Class07_Using_Transmission_lines by liamei12345

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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
3

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
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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
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 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

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
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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