[SI-LIST] Re: Why Termination at Both End ?
- From: "Yuriy Shlepnev" <shlepnev@xxxxxxxxxxxxx>
- To: "'Ihsan Erdin'" <erdinih@xxxxxxxxx>
- Date: Thu, 25 Aug 2011 16:42:45 -0700
Ihsan,
Istvan and Steve already answered your question and I just add a few things.
A resistor short-circuited by a piece of wire will have nearly zero
impedance at all frequencies (including DC) when it can be considered
lumped.
A transmission line segment transforms open-circuit impedance (infinity or
some other value) into a frequency-dependent impedance that may be nearly
zero at some frequencies.
Impedance of t-line and termination resistor impedance are connected in
parallel at the receiver end and if one of those is nearly zero at some
frequencies, the total impedance is also nearly zero. The is no difference
between actual short circuit with a wire and short-circuit with open-ended
transmission line at those frequencies. I think the frequency-dependence of
the answer is confusing part here.
It can be easily verified with simple computations as I provided below, or
simulated.
Best regard,
Yuriy
Yuriy Shlepnev
www.simberian.com
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx] On
Behalf Of Ihsan Erdin
Sent: Thursday, August 25, 2011 3:13 PM
To: shlepnev@xxxxxxxxxxxxx
Cc: steve weir; si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Re: Why Termination at Both End ?
Yuriy,
The original question is this: provided that the receiver end is an open
circuit can the impedance when observed at the driver end at a quarter
wavelength away be zero? The answer is yes if the driver end is not
terminated and no if it is. Otherwise, you have to explain how a point in
space can assume two values (i.e. zero impedance and the value of the
termination resistor) at the same time. And if it is bizarrely zero despite
the resistor there then you should also answer why Alfred didn't observe
zero volt on the resistance when he probed it but VDD/2.
However if you're talking about the impedance as seen from the load end
nobody is arguing it will assume a value independent of the line and
termination resistor. In that case it will depend on the line eigenvalues
and the length of the section. But this is irrelevant in this discussion.
Regards,
Ihsan
On Thu, Aug 25, 2011 at 5:21 PM, Yuriy Shlepnev
<shlepnev@xxxxxxxxxxxxx>wrote:
> Ihsan,
>
> A port with zero input impedance connected to a resistor will have
> zero impedance. The resistor does not change anything.
> Let's compute impedance Z of a simplified net that contains a resistor
> R connected to a transmission line segment described by complex
> propagation constant G, characteristic impedance Zo and length L, that
> is open circuited at the opposite end. From the basic theory of
> transmission lines we have:
> Z=1/(th(G*L)/Zo+1/R)
> As you can see, if G*L is exactly PI/2 (quarter of wavelength), Z is
> equal to 0 and values of Zo or R does not matter.
>
> Best regards,
> Yuriy
>
> Yuriy Shlepnev
> www.simberian.com
>
>
>
> -----Original Message-----
> From: si-list-bounce@xxxxxxxxxxxxx
> [mailto:si-list-bounce@xxxxxxxxxxxxx]
> On
> Behalf Of Ihsan Erdin
> Sent: Thursday, August 25, 2011 1:31 PM
> To: steve weir
> Cc: si-list@xxxxxxxxxxxxx
> Subject: [SI-LIST] Re: Why Termination at Both End ?
>
> Steve,
> With my questions regarding the RF theory (I'd rather call it standing
> wave theory in this particular case) I just wanted to point out some
> inconsistencies. I hope the following argument will clarify things. On
> an undisturbed line (i.e. free from any discontinuity), a short will
> be spaced at quarter wavelength from an open. Placing a termination
> (in this case at the near end), however, is mathematically equivalent
> to enforcing a boundary condition to the wave equation at that
> location. Either mathematically or physically, one will not see a
> short circuit there but only the resistor value although the location
> of the component is a quarter wavelength away from an open circuit at
> the clock freq.
>
> And finally, I didn't follow the original thread but thanks to Alfred
> for sharing this thought stimulating experience with the right info.
>
> Regards,
>
> Ihsan
>
> On Thu, Aug 25, 2011 at 12:50 AM, steve weir <weirsi@xxxxxxxxxx> wrote:
>
> > Ihsan, set-up a simulation and you can see this easily. What is
> > required is for the sum of the return wave front and the incident
> > wave front to cancel at the junction of the transmission line and
> > the source termination resistor. This will happen when:
> >
> > 1. The signal duty cycle is 50%, AND 2. The transmitter + series
> > termination matches the impedance of the transmission line, AND 3.
> > The receive end is an open circuit, AND 3.
> > The round-trip propagation delay is one phase interval.
> >
> > In this case, the incident wave launches into the Tx line with
> > 0.5*deltaVtx. At the far end the open line causes a +1 reflection.
> > After a full round trip, this +1 reflection arrives just as the
> > transmitter sends the next signal edge of opposite polarity n(2x+1)
> > = -0.5*deltaVtx. The returning wavefront identically cancels the
> > incident wavefront at the junction of the resistor and the Tx line.
> > Because the transmitter termination matches the line, there is no
> > reflection back towards the receiver. Consequently, no ISI results.
> > This is just a source terminated circuit operating properly under a
> > special condition that can be confusing to an observer.
> >
> > Going back to Sen's case, the transmit termination does not match
> > the line, and so ISI does travel back towards the receiver
> > corrupting the receive signal. If as in the case above we the round
> > trip matches the phase interval, then old opposing edges will align on
incident edges.
> > For things like clocks and other timing strobes, that's a bad, bad,
> thing.
> >
> > The RF theory is straight forward: The Tx line 1/4 wave resonance
> > seen by the composite drive circuit depends only on what is at the
> > far end of the transmission line. An open at the far-end causes a
> > reflection that sends energy back up the line. For an open, a line
> > that is 1/4 wavelength one-way the returning energy is always
> counterphase
> to the
> > driver. Using a matched series termination, this doubles the current
> > load supplied by the driver. Since the driver was matched to the
> > line before and current was vdeltaTx/(Zterm + Zline), and is now
> > vdeltaTx/(Zterm), algebraically the line looks like a short circuit.
> >
> > Steve.
> >
> > On 8/24/2011 9:16 PM, Ihsan Erdin wrote:
> > > Alfred,
> > > The RF theory explanation to this seemingly mysterious phenomenon
> > > doesn't sound convincing to me for two reasons. First, how is it
> > > possible that
> > the
> > > far end open circuit is transformed to a near end short circuit
> > > while the near end is already terminated by a resistor? Second, if
> > > it were really transformed to a short then you should have
> > > observed zero volt not the average dc voltage which is half the
> > > clock voltage swing. The only
> > plausible
> > > explanation is that the average dc voltage must be the result of
> > > the matching or almost matching near end termination-line
> > > impedance
> relation.
> > >
> > > In this case the answer has more to do with the transmission line
> > reflection
> > > theory than the standing wave theory you brought about. Let's test
> > > this argument with the numbers. Your clock has a period of about
> > > 6ns. With a
> > 50%
> > > duty cycle its bit time (i.e. high/low logic time) will be 3ns.
> > > But this
> > is
> > > pretty much the two way prop delay for a 10" long PCB trace on FR4.
> > > When your pulse reached the receiver side it doubled in amplitude
> > > (reflection from open) and started to travel back to the
> > > terminated near end. Its arrival at the near end after 3ns
> > > coincided with the following logic low sunk by the driver. So when
> > > you probed it at the near end termination
> > what
> > > you really observed was nothing but a series of back-to-back high
> > > pulses; alternating between the original signal from the driver
> > > and the far end reflection, with the latter simply filling in the
> > > original logic low
> > time.
> > > To an unsuspecting eye this whole phenomenon should look like dc
> > > at about half the voltage swing with possible glitches at the
> > > logic transition
> > times
> > > (i.e. rising/falling edges) that will look like noise.
> > >
> > > I personally didn't try this in the lab nor simulated it. This is
> > > nothing but a gedanken experiment which can be easily put to test
> > > by a 5-10 line long spice netlist. If the good old reflection
> > > theory is correct I bet you'll observe what I describe here with
> > > this interesting coincidence between the electrical line length
> > > and the
> clock
> period.
> > >
> > > Should we worry about the receiver then? No, because the reflected
> > > wave
> > will
> > > be absorbed by the near end (matching) load and the receiver will
> > > not experience this bizarre dc-looking waveform, hence no problem
> there.
> > Thus,
> > > with properly terminated lines one doesn't need to worry about the
> > > line electrical length and wavelength ratios in the constraint
> > > settings. As usual, most of our concerns for long lines should
> > > come from the less sexy copper/dielectric losses at high frequencies.
> > >
> > > Regards.
> > >
> > > Ihsan
> > >
> > > On Wed, Aug 24, 2011 at 12:40 PM, alfred1520list
> > > <alfred1520list@xxxxxxxxx>wrote:
> > >
> > >> ----- Original Message -----
> > >> From:<Peter.Pupalaikis@xxxxxxxxxx>
> > >> Sent: Wednesday, August 24, 2011 12:03 AM
> > >>
> > >>> Steve's comments reminded me of a situation we were discussing
> > >>> about a
> > >> chip
> > >>> design the other day. That is, if you do have an imperfect (or
> > purposely
> > >>> bad) termination at the receiver end (like a high-impedance),
> > >>> then try
> > >> not
> > >>> to evaluate any performance by looking at the waveform at the
> > >>> source -
> > it
> > >>> may look horrible and in the case of a clock, it is possible
> > >>> that the voltage waveform is nearly non-existent, while the
> > >>> waveform at the
> > >> receive
> > >>> end looks fine.
> > >> I can attest to this. It happened on a pretty mundane situation.
> > >> I thought I'll share so some one like myself a few years ago
> > >> would benefit.
> > >>
> > >> We have a 148.5 MHz video clock driving a relatively long trace,
> > >> around
> > 10"
> > >> long,
> > >> that's in the inner layer. As usual we only source series
> > >> terminate
> it.
> > >> So when
> > >> we have some video problems, we look at the clock line. Since we
> > >> don't have easy access to the end of the line, we looked at the
> > >> down stream side of the series termination resistor, THERE IS NO
> > >> CLOCK, just a constant DC line with a little bit of noise. How
> > >> could it have ever worked? After some series head scratching and
> > >> pondering, we realized this is exactly the situation that Peter
> > >> described.
> > >>
> > >> Looking from the RF antenna engineering point of view, this
> > >> should be obvious. When the transmission line is exactly 1/4 the
> > >> wave length, it becomes a 1/4 wave transformer that transforms
> > >> the open circuit at the far end
> > to a
> > >> short
> > >> at the near end, hence the flat line. DC is not affected by the
> > >> transformer so you see only the DC average of the waveform.
> > >>
> > >>
> > >> Best Regards,
> > >> Alfred Lee
> > >>
> > >> http://www.mds.com
> > >>
> > >> -----------------------------------------------------------------
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> > >
> >
> >
> > --
> > Steve Weir
> > IPBLOX, LLC
> > 150 N. Center St. #211
> > Reno, NV 89501
> > www.ipblox.com
> >
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> >
> >
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>
> --
> Ihsan Erdin
>
>
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Ihsan Erdin
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