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1
Differential Signaling
Introduction to Frequency Domain Analysis (3 Classes)Many thanks to Steve Hall, Intel for the use of his slidesReference Reading: Posar Ch 4.5
http://cp.literature.agilent.com/litweb/pdf/5952-1087.pdf
Slide content from Stephen Hall
Instructor: Richard Mellitz
2
Differential Signaling
Outline
Motivation: Why Use Frequency Domain Analysis2-Port Network Analysis TheoryImpedance and Admittance MatrixScattering MatrixTransmission (ABCD) MatrixMason’s RuleCascading S-Matrices and Voltage Transfer FunctionDifferential (4-port) Scattering Matrix
3
Differential Signaling
Motivation: Why Frequency Domain Analysis?Time Domain signals on T-lines lines are hard to analyze
Many properties, which can dominate performance, are frequency dependent, and difficult to directly observe in the time domain• Skin effect, Dielectric losses, dispersion, resonance
Frequency Domain Analysis allows discrete characterization of a linear network at each frequencyCharacterization at a single frequency is much easier
Frequency Analysis is beneficial for Three reasonsEase and accuracy of measurement at high frequenciesSimplified mathematicsAllows separation of electrical phenomena (loss, resonance … etc)
4
Differential Signaling
Key Concepts
Here are the key concepts that you should retain from this class
The input impedance & the input reflection coefficient of a transmission line is dependent on:Termination and characteristic impedanceDelayFrequency
S-Parameters are used to extract electrical parametersTransmission line parameters (R,L,C,G, TD and Zo) can be
extracted from S parametersVias, connectors, socket s-parameters can be used to create
equivalent circuits=
The behavior of S-parameters can be used to gain intuition of signal integrity problems
5
Differential Signaling
Review – Important Concepts
The impedance looking into a terminated transmission line changes with frequency and line length
The input reflection coefficient looking into a terminated transmission line also changes with frequency and line length
If the input reflection of a transmission line is known, then the line length can be determined by observing the periodicity of the reflection
The peak of the input reflection can be used to determine line and load impedance values
6
Differential Signaling
Two Port Network Theory
Network theory is based on the property that a linear system can be completely characterized by parameters measured ONLY at the input & output ports without regard to the content of the system
Networks can have any number of ports, however, consideration of a 2-port network is sufficient to explain the theoryA 2-port network has 1 input and 1 output port.The ports can be characterized with many parameters, each
parameter has a specific advantage
Each Parameter set is related to 4 variables2 independent variables for excitation2 dependent variables for response
7
Differential Signaling
Network characterized with Port ImpedanceMeasuring the port impedance is network is the most
simplistic and intuitive method of characterizing a network
Port 1 Port 2
Case 1Case 1: Inject current I1 into port 1 and measure the open circuit voltage at port 2 and calculate the resultant impedance from port 1 to port 2
1
2,21
port
portopen
I
VZ
Case 2Case 2: Inject current I1 into port 1 and measure the voltage at port 1 and calculate the resultant input impedance
1
1,11
port
portopen
I
VZ
2-port Network
I1 I2
+-
V1 V2+-
2- port Network
I1 I2
+-
V1 V2+-
8
Differential Signaling
Impedance MatrixA set of linear equations can be written to describe the network in terms of its port impedances
Where:
If the impedance matrix is known, the response of the system can be predicted for any input
2221212
2121111
IZIZV
IZIZV
j
iij I
VZ Open Circuit Voltage measured at Port i
Current Injected at Port j
2
1
2221
1211
2
1
I
I
ZZ
ZZ
V
V
Zii the impedance looking into port iZij the impedance between port i and j
Or
9
Differential Signaling
Impedance Matrix: Example #2
Calculate the impedance matrix for the following circuit:
Port 1 Port 2
R1R2
R3
10
Differential Signaling
Impedance Matrix: Example #2
Step 1: Calculate the input impedance
R1R2
R3I1 V1
+
-
Step 2: Calculate the impedance across the network
R1 R2
R3I1 V2
+
-
311
111
3111 )(
RRI
VZ
RRIV
31
221
3131
3131
31
312
)(
RI
VZ
RIRR
RRRI
RR
RVV
11
Differential Signaling
Impedance Matrix: Example #2
Step 3: Calculate the Impedance matrix
Assume: R1 = R2 = 30 ohmsR3=150 ohms
18030
30180MatrixZ
1803111 RRZ
3021Z
12
Differential Signaling
Measuring the impedance matrix
Question: What obstacles are expected when measuring the impedance
matrix of the following transmission line structure assuming that the micro-probes have the following parasitics? Lprobe=0.1nH Cprobe=0.3pF
Assume F=5 GHz
T-line0.1nHPort 1
Port 20.3pF
0.1nH
0.3pF
Zo=50 ohms, length=5 in
13
Differential Signaling
Measuring the impedance matrix
1062
1_ fC
Z Cprobe
32_ fLZ Lprobe
T-line
Port 2
Answer: Open circuit voltages are very hard to measure at high frequencies
because they generally do not exist for small dimensions Open circuit capacitance = impedance at high frequencies Probe and via impedance not insignificant
0.1nH
106 ohms106 ohms
Zo = 50
Without Probe Capacitance
Zo = 50
With Probe Capacitance @ 5 GHz
Z21 = 50 ohms
Z21 = 63 ohms
Port 1 Port 2
Port 1 Port 2
T-line0.1nHPort 1
Port 20.3pF
0.1nH
0.3pF
Zo=50 ohms, length=5 in
14
Differential Signaling
Advantages/Disadvantages of Impedance Matrix
Advantages:The impedance matrix is very intuitive
Relates all ports to an impedanceEasy to calculate
Disadvantages:Requires open circuit voltage measurements
Difficult to measureOpen circuit reflections cause measurement noiseOpen circuit capacitance not trivial at high frequencies
Note: The Admittance Matrix is very similar, however, it is characterized with short circuit currents instead of open circuit voltages
15
Differential Signaling
Scattering Matrix (S-parameters)Measuring the “power” at each port across a well
characterized impedance circumvents the problems measuring high frequency “opens” & “shorts”
The scattering matrix, or (S-parameters), characterizes the network by observing transmitted & reflected power waves
2-port Network2-port
NetworkPort 1 Port 2
ai represents the square root of the power wave injected into port i
R
VaP
R
VP i
i
2
a1a2
b2b1
bj represents the power wave coming out of port jR
Vb j
j
RR
16
Differential Signaling
Scattering MatrixA set of linear equations can be written to describe the network in terms of injected and transmitted power waves
Where:
Sii = the ratio of the reflected power to the injected power at port i
Sij = the ratio of the power measured at port j to the power injected at port i
2221212
2121111
aSaSb
aSaSb
jportatinjectedPower
iportatmeasuredPower
a
bS
j
iij
2
1
2221
1211
2
1
a
a
SS
SS
b
b
17
Differential Signaling
Making sense of S-Parameters – Return LossWhen there is no reflection from the load, or the line length
is zero, S11 = Reflection coefficient
50
50
1
1
1
1
021
111
o
oo
incident
reflected
aZ
Z
V
V
V
V
R
VR
V
a
bS
S11 is measure of the power returned to the source,
and is called the “Return Loss”
R=Zo
Z=-l Z=0
Zo
R=50
18
Differential Signaling
Making sense of S-Parameters – Return LossWhen there is a reflection from the load, S11 will be
composed of multiple reflections due to the standing waves
)(1
)(1)(
l
lZlZZ oin
RL
Z=-l Z=0
Zo
inZ
If the network is driven with a 50 ohm source, then S11 is calculated using the input impedance instead of Zo
50 ohms
50
50
11
in
in
Z
ZSS11 of a transmission line will exhibit periodic effects due to the standing waves
19
Differential Signaling
Example #3 – Interpreting the return loss
Based on the S11 plot shown below, calculate both the impedance and dielectric constant
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
1.0 1.5 2.0 2.5 3..0 3.5 4.0 4.5 5.0
Frequency, GHz
S11
, Mag
nit
ud
eR=50
L=5 inchesZoR=50
20
Differential Signaling
Example – Interpreting the return loss
Step 1: Calculate the time delay of the t-line using the peaks
inchpspsinchTD
inchTDpsTDTD
GHzGHzf peaks
/7.84"5/7.423/
/7.4232
176.194.2
Step 2: Calculate Er using the velocity
0.1
)/37.39/7.84
1
/1031 8
Er
minchinchps
Er
sm
Er
cv
TD
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0
Frequency, GHz
S11
, Mag
nit
ud
e
1.76GHz 2.94GHz
Peak=0.384
21
Differential Signaling
Example – Interpreting the return loss
Step 3: Calculate the input impedance to the transmission line based on the peak S11 at 1.76GHz
Note: The phase of the reflection should be either +1 or -1 at 1.76 GHz because it is aligned with the incident
33.112
384.050
5011
in
in
in
Z
Z
ZS
Step 4: Calculate the characteristic impedance based on the input impedance for x=-5 inches
9.74
)1(5050
1
)1(5050
1
33.112)5(1
)5(1
1
50
50)(
366.97.84)5(76.144
42
o
o
o
o
o
ooin
jpsGHzjLCflj
LCflj
o
olo
Z
ZZZZ
Zx
xZZ
eee
eZ
Zex
Er=1.0 and Zo=75 ohms
22
Differential Signaling
Making sense of S-Parameters – Insertion Loss
When power is injected into Port 1 with source impedance Z0 and measured at Port 2 with measurement load impedance Z0, the power ratio reduces to a voltage ratio
incident
dtransmitte
o
o
aV
V
V
V
Z
V
Z
V
a
bS
1
2
1
2
021
221
2-port Network2-port
NetworkV1
a1a2=0
b2b1
V2ZoZo
S21 is measure of the power transmitted from
port 1 to port 2, and is called the “Insertion Loss”
23
Differential Signaling
Loss free networks For a loss free network, the total power exiting the N ports must
equal the total incident power
exitincident PP If there is no loss in the network, the total power leaving the
network must be accounted for in the power reflected from the incident port and the power transmitted through network
121_1_
incident
portportdtransmitte
incident
portreflected
P
P
P
P
Since s-parameters are the square root of power ratios, the following is true for loss-free networks
1221
211 SS
If the above relationship does not equal 1, then there is loss in the network, and the difference is proportional to the power dissipated by the network
24
Differential Signaling
Insertion loss exampleQuestion: What percentage of the total power is dissipated by the
transmission line? Estimate the magnitude of Zo (bound it)
S-parameters; 5 inch microstrip
0
0.2
0.4
0.6
0.8
1
1.2
0.E+00 2.E+09 4.E+09 6.E+09 8.E+09 1.E+10 1.E+10
Frequency, Hz
Mag
nit
ud
e S(1,1)S(1,2)
25
Differential Signaling
Insertion loss example What percentage of the total power is dissipated by the transmission line ? What is the approximate Zo? How much amplitude degradation will this t-line contribute to a 8 GT/s signal? If the transmission line is placed in a 28 ohm system (such as Rambus), will
the amplitude degradation estimated above remain constant? Estimate alpha for 8 GT/s signal
S-parameters; 5 inch microstrip;
0
0.2
0.4
0.6
0.8
1
1.2
0.E+00 2.E+09 4.E+09 6.E+09 8.E+09 1.E+10
Frequency, Hz
Ma
gn
itu
de
S(1,1)
S(1,2)
26
Differential Signaling
Insertion loss exampleAnswer: Since there are minimal reflections on this line, alpha can be
estimated directly from the insertion loss S21~0.75 at 4 GHz (8 GT/s)
057.075.0 )5(21 eeS l
When the reflections are minimal, alpha can be estimated
If the reflections are NOT small, alpha must be extracted with ABCD parameters (which are reviewed later)
The loss parameter is “1/A” for ABCD parameters ABCE will be discussed later.
If S11 < ~ 0.2 (-14 dB), then the above approximation is valid
27
Differential Signaling
Important concepts demonstrated
The impedance can be determined by the magnitude of S11
The electrical delay can be determined by the phase, or periodicity of S11
The magnitude of the signal degradation can be determined by observing S21
The total power dissipated by the network can be determined by adding the square of the insertion and return losses
28
Differential Signaling
A note about the term “Loss”
True losses come from physical energy lossesOhmic (I.e., skin effect)Field dampening effects (Loss Tangent)Radiation (EMI)
Insertion and Return losses include effects such as impedance discontinuities and resonance effects, which are not true losses
Loss free networks can still exhibit significant insertion and return losses due to impedance discontinuities
29
Differential Signaling
Advantages/Disadvantages of S-parametersAdvantages:Ease of measurement
Much easier to measure power at high frequencies than open/short current and voltage
S-parameters can be used to extract the transmission line parametersn parameters and n Unknowns
Disadvantages:Most digital circuit operate using voltage thresholds. This suggest
that analysis should ultimately be related to the time domain.Many silicon loads are non-linear which make the job of
converting s-parameters back into time domain non-trivial. Conversion between time and frequency domain introduces errors
30
Differential Signaling
Cascading S parameter
While it is possible to cascade s-parameters, it gets messy.
Graphically we just flip every other matrix. Mathematically there is a better way… ABCD parameters We will analyzed this later with signal flow graphs
a1a111
b1b111
s11
s21
s12
s22
a2a211
b2b211 a1a122
b1b122
a2a222
b2b222 a1a133
b1b133
a1a133
b1b133
s111 s121
s211 s221 s112 s122
s212 s222
s113 s123
s213 s223
3 cascaded s parameter blocks3 cascaded s parameter blocks
31
Differential Signaling
ABCD Parameters
The transmission matrix describes the network in terms of both voltage and current waves
2-port Network2-port
NetworkV1
I1I2
V2
The coefficients can be defined using superposition
221
221
DICVI
BIAVV
2
2
1
1
I
V
DC
BA
I
V
02
1
02
1
02
1
02
1
2222
VIVI
I
ID
V
IC
I
VB
V
VA
32
Differential Signaling
Transmission (ABCD) Matrix
Since the ABCD matrix represents the ports in terms of currents and voltages, it is well suited for cascading elements
V1
I1I2
V2
The matrices can be cascaded by multiplication
3
3
22
2
2
2
11
1
I
V
DC
BA
I
V
I
V
DC
BA
I
V
I3
V3
1DC
BA
2DC
BA
3
3
211
1
I
V
DC
BA
DC
BA
I
V
This is the best way to cascade elements in the frequency domain.
It is accurate, intuitive and simplistic.
33
Differential Signaling
Relating the ABCD Matrix to Common Circuits
ZPort 1 Port 2 10
1
DC
ZBA
Port 1 Y Port 21
01
DYC
BA
Z1
Port 1 Port 2323
3212131
/1/1
//1
ZZDZC
ZZZZZBZZA
Z2
Z3
Y1Port 1 Port 2Y2
Y3
3132121
332
/1/
/1/1
YYDYYYYYC
YBYYA
Port 1 Port 2,oZ)cosh()sinh()/1(
)sinh()cosh(
lDlZC
lZBlA
o
o
l
Assignment 6:
Convert these to s-parameters
34
Differential Signaling
Converting to and from the S-Matrix
The S-parameters can be measured with a VNA, and converted back and forth into ABCD the MatrixAllows conversion into a more intuitive matrixAllows conversion to ABCD for cascadingABCD matrix can be directly related to several useful circuit topologies
DCZZBA
DCZZBAS
DCZZBAS
DCZZBA
BCADS
DCZZBA
DCZZBAS
oo
oo
oo
oo
oo
oo
/
/
/
2
/
)(2
/
/
11
21
12
11
21
21122211
21
21122211
21
21122211
21
21122211
2
)1)(1(
2
)1)(1(1
2
)1)(1(
2
)1)(1(
S
SSSSD
S
SSSS
ZC
S
SSSSZB
S
SSSSA
o
o
35
Differential Signaling
ABCD Matrix – Example #1
Create a model of a via from the measured s-parameters
Port 1
Port 2
153.0110.0572.0798.0
572.0798.0153.0110.0
2221
1211
jj
jj
SS
SS
36
Differential Signaling
ABCD Matrix – Example #1The model can be extracted as either a Pi or a T network
L1
CVIA
L2
The inductance values will include the L of the trace and the via barrel (it is assumed that the test setup minimizes the trace length, and subsequently the trace capacitance is minimal
The capacitance represents the via pads
Port 1
Port 2
37
Differential Signaling
ABCD Matrix – Example #1
Assume the following s-matrix measured at 5 GHz
153.0110.0572.0798.0
572.0798.0153.0110.0
2221
1211
jj
jj
SS
SS
38
Differential Signaling
ABCD Matrix – Example #1
Assume the following s-matrix measured at 5 GHz
153.0110.0572.0798.0
572.0798.0153.0110.0
2221
1211
jj
jj
SS
SS
Convert to ABCD parameters
827.00157.0
08.20827.0
j
j
DC
BA
39
Differential Signaling
ABCD Matrix – Example #1Assume the following s-matrix measured at 5 GHz
153.0110.0572.0798.0
572.0798.0153.0110.0
2221
1211
jj
jj
SS
SS
Convert to ABCD parameters
827.00157.0
08.20827.0
j
j
DC
BA
Relating the ABCD parameters to the T circuit topology, the capacitance and inductance is extracted from C & A
Z1
Port 1 Port 2
Z2
Z3
pFC
fCjZ
jC VIA
VIA
5.0
2111
0157.03
nHLLfCj
fLj
Z
ZA
VIA
35.0)2/(1
21827.01 21
3
1
40
Differential Signaling
ABCD Matrix – Example #2Calculate the resulting s-parameter matrix if the two
circuits shown below are cascaded
2-port Network
Network X 5050
Port 1 Port 2
2221
1211
XX
XXX SS
SSS
2-port Network
Network Y 5050
Port 1 Port 2
2-port Network
Network X50
Port 1
2-port Network
Network Y 50
Port 2
?XYS
2221
1211
YY
YYY SS
SSS
41
Differential Signaling
ABCD Matrix – Example #2
Step 1: Convert each measured S-Matrix to ABCD Parameters using the conversions presented earlier
XX
XXXX DC
BATS
YY
YYYY DC
BATS
Step 2: Multiply the converted T-matrices
XYXY
XYXY
YY
YY
XX
XXYXXY DC
BA
DC
BA
DC
BATTT
Step 3: Convert the resulting Matrix back into S-parameters using thee conversions presented earlier
2221
1211
XX
XXXYXY SS
SSST
42
Differential Signaling
Advantages/Disadvantages of ABCD Matrix
Advantages:The ABCD matrix is very intuitive
Describes all ports with voltages and currents
Allows easy cascading of networksEasy conversion to and from S-parametersEasy to relate to common circuit topologies
Disadvantages:Difficult to directly measure
Must convert from measured scattering matrix
43
Differential Signaling
Signal flow graphs – Start with 2 port first
The wave functions (a,b) used to define s-parameters for a two-port network are shown below. The incident waves is a1, a2 on port 1 and port 2 respectively. The reflected waves b1 and b2 are on port 1 and port 2. We will use a’s and b’s in the s-parameter follow slides
44
Differential Signaling
Signal Flow Graphs of S Parameters
“In a signal flow graph, each port is represented by two nodes. Node an represents the wave coming into the device from another device at port n, and node bn represents the wave leaving the device at port n. The complex scattering coefficients are then represented as multipliers (gains) on branches connecting the nodes within the network and in adjacent networks.”*
a1
b1
b2
a2
S L
s21
s12
s11 s22
Example
Measurement equipment strives to be match i.e. reflection coefficient is 0
See: http://cp.literature.agilent.com/litweb/pdf/5952-1087.pdf
0
011 21
1 LS
abas
45
Differential Signaling
Mason’s Rule ~ Non-Touching Loop Rule
T is the transfer function (often called gain) Tk is the transfer function of the kth forward path
L(mk) is the product of non touching loop gains on path k taken mk at time.
L(mk)|(k) is the product of non touching loop gains on path k taken mk at a time but not touching path k.
mk=1 means all individual loops
mk
mkk mk
kmkk
mkL
mkLT
))()1(1(
))()1(1(T
)(
46
Differential Signaling
Voltage Transfer function
What is really of most relevance to time domain analysis is the voltage transfer function.
It includes the effect of non-perfect loads. We will show how the voltage transfer functions for a 2 port
network is given by the following equation.
Notice it is not S21
47
Differential Signaling
Forward Wave Path
a1
b1
b2
a2
Vs
S L
s21
s12
s11 s22
Z0
ZS Z0( )
48
Differential Signaling
Reflected Wave Path
a1
b1
b2
a2
Vs
S L
s21
s12
s11 s22
Z0
ZS Z0( )
49
Differential Signaling
Combine b2 and a2
50
Differential Signaling
Convert Wave to Voltage - Multiply by sqrt(Z0)
51
Differential Signaling
Voltage transfer function using ABCD
ABCD_CHANNEL
1 s11( ) 1 s22( ) s12 s212 s21
1 s11( ) 1 s22( ) s12 s212 s21 Z0
1 s11( ) 1 s22( ) s12 s21[ ] Z02 s21
1 s11( ) 1 s22( ) s12 s212 s21
ABCD_SOURCE1
0
Zs
1
ABCD_LOAD
1
1
ZL
0
1
ZL Z01 L
1 L Zs Z0
1 s
1 s
Let’s see if we can get this results another way
52
Differential Signaling
Cascade [ABCD] to determine system [ABCD]
VOLTAGE_TRANSFER_FUNCTION ABCD_SOURCEABCD_CHANNEL ABCD_LOAD
1
0
Z01 s
1 s
1
1 s11( ) 1 s22( ) s12 s212 s21
1 s11( ) 1 s22( ) s12 s212 s21 Z0
1 s11( ) 1 s22( ) s12 s21[ ] Z02 s21
1 s11( ) 1 s22( ) s12 s212 s21
1
1
Z01 L
1 L
0
1
Simplify
21 s11 s s22 L s11 s22 s L s12 s21 s L
1 s s21 1 L
1 s22 L s11 s11 s22 L s12 s21 L
s21 Z0 1 L
Z01 s22 s11 s s11 s22 s s12 s21 s
1 s s21
12
1 s22 s11 s11 s22 s12 s21s21
53
Differential Signaling
Extract the voltage transfer function
"A" parameter which input over output transfer. We are looking for "1/A" which is output over input
21 s11 s s22 L s11 s22 s L s12 s21 s L
1 s s21 1 L
1
Simplify and re-arange
s21 1 L 1 s
2
1 s11 s s22 L s12 s21 s L s11 s22 s L
Same as with flow graph analysis
54
Differential Signaling
Cascading S-Parameter
As promised we will now look at how to cascade s-parameters and solve with Mason’s rule
The problem we will use is what was presented earlier The assertion is that the loss of cascade channel can be
determine just by adding up the losses in dB. We will show how we can gain insight about this
assertion from the equation and graphic form of a solution.
a1a111
b1b111
s11
s21
s12
s22
a2a211
b2b211 a1a122
b1b122
a2a222
b2b222 a1a133
b1b133
a1a133
b1b133
s111 s121
s211 s221
s112 s122
s212 s222
s113 s123
s213 s223
55
Differential Signaling
Creating the signal flow graph
We map output a to input b and visa versa. Next we define all the loops Loop “A” and “B” do not touch each other
11
11 11
11A1A111 B2B211 A1A122 B2B222 A1A133 B2B233
B1B111 A2A211 B1B122 A2A222 B1B133 A2A233
s12s1211
s22s2211
s21s2111
s12s1233
s21s2133
s11s1133 s22s2233
s21s2122
s11s1122 s22s2222
s12s1222
a1a111
b1b111
s11
s21
s12
s22
a2a211
b2b211 a1a122
b1b122
a2a222
b2b222 a1a133
b1b133
a1a133
b1b133
s111 s121
s211 s221
s112 s122
s212 s222
s113 s123
s213 s223
56
Differential Signaling
Use Mason’s rule
There is only one forward path a11 to b23.
There are 2 non touching looks
Mason’s RuleMason’s Ruleb6
a1
s211
s212
s213
1 s222
s111
s223
s112
s113
s221
s122
s212
s221
s112
s222
s113
11
11 11
11A1A111 B2B211 A1A122 B2B222 A1A133 B2B233
B1B111 A2A211 B1B122 A2A222 B1B133 A2A233
s12s1211
s22s2211
s21s2111
s12s1233
s21s2133
s11s1133 s22s2233
s21s2122
s11s1122 s22s2222
s12s1222
57
Differential Signaling
Evaluate the nature of the transfer function
• If response is relatively flat and reflection is relatively low– Response through a channel is s211*s212*213…
Assumption is that these are ~ 0Assumption is that these are ~ 0
b6
a1
s211
s212
s213
1 s222
s111
s223
s112
s113
s221
s122
s212
s222
s111
s223
s114
58
Differential Signaling
Jitter and dB Budgeting Change s21 into a phasor
Insertion loss in db
S Smag ej
=
i.e. For a budget just add up the db’s and i.e. For a budget just add up the db’s and jitterjitter
S211 ej 211
S212
ej 211
S213 ej 213
S211
S212
s213
ej 211 212 213
20 log S211
S212
s213 =
20 log s211 20 log s21
2 20 log s213
dbsys
n
dbi i 1 delay
n
i i 1
59
Differential Signaling
Differential S-Parameters are derived from a 4-port measurement
Traditional 4-port measurements are taken by driving each port, and recording the response at all other ports while terminated in 50 ohms
Although, it is perfectly adequate to describe a differential pair with 4-port single ended s-parameters, it is more useful to convert to a multi-mode port
Differential S-Parameters
4-port
a1a2
b1b2
S11
S22S21
S12
S33
S44S43
S34S31
S42S41
S32
S13
S24S23
S14b1
b2
b3
b4
a1
a2
a3
a4
=a3
b3
a4
b4
60
Differential Signaling
Differential S-Parameters
Matrix assumes differential and common mode stimulus
Multi-ModePort
Mu
lti-
Mo
de
Po
rt 1
adm1adm2
acm2acm1
bdm1 bdm2
bcm1 bcm2
Mu
lti-Mo
de P
ort 2
It is useful to specify the differential S-parameters in terms of differential and common mode responsesDifferential stimulus, differential responseCommon mode stimulus, Common mode responseDifferential stimulus, common mode response (aka ACCM Noise)Common mode stimulus, differential response
This can be done either by driving the network with differential and common mode stimulus, or by converting the traditional 4-port s-matrix
DS11
DS22DS21
DS12
CS11
CS22CS21
CS12CDS11
CDS22CDS21
CDS12
DCS11
DCS22DCS21
DCS12bdm1
bdm2
bcm1
bcm2
adm1
adm2
acm1
acm2
=
61
Differential Signaling
Explanation of the Multi-Mode Port
Differential Matrix:Differential Stimulus, differential responsei.e., DS21 = differential signal [(D+)-(D-)] inserted at port 1 and diff signal measured at port 2
Common mode Matrix:Common mode stimulus, common mode Response. i.e., CS21 = Com. mode signal [(D+)+(D-)] inserted at port 1 and Com. mode signal measured at port 2
Common mode conversion Matrix:Differential Stimulus, Common mode response. i.e., DCS21 = differential signal [(D+)-(D-)] inserted at port 1 and common mode signal [(D+)+(D-)] measured at port 2
differential mode conversion Matrix:Common mode Stimulus, differential mode response. i.e., DCS21 = common mode signal [(D+)+(D-)] inserted at port 1 and differential mode signal [(D+)-(D-)] measured at port 2
DS11
DS22DS21
DS12
CS11
CS22CS21
CS12CDS11
CDS22CDS21
CDS12
DCS11
DCS22DCS21
DCS12bdm1
bdm2
bcm1
bcm2
adm1
adm2
acm1
acm2
=
62
Differential Signaling
)()()()( 3414433133321223111131
4343332321313
4143132121111
031
31
0;01
111
422
SSaSSaSSaSSabb
aSaSaSaSb
aSaSaSaSb
aa
bb
a
bDS
aaaadm
dm
cmdm
Differential S-ParametersConverting the S-parameters into the multi-mode requires just a little algebra
Example Calculation, Differential Return LossThe stimulus is equal, but opposite, therefore:
2413 ; aaaa
2-port Network4-port
Network
21
3 4
Assume a symmetrical network and substitute 2413 ; aaaa
3313311111 2
1SSSSDS
14323412 ; SSSS
Other conversions that are useful for a differential bus are shown
4323412121 2
1SSSSDS 4123432121 2
1SSSSCDS
Differential Insertion Loss: Differential to Common Mode Conversion (ACCM):
Similar techniques can be used for all multi-mode Parameters
63
Differential Signaling
Next class we will develop more differential concepts
64
Differential Signaling
backup review
65
Differential Signaling
Advantages/Disadvantages of Multi-Mode Matrix over Traditional 4-portAdvantages:Describes 4-port network in terms of 4 two port matrices
DifferentialCommon modeDifferential to common modeCommon mode to differential
Easier to relate to system specificationsACCM noise, differential impedance
Disadvantages:Must convert from measured 4-port scattering matrix
66
Differential Signaling
High Frequency Electromagnetic WavesIn order to understand the frequency domain analysis, it is
necessary to explore how high frequency sinusoid signals behave on transmission lines
The equations that govern signals propagating on a transmission line can be derived from Amperes and Faradays laws assumimng a uniform plane waveThe fields are constrained so that there is no variation in the X and Y
axis and the propagation is in the Z direction
Wf
min
sm
r
1
4.39103 8
Z
X
Y
Direction of propagation
This assumption holds true for transmission lines as long as the wavelength of the signal is much greater than the trace width
For typical PCBs at 10 GHz with 5 mil traces (W=0.005”)
"005.0"59.0
67
Differential Signaling
High Frequency Electromagnetic Waves For sinusoidal time varying uniform plane waves,
Amperes and Faradays laws reduce to:
xy Ej
z
B
Amperes Law:
A magnetic Field will be induced by an electric currentor a time varying electric field
Faradays Law: An electric field will be generated by a time varying magnetic flux
yx Bj
z
E
Note that the electric (Ex) field and the magnetic (By) are orthogonal
68
Differential Signaling
High Frequency Electromagnetic Waves
If Amperes and Faradays laws are differentiated with respect to z and the equations are written in terms of the E field, the transmission line wave equation is derived
0
1
222
2
2
2
2
2
xx
xxyyx
Ejz
E
Ejjz
E
z
B
z
Bj
z
E
This differential equation is easily solvable for Ex:
zjzjx eCeCzE )(
2)(
1)(
69
Differential Signaling
High Frequency Electromagnetic WavesThe equation describes the sinusoidal E field for a plane
wave in free space
zjM
zjMx eEeEzE )()()(
Portion of wave traveling In the +z direction
Portion of wave traveling In the -z direction
Note the positive exponent is because the wave is traveling in the opposite
direction
= permittivity in Farads/meter (8.85 pF/m for free space) (determines the speed of light in a material)
= permeability in Henries/meter (1.256 uH/m for free space and non-magnetic materials)
Since inductance is proportional to & capacitance is proportional
to , then is analogous to in a transmission line, whichis the propagation delay
LC
70
Differential Signaling
High Frequency Voltage and Current Waves
The same equation applies to voltage and current waves on a transmission line
RL
z=-l z=0
If a sinusoid is injected onto a transmission line, the resulting voltageis a function of time and distance from the load (z). It is the sum of the incident and reflected values
tjzref
tjzin eeVeeVtzV ),(
Voltage wave traveling towards the load
Voltage wave reflecting off the Load and traveling
towards the source
Incident sinusoidReflected sinusoid
tje Note: is added to specifically represent the time varying Sinusoid, which was impliedin the previous derivation
71
Differential Signaling
High Frequency Voltage and Current Waves
))(( CjGLjR LCj
= Attenuation Constant (attenuation of the signal due to transmission line losses)
= Phase Constant (related to the propagation delay across the transmission line)
j = Complex propagation constant – includes all the transmission line parameters (R, L C and G)
(For the loss free case) (lossy case)
C
LG
L
CR
2
1 (For good conductors)
LC (For good conductors and good dielectrics)
The parameters in this equation completely describe the voltage on a typical transmission line
tjzref
tjzin eeVeeVtzV ),(
72
Differential Signaling
High Frequency Voltage and Current Waves
ze
)sin()cos( je j
LC
Subsequently:
)(sin)(cos
)(sin)(cos
),( )()(
ztj
ztVe
ztj
ztVe
eeVeeVtzV
refz
inz
tjzjref
tjzjin
The voltage wave equation can be put into more intuitive terms by applying the following identity:
The amplitude is degraded byThe waveform is dependent on the driving function
( ) & the delay of the linetjt sincos
73
Differential Signaling
Interaction: transmission line and a load
(Assume a line length of l (z=-l))
)(1)( leVeeVeVeVlV lin
lo
lin
lref
lin
This is the reflection coefficient looking into a t-line of length l
Zl
Z=-l Z=0
l
ol
ollol
in
lref e
ZZ
ZZe
eV
eVl
22)(
The reflection coefficient is now a function of the Zo discontinuities AND line lengthInfluenced by constructive & destructive combinations of the forward & reverse waveforms
Zo
)(l
74
Differential Signaling
)(111
)(
)(1)(
leVZ
eVeVZ
lI
leVeeVeVeVlV
lin
o
lref
lin
o
lin
lo
lin
lref
lin
This is the input impedance looking into a t-line of length l
RL
Z=-l Z=0
)(1
)(1
)(11
)(1
)(
)()(
l
lZ
leVZ
leV
lI
lVlZZ o
lin
o
lin
in
Interaction: transmission line and a loadIf the reflection coefficient is a function of line length, then the
input impedance must also be a function of length
Note: is dependent on and
)(l
l
Zin
75
Differential Signaling
Line & load interactions In chapter 2, you learned how to calculate waveforms
in a multi-reflective system using lattice diagrams Period of transmission line “ringing” proportional to the line delay Remember, the line delay is proportional to the phase constant
In frequency domain analysis, the same principles apply, however, it is more useful to calculate the frequency when the reflection coefficient is either maximum or minimum This will become more evident as the class progresses
LCjLCjj
GR
CjGLjRj
22
00
To demonstrate, lets assume a loss free transmission line
76
Differential Signaling
Line & load interactions
o
loel 2)(
The frequency where the values of the real & imaginaryreflections are zero can be calculated based on the line length
LCfljLCfleee oLClj
olj
ol
o 4sin4cos2)(2)(2
Term 1 Term 2
...5,3,18
24
nLCl
nf
nLCfl
Term 1=0Term 2 =
...3,2,14
4
nLCl
nf
nLCfl
Term 2=0Term 1 =
Note that when the imaginary portion is zero, it means the phase of the incident & reflected waveforms at the input are aligned. Also notice that value of “8” and “4” in the terms.
Remember, the input reflection takes the form
o
77
Differential Signaling
Example #1: Periodic ReflectionsCalculate:1. Line length2. RL
(assume a very low loss line)
RL
Z=-l Z=0
)(l
-2.5E-01
-2.0E-01
-1.5E-01
-1.0E-01
-5.0E-02
0.0E+00
5.0E-02
1.0E-01
1.5E-01
2.0E-01
2.5E-01
0.0E+00 5.0E+08 1.0E+09 1.5E+09 2.0E+09 2.5E+09 3.0E+09Frequency
Ref
lect
ion
Coe
ff.
Real
Imaginary
Zo=75
Er_eff=1.0
78
Differential Signaling
Example #1: SolutionStep 1: Determine the periodicity zero crossings or peaks & use the
relationships on page 15 to calculate the electrical length
psGHzff
LClTD
GHzMHzGHzLClLClLCl
ff
nn
nn
42535.2
1
)(2
1
176.158876.12
1
4
1
4
3
13
13
Imaginary
79
Differential Signaling
Example #1: Solution (cont.)
Since TD and the effective Er is known, the line length can be calculated as in chapter 2
inm
ms
ps
c
effEr
TDlength 5127.0
1031
425
_8
Note the relationship between the peaks and the electrical length
This leads to a very useful equation for transmission lines)(2
1
13
nn ffLClTD
TDFpeaks 2
1
80
Differential Signaling
Example #1: Solution (cont.) The load impedance can be calculated by observing the peak values of the reflection
When the imaginary term is zero, the real term will peak, and the maximum reflection will occur If the imaginary term is zero, the reflected wave is aligned with the incident wave and the phase term = 1
Important Concepts demonstrated The impedance can be determined by the magnitude of the reflection The line length can be determined by the phase, or periodicity of the reflection
50
2.0)1(75
75)( 2
L
L
Ll
oL
oL
R
R
Re
ZR
ZRl