Download Transmission lines

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
ab
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