[SI-LIST] Re: capacitor impedance in time domain
- From: Peter Fekete <thefekete@xxxxxxxxx>
- To: si-list@xxxxxxxxxxxxx
- Date: Sat, 29 Jan 2005 16:54:07 -0800 (PST)
Thank you Ed,
I was sure it is a textbook problem or something like that. In fact, a
simplified version of the situation is described in Hall, Hall and McCall
?High-speed Digital System Design? pages 30-31 and the same simplified case is
solved via Laplace transforms in B. Young ?Digital Signal Integrity? page 83.
I thought that this is an idealized situation without too many practical
applications, but apparently it is a starting point for certain interesting
applications like the characterization of capacitors etc.
Peter
"Dr. Edward P. Sayre" <esayre@xxxxxxxx> wrote:Dear SI Professionals:
Hats off to Peter for his excellent derivation. There is much more
information available from Tektronix when they introduced their early TDR's
and Agilent (way back when they were H-P Instrumentation Division)
regarding reflection from discrete elements. It is possible, and in fact
one of the more powerful ways to determine and characterize capacitors and
inductors to look at this problem in the time domain.
The risetime performance of the transmitted pulse is a function of both the
risetime as well as the natural time constant of the shunt or series
discrete element. One can show that for time constants on the order of the
risetime, the effect of a shunt capacitor is to delay the pulse by a
ristime which is equivalent to slowing the risetime down equivalently. The
time constant is on the order of 0.5*Zo*Cshunt. For realistic numbers such
as those encountered in a via (say 1pF), the delay is on the order of 25
pSec. For larger capacitance such as that encountered in a memory system,
the delay approximation breaks down and the problem is a full blown
transient solution directly amenable by LaPlace transform methods. In this
case, the charging of the capacitance which is the time constant portion of
the solution dominates the problem.
There is much theory around for time solutions which give far more insight
into the effects of bypass capacitance as well as via capacitance then can
be seen in the frequency domain. Again - hats off to Peter.
ed sayre
At 05:08 PM 1/28/2005 -0800, Peter Fekete wrote:
>Dear group_delay, Arpad, ?
>
>When I answered to this topic three days ago I thought that I explained
>enough in my message but apparently not.
>
>Let?s start specifying that any voltage (and current) distribution on a
>transmission line can be decomposed in two waves. One is the progressive
>(from source to load) and the other is regressive.
>
>Anywhere on the line the actual voltages and currents are superpositions
>of these, more exactly: V=V(+) + V(-) and I = I(+) - I(-); the voltages
>add and the currents are substracted
>
>For both waves the relationship between voltage and current is the well
>known V(+)=Zc* I(+) and V(-)=Zc *I(-)
>
>The way the problem is formulated (by the way, it looks like a homework
>problem) implies that we have the progressive wave V(+) defined as
>trapezoidal and we want the reflected wave ie V(-)
>
>
>
>An important detail is that the source is connected to a 50 ohm resistor
>and then to the line. Assuming the line has a 50 ohm characteristic
>impedance and the source is perfect results that the V(-) wave returning
>to the source sees a perfectly terminated line and will not give any more
>reflections.
>
>This simplifies things and also the 50 ohm resistor and 50 ohm line will
>result in half of the source voltage actually going on the line (as
>group_delay noticed).
>
>
>
>To solve the problem in time domain we don?t need to use the concept of
>impedance, although if you like you can talk about it.
>
>At the capacitor we have I=C dV/dt
>
>Note that these V and I are not the trapezoidal ones, but they are as
>mentioned V=V(+) + V(-) and I = I(+) - I(-) and only V(+) and I(+) are
>trapezoidal.
>
>This equation gives
>
>V(+)/Zc - V(-)/Zc = C ( dV(+)/dt + dV(-)/dt )
>
>V(+) is known so I?ll group the terms:
>
>
>
>dV(-)/dt + V(-)/CZc= dV(+)/dt ­V(+)/CZc
>
>
>
>the term C*Zc is the ?RC constant? mentioned in some messages = tau
>
>
>
>so the equation for the unknown V(-) is:
>
>
>
>dV(-)/dt + V(-)/tau = dV(+)/dt ­ V(+)/tau
>
>
>
>I solved this equation quickly for rectangular pulse V(+) and it took a
>while for the trapezoidal pulse but it is doable.
>
>
>
>For a rectangular pulse (rise time=0) V(-)=[1-2exp(-t/tau)]*0.5 for
>t=(0 to T) ( time origin shifted to when the pulse arrives at the
>capacitor) and V(-)= [1-exp(-T/tau)]*exp[-(t-T)/tau] for t= T to infinity
>
>Note that this gives you V(-)= - 0.5 for t=0 ie exactly the dip that takes
>the V(+)=0.5 V down to zero as group_delay noticed for zero rise time
>
>
>
>For a trapezoidal pulse it is much uglier
>
>V(-)= 0.5/tr * (t-2tau)+tau/tr * exp(-t/tau) for the up ramp t= 0 to tr
>
>And so on ?
>
>
>
>The voltage at the end of the rise time is
>
>0.5- tau/tr[1-exp(-tr/tau)]
>
>if you take the limit for tr ->0 you?ll find the previous result ­0.5
>
>however for tr different from zero it is > -0.5 and does not cancel
>completely the 0.5 of the V(+) , again as noticed by group_delay
>
> With these V(-) and V(+) propagated one can reconstruct the actual
> voltage at the capacitor and the source
>
>The problem can also be solved, as I mentioned earlier, in Laplace or
>Fourier domains, in which cases one would actually use the capacitor impedance.
>
>
>
>Hope this helped in solving the homework problem.
>
>
>
>Peter
>
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>group_delay wrote:
>arpad,
>i didn't quite understand some of the characters in your equation.
>when you write (finite dV)/(dI = 3D 0) = 3D> infinite impedance,
>what do you mean?
>
>yes, in my case the input waveform to the tline is a trapezoidal
>signal 0V to 1V. the setup is shown below.
>
>signal source -- 50 ohm resistor -- transmission line-- capacitor to
>ground.
>
>when i monitor the voltage at the junction of the resistor and the
>tline, i see it jumps to 0.5V initially which makes sense. However,
>when the reflection from the end comes back, it dips down very fast
>and then rises in an exponential fashion. The dip down looks
>parabolic (2nd order). The dip does go to zero for a zero rise time
>input, but it does not go to zero for a finite rise time signal. i
>am trying to figure out the best way to calculate the amount of this
>dip...
>
>thanks for you help.
>
>
>
>
>
>--- In si-list@xxxxxxxxxxxxxxx, "Muranyi, Arpad"
>wrote:
> > You will get full reflection, because the impedance
> > you need to use for the equation you quoted is the
> > small signal (AC) impedance. Referring to my previous
> > posting a short time ago, the constant current source
> > equivalent during the ramping portion of your trapezoid
> > waveform has a (finite dV) / (dI =3D 0) =3D> infinite impedance.
> >
> > The question I have for you is this: where do you mean
> > that your waveform is trapezoid? At the beginning of
> > the T-line, or at the end, where the capacitor is?
> > If the first, be prepared for a non trapezoid waveform
> > at the capacitor, because the T-line and the cap forms
> > an RC circuit, who's response is an exponential waveform.
> > If the ramp is faster the RC constant you will see
> > an exponential waveform, if it is slower, you will see
> > a more or less trapezoid waveform.
> >
> > I hope this helps,
> >
> > Arpad
> >
>=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
>=3D=3D=
> >
>=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
>=3D=3D=
> > =3D=3D=3D=3D=3D
> >
> >
> > -----Original Message-----
> > From: si-list-bounce@xxxx [mailto:si-list-bounce@xxxx] =
> > On Behalf Of group_delay
> > Sent: Friday, January 28, 2005 1:48 PM
> > To: si-list@xxxx
> > Subject: [SI-LIST] Re: capacitor impedance in time domain
> >
> >
> > hi all,
> > what i really want to do is find out how much waveform gets
>reflected
> > from the end of a lossless transmission line terminated with a
> > lossless capacitor, assuming the input waveform is a trapezoidal
> > signal. I know this can be computed using: gamma =3D (Zl-Zo)/
>(Zl+Zo),
> > but this requires you to calculate Zl for the time domain signal.
>If I
> > wanted to avoid it and use time domain analysis, how would I setup
>the
> > equation?
> >
> > thanks,
> > chris
> >
> >
> > --- In si-list@xxxxxxxxxxxxxxx, steve weir wrote:
> > > matthias, in the time domain we would solve the differential
> > equations for=20
> > > the network, or more likely using a computer program we would
>solve =
> > the=20
> > > difference equations over a series of discrete time steps. Now
>in
> > either=20
> > > case we could express impedance as dv/dt / di/dt. But I don't
>know =
> > how=20
> > > useful it would be towards either visualizing behavior, or
>solving the =
> >
> > > equations. Let's take the trapezoidal wave for instance. An =
> > effective=20
> > > impedance is pretty easy to come by on each: the rising, and
>falling=20
> > > portions of the waveform from the capacitance expression C =3D
> > i/dv/dt, Z =3D=20
> > > dv/dt / di/dt =3D 1/(dv/dt * C ). The flat portions are
>troublesome
> > as are=20
> > > the vertices, since dv/dt theoretically goes to zero and the
> > impedance from=20
> > > the formula jumps to an infinite value. Intuition should tell us
> > that this=20
> > > is wrong, as
> >
> > coupling capacitors routinely pass high frequency pulses.
> > >=20
> > > In the frequency domain, we have this nailed. We don't have=20
> > > discontinuities at the vertices. The vertices and flat portions
>=
> > follow=20
> > > curves formed by the frequency components, and rather than a flat
> > section=20
> > > containing DC and no HF, quite the opposite is true: the
>flatter we
> > want=20
> > > the pulse tops to be, the higher the frequency content
>required. This =
> >
> > > aligns with our intuition. But when we transform the
>representation
> > back=20
> > > to the time domain, those piecewise linear segments are now
>curved
> > solving=20
> > > the discontinuities at the vertices and eliminating the flat
>slopes
> > with=20
> > > theoretically infinite Z between the edges.
> > >=20
> > > So if someone wanted to look only at the rising and falling
>edges, an=20
> > > impedance in the time domain is reasonable, and possibly even
> > useful. But=20
> > > it really gets awkward when dealing with the whole waveform
>unless
> > we first=20
> > > perform frequency limiting operations, most easily performed in
>the=20
> > > frequency domain.
> > >=20
> > > I am not an expert on algorithms, so I really can't say from an
>error=20
> > > analysis and computational efficiency standpoint what is really
>the
> > best=20
> > > way to perform a transient analysis. But in my naivete, I would
>be=20
> > > inclined to transform everything into the frequency domain,
>compute =
> > the=20
> > > solution and transform back. In my feeble mind, this would avoid
> > some of=20
> > > the discontinuity and convergence problems in SPICE and more
>closely=20
> > > follows nature. But since people a whole lot better at math
>than I
> > have=20
> > > worked long and hard on those algorithms, I suspect either the=20
> > > computational overhead, or error build-up of my naive approach
>would =
> > be=20
> > > unacceptably high. Maybe what this world needs is a five cent,
>256 =
> > bit=20
> > > floating point, matrix solver!
> > >=20
> > > Steve.
> > >=20
> > > At 10:13 PM 1/26/2005 +0100, Matthias Bergmann wrote:
> > > >
> > > >Hello, I don`t understand why impedance should be limited to =
> > Frequency
> > > >domain. What impedance are we speaking about ? For example the
> > > >characteristicimpedance Z of a transmission line also exists in
> > time domain.
> > > >If you look along a transmission line, v(t) / i(t) have got
> > singularities
> > > >(undefined, infinite), these are called short and open ?!?!?
> > Furthermore
> > > >mostof the simulation programs use the time domain because it
>permits
> > > >non-linearities. I don`t know how what happens when your
>impulse is
> > > >trapezoidal, but if it was a rectangular and your load is a
> > capacitance, you
> > > >are answer would look like an exponential function, with your
> > reflection
> > > >co-efficient as initial value. Regards, Matthias Bergmann P.S.:
> > Yes, use
> > > >SPICE or ADS ! _m |---------+---------------------------------->
> > > >
> > > >
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> > > >
> > > >-
>list@xxxx>
> > > >-LIST] Re: capacitor impedance in time
> > >
> >
> >domain
>0;&#=
> > 160;
> > >
> >
> >60I
>;=
> > 60;
> > > > |
> > >
> > > >-----------------------------------------------------------
>------=
> > ----
> > > >-- -----------------------------------------| >I could be
> > wrong >but
> > > >tome >impedance is a concept strongly related to Frequency
>domain.
> > >>It is
> > > >meaningful just in that domain. Absolutely. If you define
>impedance =
> > as
> > > >voltage/current, then you run into great difficulties if you
>try to
> > do it in
> > > >the time domain. In general, with any complex impedance,
> > v(t)/i(t) has
> > > >singularities (undefined, infinite). I consider impedance =3D
> > v(s)/i(s) or
> > > >v(f)/i(f), which makes it a strictly frequency domain parameter.
> > Regards,
> > > >Andy
> > ------------------------------------------------------------------
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> > >
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