[SI-LIST] Re: capacitor impedance in time domain
- From: "group_delay" <group_delay@xxxxxxxxx>
- To: si-list@xxxxxxxxxxxxx
- Date: Sun, 30 Jan 2005 05:24:29 -0000
Hi Peter,
Thanks for the very insightful way of looking at the problem and
effectively solving it! It was actually not a homework problem. I am
doing a thesis on IC design and wanted to find out about the
capability of cmos buffers to drive capacitive loads, and my
knowledge of tlines is a bit rusty...been a while since my undergrad
days...
thanks again!
--- In si-list@xxxxxxxxxxxxxxx, Peter Fekete <thefekete@xxxx> wrote:
> Dear group_delay, Arpad, =85
>
> 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=3DV(+) + V(-) and I =3D I
(+) - I(-); the voltages add and the currents are substracted
>
> For both waves the relationship between voltage and current is the
well known V(+)=3DZc* I(+) and V(-)=3DZc *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=3DC dV/dt
>
> Note that these V and I are not the trapezoidal ones, but they are
as mentioned V=3DV(+) + V(-) and I =3D I(+) - I(-) and only V(+) and I
(+) are trapezoidal.
>
> This equation gives
>
> V(+)/Zc - V(-)/Zc =3D C ( dV(+)/dt + dV(-)/dt )
>
> V(+) is known so I'll group the terms:
>
>
>
> dV(-)/dt + V(-)/CZc=3D dV(+)/dt =96V(+)/CZc
>
>
>
> the term C*Zc is the "RC constant" mentioned in some messages =3D
tau
>
>
>
> so the equation for the unknown V(-) is:
>
>
>
> dV(-)/dt + V(-)/tau =3D dV(+)/dt =96 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=3D0) V(-)=3D[1-2exp(-t/tau)]
*0.5 for t=3D(0 to T) ( time origin shifted to when the pulse
arrives at the capacitor) and V(-)=3D [1-exp(-T/tau)]*exp[-(t-T)/tau]
for t=3D T to infinity
>
> Note that this gives you V(-)=3D - 0.5 for t=3D0 ie exactly the dip
that takes the V(+)=3D0.5 V down to zero as group_delay noticed for
zero rise time
>
>
>
> For a trapezoidal pulse it is much uglier
>
> V(-)=3D 0.5/tr * (t-2tau)+tau/tr * exp(-t/tau) for the up ramp t=3D 0
to tr
>
> And so on =85
>
>
>
> 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 =96
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 <group_delay@xxxx> wrote:
> arpad,
> i didn't quite understand some of the characters in your equation.
> when you write (finite dV)/(dI =3D 3D 0) =3D 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 =3D3D 0) =3D3D> 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
> >
>
=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D
> =3D3D=3D3D=3D
> >
>
=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D
> =3D3D=3D3D=3D
> > =3D3D=3D3D=3D3D=3D3D=3D3D
> >
> >
> > -----Original Message-----
> > From: si-list-bounce@xxxx [mailto:si-list-bounce@xxxx] =3D
> > 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 =3D3D (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=3D20
> > > the network, or more likely using a computer program we would
> solve =3D
> > the=3D20
> > > difference equations over a series of discrete time steps. Now
> in
> > either=3D20
> > > case we could express impedance as dv/dt / di/dt. But I don't
> know =3D
> > how=3D20
> > > useful it would be towards either visualizing behavior, or
> solving the =3D
> >
> > > equations. Let's take the trapezoidal wave for instance. An =3D
> > effective=3D20
> > > impedance is pretty easy to come by on each: the rising, and
> falling=3D20
> > > portions of the waveform from the capacitance expression C =3D3D
> > i/dv/dt, Z =3D3D=3D20
> > > dv/dt / di/dt =3D3D 1/(dv/dt * C ). The flat portions are
> troublesome
> > as are=3D20
> > > the vertices, since dv/dt theoretically goes to zero and the
> > impedance from=3D20
> > > the formula jumps to an infinite value. Intuition should tell
us
> > that this=3D20
> > > is wrong, as
> >
> > coupling capacitors routinely pass high frequency pulses.
> > >=3D20
> > > In the frequency domain, we have this nailed. We don't have=3D20
> > > discontinuities at the vertices. The vertices and flat
portions
> =3D
> > follow=3D20
> > > curves formed by the frequency components, and rather than a
flat
> > section=3D20
> > > containing DC and no HF, quite the opposite is true: the
> flatter we
> > want=3D20
> > > the pulse tops to be, the higher the frequency content
> required. This =3D
> >
> > > aligns with our intuition. But when we transform the
> representation
> > back=3D20
> > > to the time domain, those piecewise linear segments are now
> curved
> > solving=3D20
> > > the discontinuities at the vertices and eliminating the flat
> slopes
> > with=3D20
> > > theoretically infinite Z between the edges.
> > >=3D20
> > > So if someone wanted to look only at the rising and falling
> edges, an=3D20
> > > impedance in the time domain is reasonable, and possibly even
> > useful. But=3D20
> > > it really gets awkward when dealing with the whole waveform
> unless
> > we first=3D20
> > > perform frequency limiting operations, most easily performed
in
> the=3D20
> > > frequency domain.
> > >=3D20
> > > I am not an expert on algorithms, so I really can't say from
an
> error=3D20
> > > analysis and computational efficiency standpoint what is
really
> the
> > best=3D20
> > > way to perform a transient analysis. But in my naivete, I
would
> be=3D20
> > > inclined to transform everything into the frequency domain,
> compute =3D
> > the=3D20
> > > solution and transform back. In my feeble mind, this would
avoid
> > some of=3D20
> > > the discontinuity and convergence problems in SPICE and more
> closely=3D20
> > > follows nature. But since people a whole lot better at math
> than I
> > have=3D20
> > > worked long and hard on those algorithms, I suspect either
the=3D20
> > > computational overhead, or error build-up of my naive approach
> would =3D
> > be=3D20
> > > unacceptably high. Maybe what this world needs is a five cent,
> 256 =3D
> > bit=3D20
> > > floating point, matrix solver!
> > >=3D20
> > > Steve.
> > >=3D20
> > > At 10:13 PM 1/26/2005 +0100, Matthias Bergmann wrote:
> > > >
> > > >Hello, I don`t understand why impedance should be limited to =3D
> > 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 =10 =10
> 0;&#=3D
> > 160;=01
> > >
> >
> >60I
> ;=01=3D
> > 60;=10
> > > > |
> > >
> > > >-----------------------------------------------------------
> ------=3D
> > ----
> > > >-- -----------------------------------------| >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 =3D
> > 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 =3D3D
> > v(s)/i(s) or
> > > >v(f)/i(f), which makes it a strictly frequency domain
parameter.
> > Regards,
> > > >Andy
> > -----------------------------------------------------------------
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