Synchronizer Flip Flop / Metastability

To exchange flags (1 bit signals) between two unrelated clocks, a single 
synchronizer flip flop to clock that signal is used normally. However, under 
violations of setup/hold times of the flip flop, metastability can occur.

On a Virtex2 Pro, does metastability occur often? Does adding a second (or 
even third) flip flop after the synchronizer flip flop help or is that 
overcautious?

Best Regards,
Simon Heinzle

On Fri, 23 Sep 2005 13:37:11 +0200, Simon Heinzle wrote:

> To exchange flags (1 bit signals) between two unrelated clocks, a single 
> synchronizer flip flop to clock that signal is used normally. However, under 
> violations of setup/hold times of the flip flop, metastability can occur.
> 
> On a Virtex2 Pro, does metastability occur often? Does adding a second (or 
> even third) flip flop after the synchronizer flip flop help or is that 
> overcautious?
> 
> Best Regards,
> Simon Heinzle

Standard practice is two flip flops not one. It's been decades since I
looked at the math but as I recall there is a square law involved, i.e.
the probability of a double synchronizer failing is the square of a
probability of a single synchronizer failing. Someone who has looked at
this more recently then me might say that my math is wrong but in any
event there is no question that a double synchronizer is significantly
better than a single synchronizer.

Simon Heinzle wrote:
> To exchange flags (1 bit signals) between two unrelated clocks, a single
> synchronizer flip flop to clock that signal is used normally. However, under
> violations of setup/hold times of the flip flop, metastability can occur.
>
> On a Virtex2 Pro, does metastability occur often? Does adding a second (or
> even third) flip flop after the synchronizer flip flop help or is that
> overcautious?
>
> Best Regards,
> Simon Heinzle

There have been many good threads on this subject.  You may want
to Google this group for "metastability."  That being said, the
answer is of course "it depends."  The number of stages required
depends on your clock rate, the metastability characteristics of
the flip-flops, and your required failure tolerance.

Generally two stages are used, however it is possible to get away
with one if the signal is only used at the next edge of the clock
(which is almost like two stages except the second stage may have
less tolerance for metastability if there is a LUT between the two
stages).  In any case, make sure the input is synchronized only
once to avoid possible logic errors.  This may seem obvious, but
if your "single-stage" synchronizer output has many loads the
tools may duplicate the flip-flop for you unless you specifically
tell them not to.  In the case of a two-stage synchronizer, only
the second flip-flop might be duplicated as the fist stage has just
one load.  Also the routing delay from first to second stage can
become important at higher frequencies.

Just my 2 cents
Gabor

"Simon Heinzle" <sheinzle@student.ethz.ch> wrote:

>To exchange flags (1 bit signals) between two unrelated clocks, a single 
>synchronizer flip flop to clock that signal is used normally. However, under 
>violations of setup/hold times of the flip flop, metastability can occur.
>
>On a Virtex2 Pro, does metastability occur often? Does adding a second (or 
>even third) flip flop after the synchronizer flip flop help or is that 
>overcautious?

Xilinx has a application note on this general subject:

http://www.xilinx.com/bvdocs/appnotes/xapp094.pdf

While adding multiple stages can reduce the risk of a metastable
failure, it is not the only way to do so.  Both the odds of a
metastable failure and the consequences of that failure are important.

The odds can be improved by increasing the settling time.  See figure
two in the application note.  To make sure there is enough settling
time, typical designs might both put an additional timing constraint
between the synchronizer FF and the next FF or FFs, and put physical
constraint (LOC or RLOC) on these FFs to position them for minimal
delay.

The consequences can sometimes be improved by making the rest of the
design more robust.  Examples might be to design the logic to recover
from a metastable failure, to make the software check the data and
ignore/correct invalid values, etc.


-- 
Phil Hays to reply solve: phil_hays at not(coldmail) dot com  
 If not cold then hot

Simon,
Firstly, congratulations on having parents who can spell. Secondly, you may
be interested in this link:-
http://www.fpga-faq.org/FAQ_Pages/0017_Tell_me_about_metastables.htm
Lots of good stuff, complete with a link to this:-
http://www.fpga-faq.org/archives/59375.html#59399
a simple and reliable circuit posted by Rick Collins to transfer a flag from
one domain to another.
Cheers, Symon.

p.s. Isn't Google ads great? At the bottom of the above linked page, I saw
an advertising link for beach footwear! Guess which type....

"Simon Heinzle" <sheinzle@student.ethz.ch> wrote in message
news:4333e8e2$1@news1.ethz.ch...
> To exchange flags (1 bit signals) between two unrelated clocks, a single
> synchronizer flip flop to clock that signal is used normally. However,
under
> violations of setup/hold times of the flip flop, metastability can occur.
>
> On a Virtex2 Pro, does metastability occur often? Does adding a second (or
> even third) flip flop after the synchronizer flip flop help or is that
> overcautious?
>
> Best Regards,
> Simon Heinzle
>
>
>

Simon,
metastability is a problem that cannot be "solved", we can only reduce
the probability of errors due to metastability. The extra delay at the
Q output of a flip-flop (or latch) with undefined timing relationship
between D and Clk is theoretically unbounded.
This can be a surprise to digital designers who are accustomed to
deterministic behavior. Metastability is not deterministic, it is a
statistical phenomenon.
The good news is that modern CMOS flip-flops and latches recover very
fast, within a few ns, as shown in the Xilinx app note XAPP094, which
is based on actual measurements, not on theory.

In many cases you will find that the mean-time-betwen-failure is
millions or billions of years.
Just make sure that the data path from the metastable-going flip-flop
to the (single!) next synchronizing flip-flop is as fast as possible.
No extra logic, absolutely minimal routing delays.
Peter Alfke, Xilinx Applications

Thanks a lot guys!


"Simon Heinzle" <sheinzle@student.ethz.ch> wrote in message 
news:4333e8e2$1@news1.ethz.ch...
> To exchange flags (1 bit signals) between two unrelated clocks, a single 
> synchronizer flip flop to clock that signal is used normally. However, 
> under violations of setup/hold times of the flip flop, metastability can 
> occur.
>
> On a Virtex2 Pro, does metastability occur often? Does adding a second (or 
> even third) flip flop after the synchronizer flip flop help or is that 
> overcautious?
>
> Best Regards,
> Simon Heinzle
>
>
> 

Peter Alfke wrote:
> metastability is a problem that cannot be "solved", we can only
> reduce the probability of errors due to metastability.

This certainly depends on what you mean by "solved".  Even
machines intended to be completely deterministic are built
of components which have various wear and statistical failure
mechanisms.  However, engineers usually consider the problem
solved if the MTBF of the conglomeration of gears, relays, tubes,
CMOS transistors, etc. is longer than the expected operational
life of the widget by sufficient orders of magnitude.  In that
sense, one could easily consider the metastability problem
solved when the probability of the register train not resolving
is far less than that of the device getting vaporized in a
direct meteor strike.  In which case even a CMOS NAND gate would
no longer produce the correct voltage result.

> In many cases you will find that the mean-time-betwen-failure is
> millions or billions of years.

Which means the power supply, if not electromigration, background
radiation, bonding wires, or even a meteor strike, etc., would far
more likely cause any instance of the "solved" FPGA solution to
fail even earlier.

IMHO. YMMV.
-- 
Ron
rhn A.T nicholson d.O.t C-o-M

Your wear-out analogy does not apply. Metastability failure is
completely statistical and probabilistic. Even when the MTBF is a
million years, the failure can occur in the next second. Not likely,
but possible.
That's why I claim that the problem can never be solved. We can only
reduce the probability down to an acceptable level.
I really believe that this is an important distinction vs any failure
besed on wear-out.
Peter Alfke

As has been said before, it's normally safe to make sure the MTBF is (say)
double the length of time you wanna be at your present company.
Cheers, Syms.