[SI-LIST] Re: Why Termination at Both End ?
- From: Istvan Novak <istvan.novak@xxxxxxxxxxx>
- To: Ihsan Erdin <erdinih@xxxxxxxxx>
- Date: Thu, 25 Aug 2011 19:23:00 -0400
Ihsan,
The quarter-wave resonator will create zero input impedance at specific
frequencies only.
At DC the quarter-wave condition does not exists, so it wont short the
DC contents; but
it will short all frequencies where the integer multiples of the
quarter-wave condition is
satisfied. Regarding two different impedances at the same point: you
are correct, when
you connect two impedances in parallel and look at them, you get a
single impedance,
whatever is the parallel resultant of the two. However, when people
speak about two
different impedances, it usually refers to the mental exercise that you
cut the network
into two pieces at the location of observation, and look into the two
parts WITHOUT
connecting them. If you cut the network at the source, where the source
impedance
meets the transmission line, you get a driver with its impedance (say 50
ohms) and
the input of an open-terminated transmission line. The input of an
open-terminated
transmission line will create AC shorts at odd integer multiple
frequencies of the
quarter-wave resonance frequency. When you connect the two together,
the input
impedance of the transmission line behaves like the lower leg of a
discrete voltage
divider, with the driver impedance being in the upper leg. This divider
will create
zero AC voltage where the quarter-wave resonance conditions are satisfied.
Regards,
Istvan Novak
Oracle
On 8/25/2011 6:12 PM, Ihsan Erdin wrote:
> 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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