[SI-LIST] Re: Why Termination at Both End ?
- From: Jory McKinley <jory_mckinley@xxxxxxxxx>
- To: Vinu Arumugham <vinu@xxxxxxxxx>, alfred1520list <alfred1520list@xxxxxxxxx>
- Date: Wed, 24 Aug 2011 11:44:41 -0700 (PDT)
Could be a new thread.....It is important to simulate and fully understand the
potential design issue that even a simple clock net can produce. Suppose one
has to transmit a 2Ghz clock from driver to receiver. Rough 1/4 wavelength is
1000mils. So one might design around this length and place a design constraint
of say 1500mils to 2500mils for this net (3000mils would be a no no in this
scenario). Good place to start but make sure to include the parasitics of the
driver and receiver since these "filters" will have a big effect on your design
constraints. I have seen a customer case very similar where the design was
targeted for 2500mils only to find in the lab and subsequently simulations that
the parasitics shifted the 1/4 wavelength (odd multiple) right into this design
constraint basically killing the clock at the receiver.
-Jory
Alfred,
I expect the driver to source the max. possible current for the entire
duration of the clock's high pulse and sink the same for the duration of
the low pulse. At the edges I expect it to change from say 45 to -45mA
or vice versa.
The max. possible current is probably somewhat lower (~37mA) in your
case if the clock driver impedance is say 7 ohm or so.
Thanks,
Vinu
alfred1520list wrote:
> Ah, the law of unintended consequence.
>
> While the story is just interesting, Vinu's point is actually critical
> in terms of the reliability of the device that I hadn't even thought
> about.
> I believe our case with a 33 ohm series termination is ok since the
> max possible current is 1.5V/33ohmE mA which occurs only on
> the edges. It would be a different story without series termination.
>
> Thanks for point it out.
>
> Best Regards,
> Alfred Lee
>
>
> ----- Original Message ----- From: "Vinu Arumugham" <vinu@xxxxxxxxx>
> To: "alfred1520list" <alfred1520list@xxxxxxxxx>
> Cc: <si-list@xxxxxxxxxxxxx>
> Sent: Wednesday, August 24, 2011 9:59 AM
> Subject: Re: [SI-LIST] Re: Why Termination at Both End ?
>
>
>> Exactly. One thing to verify is that the clock driver spec. is not
>> violated driving this short circuit for 100% of the device's life.
>> Electromigration limits could be exceeded.
>>
>> Thanks,
>> Vinu
>>
>> alfred1520list 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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>>
>>
>
>
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From: Vinu Arumugham <vinu@xxxxxxxxx>
To: alfred1520list <alfred1520list@xxxxxxxxx>
Cc: si-list@xxxxxxxxxxxxx
Sent: Wednesday, August 24, 2011 12:43 PM
Subject: [SI-LIST] Re: Why Termination at Both End ?
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