Class07_Using_Transmission_lines by liamei12345

VIEWS: 6 PAGES: 24

									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
                                                                  3


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
  Ramp into Source Matched T- line
                                                                                      4




 Ramp function is step
  function with finite rise time
  as shown in the graph.                           RS        I1                  I2
                                                                        Z0 ,T0
     The amplitude is 0 before
     time t0                                 VS              V1
                                                                             l
                                                                                 V2
     At time t0 , it rises with
     straight-line with slope
     At time t1 , it reaches final                                T = T0l
     amplitude VA
     Thus, the rise time (TR) is
     equal to t1 - t0 .
     The edge rate (or slew rate)             VA
     is
         VA /(t1 - t0 )




                                                        t0              t1

                           Using Transmission Lines
                                                        5

Ramp into Source Matched T- line
              RS   I1                              I2
                                         Z0 ,T0
                                           l
        VS         V1                              V2


                             T = T0 l

   VA




         t0        t1
                        Using Transmission Lines
Ramp Function
                                                     6




 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
                                          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                                                                                                                                   12
                                                                                                        t max                         9              ps  10         sec
                    t min  0 ns t max 1.5 ns    imax 1000 i  0  imax           t  t min       i              ns  10       sec
                                                                                            i           imax
             Specify Rise Time Spec amplitude                            Spec Waveshape                         Spec Slew Adj Fctr
                    r  .5 ns                   A  1                            n  3                               sajf  .849
             Define Wave signal vs. time array
                          
                                  sajf  n 
                                       t i  
                           1  e  r   
                 a  A  
                  i                                b  A  a
                                                     i         i


                                                                                            Assignment: Find
                1
              0.9

                                                                                            sajf for a Gaussian
              0.8
              0.7
 Amplitude




                                                                                            and capacitive edge
              0.6
              0.5
              0.4
              0.3
              0.2
              0.1
                0
                    0       0.5       1         1.5          2        2.5      3            3.5        4            4.5          5
                                                             time in nanoseconds

                Neat tr ick to find rise time

                    t hreshold  .1 A t hreshold  .9 A        define 10 and 90% thresholds
                               1                      2
                                                                                                                                                  riset ime
                    i_thresholds  hist ( t hreshold  a)      riset ime t                                       t                                         0.5
                                                                                 i_thresholds 1 i_thresholds 0          i_thresholds 0         ns



                                                                Using Transmission Lines
Short Circuit Case
                                                                                  9




                                                                I1
                    IA
                                                                I2
  Current (A)

                0.75IA

                 0.5I A

                0.25IA                                Current
                          0   T             2T          3T       4T   Time (ns)
                                                                V1
                    VA
                                                                V2
  Voltage (V)




                0.75VA

                 0.5VA
                                                      Voltage
                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
                                      TR                                      1
                  I   A
                                                                          I
                                                                              2
Current (A)




              0.75I   A




               0.5I   A




                                            TR
                                                   Current
              0.25I   A




                          T   R

                          0       T        2T       3T                        4T   Time (ns)


                                                                                                              V1
                                                                          V   A
                                                         Voltage (V)                                          V2
                                                                       0.75V  A


                                                                                                    TR

                                  Voltage
                                                                       0.5V   A




                                                                       0.25V  A




                                                                                    0   TR     T   2T    3T    4 T Time (ns)




                                                 Using Transmission Lines
Consider the Short Circuit Case
                                           11




 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
                                                                                                                          12

Ramp into Source Matched Short T-line
                                                                     RS   I1                        I2
 Very interesting case                                                          Z0 , T0

       Interaction between rising                           VS            V1
                                                                                 L, T
                                                                                                    V2          Short
       edge and reflections
       Reflections arrive before the
       applied voltage reaches
       target amplitude                                                   TR
 Again, let us consider the
    short circuit case                              0.5VA
                                                                                            VRamp
       Let TR = 4T

                                    Voltage (V)
                                                  0.375VA
                                                                                            VStep
       The voltage at the source
                                                   0.25VA
       (V1) end is plotted
          showing comparison between
                                   0.125VA
          ramp and step
       The result is a waveform with
       three distinct slopes                                                                                  Time (ns)
                                                                 0         2T   4T         6T            8T
       The peak value is 0.25VA
   Solved with simple geometry
    and algebra
                              Using Transmission Lines
                                                                                                                                         13

Ramp into a Source Matched, Intermediate Length T-Line

   For the intermediate length                                                                 Short Circuit Case
    transmission line, let the TR =
    2T                                                                                         TR

         The reflected voltage arrives at                                     0.5VA

         the source end the instant the                                                                       VRamp




                                                  Voltage (V)
                                                                  0.375VA
                                                                                                              VStep
         input voltage has reached target                              0.25VA
         peak
   The voltage at the source (V1)                                0.125VA

    end is plotted for two cases                                                                                               Time (ns)
                                                                                       0        T       2T   3T           4T
         comparison between ramp and step
   Short circuit case                                                                              Open Circuit Case
         Negative reflected voltage arrives                                                    TR
         and reduces the amplitude until                                         VA
         zero



                                                                Voltage (V)
                                                                              0.75VA
         The result is a sharp peak of value                                   0.5VA
                                                                                                                  VRamp
         0.5VA                                                                                                    VStep
                                                                              0.25VA
   Open circuit case
         Positive reflected voltage arrives                                                0    T       2T   3T           4T Time (ns)
         and increases the amplitude to VA
         The result is a continuous, linear
         line
                                Using Transmission Lines
                                                     14

Gunning Transistor Logic (GTL)
                                                 V
      Chip (IC)      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                                              High to Low
Steady state low                                     Steady state high
                      Vtt  R(n)        Vtt
               VL                                                                               Vtt
                      Rtt  R(n)
                                              Rtt                            Zo                        Rtt
                                   Zo

 R(n)
                        Vtt                            R(n)               V  Vtt
               IL 
                    Rtt  R(n)

Switch opens                                          Switch closes
                                                                                                 Vtt
             Rtt  Zo                  Vtt                          Rtt  Zo 
  V  Vstep 1          VL                        V  Vtt  Vstep 1           
             Rtt  Zo                        Rtt                       Rtt  Zo                    Rtt


                                   Zo                                                       Zo
                                                       R(n)                           Zo
  R(n)
               Vstep  I L Zo                                       Vstep  Vtt
                                                                                   Zo  R (n)


                                    Using Transmission Lines
                                                                                                                                    16


Basics of current mode transitions - Example
         VV ( a ) _ rise  VL  Vstep  VL  I L Zo                                             70  50 
                                                             VV (b ) _ rise  18 .29 mA  50  1         0.219  1.29V
          (0.219)  (18.29mA  50)  1.13V                                                     70  50 
         1.6


                                                                                                                    1.5 V
         1.4

                                                                           V(a)                                        70 ohms
         1.2
                                                                                                      50 ohms
                                                                                                                       V(b)
                                                                                         1.5
         1.0                                                 12 Ohms               IL           18.29mA
                               V(b)                                                     70  12
 Volts




         0.8                   V(a)
                                                                                             50
                                                           VV ( a ) _ fall  1.5  1.5             0.29
                                                                                           50  12
         0.6



         0.4
                                                                                                       50  70  50 
                                                                      VV (b ) _ fall  1.5  1.5            1         0.088V
                                                                                                     50  12  70  50 
         0.2                  1.5 12
                      VL              0.219V
                              70  12
         0.0
               0          2            4              6               8              10              12
                                                  Time, ns

                                                          Using Transmission Lines
                                                                                                              17
         GTL, GTL+ BUS LOW to HIGH TRANSITION
         END AGENT DRIVING - First reflection
                                                                              Vtt
                                                                  IL 
                                                                       1
   Vtt                                    Vtt
                                                                         Rtt  R (n)
                                                                       2
          Rtt        V(A)                       Rtt                      Vtt  R (n)
                            Zs
                                                                  VL 
                                                                       1
                Zo                       V(B)                            Rtt  R (n)
R(n)                                                                   2
                                                                                             I L  Zo  Rtt
                                                                  Vdelta  I L  Zo || Rtt 
       IL = Low steady state current                                                           Zo  Rtt
       VL = Low steady state voltage
                                                                              Zo || Zs  Zo
       Vdelta = The initial voltage step launched onto the line    @ stub 
       Vinitial = Initial voltage at the driver                               Zo || Zs  Zo
       T = The transmission coefficient at the stub
                                                                  T  1   @ stub
         Notice termination was                                   Vinitial  Vdelta  VL
         added at the source                                      V ( A)  2  T  Vdelta  VL

         Why?                                                                           Rtt  Zo 
                                                                  V ( B )  T  Vdelta 1          VL
                                                                                        Rtt  Zo 
                                         Using Transmission Lines
                                                                                                   18
       GTL, GTL+ BUS HIGH to LOW TRANSITION
       END AGENT DRIVING - First reflection

                                                                   Vtt  R(n)
       Vtt                                 Vtt             VL 
                                                                  1
             Rtt        V(A)
                               Zs                Rtt                Rtt  R(n)
                                                                  2
                   Zo                     V(B)                                  Rtt || Zo
                                                           Vdelta  Vtt 
R(n)                                                                        Rtt || Zo  R(n)
                                                                     Zo || Zs  Zo
                                                           @ stub 
       IL = Low steady state current                                 Zo || Zs  Zo
       VL = Low steady state voltage
       Vdelta = Initial voltage launched onto the line     T  1   @ stub
       Vinitial = Initial voltage at the driver
                                                           Vinitial  Vtt  Vdelta
       T = The transmission coefficient at the stub
                                                           V ( A)  Vtt  2  T Vdelta
                                                                                     Rtt  Zo 
                                                           V ( B)  Vtt  T Vdelta 1        
                                                                                     Rtt  Zo 

                                       Using Transmission Lines
                                                            19


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
Driving point impedance – freq. domain
                                                      20




  Telegraphers formula
  Driving point impedance
  MathCAD and investigation

        R, L, C, G per unit length

  Zin
                                        Rdie   Cdie




                    Using Transmission Lines
                                                                                                                                                                                                                                            21

  Driving Point Impedance Example
                                                                            Physical Constants

              12                         9                           9                    6                             7 mho                        7 henry
   ps  10         sec       ns  10          sec        nH  10          henry   h  10        henry   5.96710 
                                                                                                                                    o  4.0  10                          r  1       o  r         r  4.3
                                                                                                                               m                                  m
 Propagation Constant                             Speed of light Vc  3 108
                                                                                               m
                                                                                                                                                                  a b
                                                                                                         Function for parallel circuit:                                 Cap function ZC( Cx f) 
                                                                                            sec                                                                                                                                    1
                                                                                                                                par( a  b )
   Tp d 
             1
                   r              ( f)  2  f Tp d        Tp d  2.107
                                                                              ns                                                                              ab                                                          j 2 f Cx
             Vc                                                                ft
 Input Impedance of a Transmission Line                                                            Set up Frequency Range For Plotting
                                 Zl cos    l  j Zo  sin    l                                                                                                                  ( fmax  fmin)  nf
   Zin Zl Zo    l   Zo                                                        nl  100 nf  0  nl  1 fmin  1MHz          fmax 1GHz                freq         fmin 
                                 Zo  cos    l  j Zl sin    l                                                                                                 nf                           nl

                                                                                                                                                         14 mho
 Linear Lossy Transmission Line Parameters                                               L  11
                                                                                                    nH
                                                                                                            C  4.4
                                                                                                                       pF
                                                                                                                             R  .2           G  10                           Characteristic Impedance
                                                                                                    in                 in              in                          in
                                                                                                                                                                                                      L j 2  f  R
                                                                                                                                                                                      Z0( f) 
Load Impedance Cdie  1pF Rdie  40ohm Z1( f)  par( ZC( Cdie  f)  Rdie)                                                                                                                         C j  2  f  G

 Expand  and impedances to define driving point Impecnace
                                                         1                                       1  
                                                                                                  
                                 Z1( f)  cos 2   er  len  i Z0( f)  sin 2   er   len 
                                                     f      2                                  f       2

 Zin( er  len  f)  Z0( f) 
                                               Vc                                      Vc                    
                                                         1
                                                            
                                                                                                     1
                                                                                                        
                                                                                                                 
                                                                                                                                                      80
                                                     f
                                 Z0( f)  cos 2   er 2   len  i Z1( f)  sin 2  f  er 2   len 
                                               Vc                                      Vc                                                          60
                                                                                                                             Zin r  10in freq nf
                                                                                                                                                         40


                                                                                                                                                         20
                                                                                                                                                                              2 10       4 10           6 10            8 10          1 10
                                                                                                                                                                                  8               8             8              8                  9
                                                                                                                                                              0
                                                                                                                                                                                                      freq nf



                                                                                     Using Transmission Lines
                                                                    22

Measurement – DC (low frequency)

  2 Wire Method
                                         Calibration Method
  Ohm
  Meter             I*r=ERROR            Z=(V_measure-V_short)/I
          Measure
          V
      I                          UNK


                                 Ohm
                                 Meter
                                             Measure
   4 Wire or Kelvin                          V
   measurement                         I                      UNK
   eliminates error

                       Using Transmission Lines
                                                  23

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
                                                  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

								
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