[SI-LIST] Re: Open Termination
- From: Geoff Stokes <gstokes@xxxxxxxxx>
- To: "'si_monkey2@xxxxxxxxx'" <si_monkey2@xxxxxxxxx>
- Date: Thu, 16 Sep 2004 08:38:09 +0100
Yes. In the simple (approximate) model, the transmission line has
inductance too, so V includes other terms.
> -----Original Message-----
> From: Sitar Moniker [mailto:si_monkey2@xxxxxxxxx]
> Sent: 15 September 2004 20:02
> To: Geoff Stokes
> Cc: si-list@xxxxxxxxxxxxx
> Subject: [SI-LIST] Re: Open Termination
>
>
> Let us use the capacitor equation: V = Q/C. At the open end,
> in order to make the voltage 2V, either Q get double or C get
> half. Do you see a problem with this line of thinking?
>
> Geoff Stokes <gstokes@xxxxxxxxx> wrote:
> I just forgot to say that the fringing capacitance relates to
> the fringing
> field. That varies according to the additional wave modes
> which are set up
> by the discontinuity. These change with frequency, so the fringing
> capacitance is a function of frequency, which is an added
> complication.
>
> > -----Original Message-----
> > From: Geoff Stokes
> > Sent: 15 September 2004 17:42
> > To: 'si_monkey2@xxxxxxxxx'
> > Cc: si-list@xxxxxxxxxxxxx
> > Subject: [SI-LIST] Re: Open Termination
> >
> >
> > Sitar
> >
> > Forgive me but I need to avoid spending too much time on this. The
> > microstrip metal has a surface current which is linked, by the
> > electromagnetic equations of Maxwell, to the field
> > immediately adjacent to
> > it. Microstrip's fundamental mode is "quasi-TEM" - it approximates
> > Transverse ElectroMagnetic (TEM). TEM is a wave where there
> > is no field
> > component in the direction of propagation. Also the
> electric field is
> > perpendicular to the metal surface and the magnetic field is
> > everywhere
> > perpendicular to the electric. However the field is not
> > uniform. That may
> > imply that E/H is varying (I'm not sure). The transmission
> > line is a wave
> > guide. However, the theory says that any wave motion can be
> > described by a
> > linear superposition of plane waves in various directions. For each
> > component Ei and Hi, Ei/Hi = 377 ohms. Perhaps not for the
> > total. Does it
> > matter? The current travels in the direction of propagation on the
> > underside of the signal trace, and in the reverse direction
> > on the top side
> > of the ground plane. V/I = 50 ohms or whatever, nothing to
> > do with the wave
> > impedance of free space.
> >
> > But at the open circuit, the current tapers down to nearly
> > zero (sum of
> > incident and reflected wave current). What's left flows in
> > the fringing
> > capacitance. The fringing capacitance is difficult to find,
> > but the order
> > of magnitude will be somewhere near the capacitance per unit
> > length times
> > the microstrip h. That's a small fraction of 1 pF usually.
> > To find it, you
> > have two choices:
> >
> > 1. Measure it on a network analyser or TDR analyser.
> >
> > 2. Measure it in simulation.
> >
> > Both cases require a de-embedding calculation.
> >
> > Geoff
> >
> > > -----Original Message-----
> > > From: Sitar Moniker [mailto:si_monkey2@xxxxxxxxx]
> > > Sent: 15 September 2004 17:24
> > > To: Geoff Stokes
> > > Cc: si-list@xxxxxxxxxxxxx
> > > Subject: [SI-LIST] Re: Open Termination
> > >
> > >
> > > Thank you, Geoff. To be specific, let us consider a
> > > microstrip line the load end of which is left open. There are
> > > two impedances. One is the impedance of the line (V/I) and
> > > the other, beyond the point where the line ends, is the
> > > impedance of free space (E/H) . The 377 Ohms I mentioned
> > > earlier corresponds to this (E/H). Am I right so far?
> > >
> > > If the open termination is modelled with fringing capacitance
> > > and possibly radiation loss, what is the magnitude of this
> > > impedance so we could approximate it with infinity?
> > >
> > > V and I on the line are due to propagating E and H waves.
> > > Then, what is the equivalent of a (V/I)=50 Ohms in terms of
> > > corresponding (E/H)? What is the equivalent of (E/H)=377 Ohms
> > > in terms of (V/I)?
> > >
> > > If ideal open circuit approx (Vi=Vr) is good especially at
> > > lower frequencies, why do we routinely use it at higher
> > > frequencies as well? Is there a somewhat realistic model for
> > > open termination?
> > >
> > > Geoff Stokes wrote:
> > > Sitar
> > >
> > > In addition to your modern engineering books, you also need a
> > > classical book
> > > on electricity and magnetism or microwave/rf engineering.
> > > This will show
> > > that:
> > >
> > > 1. Impedance of a wave in free medium (e.g. air) is not the
> > > impedance of a
> > > transmission line. One is E/H, the other is V/I.
> > > 2. Open termination is not infinite impedance and is
> > > sometimes modelled
> > > with fringing capacitance and possibly radiation loss.
> > > 3. For the ideal open circuit (which does not exactly exist
> > > as I just said,
> > > but sometimes it's good enough for an approximate analysis,
> > > especially at
> > > lower frequencies), the incident wave voltage, Vi amplitude
> > > is equal to the
> > > reflected wave amplitude, and in phase with it. So the total
> > > at the open is
> > > 2Vi. At some other point on the line, the phases are
> > > different, and so you
> > > will plot a standing wave.
> > >
> > > Good luck
> > > Geoff
> > >
> > > > -----Original Message-----
> > > > From: Sitar Moniker [mailto:si_monkey2@xxxxxxxxx]
> > > > Sent: 14 September 2004 22:48
> > > > To: si-list@xxxxxxxxxxxxx
> > > > Subject: [SI-LIST] Open Termination
> > > >
> > > >
> > > > In SI and HSDD books, they assume that open termination
> > > > offers infinite impedance. My understanding is open end line
> > > > is exposed to air which has about 377 ohms.
> > > >
> > > > 1) How do you get infinite impedance for unterminated line?
> > > >
> > > > 2) Is there any intuitive way to show that voltage at the
> > > > open end doubles- other than using math: ref. coeft. = +1?
> > > >
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