[SI-LIST] Re: LVDS Termination
- From: Jim G Roberts <jgroberts@xxxxxxxxxx>
- To: bmgman@xxxxxxxxxx
- Date: Fri, 11 Apr 2003 16:44:13 +0200
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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