[SI-LIST] Re: Open Termination
- From: Sitar Moniker <si_monkey2@xxxxxxxxx>
- To: "Beal, Weston" <weston_beal@xxxxxxxxxxx>, gstokes@xxxxxxxxx
- Date: Thu, 16 Sep 2004 11:04:25 -0700 (PDT)
Good point. The antenna takes the line impedance (V/I = 50, say) and transforms
into the free space impedance (E/H = 377, say) and thus facilitates radiation.
I suppose the antenna itself has some impedance between 50 and 377 to function
as transformer. I wonder what impedance value provides max radiation for the
(50,377) set.
"Beal, Weston" <weston_beal@xxxxxxxxxxx> wrote:Sitar,
Another important point that might make Geoff's wonderful explanation make more
sense (or less) is to think of an antenna as an impedance transformer. An
antenna transforms the V/I impedance of the transmission line at the driving
point of the antenna into a free space intrinsic impedance of E and H fields.
The intrinsic impedance of the medium (dieletric in most cases) is just the
ratio of E to H that the medium supports. It is independent of geometries
because it assumes that the fields are completely enclosed in the medium. On
the other hand, the impedance of a transmission line depends on the geometry of
the metal and dieletric because these cause different distributions of the
current and voltage on the metal conductors.
Regards,
Weston
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx] On
Behalf Of Geoff Stokes
Sent: Wednesday, September 15, 2004 9:42 AM
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 s
ide 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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