[SI-LIST] Re: Open Termination - Sandor
- From: Sitar Moniker <si_monkey2@xxxxxxxxx>
- To: sandord@xxxxxxxxxxxxx
- Date: Fri, 17 Sep 2004 08:53:24 -0700 (PDT)
From your explanation in previous mail, what I understand is the E field
(though you called it energy wave) rolls down the line happily and suddenly
finds itself standing in a situation where it can't jump out into the air
(infinite impedance discontinuity) nor just sit there and roadblock junior E
fields rushing from behind. The clever E field turns around (reflection) and
hugs the oncoming E field (superposition). Thus, we have total voltage at the
open equal to 2V and a E field standing there corresponding to 2V.
By now, the E field guy standing at the open end prefers to stand looking
toward source end(reflection mode) and use superposition to his advantage. My
question is why there are no more superpositions to make it 3V, 4V etc. even
though the driver is continuously pumping a constant V and more and more E
field guys approaching the open end.
What happens if we make the source end also open (by yanking the source) the
instant the voltage at the load end turned 2V?
Open end case is as ideal as it gets. So, let us keep everything else related
to this as ideal as they could be. The answers may be trivial to the experts
but I don't know how to get a hang of these concepts.
Sandor Daranyi <sandord@xxxxxxxxxxxxx> wrote:If you imagine the wave reflected
at the open end, the
direction of travel may be reversed but it still sees
the same line impedance, so the voltage amplitude
won't change. After superposition, 1+1=3D2. In real life
of course, there are always losses, so it will be
less, but the reflection (or rather its effect on the
signal) is something that is really easy to observe
with an oscilloscope. As always, in real life it's a
bit more complicated because there may be some ringing
on the signal and perhaps several reflections, due to
other "imperfections" beside an unterminated line.
These imperfections are anything that can make the
transmission line nonideal. E.g. when it has
stubs/branches hanging off it, finite impedances
connected to it (a device inputs perhaps) somewhere
along the line, line geometry changes etc. In general:
impedance variations.
Now you could say, yes, but what happens when the
wavefront reaches the beginning of the transmission
line and gets reflected again, wouldn't that result in
a higher voltage on the line? Well, the thing to note
here is that there is a driver hanging off the source
end. In ideal case, the driver will be a voltage
source or maybe a current source. If it's a current
source, it has infinite impedance and you do have a
reflection again, so you end up with higher and higher
voltages but going back to looking at the energy, it
is not surprising, since the current source will keep
pumping energy in there! If you have an ideal voltage
source driving, it'll have zero impedance. Just like
infinite impedance, zero impedance can not "swallow"
energy, so it will be reflected back, but the
difference is that voltage is reflected back with the
opposite phase in this case.
If you are interested in a more practical case, the
basic things to consider are that real life drivers
have finite source impedances, they have
nonlinearities and limited V & I ranges. =20
So, with all the signal integrity problems they can
cause, are reflections always bad? Not necessarily.
For example PCI uses reflected wave switching which
allows the use of relatively weak drivers.
Cheers,
Sandor
-----Original Message-----
From: Sitar Moniker [mailto:si_monkey2@xxxxxxxxx]=20
Sent: Friday, 17 September 2004 5:04
To: sandord@xxxxxxxxxxxxx
Cc: si-list@xxxxxxxxxxxxx
Subject: Re: [SI-LIST] Re: Open Termination
Thank you. It definitely helps to understand in terms of energy. As you men=
tioned, due to voltage superposition, the voltage at the open end gets do=
ubled. Extending this idea, will there be an occasion (ideal or otherwise=
) when the open end voltage becomes 3X or 4X. If so, under what condition=
s? If not, what limits this from happening?
Sandor Daranyi wrote:
Hello Sitar,
Others have already given good information regarding your questions but sin=
=3D
ce you asked for something intuitive, maybe I can add something worthwhil=
=3D
e. The warning is that intuitive explanations can often be oversimplific=3D
ations.
> -----Original Message-----
> From: Sitar Moniker
> Sent: Wednesday, 15 September 2004 7:48
> To: si-list@xxxxxxxxxxxxx
> Subject: [SI-LIST] Open Termination
>=3D20
> 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.
>=3D20
> 1) How do you get infinite impedance for unterminated line?
The easiest way to see what happens is in terms of energy. (Looking at what=
=3D
happens to energy is a very useful techni que when trying to understand s=
=3D
omething like this.) The driver pumps energy into the transmission line. =
=3D
The resulting energy wave travels down the line at close to the speed of =
=3D
light and eventually reaches the end. The energy COULD continue to travel=
=3D
in air but you would need a coupling mechanism between the transmission =3D
line and free air. If there is no coupling, it is irrelevant what would h=
=3D
appen in air, or indeed that it's 377ohms. If energy can't be transferred=
=3D
, then by definition the impedance must be infinite. If it weren't, there=
=3D
would be a finite impedance and energy could be transferred.
Nothing is perfect in the real world, so there can be some little coupling =
=3D
e.g. due to losses and radiation from the transmission line, so the menti=
=3D
oned infinite impedance is only an approximation, but for most practical =
=3D
cases, quite a valid one.
Of course, if you have an antenna at the end of the transmission line, sudd=
=3D
enly you do have a coupling mechanism and the story changes. Everything g=
=3D
ets a bit trickier though, because now you would have things like radiati=
=3D
on efficiency and the antenna itself would behave like an impedance trans=
=3D
former between the transmission line and free air.
> 2) Is there any intuitive way to show that voltage at the open end
> doubles- other than using math: ref. coeft. =3D3D +1?
From the previous answer, you may already see what happens here. When the f=
=3D
ace of the energy wavefront reaches the end of the transmission line and =
=3D
more energy still keeps arriving, it won't jump into air and it won't pil=
=3D
e up there, either. Instead, it will get reflected and the reflected wave=
=3D
will get superimposed on the arriving part of the wave. Why twice the vo=3D
ltage? Due to voltage superposition, which should sound familiar from bas=
=3D
ic electronic theory (a quick "voltage superposition" google search threw=
=3D
up: http://www.allaboutcircuits.com/vol_1/chpt_10/6.html).
Sandor
------
Sandor Daranyi
Snr Design Engineer
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