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Using Transmission Lines III – class 7 Purpose – Consider finite transition time edges and GTL. Acknowledgements: Intel Bus Boot Camp: Michael Leddige Agenda 2 Source Matched transmission of signals with finite slew rate Real Edges Open and short transmission line analysis for source matched finite slew rates GTL Analyzing GTL on a transmission line Transmission line impedances DC measurements High Frequency measurements Using Transmission Lines Introduction to Advanced Transmission Line Analysis Propagation of pulses with non-zero rise/fall times Introduction to GTL current mode analysis Now the effect of rise time will be discussed with the use of ramp functions to add more realism to our analysis. Finally, we will wrap up this class with an example from Intel’s main processor bus and signaling technology. Using Transmission Lines 3 Ramp into Source Matched T- line Ramp function is step function with finite rise time as shown in the graph. The amplitude is 0 before time t0 At time t0 , it rises with straight-line with slope At time t1 , it reaches final amplitude VA Thus, the rise time (TR) is equal to t1 - t0 . The edge rate (or slew rate) is VA /(t1 - t0 ) I1 RS Z0 ,T0 l V1 VS T = T0 l VA t0 Using Transmission Lines t1 I2 V2 4 5 Ramp into Source Matched T- line RS VS I1 Z0 ,T0 l V1 T = T0l VA t0 t1 Using Transmission Lines I2 V2 Ramp Function Ramp function is step function with finite rise time as shown in the graph. The amplitude is 0 before time t0 At time t0 , it rises with straight-line with slope At time t1 , it reaches final amplitude VA Thus, the rise time (TR) is equal to t1 - t0 . The edge rate (or slew rate) is VA /(t1 - t0 ) Using Transmission Lines 6 7 Ramp Cases When dealing with ramps in transmission line networks, there are three general cases: Long line (T >> TR) Short line (T << TR) Intermediate (T ~ TR) Using Transmission Lines 8 Real Edges Set up time array tmin 0 ns tmax 1.5 ns imax 1000 i 0 imax Specify Rise Time Spec amplitude r .5 ns A 1 imax Spec Waveshape b A a i i ns 10 sec Spec Slew Adj Fctr sajf .849 i Assignment: Find sajf for a Gaussian and capacitive edge 1 0.9 0.8 0.7 Amplitude i 12 ps 10 9 n 3 Define Wave signal vs. time array sajf n t i r a A 1 e i 0.6 0.5 0.4 0.3 0.2 0.1 0 t tmin tmax 0 0.5 1 1.5 2 2.5 3 time in nanoseconds 3.5 4 4.5 5 Neat trick to find rise time threshold .1 A threshold .9 A 1 2 i_thresholds hist( threshold a) define 10 and 90% thresholds risetime t i_thresholds1 i_thresholds0 Using Transmission Lines t i_thresholds0 risetime ns 0.5 sec Short Circuit Case 9 I1 I2 Current (A) IA 0.75IA 0.5I A Current 0.25IA 0 T 2T 3T Time (ns) V1 V2 VA Voltage (V) 4T 0.75VA Voltage 0.5VA 0.25VA Next step 0 T 2T 3T 4T Time (ns) Replace the step function response with one modified with a finite rise time The voltage settles before the reflected wave is encountered. Using Transmission Lines 10 Open Circuit with Finite Slew Rate I I TR 1 A I 0.75I A 0.5I A 0.25I TR Current A T 0 R T 2T 3T 4T Time (ns) V1 V A Voltage Voltage (V) Current (A) 2 V2 0.75V A TR 0.5V A 0.25V A 0 TR Using Transmission Lines T 2T 3T 4 T Time (ns) Consider the Short Circuit Case Voltage and current waveforms are shown for the step function as a refresher Below that the ramp case is shown Both the voltages and currents waveforms are shown with the rise time effect For example I2 doubles at the load end in step case, instantaneously in the ramp case, it takesTR Using Transmission Lines 11 Ramp into Source Matched Short T-line RS Very interesting case Interaction between rising edge and reflections Reflections arrive before the applied voltage reaches target amplitude short circuit case Let TR = 4T The voltage at the source (V1) end is plotted Voltage (V) Again, let us consider the I1 V1 VS I2 Z0 , T0 L, T Short V2 TR 0.5VA VRamp VStep 0.375VA 0.25VA showing comparison between0.125V A ramp and step The result is a waveform with three distinct slopes The peak value is 0.25VA 0 Solved with simple geometry and algebra Using Transmission Lines 2T 4T 6T 8T Time (ns) 12 Ramp into a Source Matched, Intermediate Length T-Line For the intermediate length Short Circuit Case transmission line, let the TR = 2T The voltage at the source (V1) TR 0.5VA Voltage (V) The reflected voltage arrives at the source end the instant the input voltage has reached target peak VRamp VStep 0.375VA 0.25VA 0.125VA end is plotted for two cases T 0 comparison between ramp and step Short circuit case Open circuit case 2T 3T 4T Time (ns) Open Circuit Case TR VA Voltage (V) Negative reflected voltage arrives and reduces the amplitude until zero The result is a sharp peak of value 0.5VA 13 0.75VA VRamp VStep 0.5VA 0.25VA Positive reflected voltage arrives and increases the amplitude to VA The result is a continuous, linear line Using Transmission Lines 0 T 2T 3T 4T Time (ns) 14 Gunning Transistor Logic (GTL) Chip (IC) V Chip (IC) Voltage source is outside of chip Reduces power pins and chip power dissipation “Open Drain” circuit Related to earlier open collector switching Can connect multiple device to same. Performs a “wire-or” function Can be used for “multi-drop bus” Using Transmission Lines Basics of GTL signaling – current mode transitions 15 Low to High Steady state low VL High to Low Steady state high Vtt R(n) Rtt R(n) Vtt Zo R(n) Vtt Rtt R(n) Vtt IL Rtt R(n) Switch opens Vtt Rtt Zo Vstep I L Zo V Vtt Switch closes Rtt Zo V Vstep 1 VL Rtt Zo R(n) Rtt Zo Vtt Rtt Zo V Vtt Vstep 1 Rtt Zo Rtt Zo R(n) Using Transmission Lines Vstep Vtt Zo Zo R(n) 16 Basics of current mode transitions - Example VV ( a ) _ rise VL Vstep VL I L Zo 70 50 VV (b ) _ rise 18.29mA 50 1 0.219 1.29V 70 50 (0.219) (18.29mA 50) 1.13V 1.6 1.5 V 1.4 V(a) 50 ohms 1.2 12 Ohms Volts 1.0 V(b) V(a) 0.8 1.5 IL 18.29mA 70 12 VV ( a ) _ fall 1.5 1.5 0.6 0.4 VL 1.5 12 0.219V 70 12 50 70 50 1 0.088V 50 12 70 50 0.0 0 2 4 6 8 10 Time, ns Using Transmission Lines V(b) 50 0.29 50 12 VV (b ) _ fall 1.5 1.5 0.2 70 ohms 12 17 GTL, GTL+ BUS LOW to HIGH TRANSITION END AGENT DRIVING - First reflection IL Vtt Vtt V(A) Rtt Zo Zs Rtt V(B) R(n) IL = Low steady state current VL = Low steady state voltage Vdelta = The initial voltage step launched onto the line Vinitial = Initial voltage at the driver T = The transmission coefficient at the stub Vtt 1 Rtt R(n) 2 Vtt R(n) VL 1 Rtt R(n) 2 Vdelta I L Zo || Rtt I L Zo Rtt Zo Rtt Zo || Zs Zo @ stub Zo || Zs Zo T 1 @ stub Notice termination was added at the source Vinitial Vdelta VL Why? Rtt Zo V ( B) T Vdelta 1 VL Rtt Zo V ( A) 2 T Vdelta VL Using Transmission Lines GTL, GTL+ BUS HIGH to LOW TRANSITION END AGENT DRIVING - First reflection V(A) Rtt Zo Zs Vtt R(n) 1 Rtt R(n) 2 Rtt || Zo Vdelta Vtt Rtt || Zo R(n) Zo || Zs Zo @ stub Zo || Zs Zo T 1 @ stub VL Vtt Vtt Rtt V(B) R(n) IL = Low steady state current VL = Low steady state voltage Vdelta = Initial voltage launched onto the line Vinitial = Initial voltage at the driver T = The transmission coefficient at the stub Vinitial Vtt Vdelta V ( A) Vtt 2 T Vdelta Rtt Zo V ( B) Vtt T Vdelta 1 Rtt Zo Using Transmission Lines 18 Transmission Line Modeling Assumptions All physical transmission have non-TEM characteristic at some sufficiently high frequency. Transmission line theory is only accurate for TEM and Quasi-TEM channels Transmission line assumption breaks down at certain physical junctions Transmission line to load Transmission line to transmission line Transmission line to connector. Assignment Electrically what is a connector (or package)? Electrically what is a via? I.e. via modeling PWB through vias Package blind and buried vias Using Transmission Lines 19 Driving point impedance – freq. domain Telegraphers formula Driving point impedance MathCAD and investigation R, L, C, G per unit length Zin Rdie Using Transmission Lines Cdie 20 21 Driving Point Impedance Example Physical Constants 12 ps 10 sec 9 ns 10 Propagation Constant Tpd 1 Vc r sec 9 nH 10 Speed of light ( f) 2 f Tpd Zo Zl cos l j Zo sin l Zo cos l j Zl sin l Linear Lossy Transmission Line Parameters h 10 7 mho henry 5.967 10 8 m Vc 3 10 Tpd 2.107 Input Impedance of a Transmission Line Zin Zl Zo l 6 henry 7 henry o 4.0 10 r 1 m o r r 4.3 a b Function for parallel circuit: Cap function ZC( Cx f) par( a b ) sec ab ns ft m 1 j 2 f Cx Set up Frequency Range For Plotting nl 100 nf 0 nl 1 fmin 1MHz L 11 nH in C 4.4 pF in R .2 fmax 1GHz in freq nf 14 mho G 10 in fmin ( fmax fmin ) nf nl Characteristic Impedance Z0( f) Load Impedance Cdie 1pF Rdie 40ohm Z1( f) par( ZC( Cdie f) Rdie) L j 2 f R C j 2 f G Expand and impedances to define driving point Impecnace 1 1 f f 2 2 Z1( f) cos 2 er len i Z0( f) sin 2 er len Vc Vc Zin ( er len f) Z0( f) 1 1 f f 2 2 len Z0 ( f ) cos 2 er len i Z1 ( f ) sin 2 er Vc Vc 80 Zin r 10in freq nf 60 40 20 0 Using Transmission Lines 2 10 8 4 10 8 6 10 freq nf 8 8 10 8 1 10 9 22 Measurement – DC (low frequency) 2 Wire Method Calibration Method Ohm Meter I Measure V Z=(V_measure-V_short)/I I*r=ERROR 4 Wire or Kelvin measurement eliminates error UNK Ohm Meter I Measure V Using Transmission Lines UNK High Frequency Measurement At high frequencies 4 wires are impractical. The 2 wire reduces to a transmission line The Vshort calibration migrates to calibration with sweep of frequencies for selection of impedance loads. Because of the nature of transmission lines illustrated in earlier slides Vector Network Analyzers (VNAs) used this basic method but utilized s-parameters More later on s parameters. Using Transmission Lines 23 24 Assignment Find driving point impedance vs. frequency of a short and open line (a) Derive the equation (b) given L=10inch, Er=4, L=11 nH/in, C=4.4 pF/in, R=0.2 Ohm/in, G=10^(-14) Mho/in, plot the driving point impedance vs freq for short & open line. (Mathcad or Matlab) (c) Use Pspice to do the simulation and validate the result in (b) Using Transmission Lines