[SI-LIST] Re: AC Termination : Capacitor Value
- From: wolfgang.maichen@xxxxxxxxxxxx
- To: joepaulm@xxxxxxxxx
- Date: Tue, 15 Jan 2008 12:52:44 -0800
Joe,
I am assuming you mean termination with a 50 Ohm resistor connected to a
capacitor whose other end goes to ground? I.e.
transmission line --- receiver
|
|
R
|
C
|
GND
This type of termination avoids static current flowing when there is no
activity on the line. The price you pay is that it cannot terminate slow
(DC) signals. That means this scheme will only work with DC balanced
signals (usually means an equal number of 1's and 0's in the data stream.
Examples are 50% duty cycle clock signals, Manchester encoded data, or
8b/10b encoded). The line will then settle to the average level on the
line.
Signal rise/fall times will NOT be increased because of the capacitor -
for such short time intervals (assuming sufficient capacitance) the
capacitor looks like a perfect short to ground, leaving you with a matched
50 Ohm termination. BUT the levels at the receiver will drift around if
the signal stays low or high for too long. So the important figure of
merit is the "running disparity", i.e. the maximum number of excessive 1's
or 0's occuring in your data stream over any subsection of your pattern
("excessive" meaning not being balanced by preceding bits of the opposite
polarity). Have a look at Howard Johnsons "Digital Signal Propagation"
book which takes more about that concept. In a nutshell, your RC time
constant should be much longer than this time scale. E.g. 8b/10b encoded
data has a maximum running disparity of +/-2. At e.g. 1 Gb/sec data rate
that means RxC should be much longer than 2 x 1ns = 2ns. Let's e.g. take a
factor of 10x, i.e. 20ns. With R=50 Ohms and a line impedance of Zo=50
Ohms (i.e. effective source impedance of 100 Ohms since R and Zo act in
series) you'll need at least 0.2nF of capacitance.
Should you use much more to be on the safe side? Depends. First, larger
capacitors tend to have larger inductive parasitics, which is not good for
high-speed termination. Even more important, after start of the data
transmission the line needs time to settle to the average signal level,
meaning the receiver will need to throw away the first few (or many) data
bits before it can expect valid data. That's also the reason SerDes
transmission schemes need to keep sending "idle" patterns when there is no
data to transmit, to avoid drifting away from the settled levels. The
settling time is given by the time constant, (Zo+R)xC. So your data
transmission needs to run sufficient "warmup cycles" to let the levels
settle out. Make C too large and the system will take "forever" to settle.
That gives you an upper bound. Summing up:
(R+Zo) x C >> RD x Tbit
(RD = running diparity, Tbit = bit interval)
(R+Zo) x C << maximum tolerable warmup time
So for the present example, maybe 10nF would be a good choice.
Wolfgang
Joe Paul M <joepaulm@xxxxxxxxx>
Sent by: si-list-bounce@xxxxxxxxxxxxx
01/15/2008 12:19 PM
Please respond to
joepaulm@xxxxxxxxx
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Subject
[SI-LIST] AC Termination : Capacitor Value
Is there a thumb rule for finding Capacitor value for AC Parallel
termination (R-C) for single ended lines ?
I came across people who feel that higher capacitor value for R-C
termination , will increase capacitive loading on line and hence
increase rise/fall time . Elementary SI knowledge ,makes me feel that
this is not true. Am I missing something ?
Joe
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