interconnect ii – class 22
Post on 31-Dec-2015
37 Views
Preview:
DESCRIPTION
TRANSCRIPT
12/4/2002
Interconnect II – Class 22
Prerequisite Reading - Chapter 4
12/4/2002Interconnect II
2
Key Topics:
Frequency Content of Digital WaveformsFrequency Envelope Incorporating frequency domain effects into
time domain signals
Effects of Frequency Domain Phenomena on Time Domain Digital Signals
12/4/2002Interconnect II
3
Square wave: Y = 0 for - < x < 0 and Y=1 for 0 < x <
Y = 1/2 + 2/pi( sinx + sin3x/3 + sin5x/5 + sin7x/7 … + sin(2m+1)x/(2m+1) + …) 1 2 3 4 5May do with sum of cosines too.
1
0- 2 3
1
1 + 2 1 + 2 + 3
1 + 2 + 3 + 4 1 + 2 + 3 + 4 + 5
Decomposing a Digital Signal into Frequency Components
• Digital signals are composed of an infinite number of sinusoidal functions – the Fourier series
The Fourier series is shown in its progression to approximate a square wave:
12/4/2002Interconnect II
4
• The amplitude of the the sinusoid components are used to construct the “frequency envelope” – Output of FT
Frequency Content of Digital Signals
1 3 5 7 9 …...Harmonic Number
20dB/decade
40dB/decade
rT
35.0
Pw
T
T
1
Tr
12/4/2002Interconnect II
5
Estimating the Frequency Content• Where does that famous equation come from?
• It can be derived from the response of a step function into a filter with time constant tau
TrF
35.0
)1( /tinput eVV
• Setting V=0.1Vinput and V=0.9Vinput allows the calculation of the 10-90% risetime in terms of the time constant
195.2105.03.2%10%90%9010 ttt
• The frequency response of a 1 pole network is
dBdB F
F3
3 2
1
2
1
• Substituting into the step response yields
dBdB FFt
33%9010
35.009.1
12/4/2002Interconnect II
6
Estimating the Frequency Content
This equation says:The frequency response of the network with
time constant tau will degrade a step function to a risetime of t10-90%
The frequency response of the network determines the resulting rise time ( or transition time)
The majority of the spectral energy will be contained below F3dB
• This is a good “back of the envelope” way to estimate the frequency response of a digital signal.
• Simple time constant estimate can take the form L/R, L/Z0, R*C or Z0*C.
dBFt
3%9010
35.0
Edge time factor
12/4/2002Interconnect II
7
• The frequency dependent effects described earlier in this class can be applied to each sinusoidal function in the seriesDigital signal decomposed into its sinusoidal
componentsFrequency domain transfer functions applied to
each sinusoidal componentModified sinusoidal functions are then re-
combined to construct the altered time digital signal
• There are several ways to determine this responseFourier series (just described)Fast Fourier transform (FFT)
Widely available in tools such as excel, Mathematica, MathCad…
Examining Frequency Content of Digital Signals
12/4/2002Interconnect II
83 Method of Generating a Square Wave
Ramp pulsesUse Heavy Side functionUsed for first pass simulations
Power Exponential PulsesRealistic edge that can match silicon performanceUsed for behavioral simulation that match silicon performance.
Sum of CosinesText book identity.Used to get a quick feel for impact of frequency dependant phenomena on a wave.
12/4/2002Interconnect II
9
Ramp Square WaveRamp Pulse Train
RUe tt( ) tt tt( ) tt1
tt
1
onepulse tt k( ) RUe tt period k( ) RUe tt pw k period( )( )
Rpt t( )
0
number_of_pulses
k
onepulse t k( )
0 50 100 150
0
1
Rpt ti
ti
ns
12/4/2002Interconnect II
10
Power Exponential Square WavePower Exponetial Pulse Train
edge tt( ) Va 1 e tt( )
2.5
Power Exponential Edge
Ppt tt( ) edge tt( ) tt( )
onepulse tt k( ) Ppt tt period k( ) Ppt tt pw k period( )( )
Ppt t( )
0
number_of_pulses
k
onepulse t k( )
0 50 100 150
0
1
Ppt ti
ti
ns
12/4/2002Interconnect II
11
Sum Cosine Square WaveSum Cosine Pulse Wave
Establish Fourier SpectrumC
period
2Tedge a
C
period
freq n( )n 2period
n 1 3 10
Define pulse train of fourier coef i.e. pulse = sum of cosines
Fpt tt( )1
2
4
2
1 2 a( ) n
1
n2
cos n a cos freq n( ) tt( )
Va
0 50 100 1500.5
0
0.5
1
1.5
Fpt ti
ti
ns
12/4/2002Interconnect II
12Applying Frequency Dependent Effects to Digital Functions
FT
Multiply
Inverse FFT
FFT
Frequency
Att
enua
tion
(V
2/V
1)
Frequency
Vol
ts
Time
Vol
tsTime
No AC lossesWith AC losses
AC losses will degrade BOTH the amplitude and the edge rate
Input signal into lossy t-line Spectral content of waveform
Loss characteristics if t-line
Time domain waveform with frequency dependent losses
riseTF
35.0
12/4/2002Interconnect II
13
Assignment
Use MathCad to create a pulse wave with
Sum of sine wavesSum of rampsSum of realistic edge waveforms
Exponential powers
Use MathCad to determine edge time factor for exponential and Gaussian wave,
10% - 90%20% - 80%
12/4/2002Interconnect II
14
Edge Rate Degradation due to filtering
This equation says:
• If a step response is driven into a filter with tine constant tau, the output edge rate is t10-90%
• However, realistic edge rates are not step functions• RSS the input edge rate with the filter response
195.2105.03.2%10%90%9010 tttRemember this equation from a few slides ago?
Zo=50 C=5pF
tr = 300ps
pspspsTTt
pspFCZot
outrout 625)548()300(
548)550(195.2)(195.2195.2
2222
%9010
Input edge
tout=Output edgeExample:
12/4/2002Interconnect II
15
Key Topics:
Serpentine tracesBends ISITopology
Additional Effects
12/4/2002Interconnect II
16
Effects of a Serpentine Trace
• Serpentine traces will exhibit 2 modes of propagation• Typical “straight line” mode• Coupled mode via the parallel sections• Causes the signal to “speed up” because a portion of
the signal will propagate perpendicular to the serpentine
• ”Speed up” is dependent on the spacing and the length
Lp
S
-0.15
-0.05
0.05
0.15
0.25
0.35
0.45
0.55
0.65
0.0E+
00
5.0E-10
1.0E-09
1.5E-09
2.0E-09
2.5E-09
3.0E-09
3.5E-09
4.0E-09
Time, s
Vo
lts
S=5S=15
12/4/2002Interconnect II
17
Modeling Serpentines
Assignment – Find a the uncoupled trace length that matches the delay of the serpentine route below
Use Maxwell Spice/2D modeling of serpentine vs. equal length wave.
10 portTransmission Line Spice
Model Couple length=2 inches
Trace route on PWB
5 mil 2 port Tline Model
1” 2 port Tline model
1 oz copper5 mil space5mil width5 mil distance to ground
planeSymmetric striplineUse 50 ohm V source w/ 1ns
rise time (do for ramp and Gaussian)
12/4/2002Interconnect II
18
Rules of Thumb for Serpentine Trace
• The following suggestions will help minimize the effect of serpentine traces• Make the minimum spacing between parallel section (s)
at least 3-4H, this will minimize the coupling between parallel sections
• Minimize the length of the parallel sections (Lp) as much as possible
• Embedded microstrips and striplines exhibits less serpentine effects than normal m9ictrostirpsd
12/4/2002Interconnect II
19
Effects of bends• Virtually every PCB design will exhibit bends• The excess area caused by a 90o bend will increase
the self capacitance seen at the bend• Empirically inspired model of a 90o bend is simply 1
square of excess capacitanceWCC
bendo 11_90
• Measurements have shown increased delays due to the current components “hugging” the corner increasing the mean length
• 2 rights do not necessarily equal a left and a right, especially for wide traces
• 45o bends, round and chamfered bends exhibit reduced effects
Capacitance of 1 extra square
12/4/2002Interconnect II
20
Inter Symbol Interference• Inter symbol interference (ISI) is reflection noise that
effects both amplitude and timing The nature of this interference is cause by a signal not settling
to a steady stated value before the next transition occurs. Can have an effect similar to crosstalk but has completely
different physics
Waveform beginning transition from low to high with unsettled noise cased by reflections.
Different starting point due to ISI
Volts
Time
Ideal waveform beginning transition from low to high with no reflections or losses
Timing difference
Receiver switching threshold
12/4/2002Interconnect II
21
Inter Symbol Interference
• ISI can dramatically affect the signal quality Depending on the switching rate/pattern, significant differences in waveform
shape can be realized – one or two patterns won’t produce worst case If the designer does not account for this effect, switching patterns that are
unaccounted for result in latent product defects.
-2
-1
0
1
2
3
4
0.E
+00
1.E
-09
2.E
-09
3.E
-09
4.E
-09
5.E
-09
6.E
-09
7.E
-09
8.E
-09
9.E
-09
1.E
-08
Time, s
Vo
lts
Ideal 400 MHz waveform
400 MHz switching200 MHz switching
12/4/2002Interconnect II
22
Topology – the Key to a sound design
• What about the case where there is more than one receiver, or more than one driver (e.g., a Multi-processor FSB)
Vs Zo3
Zo2
0-2V Zo1
Rs=ZoReceiver 1
Receiver 2
L2L3
(L1=L2)
• There will be an impedance discontinuity at the junction The equivalent input impedance looking into the junction will
be the parallel combination of Zo2 and Zo1
132
132
||
||
ooo
ooo
ZZZ
ZZZ
• This model can be simplified and solved with lattice diagrams Valid when L1=L2
Vs Z=Zo2||Zo30-2V Zo1
Rs=Zo L2
L1
L1
12/4/2002Interconnect II
23
Topology – the Key to a sound design
• Now, consider the case where L2 and L3 are NOT Equal
VsZo3
Zo2
0-2VZo1
Rs=ZoReceiver 1
Receiver 2
L2L3
The reflections from the receiver discontinuities will not arrive at the same time; the 2 segment simplification is not applicable
This topology will ring with a frequency dependant in L2 and L3
This topology can be solved with a multi-segment lattice diagram0.0
0.5
1.0
1.5
2.0
2.5
0.0
Vo
ltag
e at
rec
eive
r
2.0 4.0 6.0 8.0 10.0
Time, ns
12/4/2002Interconnect II
24
Topology – the Key to a sound design
A
B
01 2 33T 2T
C
9
20
9
2
9
2
3
2
3
2
3
2
3
2'
3
4
3
2
3
2'
9
16
9
2
9
2
3
2
3
2
9
2
9
2
3
2
3
29
8
9
2
9
2
3
2
3
23
4
3
2
3
23
21
3
1
2
2
12
32
32
B
A
C
B
A
TT
ZoZo
ZoZoRsZo
ZoVinitial
B’
A’
In J R1 R2
Vs Zo
Zo
0-2VZo
Rs=ZoR1
R 2
JIn
top related