[SI-LIST] Re: Why Termination at Both End ?
- From: Vinu Arumugham <vinu@xxxxxxxxx>
- To: alfred1520list <alfred1520list@xxxxxxxxx>
- Date: Wed, 24 Aug 2011 09:59:13 -0700
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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