stripline: WC = 6 mils, TD 1 = TD 2 = 4 mils, TC = 1 mil, er = 4 Write Math Cad Program to calculate Z 0, Td, L & C for each case. What factors cause the errors with the parallel plate approximation? Transmission Lines Class 6 28

Transmission line equivalent circuits and relevant equations § Basic pulse launching onto transmission lines § Calculation of near and far end waveforms for classic load conditions Transmission Lines Class 6 29

Review: Voltage Divider Circuit 30 § Consider the simple circuit that contains source voltage VS, source resistance RS, and resistive load RL. RS RL VS § The output voltage, VL is easily calculated from the source amplitude and the values of the two series resistors. VL = VS Why do we care for? Next page…. Transmission Lines Class 6 RL RL + RS VL

Solving Transmission Line Problems The 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: 1. Determination of launch voltage & final “DC” or “t =0” voltage 2. Calculation of load reflection coefficient and voltage delivered to the load 3. Calculation of source reflection coefficient and resultant source voltage These are the steps for solving all t-line problems. Transmission Lines Class 6 31

Determining Launch Voltage 32 TD Vs 0 Rs A B Zo Vs Rt (initial voltage) t=0, V=Vi Vi = VS Z 0 + RS Vf = VS Rt 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

Voltage Delivered to the Load TD Vs Rs A Zo Vs 0 B Rt (initial voltage) t=0, V=Vi r. B = t=2 TD, r. A(r. B)(Vi ) V=Vi - +Zo Rt+ r. B(Vi) Rt + Zo (signal is reflected) t=TD, V=Vi +r. B(Vi ) Vreflected = r. B (Vincident) VB = Vincident + Vreflected Step 2: Determine VB in the circuit at time t = TD l l The transient behavior of transmission line delays the arrival of launched voltage until time t = TD. l VB at time 0 < t < TD is at quiescent voltage (0 in this case) l VB = Vincident + Vreflected at time t = TD Voltage wavefront will be reflected at the end of the t-line Transmission Lines Class 6 33

Voltage Reflected Back to the Source Vs 0 Rs A Vs B Zo r. A r. B Rt TD (initial voltage) t=0, V=Vi (signal is reflected) t=2 TD, V=Vi + r. B (Vi) + r. A(r B )(Vi ) Transmission Lines Class 6 t=TD, V=Vi + r. B (Vi ) 34

Voltage Reflected Back to the Source r. A - Zo Rs = Rs + Zo Vreflected = r. A (Vincident) VA = Vlaunch + Vincident + Vreflected Step 3: Determine VA in the circuit at time t = 2 TD l l The transient behavior of transmission line delays the arrival of voltage reflected from the load until time t = 2 TD. l VA at time 0 < t < 2 TD is at launch voltage l VA = Vlaunch + Vincident + Vreflected at time t = 2 TD Voltage wavefront will be reflected at the source In the steady state, the solution converges to VB = VS[Rt / (Rt + Rs)] Transmission Lines Class 6 35

Problems Solved Homework Consider the circuit shown to the right with a resistive load, assume propagation delay = T, RS= Z 0. Calculate and show the wave forms of V 1(t), I 1(t), V 2(t), and I 2(t) for (a) RL= and (b) RL= 3 Z 0 Transmission Lines Class 6 36

Step-Function into T-Line: Relationships § Source matched case: RS= Z 0 V 1(0) = 0. 5 VA, I 1(0) = 0. 5 IA GS = 0, V(x, ) = 0. 5 VA(1+ GL) § Uncharged line V 2(0) = 0, I 2(0) = 0 § Open circuit means RL= GL = / = 1 V 1( ) = V 2( ) = 0. 5 VA(1+1) = VA I 1( ) = I 2 ( ) = 0. 5 IA(1 -1) = 0 Solution Transmission Lines Class 6 37

Step-Function into T-Line with Open Ckt § At t = T, the voltage wave reaches load end and doubled wave travels back to source end V 1(T) = 0. 5 VA, I 1(T) = 0. 5 VA/Z 0 V 2(T) = VA, I 2 (T) = 0 § At t = 2 T, the doubled wave reaches the source end and is not reflected V 1(2 T) = VA, I 1(2 T) = 0 V 2(2 T) = VA, I 2(2 T) = 0 Solution Transmission Lines Class 6 38

39 Waveshape: Step-Function into T-Line with Open Ckt This is called “reflected wave switching” Solution Transmission Lines Class 6

40 Problem 1 b: Relationships § Source matched case: RS= Z 0 V 1(0) = 0. 5 VA, I 1(0) = 0. 5 IA GS = 0, V(x, ) = 0. 5 VA(1+ GL) § Uncharged line V 2(0) = 0, I 2(0) = 0 § RL= 3 Z 0 GL = (3 Z 0 -Z 0) / (3 Z 0 +Z 0) = 0. 5 V 1( ) = V 2( ) = 0. 5 VA(1+0. 5) = 0. 75 VA I 1( ) = I 2( ) = 0. 5 IA(1 -0. 5) = 0. 25 IA Solution Transmission Lines Class 6

41 Problem 1 b: Solution § At t = T, the voltage wave reaches load end and positive wave travels back to the source V 1(T) = 0. 5 VA, I 1(T) = 0. 5 IA V 2(T) = 0. 75 VA , I 2(T) = 0. 25 IA § At t = 2 T, the reflected wave reaches the source end absorbed V 1(2 T) = 0. 75 VA , I 1(2 T) = 0. 25 IA V 2(2 T) = 0. 75 VA , I 2(2 T) = 0. 25 IA Solution Transmission Lines Class 6

Waveshapes for Problem 1 b 42 Note that a properly terminated wave settle out at 0. 5 V Solution Transmission Lines Class 6

Transmission line step response § 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 43

44 Lattice Diagram Analysis – Key Concepts The lattice diagram is a tool/technique to simplify the accounting of reflections and waveforms § Diagram shows the boundaries § § (x =0 and x=l) and the reflection coefficients (GL and GL ) Time (in T) axis shown vertically Slope of the line should indicate flight time of signal Particularly important for multiple reflection problems using both microstrip and stripline mediums. Vs 0 Vs Zo V(source) Rs TD = N ps V(load) Time V(source) N ps a A A’ b 2 N ps 3 N ps c § B’ B d § Calculate voltage amplitude for each successive reflected wave Total voltage at any point is the sum of all the waves that have reached that point Rt rload rsource 0 V(load) 4 N ps 5 N ps Transmission Lines Class 6 C’ e

Lattice Diagram Analysis – Detail r r source load V(load) V(source) 0 45 Vlaunch 0 Time Vlaunch N ps Vlaunch rload Vlaunch(1+rload) 2 N ps Time Vlaunch rloadrsource Vlaunch(1+rload rsource) 3 N ps Vlaunch r 2 loadrsource Vlaunch(1+rload+r 2 loadrsource+ r 2 loadr 2 source) 4 N ps Vlaunch r 2 loadr 2 source 0 V(load) V(source) Zo Vs Rs TD = N ps Vs Rt 5 N ps Transmission Lines Class 6

Transient Analysis – Over Damped Transmission Lines Class 6 46

Transient Analysis – Under Damped 2 v 0 Zo Zs TD = 250 ps Vs rsource = -0. 3333 Time Assume Zs=25 ohms Zo =50 ohms Vs=0 -2 volts V(load) V(source) 0 1. 33 v 500 ps 1. 33 v 1000 ps rload = æ 50 ö Zo = (2) ç ÷ =1. 3333 + + Zs Zo è 25 50 ø Zs - Zo 25 -50 = = -0. 33333 Zs + Zo 25 + 50 Zl - Zo ¥ -50 = =1 + ¥ + Zl Zo 50 2. 66 v Response from lattice diagram 3 -0. 443 v 2. 5 1. 77 v 2000 ps rsource = 0 v 1. 33 v -0. 443 v 1500 ps 2. 22 v rload = 1 Vinitial =Vs 0. 148 v Volts V(source) 2 1. 5 Source 1 0. 5 2500 ps 1. 92 Load 0 0. 148 v 0 250 500 750 1000 1250 1500 1750 2000 2250 Time, ps 2. 07 Transmission Lines Class 6 47

Two Segment Transmission Line Structures Transmission Lines Class 6 48

Previous examples are the Assignment Consider the two segment preparation transmission line shown to the right. Assume RS= 3 Z 01 and Z 02= 3 Z 01. Use Lattice diagram and calculate reflection coefficients at the interfaces and show the wave forms of V 1(t), V 2(t), and V 3(t). Check results with PSPICE Transmission Lines Class 6 49