Re: [SI-LIST] : LVDS questions

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From: D. C. Sessions ([email protected])
Date: Tue Feb 01 2000 - 12:39:18 PST

Tom Zimmerman wrote:

> 1. My basic understanding of LVDS (which may be incorrect) is that a
> constant current of several mA is supplied to one of the output pads (pads
> are terminated in approx. 100 ohms) in order to supply a voltage
> differential across the load of several hundred mv. In order to establish a
> common mode voltage (approx. 1.2 V), the other pad must effectively be
> connected to a voltage source. Is this really the way LVDS is done?

No. LVDS requires that the difference between Vos (output common mode) and
Vod (output differential) for one and zero be kept pretty small. In other
words, signaling must be balanced.

> This
> configuration seems inherently unbalanced, since one output pad is low
> impedance, the other high impedance. It seems that this would reduce common
> mode noise rejection. Wouldn't a balanced configuration be more desirable??

Indeed, and that's why it is. Oh, and the Rs is also constrained.

> 2. How does LVDS compare to other schemes such as PECL or CML (is CML just
> essentially 2 open collectors which switch a current to one output pad or
> the other??). Is there some consensus as to a superior scheme, or does each
> scheme have its own advantages and disadvantages depending on the
> application?

CML has the extremely nice properties of

1) really, really, low differential skew (it's easy to balance the drive)
2) simple receiver (the common-mode issue is easy for current drive)

and the truly ugly property of

3) hideous power dissipation.

PECL, on the other hand, has nasty habits of its own. My unfavorite is
its reference to the positive rail, which is naturally the one that keeps
moving as supplies migrate downward.

Bothe LVDS and PECL, being push/pull, have problems with crossover noise.
Both have the additional nasty of common-mode points offset from EITHER
supply, which makes doing them in advanced (read low-voltage) processes
ugly to impossible.

> 3. Some time ago I was asked to provide low-level differential LVDS-like
> digital outputs on a chip I designed. I ended up
> simply switching the output pads between 2 equal valued on-chip resistors
> (hundreds of ohms), one connected to ground, the other to a positive low
> impedance voltage source. In other words, this scheme is like LVDS except
> it uses 2 equal valued current sources of rather low output impedance (the
> resistors). The common mode output voltage then lies at half the positive
> voltage source value. I was attracted to this scheme (perhaps naively)
> because of its inherent simplicity and its balanced nature. The users of
> this chip report that it works just fine and that they are pleased with the
> performance. I would like to know what the disadvantages of this approach
> might be (since I am considering using it again). Is this a silly thing to
> do, and if so, why? Isn't a balanced output preferable to unbalanced
> (if LVDS is indeed unbalanced)?

I've seen LVDS drivers done this way. Assuming that you can get the common-mode
point into the rather narrow allowed range I don't know of any reason why it
shouldn't work.

D. C. Sessions
[email protected]

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