> So, Ivan's comments about ONLY a resistor can absorb is correct when
talking
> about conventional "hard output" CMOS chips.  Due to these transmission
line
> problems, many newer chips are putting active resistors on critical output
pins.

Yes, I suppose I should have said that "only a R +/- jX impedance, where R
is non-zero, can absorb".  However, it seemed more clear and direct to say
that "only a resistor can absorb".

Best regards,
Ivan Baggett
Bagotronix Inc.
website:  www.bagotronix.com


----- Original Message -----
From: "Jon Elson" <[EMAIL PROTECTED]>
To: "Protel EDA Forum" <[EMAIL PROTECTED]>
Sent: Monday, August 06, 2001 7:01 PM
Subject: Re: [PEDA] 5/5 Manufacturing Issues


>
>
> Tim Hutcheson wrote:
>
> > Actually I was thinking the authors meant that the driver absorbs about
5/6
> > of the signal at frequency, while the resistor "absorbs" the remaining
1/10,
> > at any frequency. Is that incorrect?
>
> That is correct.
>
> > Keep in mind that only a resistor can "absorb".  A PCB trace cannot
absorb
> > anything.  If you really need a reflection absorbed at the source, you
> > really need a resistor to absorb it.
>
> Yes, but!
>
> A resistor can be made many ways.  One way is to bias a transistor into
its
> linear region.  Another way is to put an amplifer in series with a sense
> resistor,
> and have a feedback loop adjust output voltage so that a specific voltage
ratio
> is seen across the sense resistor vs. the output pin.  This second method
is
> probably what your chip is doing.  The first method seems to be what some
> other special drivers do.
>
> The typical CMOS output driver has an output impedance around 1-2 Ohms
> in the normal current range of the chip.  For short circuit protection,
the
> output impedance goes up with higher current.  This can actually be a
> point to consider, that the pin is not ALWAYS a 50 Ohm load on the line,
> only within a certain domain.
>
> So, Ivan's comments about ONLY a resistor can absorb is correct when
talking
> about conventional "hard output" CMOS chips.  Due to these transmission
line
> problems, many newer chips are putting active resistors on critical output
pins.
>
> Any digital chip with a 50 Ohm output impedance is definitely specialized
for
> this sort
> of problem.
>
> Jon
>

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