From: S. Weir ([email protected])
Date: Mon Jun 04 2001 - 20:06:30 PDT
Chris,
I don't see how you can justify the surprise introduction of a 20 ohm
driver resistance into the discussion. That reduces the incident amplitude
seen by the load at the far end similarly. Is it appropriate to compare a
circuit that provides 70% (Rdvr=20, Zl=50, Rload=50), against one that
provides 100%, (Rdvr=20, Rser=30, Zl=50, Rload=open)?
For your analogy of a tight "U" shaped trace, I think that model is
great. But the radiating cross-sectional area does not increase. The
whole point is that we are assuming TEM, and so the wave is polarized. The
scalar formula you are using holds only for the perpendicular cross-section
which is just the original Length * Height. Put another way, if we have
two traces, A and B where the driver for A is on the left, and B is on the
right, and they are close to each other, driving each line alternately will
not increase E over what we get by driving one line alone.
Regards,
Steve.
At 06:13 PM 6/4/01 -0700, you wrote:
>Steve,
>
>I like your power analogy except that it assumes each I(f) is halved which
>is not the case. If a driver of 20 ohms feeds a 50 ohm line with/without a
>30 ohm series terminator, the current is reduced to .71 (not .5) its
>original value. Squaring that is .5 which is multipled by twice the time
>ending up back at one. Seems like we agree to me!
>
>For reference:
>
> k * I(f) * f^2 * A
> E(f) = --------------------
> r
>
> where k is a constant,
> I(f) if current at a given frequency
> f is frequency,
> A is loop area (sep distance times length)
> r is antenna distance.
>
>Now in attempting to map the series/parallel question into this formula I
>reasoned to double the area in the sense that the wave (albreit with less
>current than parallel case) travels down and then back. It happens to be at
>the same location. I don't see how this thinking is drastically different
>than creating a an actual route with twice the length (and loop area) that
>folds back and runs along side itself to terminate near its source (with
>adjusted feed current and end termination). This reasoning implies that the
>reduction in current must outweigh the increase in loop area to be
>effective.
>
>If you don't like messing with "A", then another way to map this problem
>into the simple formula is to look at the Fourier equivalent I(f) of the
>signal resulting from each of the cases assuming some periodic switching
>frequency (the formula is for sine wave of freq "f"). In the parallel case,
>the signal content is that of a gorgeous trapezoidal waveform--each segment
>sees a trapezoidal blip travel down it per cycle. However in the series
>termination, it looks much different. The energy is not distributed among
>harmonics in the same proportions, each small segment sees something
>different than the others over a given cycle. Aren't the odds spectacular
>that some of these small segments will sing in chorus at some frequency at
>some angle and some distance away (at your FCC test).
>
>This is all just hearsay anyway until someone publishes some results.
>
>Best Regards,
>
>Chris Rokusek
>Innoveda
>
>
> > -----Original Message-----
> > From: [email protected]
> > [mailto:[email protected]]On Behalf Of S. Weir
> > Sent: Monday, June 04, 2001 1:52 PM
> > To: [email protected]
> > Subject: RE: [SI-LIST] : Antenna Problem on the Board
> >
> >
> > Chris,
> >
> > I really don't follow your reasoning:
> >
> > At 11:16 AM 6/4/01 -0700, you wrote:
> > >Richard,
> > >
> > >Your first paragraph sounds good to me.
> > >
> > >Something else that to consider is that with parallel
> > termination, the wave
> > >flows down the line without a reflection but with source termination the
> > >wave has to travel _twice_ as far before it is absorbed. This
> > seems loosely
> > >like doubling the loop area. Sounds like a good case for
> > >simulation/measurement.
> > >
> > >Chris Rokusek
> > >Innoveda
> >
> > The current imparted is one half for each direction, Vdelta/(Rt + Zl)
> > versus Vdelta/(Zl) so we have one quarter the power for twice the
> > time. This says one-half the total energy, and the peak amplitude is one
> > quarter. With the same geometries for each, what condition would
> > ever give
> > rise to the series termination radiating more than the parallel
> > termination?
> >
> > Regards,
> >
> > Steve.
> >
> >
>
>
>
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