Transcript Introduction to Radar Systems

High-Speed Digital Circuit Design
Chris Allen ([email protected])
Course website URL
people.eecs.ku.edu/~callen/713/EECS713.htm
1
Outline
Syllabus
Instructor information, course description, prerequisites
Textbook, reference books, grading, course outline
Preliminary schedule
Introductions
What to expect
First assignment
Review of circuit fundamentals
When are HSD design techniques needed?
Transient response of reactive circuits
Measuring device reactance
Impact of via inductance
Crosstalk from mutual reactance
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Syllabus
Prof. Chris Allen
Ph.D. in Electrical Engineering from KU 1984
10 years industry experience
Sandia National Labs, Albuquerque, NM
AlliedSignal, Kansas City Plant, Kansas City, MO
Phone: 785-864-8801
Email: [email protected]
Office: 3024 Eaton Hall
Office hours: TR: 10 to 10:50 AM
Course description
Basic concepts and techniques in the design and analysis of highfrequency digital and analog circuits. Topics include: transmission lines,
ground and power planes, layer stacking, substrate materials,
terminations, vias, component issues, clock distribution, cross-talk,
filtering and decoupling, shielding, signal launching.
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Syllabus
Prerequisites
Electronic Circuits I (EECS 312) Non-linear circuit elements, MOSFETs, BJTs, diodes,
digital circuits and logic gates
Electromagnetics II (EECS 420) (recommended) transmission line theory
Textbook
High-Speed Digital Design
by H. W. Johnson and M. Graham
PTR Prentice-Hall, 1993, ISBN 0133957241
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Syllabus
Reference books
High-Speed Digital Propagation –
Advanced Black Magic
by H. W. Johnson and M. Graham
Prentice Hall PTR , 2003, ISBN 013084408X
High-Speed Digital System Design –
A Handbook of Interconnect Theory and Design Practices
by S. H. Hall, G. W. Hall, J. A. McCall
Wiley-IEEE Press, 2000, ISBN 0471360902
Digital Transmission Lines –
Computer Modeling and Analysis
by K. D. Granzow
Oxford University Press, 1998, ISBN 019511292X
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Grades and course policies
The following factors will be used to arrive at the final
course grade:
Homework & quizzes
Design project
Midterm exam
Final exam
15 %
15 %
35 %
35 %
Grades will be assigned to the following scale:
A
B
C
D
F
90 - 100 %
80 - 89 %
70 - 79 %
60 - 69 %
< 60 %
These are guaranteed maximum scales and may be revised downward at the
instructor's discretion.
Read the policies regarding homework, exams, ethics, and
plagiarism.
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Outline and schedule
Tentative Course Outline (subject to change)
Review of circuits, electronics and traveling wave theory
(Thevenin and Norton equivalents, device capacitance and inductance, current sourcing and
sinking transmission line impedance, source and load impedance, reflections)
Measurement Issues
(requirements and specifications, design for test, test equipment, special fixtures)
Properties of high-speed gates
(circuit families and their characteristics, propagation delay, rise/fall times, input impedance, output
impedance, sensitivity to electro-static discharge (ESD), heat dissipation)
Transmission lines
(microstrip, stripline, coplanar, multiwire)
Ground and power planes
(number of planes, placement, characteristics)
Substrate materials
(printed wiring boards (PWBs), multi-chip modules (MCMs))
Thermal issues
(junction temperature, thermal resistance, thermal vias, cooling options)
Packaging technologies
(packaged parts on PWB (through hole and surface mount), bare die, chip-on-board, multi-chip
modules)
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Outline and schedule
Tentative Course Outline (continued)
Routing issues
(fanout limits, stubs, daisy chaining, testability)
Terminations and vias
(termination options)
Clock distribution
(timing skew, fanout, fine adjustments)
Cross-talk
(analysis, design rules, consequences)
Filtering and decoupling
(techniques and requirements)
Shielding and grounding
(electromagnetic interference (EMI) and electromagnetic compatibility (EMC))
Signal launching
(connection between boards, impedance matching)
Special high-speed circuit design techniques
(pipelining and latency, multiplexing)
Future trends
(chip speed, complexity, number of I/O, optical interconnection (die and board level), chip
stacking)
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Overview
Many of the topics discussed are covered adequately in the text
Other topics to be discussed will use additional resources including
manufacturer’s application notes, product data sheets, and other texts.
High-frequency circuit design is both an art and a science;
hence the Black Magic reference in the title.
The text presents general design rules frequently without deriving them;
these result from past experience and were learned the hard way.
What is meant by high frequency or high speed?
These are relative terms.
What we mean is typically signals with fundamental frequency components
> 100 MHz although lower frequency signals may qualify in some cases.
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Applications
Numerous systems use high-speed digital signals.
Examples include:
Radar systems
GSa/s analog-to-digital converters (ADCs) and digital-to-analog
converters (DACs); systems with wide signal bandwidths;
computations measured in GFLOPS
Communication systems
channel rates of 40 Gb/s are common today over long-distance
optical fiber
Computers
supercomputers and cluster computing; performance measured in
operations per second (OPS); fine-resolution displays
Giga-sample per second: GSa/s
Giga-bit per second: Gb/s
Giga-floating operations per second: GFLOPS
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Applications
Many of the high-speed digital design techniques apply to
both analog and digital signals. Examples include passive
component selection and interconnection techniques (e.g.,
transmission lines), but not active component design.
A major difference is that analog systems often have a
bandwidth that is a small fraction of the carrier frequency
whereas in digital systems generally require signal
frequencies extending from DC to 2 or more times the
highest clock frequency.
The key issue is preservation of signal integrity.
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Outline and schedule
Class Meeting Schedule
August: 25, 27
September: 1, 3, 8, 10, 15, 17, 22, 24, 29
October: 1, 6, 8, (no class on 13th), 15, 20, 22, 27, 29
November: 3, 5, 10, 12, 17, 19, 24, (no class on 26th)
December: 1, 3, 8, 10
Final exam scheduled for Wednesday, Dec 16, 10:30 a.m. to 1:00 p.m.
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Course website
URL: people.eecs.ku.edu/~callen/713/EECS713.htm
Contains –
Syllabus
Class assignments
Some supplemental course material
Project information (when issued)
Powerpoint files used in class presentations
• continually updated to correct errors or enhanced
• file contents typically span many presentations (class sessions)
• max slide count ~ 100
Links to recorded presentations (audio and Powerpoint)
Special announcements (when issued)
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Introductions
Name
Major
Specialty
What you hope to get from of this experience
(Not asking what grade you are aiming for )
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What to expect
Course is being webcast, therefore …
Most presentation material will be in PowerPoint format 
Presentations will be recorded and archived (for duration of semester)
• Not 100% reliable (occasionally recordings fail due to a variety of causes)
Student interaction is encouraged
Remote students must activate microphone before speaking
Please disable microphone when finished
Homework assignments will be posted on website
Electronic homework submission logistics to be worked out
We may have guest lecturers later in the semester
To break the monotony, we’ll take a couple of 1- to 2minute breaks during each class session
(roughly every 15 to 20 min)
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High-speed digital circuit design
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Your first assignment
Send me an email (from the account you check most often)
To: [email protected]
Subject line: Your name – EECS 713
Tell me a little about yourself and what knowledge you
hope to gain from this experience
Attach your ARTS form (or equivalent)
ARTS: Academic Requirements Tracking System
Its basically an unofficial academic record
I use this to get a sense of what academic experiences you’ve had
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Review
Thevenin and Norton equivalent circuits
Complex circuits can be modeled in terms of
VT and ZT (Thevenin) or IN and ZN (Norton)
Note that ZT = ZN
This concept applies to
both analog and digital devices
both inputs and outputs
So we can determine the impedance (Z)
of a source (driver) or a load
18
Review
Impedances are generally complex: Z = R + jX
R is real part (resistive)
X is reactive part (inductive or capacitive)
Of particular interest is the capacitive nature of the device
as this often determines the circuit’s frequency response
(time constant)
Another important parameter is the drive device’s current
sourcing and sinking capacity (IN)
These (sourcing / sinking) are not necessarily equal depending on
the circuit design
19
Review
Transmission lines
Characteristic impedance, Zo = V / I
Velocity of propagation, vp < c
Propagation delay, d = ℓ / vp where ℓ is the line length
Impedance mismatches between the transmission line and the
source impedance (ZS) or load impedance (ZL) it connects will
reflect part of the impinging wave resulting in distortions in the
voltage and current along the transmission line
20
When are HSD design techniques needed?
High-speed design (HSD) techniques (to be explored later)
should be applied when the circuit trace length (the line
length) ℓ is greater than about a quarter of the length of
the rising edge, l
ℓ>l/4
Where
rise time
Tr
l

propagatio n delay D
and
propagation delay = (propagation velocity)-1
21
When are HSD design techniques needed?
What is rise time ?
Rise time – or fall time (usually assumed to be equal) – can be
defined in a variety of methods
Rise time, Tr, is defined as the time required for a signal to change
from a specified low value to a specified high value.
Typically these values are 10% and 90% of the step height.
Others may use 20% and 80% of the step height
or the step height divided by the center or maximum slope
(See Appendix B in the text for more details)
22
When are HSD design techniques needed?
Why use rise time, Tr, and not the clock frequency, fCLK?
Consider the circuit
In the time domain the output signal appears as
23
When are HSD design techniques needed?
In the frequency domain the output signal appears as
The knee frequency,
Fknee, which is inversely
related to the rise time,
is much higher than
the clock frequency,
FCLK.
FCLK  Fknee 
1
2 Tr
Most energy in digital pulses concentrates below the knee frequency.
Any circuit with a flat frequency response up to the Fknee frequency will
pass a digital signal practically undistorted.
24
When are HSD design techniques needed?
What is propagation delay ?
1 ns = 10-9 s
Signal delay is inversely related to signal velocity
1 ps = 10-12 s
In free space, signals travel at speed of light, vp = c
v p  c  1 oo
1 in = 2.54 cm
c = 3 x 108 m/s = 30 cm/ns = 11.8 in/ns = 0.0118 in/ps
In free space, delay is Dfreespace = 1/speed = 84.7 ps/in
Propagation through a medium is slower than in free space.
In non-free space v p  c  r  r
where r is the relative permeability and
r is the relative permittivity of the medium
Typically, r = 1 so that
vp  c
r  c n
where n = refractive index =  r
25
When are HSD design techniques needed?
What is propagation delay ?
Consider a stripline trace in FR-4 (r = 4.5, n = 2.12)
Stripline transmission
line geometry
Since the field lines are confined entirely within
the glass-epoxy dielectric, the propagation
velocity is simply
11.8 in ns
 5.56 in ns 47% of c 
2.12
D  180 ps in
vp 
Trace – A line or "wire" of
conductive material such as
copper, silver or gold, on the
surface of or sandwiched
inside a PCB, printed circuit
board.
FR-4 – Flame-Retardant
industrial laminate having a
substrate of woven-glass
fabric and resin binder of
epoxy. FR-4 is the most
common dielectric material
used in the construction of
PCBs in the USA
26
When are HSD design techniques needed?
What is propagation delay ?
Now consider a microstrip trace on FR-4
Microstrip transmission line geometry
Here the field lines are not confined within the glass-epoxy
dielectric, so the effective permittivity is between that of air (r = 1)
and that of FR-4 (r = 4.5).
Therefore the propagation velocity is between c and 0.47c.
So for microstrip lines
85 ps/in (free space) < D < 180 ps/in (stripline).
Furthermore, the propagation velocity may be frequency dependent
leading to signal dispersion and pulse distortion.
27
When are HSD design techniques needed?
What is l ?
l is the length of the rising edge and l = Tr /D.
Consider a circuit with Tr = 800 ps
(GaAs technology)
GaAs – An alloy of
gallium and arsenic
compound (GaAs)
that is used as the
base material for
chips.
Fknee for this signal will be (1600 ps)-1 or 625 MHz
The signal propagates on a stripline transmission line
Alumina – A ceramic
fabricated on alumina (r = 8.7).
used for insulators in
Find l
ps
ps
D  85  8.7  251
in
in
800 ps
l
 3.2 in  8.13 cm
251 ps in
So l/4 = 0.8 in or ~ 2 cm
substrates in thin film
circuits. It can
withstand
continuously high
temperatures and
has a low dielectric
loss over a wide
frequency
range. Aluminum
oxide (Al2O3).
Therefore if the circuit length ℓ > 2 cm HSD design techniques
should be followed.
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When are HSD design techniques needed?
Failure to follow HSD design techniques may result in
Signal reflections
Cross talk
Interference
Ringing
29
Transient response of reactive circuits
Reactances are not usually considered in low-speed digital
circuit designs.
Think about low-speed behavior as essentially DC
Reactive circuit elements of interest include:
capacitance – load, distributed, parasitic
inductance – load, distributed, parasitic
mutual capacitance – capacitive coupling
mutual inductance – inductive coupling
30
Transient response of reactive circuits
Whatever the nature of the reactance,
we can treat it as a lumped quantity
being driven by a signal generator.
Consider the circuit loaded by a capacitor, driven by a step function.
Predict the resulting output voltage (Vo(t)), current (I), and short-term
impedance (i.e., Vo(t)/I(t)) as the circuit responds to the stimulus.
• What will the output voltage be immediately after the step function?
• What will be the steady-state output voltage?
• What will the current be immediately after the step function?
• What will be the steady-state current?
31
Transient response of reactive circuits
Time-domain viewpoint
Capacitor voltage cannot change instantaneously
• Over short-time intervals capacitors behave as ideal voltage sources.
• For modeling purposes, capacitors are modeled as a short circuits.
Long after the transient, the capacitor current goes to zero
• For modeling purposes, capacitors are modeled as open circuits.
Inductor current cannot change instantaneously
• Over short-time intervals an inductor behaves as a current source.
• For modeling purposes, inductors are modeled as open circuits.
Long after the transient, the inductor voltage goes to zero
• For modeling purposes, inductors are modeled as short circuits.
Frequency-domain viewpoint
At high frequencies, capacitors behave as shorts, inductors as opens
At low frequencies, capacitors behave as opens, inductors as shorts
32
Transient response of reactive circuits
Transient response to step function.
33
Transient response of reactive circuits
Consider the circuit loaded by a perfect inductor,
driven by a step function.
Predict the resulting output voltage (Vo(t)), current (I), and short-term
impedance (i.e., Vo(t)/I(t)) as the circuit responds to the stimulus.
•What will the output voltage be immediately after the step function?
•What will be the steady-state output voltage?
•What will the current be immediately after the step function?
•What will be the steady-state current?
34
Transient response of reactive circuits
Transient response to step function.
35
Transient response of reactive circuits
Homework #1
Following a similar procedure, predict the transient response for the
four load conditions shown.
See the course website for homework assignment details.
36
Measuring device reactance
Sometimes is it necessary to measure reactance
(capacitance & inductance) of devices
• circuit structures (traces)
• packages
• leaded components
Why not apply EM analysis instead ?
• too many unknowns (r, internal geometry, material )
• complex geometries, difficult to measure or model
• cost in $, time, resources
Building a small test fixture may be
• more efficient and accurate
• relatively simple to fabricate
37
Measuring device reactance
Test equipment requirements
While specialized test equipment exists to characterize inductance,
capacitance, resistance, etc., such instruments may not be
available.
More common instrumentation that may be used include
• Pulse generator with a small Tr value
• Oscilloscope with a wide bandwidth
BW3dB
Tr
100 ps
800 ps
2 ns
0.35

Tr
see page 85 in text
BW
3.5 GHz
440 MHz
175 MHz
BW3dB – bandwidth over which signal power falls by 50% (3 dB)
38
Measuring device capacitance
Capacitance test fixture design
Simplified test circuit
Thevenin equivalent circuit connected to unknown capacitive load,
Z, also known as the Device Under Test (DUT)
The output voltage from simple, first-order RC circuit is


Vo t   V 1  e  t  , for t  0
where  is the circuit’s time constant,  = Rs C.
If Rs is known, the unknown capacitance, C, can be estimated by
observing the rise time  of Vo
39
Measuring device capacitance
If the anticipated DUT capacitance is a few pF,
what value of Rs is required to provide a measurable  ?
Example:
assume pulse generator’s Tr = 800 ps (BW = 0.35/800 ps = 440 MHz)
assume oscilloscope’s BW  440 MHz
oscilloscope’s smallest useful time scale ~ 500 ps / div
therefore we need   500 ps
If we assume C = 1 pF and set  = 500 ps, then
Rs =  / C  500 ps / 1 pF
or
Rs  500 
40
Measuring device capacitance
Suggested capacitance test fixture design
Determine values for R1 and R2 so that Rs  500 
41
Measuring device capacitance
Capacitance test fixture design insights
To reduce reflection which would corrupt the measurement –
the cables have Zo = 50 
the oscilloscope has internal termination of 50 
the test fixture has termination of 50 
the pulse generator has source termination of 50 
Coaxial cables enter/exit from opposite sides to reduce direct feedthrough
R1 provides isolation between the source and the DUT
R2 acts as a voltage divider with the oscilloscope’s 50- termination
42
Measuring device capacitance
Suggested value for R2 is 1000  (1 k)
This keeps the scope from loading the capacitor,
i.e., without R2, Rs ~ 50 
For R2 = 1000 , what should R1 be to make Rs  500  ?
To answer this, analyze the circuit as seen by the DUT
43
Measuring device capacitance
The resistance between the DUT terminals written as
Rs = (1000 + 50) // (R1 + 50 // 50)  500 
or
Rs = (1050) // (25 + R1)  500 
or
1/1050 + 1/(25 + R1)  1/500
or
R1  930 
With 5% resistors
choices are
910  or 1000 
Therefore
R1 = 1000 
and
Rs = 519 
44
Measuring device capacitance
We can test our setup by shorting the DUT test points
ideal response should be zero
Circuit analysis needed to
predict the range of Vo
select the power rating for resistors
using steady-state analysis when V = 1 V
45
Measuring device capacitance
In steady state, treat DUT as open circuit
Resistance seen by source
R = 50 + 50//2050 or
R = 98.8 
Source current, I1, is
I1 = 1 V / 98.8 
I1 = 10.1 mA
Node voltages and branch currents are
VA = 1 – (50) (I1) = 494 mV
I2 = 494 mV / 50 = 9.88 mA
I3 = VA / 2050 = 241 A
VB = (1050) (I3) = 253 mV
VC = 12 mV
Note that VB/VC = 21
a 21:1 voltage reduction
46
Measuring device capacitance
Capacitance test fixture design
Now find the power dissipation in various resistors
In the pulse generator’s Rs, Pdiss = RS(I1)2 = 5 mW
In the test fixture’s 1-k resistors, Pdiss = 1k(I3)2 = 58 W
In the test fixture’s 50- resistor, Pdiss = 50(I2)2 = 5 mW
Therefore 1/8-W resistors can be used in the test fixture
47
Measuring device capacitance
Capacitance test fixture application
The Thevenin equivalent circuit for this test fixture is
Vmeas = Vo / 21
With the a component in the DUT, find the rise time 

Vo  Vs 1  e  t 

when t = , Vo = Vs(1 – e-1) = 63.2% of Vs or
Vo = 160 mV and Vmeas = 7.58 mV
 corresponds to the point where ΔV = 7.58 mV
C =  / 519 
For example: If  = 15 ns, then C = 29 pF
48
Measuring device inductance
Similar to the discussion on capacitance measurement, we
can design a fixture for measuring device inductance.
Setup designed to measure inductances as low as a few nH.
Thevenin equivalent circuit connected to unknown inductive load, Z,
also known as the Device Under Test (DUT).
We know that the current through an inductor cannot change
instantaneously, and
t 
 V e
Vo t   
 0,
, for t  0
for t  0
where the time constant,  = L/Rs
For L = 1 nH (10-9 H) and a desired  of 500 ps
requires Rs ~ 2  (i.e., a small Rs value is desired)
49
Measuring device capacitance
Suggested inductance test fixture design
Determine values for R1 and R2 so that Rs < 2 
50
Measuring device capacitance
Design requirements
• R1 + R2 = 50 
at t = 0 inductance behaves as open circuit
• Rs = 50 // R2 // (R1 + 50)
first 50  term represents oscilloscope termination
second 50  term represents pulse generator termination
Notice: There is no resistor between
the DUT and the oscilloscope
 no voltage reduction, Vo = Vmeas
Also notice: Only R1 isolates the
pulse generator from the DUT
 the 50-  back termination is very
important to manage reflections
51
Measuring device inductance
Begin by letting R1 = 39  (from author’s example)
Therefore R2 = 10   Rs = 7.6 
[does not meet 2- requirement]
Voltage applied across DUT will be
Vs (10//50) / (50 + 39 + 10//50) = 8.6% of Vs
52
Measuring device inductance
Try again, let R1 = 47 
Therefore R2 = 2.2   Rs = 2.06  [meets 2- requirement]
Voltage applied across DUT will be
Vs (2.2//50) / (50 + 47 + 2.2//50) = 2.1% of Vs
53
Measuring device inductance
Now consider the effects of testing a component with both
capacitance and inductance
Example in text describes a DUT is a 1 long trace, 0.010 (10 mils)
wide with a 0.035 (35 mils) diameter via to ground.
1
 Cross-section view of
Oblique view of 
microstrip trace and via....
microstrip trace and via.
10 mils
35 mils
Estimated characteristics for this structure are
C = 2 pF, L = 9 nH
At the leading pulse’s leading edge (Tr = 800 ps) the max frequency
of interest is Fknee = 0.5/Tr = 625 MHz
The capacitive reactance, Xc = (2  f C)-1 and for f = Fknee
Xc(f=Fknee) = Tr/( C) = 800 ps /( 2 pF) = 127 Ω
The inductice reactance, XL = 2 f L and for f = Fknee
XL(f=Fknee) =  L / Tr =  9 nH / 800 ps = 35 Ω
54
Measuring device inductance
Equivalent circuit for via using  transmission line model

Now, Xc = 2Tr/( C) = 254 Ω
The instantaneous impedance, Vo(t)/I(t) at t=0+, is
XC // XL or 256 Ω // 35 Ω = 30.8 Ω
So the error in the measurement of XL is
(35 - 30.8) / 35 = 0.12
 12% error due to parasitic capacitance
55
Measuring device inductance
Now find the inductance from the observed waveform
Vo t   V e  t R s
at time t1, Vo t1   V1  V e  t1 R s
L
L
find time t2 on waveform so that Vo t 2   V1 e 1  Vo t1  0.37
Vo t 2   V e  t 2 R s
L
 V e  t1 R s
L
e 1
 t2 Rs/L = 1+ t1 Rs/L
t2 – t1 = L/Rs 
L = Rs (t2 – t1)
56
Measuring device inductance
The author also presents an alternative technique for
finding L from the observed Vo(t) using the area under the
curve.
An advantage of this approach is that it is less sensitive to
noise (measurement error) and less sensitive to the rise
time.
A disadvantage is that it requires integration of the
waveform, however this may be achieved either using
built-in math functions in modern oscilloscopes or
by exporting the captured the waveform for external
analysis.
57
Impact of via inductance
Now consider the impact of via inductance on circuit
performance. Consider the circuit
A 50- transmission line terminated
with 50- load resistor connected to
ground through a via.
Physically this may look like
An equivalent circuit is
58
Impact of via inductance
Instantaneous impedance at load end is
(50 + XL // XC) // 10k
where XC = Tr / ( C) and XL =  L / Tr
For Tr = 3 ns, XC = 954  and XL = 9.42 , XL // XC = 9.3 
The effective impedance to ground is 59 
resulting impedance mismatch produces a small reflection
Reflection = (ZL – Zo) / (ZL + Zo) = 8%
59
Impact of via inductance
Now consider the impact of via
inductance on a circuit where
a bus of 8 lines use a single
ground via for all 8 termination
resistors.
Physically this may look like
An equivalent circuit is
60
Impact of via inductance
Now the find the instantaneous impedance at load end for
Tr = 3 ns. We know that XL // XC = 9.3 
Consider the case when all 8 lines transition low to high
The effective transmission line impedance is 6.25 
and the effective impedance to ground is 15.55 
resulting a significant impedance mismatch
Reflection = (ZL – Zo) / (ZL + Zo) = (15.55 – 6.25) / (15.35 + 6.25) = 43%
61
Crosstalk from mutual reactance
Consider the circuit
In the absence of coupling
the signals A and B operate
independently, i.e., no crosstalk.
Crosstalk between circuits can occur through mutual
capacitance or through mutual inductance.
62
Crosstalk from mutual capacitance
Mutual capacitance involved electric field interaction
between circuits A and B.
The magnitude of the mutual
capacitance depends on the
circuit geometry and on the
material properties.
We know IM = CM dV/dt = CM d(VA-VB)/dt
Consider the case where VA = VB
static case: both high or both low and unchanging
dynamic case: both transitioning low-to-high or high-to-low
IM = 0, no capacitive current flows
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Crosstalk from mutual capacitance
Now consider the case where VA ≠ VB
signal A transitioning from low-to-high
while signal B is static low
d VA
 d VA d VB 
IM  CM 

 CM

dt 
dt
 dt
Now
so
0
dVA/dt ≈ ΔV/Tr
IM = CM ΔV/Tr
IM represents additional current flowing in circuit B
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Crosstalk from mutual capacitance
IM = CM ΔV/Tr flows into circuit B
VB = IM RB // RSB
if RSB >> RB, then VB = IM RB
if RSB = RB, then VB = IM RB / 2
This voltage can be thought of as crosstalk
Crosstalk 
unwanted signal
pure signal
Unwanted signal = IM RB // RSB = CM (RB // RSB) ΔV/Tr
Pure signal in B is ΔV
Crosstalk 
C M R B // R SB
TR
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Crosstalk from mutual capacitance
The resulting crosstalk
for capacitive coupling
appears as
Can we measure CM ?
For a known ΔV, Tr, RB << RSB
IM  CM
d VA
dt
Integrate both sides to get
d VA
 I M d t  CM  d t dt  C M V
Area  R B  I M d t  R B C M V
so
Area
CM 
V R B
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Crosstalk from mutual inductance
Interaction between circuits can also result through
magnetic field coupling, this is
called mutual inductance.
The magnitude of mutual inductance
depends on the geometry of the circuit.
Recall, Vx  L M
d I A  I B 
dt
assume d I A  0 , d I B  0
dt
dt
If reactance of inductance << RA and RA << RSA
1 d VA
1 V
then d I A
dt

RA
dt

R A Tr
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Crosstalk from mutual inductance
So we have Vx  L M
V
R A Tr
Recalling our definition of crosstalk Crosstalk 
we have Crosstalk 
unwanted signal
pure signal
Vx
LM

V R A Tr
The resulting crosstalk
for inductive coupling
may appear as
The circuit geometry will determine polarity of inductive
crosstalk, so polarity of signal induced into circuit B could
be negative.
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Crosstalk from mutual inductance
Can we measure LM ?
For a given ΔV, Tr, RA << RSA we know
Vx  L M
d IA
dt
Integrate both sides to get
d IA
LM
 Vx dt  L M  d t dt  R A
d V
 dt
L M V
 Vx dt  R A
L M V
Area   Vx dt 
RA
so
Area R A
LM 
V
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Crosstalk from mutual inductance
Actually the induced voltage pulse is split across the
inductor.
So half of Vx is seen at RB initially
and the area under the curve must
be doubled to correctly estimate LM.
So L M 
2 Area R A
V
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Crosstalk from mutual reactance
Note that the sign of Vx can be positive or negative
depending on the dot convention which depends on the
circuit geometry.
Mutual inductance
may produce this 
in circuit B
or this 
Whereas capacitively
generated signals in B
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Crosstalk from mutual reactance
How to reduce crosstalk?
Recall that the source is coupled fields
• electric fields  capacitance
• magnetic fields  inductance
techniques to reduce field coupling  reduced crosstalk
Examples include
• increasing separation between circuits
• reduce the loop area for inductive coupling
• reduce the area for capacitive coupling
Other approaches
• introduce isolating structures (guard traces or fence of ground vias)
• increase Tr if possible
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