[SI-LIST] TDR impedance measurement and rise time
- From: "Howard Johnson" <howie03@xxxxxxxxxx>
- To: <si-list@xxxxxxxxxxxxx>
- Date: Tue, 1 Sep 2009 10:08:21 -0700
Dear Mick Zhou,
Regarding your note below about equations [3.87] and [3.88] in "High-Speed
Signal Propagation: Advanced Black Magic", I shall attempt to explain here
what you may be missing when you read those equations.
Those of you who do not derive intense feelings of satisfaction at having
your Laplace transforms turn out just right may want to step out for a cup
of tea while I finish off the "hard part" of this message and skip to the
"CONCLUSIONS" at the end.
First, the context of equations [3.87] and [3.88] has to do with TDR
testing, which occurs with a STEP excitation. The time-domain pictures at
the top of the page show STEP waveforms, not IMPULSES. I thought that was
made clear, but from your comments I see that there may be some confusion
about that point. The notation is as follows:
Characteristic impedance as a function of frequency is noted as Z[sub-C]
(expression [3.87]).
The step response of Z[sub-C] as a function of time is noted as
Z[sub-c](t) (expression [3.88]).
The idealized measurement setup for taking the step response
of Z[sub-C] applies a step of current to the front end of an
infinitely long transmission line.
In your note below you stated, "I checked Fourier transform and Laplace
transform pairs, 1/(j*omega) <--> 1, not t. So the rest is incorrect.., "
My response to that is that the step response [3.88] is not (and should not
be) the Laplace transform of frequency-domain expression [3.87]. To go from
frequency domain [3.87] to step response [3.88] you may take either of two
paths
PATH A: Multiply the frequency-domain expression [3.87] times the Laplace
transform of a step function (1/jw) and then transform the expression.
Omitting, for the moment, the scaling factor Z[sub-0], the multiplication
turns "1" into "1/jw" and turns "1/jw[tau]" into "((1/jw)-squared)x(1/tau)".
The first of those two terms becomes, in the time-domain, a unit step and
the second becomes a linear ramp with slope 1/[tau].
PATH B: Transform expression [3.87] directly into the time domain, and
then integrate it over time to form the step response. In this case the
first term "1" becomes, in the time domain, a unit impulse, and the
"1/jw[tau]" becomes a step with amplitude 1/[tau]. When integrated, the unit
impulse becomes a step with unit height and the step becomes a linear ramp
with slope 1/[tau].
In either case equation 3.88 is correct.
Istvan is correct that the approximation taken in [3.87] does not continue
down to zero frequency. What that means, in the time domain, is that the
time-domain step response does not continue up forever on a linear slope. In
practice, at times greater than [tau] the step response continues to
increase, but at a gradually slowing rate. Equation [3.87] takes into
account only the first two terms of the Taylor's series for [3.86]. The
other terms are extremely low-pass limited and all have initial time-domain
slopes of zero. That's why you can ignore them for the analysis of behavior
at times less than [tau].
It is helpful to have a physical model in mind when working with this
problem. Apply a step current of 1 Amp to the front end of an infinitely
long transmission line. For the case of times shorter than [tau], imagine
your step edge propagating uniformly without significant dispersion or loss
of signal amplitude. At any time "t", the voltage at the source equals the
accumulated voltage drop across the length of the cable up to the present
location of the propagating edge, plus the step voltage at the propagating
edge itself. If the current in the transmission line at all points from the
source up to the present location of the propagating edge equals precisely
one Amp, the accumulated voltage drop must then simply equal one amp
multiplied times the total accumulated resistance of the cable up to that
point. That's why the step response of Z[sub-C] trends up linearly.
The linear-increase principle exhibits itself directly in the construction
of the constant [tau]. Ignoring for now the skin-effect, the total
accumulated ressitance after time "t" should equal the time "t" multiplied
by the length of cable traversed per unit time, further multiplied by the
transmission-line resistance per unit length. That makes
(t)(1/sqrt(LC)(Z[sub-DC]). Check this against the linear term in
expression [3.88], which is Z[sub-0](t/[tau]). Plugging in the defined
value of [tau]=2L/R[sub-DC] and remembering that Z[sub-0] equals sqrt(L/C) I
hope you can see that these two expressions are the same except for a factor
of two.
The factor of two can be explained if you take into account the slight
attenuation in signal amplitude that happens as your signal edge propagates.
That attenuation REDUCES the front-end step response by a factor precisely
half as great as the INCREASE in front-end step response created by the
accumulated voltage drop.
I brought up the "accumulated voltage drop" argument because it shows
clearly why (and under what conditions) the step response will NOT continue
to increase linearly.
(1) If the signal amplitude does not decrases linearly with time. That
happens in systems with large amounts of attenuation. For example, if you
lose 1% of signal amplitude per inch in the first ten inches of propagation
the signal is down about 10% at that point. Going another 100 inches leaves
the signal small indeed, but still more than zero. Usually, in a TDR test,
the person developing the test will ENSURE that there very little signal
attenuation occurs as part of the test setup.
(2) If the signal does not propagate with uniform speed. That happens in
system with large amounts of signal dispersion. For example, an R-C
transmission line (something very familiar to those of you accustomed to
on-chip design), introduces a lot of signal dispersion. If you attempted a
"TDR test" on an R-C mode transmission line you might gain the impression
that the characteristic impedance of this type of line varies substantially
with freuqency (it does). Telephone engineers know this fact quite well. The
characteristic impedance of ordinary twisted-pair phone cable is about 600
ohms at 1 KHz, with a slope of about -3dB per octave. Section 8.3.3. of
"High-Speed Signal Propagation" explores the challenge of matching that
impedance for the purpose of making a hybrid circuit for bidirectional
transmission.
CONCLUSIONS:
------------
So, when does any of this matter?
I brought up the subject of the upward-trending slope in TDR measurements in
my book due, in part, to inquiries from Kodak regarding the use of thin-film
circuits. These circuits, because the conductors are extremely thin, have
enormous DC resistance and so the upward-trend in TDR measurements became a
source of considerable concern. Thin-film circuits also have a lot of signal
attenuation.
Ribbon conductors with printed conductors would have a similar effect.
Silicon wafers used as large substrates to make supercomputer pcbs have a
similar effect.
If you are doing ordinary electrodeposited copper pcbs with trace widths 3-6
mils you will probably see an upward tilt to the TDR curves of about 6-12
ohms per meter (with the fatter trace having less of a tilt). If you are
trying to measure the impedance uniformity of a 1-meter backplane, that
effect can appear troubling.
A tilt of twelve ohms per meter equals about 0.3 ohms per inch. If you have
a 12-inch test coupon, measurements taken at the front, end, or center of
that coupon could therefore return characteristic impedances of 48.2, 50,
and 51.8 ohms. That is not a huge effect, but the uncertainty subtracts
from all your impedance-related tolerances. In a board where accuracy
counts, therefore, I suggest that you develop a standard way to measure
impedance for your company so that, at least, all your pcb vendors do it the
same way. I like a 12-in. coupon, using a 1-ns edge, measured at a point 2
ns from the midpoint of the leading edge (that's six inches from the start).
Changing from a 1-ns edge to a 100-ps edge modifies the measurement to the
extent that the lingering effects of the different edges as measured 2 ns
later could be microscopically different, and clutters the measurement with
a higher degree of signal reflections emanating from the test board
interface connector. I like the slower edge. It averages the impedance over
a longer (6-inch) chunk of trace.
LAST NOTE: Skin-effect and dielectric losses complicate the discussion
further. Here's an example from one of my EDN articles:
"Characteristic Impedance of Lossy Line"
www.sigcon.com/Pubs/edn/LossyLine.htm
Best regards,
Dr. Howard Johnson, Signal Consulting Inc.,
tel +1 509-997-0505, howie03@xxxxxxxxxx
www.sigcon.com -- High-Speed Digital Design seminars, publications and films
P.S. -- I'll be teaching public classes in Phoenix Nov. 2-5, 2009.
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx] On
Behalf Of Istvan Novak
Sent: Saturday, August 29, 2009 2:42 PM
To: Mick zhou
Cc: Istvan Nagy; Yuriy Shlepnev; si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Re: TDR impedance measurement and rise time
Hi Mick,
Not sure what you want to say with this... The note for [3.87] states that
the approximation used to get [3.87] (from [3.86]) is valid only well above
the wlc corner frequency. When you apply Fourier or Laplace transform, they
require (or rather use) an infinite frequency range, so to me it means that
you can apply the transformations to [3.86] if you wish, but not to [3.87].
Regards,
Istvan Novak
SUN Microsystems
Mick zhou wrote:
> All,
>
> This discussion has been there for a while. I reviewed the section in
> Johnson's book (Advanced Black Magic) about the slope in Zc(t). From
> [3.87] to [3.88], I checked Fourier transform and Laplace transform
> pairs, 1/(j*omega) <-->1, not t. So the rest is incorrect, unless we
> go to the second order. Can somebody help double check?
>
> Thanks,
>
> Mick
>
> On Wed, May 6, 2009 at 11:12 AM, Mick zhou <mick.zhou@xxxxxxxxx
> <mailto:mick.zhou@xxxxxxxxx>> wrote:
>
> Istvan,
>
> I cannot agree with you more. That's why I emphasized "general
> cases".
> There are also elements that Laplace transformation can be
> performed correctly, for example ideal L and C.
>
> Yep, fortunately we live in "Newton's world" most likely.
>
> Best regards,
>
> Mick
>
>
>
> On Wed, May 6, 2009 at 9:18 AM, Istvan Novak <istvan.novak@xxxxxxx
> <mailto:istvan.novak@xxxxxxx>> wrote:
>
> Hi Mick,
>
> I see two related, but independent statements/questions in
> your message. The answer to "How can a
> complex value (f-domain) equals a real value (TDR, t-domain)
> in general cases?" the answer is: it is
> doable without loss of accuracy, if done correctly. The
> sentence you quote "Severe degrees of tilt
>
> make it very difficult to define one correct measurement
> procedure that is best for all appications"
> may apply to some laminate materials, but a) it has nothing to
> do with the complex or real nature of
> the data, and b) luckily today the typical materials used by
> the PCB industry dont fall into this
> category, not at least in the frequency range of common
> interest for digital people.
>
>
> Regards,
>
> Istvan Novak
> SUN Microsystems
>
>
>
>
>
>
>
> Mick zhou wrote:
>
> Istvan,
> Well, I should say fortunately we have no serious problems
> in most
> interconnect design practices since we intend to make them
> low loss etc.
>
> How can a complex value (f-domain) equals a real value
> (TDR, t-domain) in
> general cases? In High Speed Signal Propagation: Advanced
> Black Magic by
> H.Johnson, sect. 3.6.3, the authors touched the surface of
> the difficulty by
> concluding " Severe degrees of tilt make it very difficult
> to define one
> correct measurement procedure that is best for all
> appications" (p.172).
> Actually I think it is a fundamental problem: how can we
> define Z in
> t-domain that is compatible with Z in f-domain
> scientifically in general
> cases? f-dependent, nonlinear etc.
>
> What we have done is to map Z(f)=V(f)/I(f) into
> Z(t)=V(t)/I(t), and so
> reflection etc. Obviously, these definitions are not
> compatible even
> mathematically. Laplace transformation is not satisfied in
> general cases.
>
> My point is the current TDR theory has limitations that
> make some of our
> interpretations (struggles) meaningless. However,the
> difficulty does not
> stop our engineering work until it is absolutely necessary
> to correct the
> theory. We don't employ relativity to solve most of our
> engineering
> problems, but it is good to know the limitations of
> Newton's theory to avoid
> unnecessary struggles if we run outside of the territory.
> Cheer!
>
> I don't think it is easy to solve this difficulty by
> emails in this list
> unless we want to confuse people more. I'd leave it to
> theorists again.
>
> Best regards,
>
> Mick
>
>
> 2009/4/23 Istvan Nagy <buenos@xxxxxxxxxxx
> <mailto:buenos@xxxxxxxxxxx>>
>
>
>
> Hi
>
> So, does it mean that we can not do anything useful
> about frequency
> dependent impedance control on digital boards?
> Impedance can vary 5% from 100MHz (analog VGA,
> reference clocks) to few GHz
> (PCI-express, SATA, XAUI), so it can cause a problem.
> Or that is the maximum
> accuracy what we can get?
> 5% unaccuracy is 5% extra mismatch for the
> termination, if we have other
> sources of a mismatch already (component tolerance).
> Isn't 5% bad, or is it
> acceptable?
>
> Another aspect is what single frequency to substitute
> for a digital signal
> for impedance/trace_width calculations/simulations?
> I thought it would be the knee frequency based on the
> signal's rise time,
> but i am not shure anymore.
> For 8b10b encoded signals, there should be a lower
> frequency (data_rate/10)
> limit in the signal's spectrum, since maximum 5 zeroes
> or ones can follow
> each other.
> Where do we need best matching for terminations, at
> the highest frequency
> components, or at the mean of the spectrum, or at the
> highest peak...?
> I was trying to do some simulations with different bit
> patterns in QUCS and
> cadence SigExplorer, then do FFT, but the result looks
> mostly meaningless
> garbage with some negative slope.
> Anyway, how does the spectrum looks like for real data
> signals, especially
> at the lower end of the spectrum?
>
> How does the TDR determine the impedance? Does it
> measure the reflected
> signal voltage peak?
> And at what frequency? if we check the impedance
> characteristics from DC to
> infinite Hz, the impedance varies a lot. In theory, if
> both a simulation and
> a TDR measurement gives a number, then at what
> frequency should they be
> equal, and why?
>
> regards,
> Istvan
>
>
> ----- Original Message ----- From: "Mick zhou"
> <mick.zhou@xxxxxxxxx <mailto:mick.zhou@xxxxxxxxx>>
> To: "Yuriy Shlepnev" <shlepnev@xxxxxxxxxxxxx
> <mailto:shlepnev@xxxxxxxxxxxxx>>
> Cc: "Istvan Nagy" <buenos@xxxxxxxxxxx
> <mailto:buenos@xxxxxxxxxxx>>; <si-list@xxxxxxxxxxxxx
> <mailto:si-list@xxxxxxxxxxxxx>>
> Sent: Monday, April 20, 2009 11:21 PM
>
> Subject: [SI-LIST] Re: TDR impedance measurement and
> rise time
>
>
> Yurily,
>
>
> Nice study.
> I'd like to bring it deeper if not re-invent the
> wheels.
>
> Except some practical issues, I think there is a
> fundamental issue
> that is the definition of Z in t-domain and
> f-domain. The same
> formula rho=(ZL-Z0)/(ZL+Z0) (or its V(t) form) is
> simply used in both
> t- and f-domains. It does not matter if Z is
> f/t-independent,
> otherwise it is questionable Unfortunately, it is
> the foundation of
> most TDR algorithms so far. You can simply apply
> Fourier
> transformation, convolution must be involved even
> we assume Z0 is a
> constant. I don't know there is a good solution so
> far until we make
> necessary corrections in the math.
>
> We may conclude that one to one match from
> f-domain to t-domain is
> meaningless in general cases. That's probably the
> root cause of many
> confusions. We can always find a point we like to
> have a "match".
> For weak f-/t- dependent, it should be OK.
> Fortunately, most cases in
> out community are weak f-/t- dependent? We don't
> need to worry as much
> as we should?
>
> Thanks,
>
> Mick
>
>
>
>
>
>
> 2009/4/8 Yuriy Shlepnev <shlepnev@xxxxxxxxxxxxx
> <mailto:shlepnev@xxxxxxxxxxxxx>>:
>
>
>
> Hi Istvan,
>
> Looking through this thread, I finally decided
> to spend a couple of hours
> and to do a simple numerical TDR experiment
> with a broad-band model of a
> micro-strip line segment, to see at least
> theoretical effect of the rise
> time and to correlate frequency-dependent
> characteristic impedance of the
> line with the values that can be observed on
> TDR. The results of this
> simple
> experiment are available as App. Note #2009_04 at
> http://www.simberian.com/AppNotes.php (no
> registration required). The
> conclusion is that the observed TDR impedance
> depends on the rise time
> and
> can be correlated with the characteristic
> impedance at different
> frequency
> bands (well, at least theoretically).
>
> Best regards,
> Yuriy Shlepnev
> www.simberian.com <http://www.simberian.com>
>
>
> -----Original Message-----
> From: si-list-bounce@xxxxxxxxxxxxx
> <mailto:si-list-bounce@xxxxxxxxxxxxx>
> [mailto:si-list-bounce@xxxxxxxxxxxxx
> <mailto:si-list-bounce@xxxxxxxxxxxxx>]
> On
> Behalf Of Istvan Nagy
> Sent: Tuesday, April 07, 2009 1:34 PM
> To: si-list@xxxxxxxxxxxxx
> <mailto:si-list@xxxxxxxxxxxxx>
> Subject: [SI-LIST] Re: TDR impedance
> measurement and rise time
>
> Hi
>
>
> Peter from LeCroy wrote:
> "short impedance discontinuities... if you
> limit the frequency content
> ...,
> the bumps get smeared out by the slower
> risetime and they don't look so
> bad"
>
> - i think for these Test Coupon measurements
> is the point not to measure
> a
> real PCB trace with the lots of
> discontinuities, but to get the impedance
> based on the cross section. otherwise we would
> need different trace
> widths
> for every trace segment and we would need
> real-time 3D simulationd during
> PCB layout design.
>
> Exploring discontinuities on a real PCB (not
> on a test coupon) is is
> another
>
> story. I was asking about the measurements for
> the test coupons (maybe I
> forgot to mention). Normally (our) boards have
> hundreds of controlled
> impedance interconnects, those at the first
> place should be correct based
> on
>
> the cross section and test coupons. The rest
> is design practices, to make
> shure we dont deviate too much with
> discontinuitise. Of course its
> probably
> nice to characterise a full board, but in
> short development cycles, it
> wouldn't work very well. but i dont know,
> maybe it would...
>
> "Howard Johnson had an excellent video "
> - if anyone knows where to find it, i would
> appreciate...
>
>
> Jeff Loyer wrote:
> "The TDR will report the same characteristic
> impedance of your trace
> regardless of risetime"
>
> - which impedance? the impedance at 1 GHz? or
> at 10 GHz? or at xxx GHz?
> The characteristic impedance of a PCB trace
> depends on the frequency,
> since
> Er and the loss tangent are frequency
> dependent, and there is skin effect
> and others... so Z0(1GHz) is not equal to
> Z0(xxxGHz). So if a signal
> (lets
> simplify it) is at xxx GHz, then its
> terminations should be best matched
> at
> xxx GHz, and not at yyyGHz, so the board
> impedance should be correct at
> xxx
> GHz, and not at yyyGHz.
>
>
> Rob Sleigh wrote:
> "Yes, it's a very common practice to
> characterize a PDB with a TDR whose
> rise time is similar to the signal's rise
> time. It's up to the designer
> to
> decide, but usually pick a faster rise time
> than the system rise time to
> provide yourself with some margin."
>
> -most of the PCB manufacturers we talked to,
> they never asked about
> rise_time or frequency information of our
> signals, and when we tried to
> provide these to them they said they have
> deleoped their super-duper test
> setup which is based on tonns of measurements
> and it is accurate, and
> they
> dont care about our signal's frequency or rise
> time, and we should just
> pay
> and shut up... We tried In europe, north
> america and china. And the best
> what they say is they compensate for
> frequencies up to 10GHz, without
> knowing anything about our signal's freq/Tr.
> The last one said they can't or don't change
> rise times on their TDR...
>
>
> Kihong (Joshua) Kim wrote:
> "maximum frequency that may capture the
> bandwidth of imformation in
> digital
> world."
>
> - I was trying to estimate rise times and
> bandwidth. Especially at the
> receiver. I can't explain why it would be
> better than at the transmitter
> if
> they are both matched terminated to Z0, but I
> have a feeling like that...
> Normally at the receiver we have slower rise
> times. For example for PCIe
> and
>
> SATA, the signal looks sinusoid, not that
> rectangular as at the
> transmitter.
>
> So at a pattern 1010101010 the frequency would
> be fÚta_rate/2. For
>
> other
> interfaces, like DDR2/3, we can get rise times
> from simulation. So, I
> would
> provide these to the PCB manufacturer to
> calculate trace widths and
> verify
> by TDR/test-coupon measurements.
>
>
>
>
> regards,
> Istvan Nagy
> CCT, UK
>
>
> ----- Original Message -----
> From: "Kihong Joshua Kim" <joshuakh@xxxxxxxxx
> <mailto:joshuakh@xxxxxxxxx>>
> To: "Nagy István" <buenos@xxxxxxxxxxx
> <mailto:buenos@xxxxxxxxxxx>>
> Cc: <si-list@xxxxxxxxxxxxx
> <mailto:si-list@xxxxxxxxxxxxx>>
> Sent: Tuesday, April 07, 2009 4:51 PM
> Subject: [SI-LIST] Re: TDR impedance
> measurement and rise time
>
>
> Nagy,
>
>
> Couple of TDR measurements experience for
> real boards with known trace
> models and physical data will give you
> good sense of what TDR means.
> However, if you do not have time to build
> sample boards nor have TDR
> equipment...here is my help.
>
> Risetime conversion to frequency needs to
> be dealt with in-depth
> understanding of the topic. The quick rule
> of thumb equation mentioned
> in one of threaded mails is the maximum
> frequency that may capture the
> bandwidth of imformation in digital world.
> This is weird part because
> one
> might has question on why I am talking
> about digital bandwith when
> others
> discuss about analog nature of signal
> (rise time). Some excercise to
> uderstand Fourier analysis would give you
> an idea about what it meant.
>
> Anyhow, to get out of math and get the
> real sense of TDR with variety of
> sample boards.
> I had developed couple of years ago a
> virtual TDR head (IBIS TDR
> model) working just fine in any IBIS
> simualtion tools and I found out
> the
> paper in the internet (wow!). You could
> try sample boards as long as you
> have real board file and connector models
> and etc....
>
> If you google key words for IBIS TDR or
> TDR IBIS, you will find it
> easily.
> But just in case I attached here...
>
>
>
>
>
>
http://www.cadence.com/rl/Resources/conference_papers/stp_TDR_in_IBIS_Kim.pd
> f
>
>
>
> Regards,
>
> Kihong (Joshua) Kim
> http://www.linkedin.com/in/joshuakh
>
>
>
> On Tue, Apr 7, 2009 at 10:39 AM, Loyer,
> Jeff <jeff.loyer@xxxxxxxxx
> <mailto:jeff.loyer@xxxxxxxxx>>
> wrote:
>
> Concerning measuring Z0:
>
>
> The TDR will report the same
> characteristic impedance of your trace
> regardless of risetime, assuming your
> trace is long enough and there
> aren't
> significant variations in impedance
> along its length.
>
> Typically, we have very similar 6"
> coupons for all our controlled
> impedances. The board manufacturer
> will typically measure them with an
> HVM-compatible TDR, probably about 200
> ps risetime. We verify the
> impedances with our ~17ps TDR.
>
> For simulations, on the other hand,
> you'll probably want a risetime
> faster
> than the projected risetime of your
> device (I'd guess about 2x; I don't
> remember seeing it quantified). I
> typically see folks just go with the
> risetime of the equipment, ~17ps, and
> ensure simulation match those
> measurements. They may be a little
> conservative, but probably less work
> in
> the long run than trying to exactly
> justify any particular risetime.
>
> The advantages/disadvantages I can
> think of off-hand for fast risetimes
> are:
> 1) fast R.T. = resolution of finer
> features (discontinuities).
> Unfortunately, this can also
> erroneously lead you to believe you need
> to
> fix things that are "invisible" at
> your risetime of interest. Filtering
> to
> your risetime of interest can help you
> decide whether a discontinuity
> is
> significant or not.
> 2) fast R.T. = smaller probing
> geometries. It doesn't make sense to try
> to
> maintain a 15 ps risetime through a
> launch structure with 30 mil vias
> spaced
> 100 mils apart (such as might be used
> for manufacturing testing).
> Living
> with slower risetimes can allow you to
> adopt much more HVM-friendly
> launch
> structures, including pogo-pinned
> probe connections.
> 3) fast R.T. = less ESD protection.
> It's very easy to damage a TDR head
> from static discharge - HVM-compatible
> TDR machines with slower
> risetimes
> have ESD protection.
>
> If the scope or post-processing
> software doesn't have the ability to
> slow
> your risetimes, you can buy filters
> from Picosecond Pulse labs (buy a
> filter
> at 0.35/RT). They also sell hardware
> to put out very fast risetimes.
>
> Jeff Loyer
>
> -----Original Message-----
> From: si-list-bounce@xxxxxxxxxxxxx
> <mailto:si-list-bounce@xxxxxxxxxxxxx>
> [mailto:
> si-list-bounce@xxxxxxxxxxxxx
> <mailto:si-list-bounce@xxxxxxxxxxxxx>]
> On Behalf Of Nagy István
> Sent: Tuesday, April 07, 2009 4:59 AM
> To: si-list@xxxxxxxxxxxxx
> <mailto:si-list@xxxxxxxxxxxxx>
> Subject: [SI-LIST] TDR impedance
> measurement and rise time
>
> hi
> If we measure PCB test coupons with a
> TDR to determine characteristic
> impedance, should we set the rise time
> to be the same as the signal's
> rise
> time? is it possible to set it at all?
>
> what i found on the internet, the TDR
> manufacturers try to make rise
> time
> to be as low as possible, like
> 15ps..., and thats it.
>
> If the rise time is always 15ps, then
> i think it will always measure
> the
> impedance on a very high frequency,
> 2/t_rise or something, so several
> gigahertz. Usually on a board we have
> different signals, some are
> running
> 100MHz analog, some other are 800MT/s
> digital, or 2.5Gb/s digital.
> shouldn't we do different setups for
> these, to get impedances on the
> signal's operating frequency?
>
> Someone from a Fab told me, that the
> "TDR is not frequency dependent".
> so
> they dont take the signal's frequency
> into account.
>
> what is the correct handling of
> signaling frequency for impedance
> measurements, and simulations?
>
> regards,
>
> Istvan Nagy
> CCT
>
>
>
>
>
>
>
>
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