Join Teledyne LeCroy for a discussion of what S-parameters are and why we should care about them. As serial data rates move into the multi-gigabit domain, S-parameters play an important role in understanding system performance. We will uncover the four main patterns found in s-parameters and learn what they can tell us about our interconnects.
3. Characterize an Interconnect by How “Precision” Reference Signals Scatter Off a DUT
Incident Wave
Reflected Wave
Transmitted wave
TDR TDT
S11 S21
Time Domain t
t
Frequency Domain
waveform out from a port
waveform in to a port
parameter =
Measured in time and frequency domains
Displayed in frequency domain
Displayed in time domain
Company Confidential Dec 2013 3
4. S-Parameters From Many Sources: Measurement, Circuit Simulation, EM Simulation
Measurement
Electromagnetic simulation
Teledyne
LeCroy SPARQ
Circuit simulation
Polar Instruments 2D field solver Simbeor 3D Field Solver
Company Confidential Dec 2013 4
5. The “Ends” as “Ports”
incident
50W Z0 = 50W
~ reflected
50 W source impedance
Port as a coax cable
S reflected sine wave
11
= =
incident sine wave
• Has magnitude and phase
• Can be described as a complex number: real and imaginary
parts
( ) ( )
Mag S = Amplitude V
reflected
( ) incident
11 Amplitude V
Phase(S11) = Phase(Vreflected) - Phase(Vincident)
Company Confidential Dec 2013 5
6. 2 Port S-Parameters
S11: Return loss
It is the reflected signal
Impedance mismatch from 50 Ohms
throughout the interconnect
A little about losses
S21: Insertion loss is |S21 in dB|
It is the transmitted signal
Sensitive to losses
Sometimes sensitive to impedance
mismatches throughout the interconnect
~
50W Z0 = 50W
V
source
Magnitude
and phase
Detector
DUT
magnitude/
phase
detector
50W
Z0 = 50W
1 S11
VSWR
+
1 S11
=
-
Transparent interconnect:
S11: large, negative dB
S21: small, negative dB
Company Confidential Dec 2013 6
7. Connection Between Insertion and Return Loss
1 S S2 losses
= 2
+ +
11 21
Vincident
Vtransmitted
Vreflected How is Vtransmitted related to Vincident and Vreflected?
There is no such thing as conservation of voltage
There is conservation of energy
Energy in a wave ~ V2
2
21 11 In a lossless interconnect S = 1- S
Sources of loss:
- conductor
- dielectric
- coupling (cross talk)
- radiation
S11 + S21 = 1
How large does S11 have to be to affect S21? What DZ is this?
Company Confidential Dec 2013 7
8. For Lossless Interconnect:
Insertion Loss from Return Loss
WWhheenn ppllootttteedd iinn ddBB,, rreellaattiioonnsshhiipp bbeettwweeeenn
0.0
-0.5
-1.0
-1.5
-2.0
-2.5
-3.0
-3.5
-4.0
-4.5
-5.0
SS1111 aanndd SS2211 llooookkss uunnuussuuaall
-45-40--50 35-30-25-20-15-10 -5 0
dB(S11_sim)
dB(S21_sim)
2
21 11 S = 1- S
S11 of < -10 dB yields S21 > -0.5 dB
Company Confidential Dec 2013 8
9. Four Important Patterns in S11, S21
(also applies to differential S-parameters)
Ripples in S11, sometimes in S21: reflections
Monotonic drop in S21: losses
Broad dips: ¼ wave stub resonances
Sharp dips (hi Q), coupling to resonances
Insertion loss
return loss
Insertion loss
Company Confidential Dec 2013 9
10. How Return, Insertion Loss Gets its Ripples
When Len << ¼ l
Reflections from front and back, 180 deg out
of phase
No net reflection, all transmitted waves in
phase and add
S11 large negative dB, S21 nearly 0 db
Extrapolating to DC: phase S21 = 0
Z0 < 50 Ohms
50 W 50 W
When Len << ¼ l
½
0
0 0
min S11 max S21
0 0
S21
S11
At low frequency
ALL interconnects
are transparent
Company Confidential Dec 2013 10
11. When Len = ¼ l, S21 Lowest, S11 Highest
When Len = (½ x n + ¼ ) l
Max S11, min S21
Z0 < 50 Ohms
min S11 max S21
min S21
When Len << ¼ l
When Len = 1/4 l
½
¼ + 0 ¼
max S11
½
½ + 0 ¾
½
0
0 0
0 0
S21
S11
Company Confidential Dec 2013 11
12. When Z0 ≠ 50 Ohms, Multiple Reflections From The Terminations Cause Ripples
When Len = n x ½ l
Reflected waves from front and back subtract
Minimum reflected signal
Transmitted waves all in phase, S21 max
As frequency increases, insertion, return loss
increase, decrease
Longer distance between reflections, shorter the frequency
between high and low
Larger the impedance difference, the larger the modulation
Z0 < 50 Ohms
min S11 max S21
min S21
When Len << ¼ l
When Len = 1/4 l
½
¼ + 0 ¼
max S11
½
½ + 0 ¾
½
0
0 0
0 0
When Len = ½ l
½
1
½ + 0 ½
min S11 max S21
1 + 0 1½
S21
S11
Company Confidential Dec 2013 12
13. Impact of Other Features
Uniform 46 Ohm
lossy line with
TDR response with connectors
connectors
Uniform 46 Ohm lossless line
Uniform 46 Ohm lossy line
Connector discontinuities
at the ends generally
increase S11 with
frequency
Simulated examples
Company Confidential Dec 2013 13
14. Measured Examples: What do you See?
50 ohm 4 inch MS with capacitive
Company Confidential Dec 2013 14
launches
16. Measured Examples: What do you See?
S11
S21
90 ohm 4 inch MS
40 dB full scale,
20 GHz full scale
50 ohm 0.5 inch MS
50 ohm 6 inch MS
50 ohm 3 inch PTFE MS
Company Confidential Dec 2013 16
17. Attenuation and Insertion Loss
In real interconnects,
amplitude drops off
exponentially with distance
d
( ) dB/len
nepers/len
d d 20
out in in V d V e V 10
a
-a - = =
S21 V
æ ö
S21 in dB 20 x log V dB / in x d attenuation
= ç ¸= -a =
V
è ø
SS2211 iiss aatttteennuuaattiioonn wwhheenn tteerrmmiinnaattiioonnss aarree mmaattcchheedd aanndd tthheerree iiss nnoo
ccoouupplliinngg oouutt ooff tthhee ttrraannssmmiissssiioonn lliinnee
out
in
V
=
[ ] out [ ]
in
Company Confidential Dec 2013 17
18. Estimating Attenuation per Length in 50 Ohm Interconnects
S dB 1 S per Length atten dB / in f 2.3 x f x Df x Dk
= = = -æ + ö çè ø¸
dB/inch
w = line width in mils
f in GHz
Dk = dielectric constant
Df = dissipation factor
[ ]
[ ] 21 [ ]
21 Len
Len in w
w = 5 mils
Dk = 4
Df = 0.02
Measured attenuation
from two different
transmission lines
Simple estimate matches
measured behavior ok
( ) Len GHz atten ~ 2.3 x Df x Dk ~ 0.1dB / inch / GHz - -
Company Confidential Dec 2013 18
19. The ¼ Wave Stub Resonance
TD, Lenstub 2 x TD
When 2 x TD = ½ cycle, minimum received signal
TD = ¼ cycle: the quarter wave resonance
TD Len
TD 1 1
4 f
res
=
in
nsec
6
=
1.5
Len
f 1 1
= res
=
4 TD
f in GHz Len in inches
Company Confidential Dec 2013 19
20. ¼ Wave Resonance in the Frequency Domain
Lenstub
f 1 1
1.5
Len
=
res
4 TD
=
Stub 0.5 inches long
fres = 3 GHz
Stub length = 0.5 inches
What is worst case Bit
Rate (BR) for this stub? If Nyquist = 3 GHz,
no eye at BR = 6 Gbps
Where does the energy go?
S21 S11
Company Confidential Dec 2013 20
21. Measured Example: ~3/8” Stub
f 1 1
4 TD
1.5
Len
=
=
res
fres = 3.9 GHz
Stub 0.385 inches long
Company Confidential Dec 2013 21
22. Measured Insertion Loss with and without Via Stub in a ¼ inch Thick Backplane
No via stub
in the
channel
0.2 inch via stub
f 1.5 1.5 7.5 GHz
res
= = =
Len 0.20
SDD11
SDD21
Company Confidential Dec 2013 22
23. Coupling to Hi-Q Resonances
What are hi-Q resonators?
Any floating metal
Plane cavities
Voltage between the two planes
Company Confidential Dec 2013 23
24. Via to Cavity Coupling in 4 Layer Board
3.25 inches
3 GHz
Len
f 12 GHz res = =
Dk 2 x Len
1.187 inches
0.8 inches
Plane resonances expected:
Len = 3.25 in fres = 0.92 GHz
Len = 1.187 in fres = 2.5 GHz
Len = 0.8 in fres = 3.75 GHz
0.8 inches
S21 no return vias
Company Confidential Dec 2013 24
25. Introducing the SPARQ (“S-Parameters Quick”)
40 GHz frequency range on up to 4 ports, 30GHz to 12
Low cost for performance
Single-button-press operation
Built-in automatic OSLT calibration
Fast calibration and measurement time
Small footprint and Portable (4 port: 13” x 13” x 7”, 17 lbs)
Editor's Notes
I’ll start off by giving some background information about the high speed serial data channel and the testing challenges posed by these high speed links. Then I’ll provide some more in-depth information about what can be done to better model and predict the behavior of a link from a signal integrity standpoint and show how these can be used to solve real design problems.
In TDR measurements, a time domain measurement, …
a fast rising step edge is sent into the DUT and the…
reflected signal measured. In addition, we can look at the signal that is transmitted through the DUT.
This is the time domain transmitted signal, TDT. If we look at the signal incident into the DUT, it can be thought of as being composed of a series of sine waves, …
each with a different frequency, amplitude and phase. Each sine wave component will interact with the DUT independently. When a sine wave reflects from the DUT, …
the amplitude and phase may change a different amount for each frequency. This variation gives rise to the particular reflected pattern. Likewise, the…
transmitted signal will have each incident frequency component with a different amplitude and phase.
There is no difference in the information content between the time domain view of the TDR or TDT signal, or the frequency domain view. Using Fourier Transform techniques, the time domain response can be mathematically transformed into the frequency domain response and back again without changing or losing any information. These two domains tell the same story, they just emphasize different parts of the story.
TDR measurements are more sensitive to the instantaneous impedance profile of the interconnect, while frequency domain measurements are more sensitive to the total input impedance looking into the DUT. To distinguish these two domains, we also use different words to refer to them. Time domain measurements are TDR and TDT responses, while frequency domain reflection and transmission responses are referred to as S (scattering) parameters. S11 is the reflected signal, S21 is the transmitted signal. They are also often called the return loss and the insertion loss.
Depending on the question asked, the answer may be obtained faster from one domain or the other. If the question is, what is the characteristic impedance of the uniform transmission line, the time domain display of the information will get us the answer faster. If the question is what is the bandwidth of the model, the frequency domain display of the information will get us the answer faster.
We can go back and forth between the time domain and frequency domain without introducing or loosing any information. We can also generate 4-Port S-Parameters from simulation using ADS, for example. Either of these three methods can be used to generate the single ended S parameters.
All the information is contained in these terms. The single ended S parameters can be converted into the differential S-Parameters. It will always be easier to measure the terms as single ended elements, and then convert them mathematically. The rf signal sources are easier to manufacture and the calibration sources are much more straightforward as single ended elements, with the burden placed on the software to do the conversion.
We can convert the single ended S parameters in the frequency domain into the balanced S parameters in the frequency domain or in the time domain. Once we have any one of these forms of the 4 port S parameters, we can convert them interchangeably into every other form as well. We should not make a distinction between the information in any one form- they are all equivalent. Independent of how we get the S parameters, we want to be able to seamlessly change them into the form that that will get us to the answer the fastest. All of these conversion operations can be done with one button click using the LeCroy SPARQ software..
The port of a device under test is the interface of the interconnect to the S parameters. Any interconnect will have ends to it that can interact with signals. As part of the S parameter formalism, we call these ends, ports.
Here is an example of an interconnect, a transmission line. It has en end, or a port
The most important thing to keep in mind about a port has two conductors at the end, the signal and the return. This is an absolute requirement. Sometimes we get lazy in drawing diagrams and just draw the signal connection. There is always a return connection at the port. Think of a port connection as a small coax connector, with a signal and return conductor.
Sine wave are incident on their port, and sine waves can come back out
Through this port, as a reflected signal. Obviously, the properties of the reflected signal will depend on
How the device under test is terminated. When we have just 1 port we are monitoring, we have to know how the other ports are terminated to be able to unambiguously interpret the results.
We will typically use a network analyzer to measure the S parameters of a device. We sometimes call this a vector network analyzer, or VNA The connections to the front of the network analyzer we also call the ports. In this example, the network analyzer has two ports,
port 1 and
port 2. Inside each port is a generator of sine waves
with a precision 50 ohms source impedance and a high quality 50 ohm cable connecting the electronics to the front connector. Internal to the instrument is the electronics that will measure the incident sine wave and any sine wave that reflects back from the front connector.
Now we are ready to look at the return and insertion loss of a transmission line, which we mean in this context as any interconnect with two ports and a direct connection from one end to the other.
Here is the configuration in which return and insertion loss is measured with a network analyzer. The source and load impedance on the ends is 50 ohms.
We send a sine wave on port 1 and look at how much reflects back from port 1 to get the return loss.
The only feature of the interconnect that influences the return loss is the miss match to 50 ohms, not just at the front end, but
everywhere along the line and at the far end, where port 2 is connected to the other channel of the VNA.
To a lesser extent the losses in the line might influence the return loss. After all, with a lot of losses, the reflections from the far end may not make it back to port 1. The insertion loss is about how much makes it through the interconnect
As the transmitted signal. To first order, it is related to
how much doesn&apos;t reflect bas as return loss. However, for long lines, insertion loss is also dominated by
The losses in the interconnect. In general, higher frequencies will be attenuated more than low frequencies due to skin depth effects and dielectric loss, so insert loss general gets worse at higher frequencies.
-20 dB is 10% energy reflected back, has almost no impact on S21!
-10 dB is 30% energy reflected back, has only 0.5 dB impact on S21
These are the most important principles. There’s a forth principle that’s important, and that’s about the length of the socket. We haven’t said anything about how long the socket is and what the impact there is. And I want to spend just a couple of minutes talking about that and then we’ll talk about what’s not important, and then we’ll talk about power integrity. So it’s important to think about time delay and the length of the socket. We said at the beginning, what’s the speed of a signal through a material? It’s 6 inches per nanosecond. And Torlon dielectrics is roughly four. It’s a little bit less, like 3.7 or so. So 6 inches per nanosecond is a pretty good estimate. It’s about 150 mm per nanosecond is the speed in most polymers. The time delay, the time it takes to go from one end to the other is really the length divided by the speed. So here’s easy numbers to remember. If you’re dealing with, how many of you like inches versus millimeters, or mils versus millimeters? How many of you like millimeters versus mils? The millimeter guys are a little more vocal. So I’ve got them both here; 160 picoseconds per inch is the wiring delay or 6.7 picoseconds per millimeter wiring delay. So you tell me the length, I’ll tell you what the wiring delay or the time delay is. So here’s our transmission line. It’s got some; there are two terms that define it. We said there was the characteristic impedance. That’s the instantaneous impedance a signal would see each step along the way inside. So it’s characteristic impedance. And there’s a time delay. And here I use the two different units if you want to be bilingual. If it’s 1 mm long, what’s the time delay? About 7 picoseconds. If it’s 3 mm long we’re talking about 20 picoseconds as the time delay. Okay. So 20 picoseconds.
So here’s 30 and here’s 50. So it goes from here to here. What’s the phase shift? It’s the second guy minus the first. Fifth minus 30, what’s that? Plus 20. What’s the phase shift? Positive. It’s zero phase shift. If this is 30 ohms and there are 50 on both sides, I get exactly the phase shift on this side as this side. If this is zero phase shift on the reflection, this is zero phase shift on the reflection. If this is -1, or a minus number, that’s going to be 180° over here, it’s going to be 180° over here because they’re both this impedance is the first, this one is the second. So when we come out over here, the signal that reflects, there’s no phase shift going this way compared to this guy. It hits the end. What happens over here? It gets another reflection. What’s the phase shift over here? No phase shift if there’s none over here. So we’re going from down and back, what’s the condition. If this is really, really, really, really short compared to a wavelength, what’s the phase of this reflective wave compared to this reflective wave? They’re about the same. And so what’s going to happen to the signals coming out here? They’re going to add. That means there’s going to be pretty close to a magnitude of 1. That’s going to be the largest insertion loss. It’s going to be close to 0 dB. So at low frequency, whatever the reflections are, I’m going to have basically this guy’s reflecting coming back here is going to add to this guy. The sum of those, I’m not going to have any loss because what happens here? I get a signal reflecting here. Let’s see, if I don’t get it going through, where is it going? If mean if I don’t have an insertion loss of 1, a magnitude of 1, where is the signal, if it’s not going through, where is it ending up? It’s heading back. Let’s look at the reflected signal. Suppose this is 30 ohms and 50 ohms, what’s the sine of this reflected wave? It’s a negative, that’s right. So 180° phase shift goes through. What’s the sine of the reflection over here? Here we’re going from a high to a low. That gave us a reflection. Here we go from a low to a high. No phase shift. And so the reflective signal has no phase shift, so it heads back with no phase through the transmission. So this is 180° out of phase from this. If they’re really, really, really short, so there’s no phase sift going down and back, we have a phase shift here and none here, what’s this wave compared to this wave? 180° out of phase. When they add up, what do you get? Zero. So when it’s really, really, really short, no matter what the impedance of this guy is, the reflected waves are always, by definition, have to be 180° out of phase. If there’s very little phase shift through the path because they’re really short, they have to add up to give zero. So if nothing comes back, where does it all go? It has to go through. And so by basic fundamental definition, if this guy is short compared to a wavelength, so I’m transmitting down and back. We don’t have very much of a cycle added; the reflected waves are all out of phase. The transmitted waves are all in phase. So what’s the insertion loss going to be? If nothing comes back, it all goes through, the insertion loss has to be 1 or 0 dB.
difference, and I’m kind of assuming that we’re kind of close to 50 ohms, in the range of 30 to 80 ohms. So as long as we’re not too far…I mean clearly if this is a short, how much gets transmitted through? None of it, right? It’s a short. Right? So we’re assuming that it’s, you know, the reflection coefficient here is less than 50%. The reflective signals coming back, if they’re 180° out of phase, they cancel out the reflective wave or the transmitted wave, they’re in phase all that comes through. The two waves in coming through, they were in phase and this is really short. Suppose this is exactly a quarter of a wavelength in length. Let’s see. We go through here. We get some wave coming out. We go down and back. If that’s exactly a quarter wave and this is a quarter of a wave, what’s the total path length going down and back? A half a wave. So if this comes out and this transmitter wave comes out and they’re exactly a half a wave apart, that’s 180° phase shift. What’s going to happen to the two amplitudes here? They’re going to subtract. They’re 180° out of phase, and so that’s going to be when this length here is a quarter of a wave, that’s going to be the case of the lowest insertion loss. So whatever the impedance of this guy is, at low frequency, when the wavelength is long and this structure is short compared to a quarter of a wave, the insertion loss is always going to be 0 dB. That’s why at a low enough frequency, however long or short the interconnect is, at a low enough frequency the insertion loss is always going to be 0 dB.
These are the most important principles. There’s a forth principle that’s important, and that’s about the length of the socket. We haven’t said anything about how long the socket is and what the impact there is. And I want to spend just a couple of minutes talking about that and then we’ll talk about what’s not important, and then we’ll talk about power integrity. So it’s important to think about time delay and the length of the socket. We said at the beginning, what’s the speed of a signal through a material? It’s 6 inches per nanosecond. And Torlon dielectrics is roughly four. It’s a little bit less, like 3.7 or so. So 6 inches per nanosecond is a pretty good estimate. It’s about 150 mm per nanosecond is the speed in most polymers. The time delay, the time it takes to go from one end to the other is really the length divided by the speed. So here’s easy numbers to remember. If you’re dealing with, how many of you like inches versus millimeters, or mils versus millimeters? How many of you like millimeters versus mils? The millimeter guys are a little more vocal. So I’ve got them both here; 160 picoseconds per inch is the wiring delay or 6.7 picoseconds per millimeter wiring delay. So you tell me the length, I’ll tell you what the wiring delay or the time delay is. So here’s our transmission line. It’s got some; there are two terms that define it. We said there was the characteristic impedance. That’s the instantaneous impedance a signal would see each step along the way inside. So it’s characteristic impedance. And there’s a time delay. And here I use the two different units if you want to be bilingual. If it’s 1 mm long, what’s the time delay? About 7 picoseconds. If it’s 3 mm long we’re talking about 20 picoseconds as the time delay. Okay. So 20 picoseconds.
So here’s 30 and here’s 50. So it goes from here to here. What’s the phase shift? It’s the second guy minus the first. Fifth minus 30, what’s that? Plus 20. What’s the phase shift? Positive. It’s zero phase shift. If this is 30 ohms and there are 50 on both sides, I get exactly the phase shift on this side as this side. If this is zero phase shift on the reflection, this is zero phase shift on the reflection. If this is -1, or a minus number, that’s going to be 180° over here, it’s going to be 180° over here because they’re both this impedance is the first, this one is the second. So when we come out over here, the signal that reflects, there’s no phase shift going this way compared to this guy. It hits the end. What happens over here? It gets another reflection. What’s the phase shift over here? No phase shift if there’s none over here. So we’re going from down and back, what’s the condition. If this is really, really, really, really short compared to a wavelength, what’s the phase of this reflective wave compared to this reflective wave? They’re about the same. And so what’s going to happen to the signals coming out here? They’re going to add. That means there’s going to be pretty close to a magnitude of 1. That’s going to be the largest insertion loss. It’s going to be close to 0 dB. So at low frequency, whatever the reflections are, I’m going to have basically this guy’s reflecting coming back here is going to add to this guy. The sum of those, I’m not going to have any loss because what happens here? I get a signal reflecting here. Let’s see, if I don’t get it going through, where is it going? If mean if I don’t have an insertion loss of 1, a magnitude of 1, where is the signal, if it’s not going through, where is it ending up? It’s heading back. Let’s look at the reflected signal. Suppose this is 30 ohms and 50 ohms, what’s the sine of this reflected wave? It’s a negative, that’s right. So 180° phase shift goes through. What’s the sine of the reflection over here? Here we’re going from a high to a low. That gave us a reflection. Here we go from a low to a high. No phase shift. And so the reflective signal has no phase shift, so it heads back with no phase through the transmission. So this is 180° out of phase from this. If they’re really, really, really short, so there’s no phase sift going down and back, we have a phase shift here and none here, what’s this wave compared to this wave? 180° out of phase. When they add up, what do you get? Zero. So when it’s really, really, really short, no matter what the impedance of this guy is, the reflected waves are always, by definition, have to be 180° out of phase. If there’s very little phase shift through the path because they’re really short, they have to add up to give zero. So if nothing comes back, where does it all go? It has to go through. And so by basic fundamental definition, if this guy is short compared to a wavelength, so I’m transmitting down and back. We don’t have very much of a cycle added; the reflected waves are all out of phase. The transmitted waves are all in phase. So what’s the insertion loss going to be? If nothing comes back, it all goes through, the insertion loss has to be 1 or 0 dB.
difference, and I’m kind of assuming that we’re kind of close to 50 ohms, in the range of 30 to 80 ohms. So as long as we’re not too far…I mean clearly if this is a short, how much gets transmitted through? None of it, right? It’s a short. Right? So we’re assuming that it’s, you know, the reflection coefficient here is less than 50%. The reflective signals coming back, if they’re 180° out of phase, they cancel out the reflective wave or the transmitted wave, they’re in phase all that comes through. The two waves in coming through, they were in phase and this is really short. Suppose this is exactly a quarter of a wavelength in length. Let’s see. We go through here. We get some wave coming out. We go down and back. If that’s exactly a quarter wave and this is a quarter of a wave, what’s the total path length going down and back? A half a wave. So if this comes out and this transmitter wave comes out and they’re exactly a half a wave apart, that’s 180° phase shift. What’s going to happen to the two amplitudes here? They’re going to subtract. They’re 180° out of phase, and so that’s going to be when this length here is a quarter of a wave, that’s going to be the case of the lowest insertion loss. So whatever the impedance of this guy is, at low frequency, when the wavelength is long and this structure is short compared to a quarter of a wave, the insertion loss is always going to be 0 dB. That’s why at a low enough frequency, however long or short the interconnect is, at a low enough frequency the insertion loss is always going to be 0 dB.
These are the most important principles. There’s a forth principle that’s important, and that’s about the length of the socket. We haven’t said anything about how long the socket is and what the impact there is. And I want to spend just a couple of minutes talking about that and then we’ll talk about what’s not important, and then we’ll talk about power integrity. So it’s important to think about time delay and the length of the socket. We said at the beginning, what’s the speed of a signal through a material? It’s 6 inches per nanosecond. And Torlon dielectrics is roughly four. It’s a little bit less, like 3.7 or so. So 6 inches per nanosecond is a pretty good estimate. It’s about 150 mm per nanosecond is the speed in most polymers. The time delay, the time it takes to go from one end to the other is really the length divided by the speed. So here’s easy numbers to remember. If you’re dealing with, how many of you like inches versus millimeters, or mils versus millimeters? How many of you like millimeters versus mils? The millimeter guys are a little more vocal. So I’ve got them both here; 160 picoseconds per inch is the wiring delay or 6.7 picoseconds per millimeter wiring delay. So you tell me the length, I’ll tell you what the wiring delay or the time delay is. So here’s our transmission line. It’s got some; there are two terms that define it. We said there was the characteristic impedance. That’s the instantaneous impedance a signal would see each step along the way inside. So it’s characteristic impedance. And there’s a time delay. And here I use the two different units if you want to be bilingual. If it’s 1 mm long, what’s the time delay? About 7 picoseconds. If it’s 3 mm long we’re talking about 20 picoseconds as the time delay. Okay. So 20 picoseconds.
So here’s 30 and here’s 50. So it goes from here to here. What’s the phase shift? It’s the second guy minus the first. Fifth minus 30, what’s that? Plus 20. What’s the phase shift? Positive. It’s zero phase shift. If this is 30 ohms and there are 50 on both sides, I get exactly the phase shift on this side as this side. If this is zero phase shift on the reflection, this is zero phase shift on the reflection. If this is -1, or a minus number, that’s going to be 180° over here, it’s going to be 180° over here because they’re both this impedance is the first, this one is the second. So when we come out over here, the signal that reflects, there’s no phase shift going this way compared to this guy. It hits the end. What happens over here? It gets another reflection. What’s the phase shift over here? No phase shift if there’s none over here. So we’re going from down and back, what’s the condition. If this is really, really, really, really short compared to a wavelength, what’s the phase of this reflective wave compared to this reflective wave? They’re about the same. And so what’s going to happen to the signals coming out here? They’re going to add. That means there’s going to be pretty close to a magnitude of 1. That’s going to be the largest insertion loss. It’s going to be close to 0 dB. So at low frequency, whatever the reflections are, I’m going to have basically this guy’s reflecting coming back here is going to add to this guy. The sum of those, I’m not going to have any loss because what happens here? I get a signal reflecting here. Let’s see, if I don’t get it going through, where is it going? If mean if I don’t have an insertion loss of 1, a magnitude of 1, where is the signal, if it’s not going through, where is it ending up? It’s heading back. Let’s look at the reflected signal. Suppose this is 30 ohms and 50 ohms, what’s the sine of this reflected wave? It’s a negative, that’s right. So 180° phase shift goes through. What’s the sine of the reflection over here? Here we’re going from a high to a low. That gave us a reflection. Here we go from a low to a high. No phase shift. And so the reflective signal has no phase shift, so it heads back with no phase through the transmission. So this is 180° out of phase from this. If they’re really, really, really short, so there’s no phase sift going down and back, we have a phase shift here and none here, what’s this wave compared to this wave? 180° out of phase. When they add up, what do you get? Zero. So when it’s really, really, really short, no matter what the impedance of this guy is, the reflected waves are always, by definition, have to be 180° out of phase. If there’s very little phase shift through the path because they’re really short, they have to add up to give zero. So if nothing comes back, where does it all go? It has to go through. And so by basic fundamental definition, if this guy is short compared to a wavelength, so I’m transmitting down and back. We don’t have very much of a cycle added; the reflected waves are all out of phase. The transmitted waves are all in phase. So what’s the insertion loss going to be? If nothing comes back, it all goes through, the insertion loss has to be 1 or 0 dB.
difference, and I’m kind of assuming that we’re kind of close to 50 ohms, in the range of 30 to 80 ohms. So as long as we’re not too far…I mean clearly if this is a short, how much gets transmitted through? None of it, right? It’s a short. Right? So we’re assuming that it’s, you know, the reflection coefficient here is less than 50%. The reflective signals coming back, if they’re 180° out of phase, they cancel out the reflective wave or the transmitted wave, they’re in phase all that comes through. The two waves in coming through, they were in phase and this is really short. Suppose this is exactly a quarter of a wavelength in length. Let’s see. We go through here. We get some wave coming out. We go down and back. If that’s exactly a quarter wave and this is a quarter of a wave, what’s the total path length going down and back? A half a wave. So if this comes out and this transmitter wave comes out and they’re exactly a half a wave apart, that’s 180° phase shift. What’s going to happen to the two amplitudes here? They’re going to subtract. They’re 180° out of phase, and so that’s going to be when this length here is a quarter of a wave, that’s going to be the case of the lowest insertion loss. So whatever the impedance of this guy is, at low frequency, when the wavelength is long and this structure is short compared to a quarter of a wave, the insertion loss is always going to be 0 dB. That’s why at a low enough frequency, however long or short the interconnect is, at a low enough frequency the insertion loss is always going to be 0 dB.
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