[SI-LIST] Re: capacitor impedance in time domain
- From: Steve Corey <steven.corey@xxxxxxxxxxxxxx>
- To: si-list@xxxxxxxxxxxxx
- Date: Sat, 29 Jan 2005 23:54:19 -0800
In my opinion, the easiest way to solve the problem is to think of what
you would expect when TDR'ing an ideal 50-ohm T-Line of delay Td
terminated by an ideal capacitance C, using an ideal 50-ohm TDR with a
total voltage step of vstep. The capacitor initially appears as a short
circuit (voltage goes to zero, gamma=-1) and finally appears as an open
circuit (voltage goes to vstep, gamma=+1). Now think of the view from
the perspective of the capacitor. When it looks up the line, all it
sees is 50 ohms to ground, tline energy storage notwithstanding -- the
capacitor doesn't care if it's a terminated tline or a resistor. As a
result, the controlling time constant to charge the capacitor is
(C*50ohms). When we mix all this together, we come up with a TDR
waveform that starts at zero, steps up to vstep/2, is flat for 2*Td,
steps instantaneously to zero, then "decays" up to vstep with a time
constant given by C*50ohms.
We review this case, and a number of related cases, in our TDR quick
guide, although the pictures do include parasitic effects such as finite
risetime and a nonideal capacitor:
http://www.tdasystems.com/library/appnotes/tdr_iconnect_quick_guide.PDF
For a actual measurements on a power/ground plane pair, you can take a
look at figure 4 in the following app note:
http://www.tdasystems.com/library/appnotes/PDNA-0703.pdf
Regarding Steve's earlier post -- replacing jw with dv/dt to map Z =
1/jwC to Z = 1/(dv/dt * C) is really no different from a frequency
domain analysis since the latter expression is only valid for a
single-frequency sinusoidal v. Looking for flat and steep places in a
non-sinusoidal v and applying that relationship will not give valid results.
-- Steve
-------------------------------------------
Steven D. Corey, Ph.D.
Time Domain Analysis Systems, Inc.
"The Interconnect Analysis Company."
http://www.tdasystems.com
email: steven.corey@xxxxxxxxxxxxxx
phone: (503) 246-2272
fax: (503) 246-2282
-------------------------------------------
Steve Weir wrote:
> Alan, well if we short a transmission line at the far end with a capacitor,
> then Z still equals 1/( dv/dt * C ). Hopefully this is obvious as a
> reexpression the sinewave response Z = 1/jwC. With a big "perfect"
> capacitor, this will just look like a shorted line with a -1 reflection
> coefficient. As the capacitor gets smaller, the pulse top reflection will
> decay, and with further reductions in C the edge reflections will decay.
>
> The problem with viewing this as an RC is the stored energy in the line.
>
> Regards,
>
>
> Steve
> At 03:04 PM 1/28/2005 -0800, Alan Hilton-Nickel wrote:
>
>>Steve, I don't think you and Chris are talking about the same topology. My
>>understanding is that Chris is looking at
>>"a lossless transmission line terminated with a lossless capacitor", which
>>to me looks like:
>>
>>driver => txline => capacitor => GND
>>
>>I believe Arpad is correct in looking at this topology as an RC
>>connection, especially in the lossless case.
>>
>>Alan
>>
>>Steve Weir wrote:
>>
>>
>>>Arpad, this would be a good time to use Jon Powell's si-draw tool. But
>>>let me try this without:
>>>
>>>driver => txline => coupling cap => txline => terminator
>>>
>>>Now, let's suppose the driver has a Tr of 100ps, a Tf of 100ps, and a
>>>pulse width of 2.4ns, an output impedance of 0.0001 ohms, and a drive
>>>level of 2.5V. Let's try three different values of coupling capacitors,
>>>all in the same 0402 body and 50 ohm txlines.
>>>
>>>10pF
>>>1nF
>>>100nF
>>>
>>>Now, what sort of wave form do you expect to see at each point?
>>>
>>>I know what my answers are: 10pF lots of reflection towards the source
>>>and droop towards the terminator. 1nF and 100nF very little reflection
>>>or droop. Plug it into your favorite simulator and see what you get.
>>>
>>>Regards,
>>>
>>>
>>>Steve
>>>At 02:03 PM 1/28/2005 -0800, 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@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx] =
>>>>On Behalf Of group_delay
>>>>Sent: Friday, January 28, 2005 1:48 PM
>>>>To: si-list@xxxxxxxxxxxxx
>>>>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 <weirsp@xxxx> 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          &#=
>>>>>
>>>>
>>>>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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>>>>>
>>>>>            &=
>>>>>
>>>>
>>>>#160;
>>>>
>>>>
>>>>
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>>>>>>
>>>>>
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>>>>
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