[SI-LIST] Re: LVDS Termination
- From: "Dr. Howard Johnson" <howiej@xxxxxxxxxx>
- To: <si-list@xxxxxxxxxxxxx>
- Date: Mon, 14 Apr 2003 11:21:00 -0700
Dear Mr. Brown,
You may indeed have missed some good articles in your quick
scan of my site
www.sigcon.com. Two articles are particularly relevant to
the issue of differential coupling:
"Differential Trace Impedance", and
"Ribbon Cable Impedance"
To find them, look under "archives" for the "key word
index", and
try the keyword "differential".
The content of those articles generally agrees with the
writeup you have provided below.
Best regards,
Dr. Howard Johnson, Signal Consulting Inc.,
tel +1 509-997-0505, howiej@xxxxxxxxxx
http:\\sigcon.com -- High-Speed Digital Design articles,
books, tools, and seminars
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx
[mailto:si-list-bounce@xxxxxxxxxxxxx]On Behalf Of Mike Brown
Sent: Friday, April 11, 2003 4:50 PM
To: Jim G Roberts
Cc: si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Re: LVDS Termination
Jim,
"The two lines in parallel, . . ." is missing one word to be
correct.
That word is "isolated".
When is 50 ohms in parallel with 50 ohms not equal to 25
ohms? When the
50 ohm impedances are half the differential impedance of a
pair of
coupled lines. The impedance depends on the excitation.
Let's look
into an ideal symmetrical line.
o ---- Cm ---- o <-- line conductor
| |
Cg Cg
| |
gnd gnd
If the lines are driven differentially, current flows in
both the ground
capacitance Cg and the line-to-line capicatance Cm, and both
figure into
the Zdiff calculation. If the lines are driven in parallel
(common
mode, CM, excitation), there's no CM current in Cm because
both sides
of Cm operate at the same potential (by definition of the CM
drive
stimulus). If there is no current flow through an element
it may be
removed from the circuit being analyzed. Less C => higher
Z.
If the lines are loosely coupled (Cm ~ 0) then your model is
correct -
the parallel impedance is practically 25 ohms, and the Rcm
value is
practically 0. If Cm is other than very small, then the CM
impedance is
greater than Zdiff/2.
If asymmetry is introduced into the lines, common-mode
excitation
results in differential-mode noise, and vice versa. I'm not
particularly concerned about high frequency Vcm spikes from
a signalling
viewpoint (within the spec Vcm range of the receiver), but I
am
concerned about the resultant differential noise. I also
understand the
EMI threat of CM noise. That's two reasons why CM
termination is important.
Dr. Johnson's HSDD book, and his web site articles (see
Charles Grasso,
below), seem to ignore the coupled line effects on the CM
impedance
(unless I missed the article in my admittedly quick scan of
the site).
This is possibly because the achievable coupling in a PWB is
low, hence
the practical CM impedance in a PWB differs little from
Zdiff/2. That's
not necessarily the case if the transmission medium is not a
PWB.
Whether or not there's an added Rcm in the ground return
path, there
must be at least an AC ground return path from the center of
the
differential termination in order to terminate CM noise.
Regards
Mike
Jim G Roberts wrote:
>Hi Mike,
> I think I am missing something here.
>The two lines in parallel, in accordance with your sketch,
will have a
>common-mode impedance of 25ohm [50/2].
>With Rcm at zero you will always terminate the common
signals at the
receiver
>end however they are generated.
>More often high frequency Vcm spikes, etc. are generated by
unequal length
of
>lines. Of course crosstalk and power supply ripple can
induce unwanted
signals.
>Can you explain why you would need a different impedance
for the
common-mode ac
>signals.
>--
>Regards, __________ James G Roberts
> /___ ____ | jrobert@xxxxxxxxxxxxxxxxxxxxx
> Jim __ / /___/ / jgroberts@xxxxxxxxxx
> / /_/ /---| | Room: BE436, Hilversum
> \____/ /_/ Tel: +31 35 687 4308 Fax: 5976
>
>Mike Brown wrote:
>
>
>
>>Justin,
>>
>>Perhaps a sketch will help, just in case you are thinking
about a
>>different topology than I (differential sources omitted):
>> (Rdiff)
>> +---- Line + --- 50 ---+
>>Gnd --- Vcm ---+ +--- Rcm ---
Cap --- Gnd
>> +---- Line - --- 50 ---+
>>
>>The cap charges to Vcm (average); the termination
impedance of the
>>common-mode driven lines for AC is Rdiff/2 + Rcm, where
that sum is
>>ideally the CM impedance. With no AC CM, no current flows
in the CM
>>termination. The requirement is that (25+Rcm)*Cap >>
period of the
>>common mode noise. The cap is thus a near short at the
frequencies of
>>concern.
>>
>>In widely-spaced lines, Rcm approaches zero, because the
CM impedance
>>approaches Zdiff/2. This is the typical center-tapped
termination. If
>>the lines are coupled, the mutual C, Cm, (too many
CMs/Cm's in this
>>discussion!) impacts the differential Z, but drops out of
the
>>calculation of Zcm. For common mode excitation, the
voltage at each
>>terminal of Cm is equal, hence no current flows in Cm. If
there's no
>>current flow through the component, the component can be
excluded from
>>the analysis. The result is that Zcm is greater than
Zdiff/2, and the
>>need for Rcm for ideal termination.
>>
>>HTH
>>
>>Mike
>>
>>Tabatchnick, Justin wrote:
>>
>>
>>
>>>Mike;
>>>
>>>I don't quite understand your approach , it seems to me
that you will
>>>still get large reflection because your load impedance is
now complex
>>>with the addition of the cap and the real part is still a
large
>>>impedance. Can you elaborate more ?
>>>
>>>Thanks
>>>
>>>Justin
>>>
>>>-----Original Message-----
>>>From: Mike Brown [mailto:bmgman@xxxxxxxxxx]=20
>>>Sent: Thursday, April 10, 2003 8:08 PM
>>>To: kenneth.w.siders@xxxxxxxx
>>>Cc: 'si-list@xxxxxxxxxxxxx'
>>>Subject: [SI-LIST] Re: LVDS Termination
>>>
>>>Think about driving the 2 lines in parallel, with only a
high-Z path to
>>>
>>>
>>ground at the receiver input (the condition without the
cap). Without the
>>capacitor, the CM reflection coefficient approaches 1.
With the cap, it
>>approaches zero. With an appropriate resistor in series
with the cap at
the
>>CT of the 50 ohms, it can be made to be zero.
>>
>>
>>>The obvious drawback is parts count and board real
estate.
>>>
>>>Regards
>>>Mike
>>>
>>>Siders, Kenneth W wrote:
>>>
>>>
>>>>Hello Group,
>>>>My question concerns terminating an LVDS receiver with
two 50 ohm
resistors
>>>>
>>>>
>>in series with a center tap capacitor to ground across the
inputs of the
>>receiver to achieve better common mode noise rejection?
What type of
>>improvement if any will I see. Also, are there any draw
back with this
type
>>of termination.
>>
>>
>>>>Regards,
>>>>Ken
>>>>
>>>>Ken Siders
>>>>Sr. Design Engineer
>>>>Lockheed Martin MFC
>>>>5600 Sand Lake Rd.
>>>>MP 916
>>>>Orlando, FL 32819-8907
>>>>Phone: (407) 356-4891
>>>>Pager: (888) 363-3411
>>>>Fax: (407) 356-8944
>>>>
>>>>
>>>>Email: kenneth.w.siders@xxxxxxxx
>>>>
>>>>
>
>
>
>
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