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Old 05-25-2006, 03:49 PM
Tomasz Dziecielewski
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Default Metastability question (newbie)

Hello,

Please excuse me posting a presumably lame question here, but despite
a rather thorough search I can't find answers to two bothering
questions.
1. I understand that when, say, a D flop's input changes along with
the arrival of the active clock edge, the flop is likely to go
metastable. But assuming it does, what will happen when no timing
violations occur on the next active clock edge (i.e. the flop's input
is ready and steady). Will the flop remain metastable or will its
output settle to the valid input?
2. What happens if a metastable flop's output is presented to the
following (in a chain) flop's input? Will it go metastable too? Or is
its action undefined?

Regards,
Tomasz Dziecielewski
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Old 05-25-2006, 04:34 PM
Phil Hays
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Default Re: Metastability question (newbie)

Tomasz Dziecielewski <[email protected]> wrote:

>Please excuse me posting a presumably lame question here, but despite
>a rather thorough search I can't find answers to two bothering
>questions.
>1. I understand that when, say, a D flop's input changes along with
>the arrival of the active clock edge, the flop is likely to go
>metastable. But assuming it does, what will happen when no timing
>violations occur on the next active clock edge (i.e. the flop's input
>is ready and steady). Will the flop remain metastable or will its
>output settle to the valid input?


The odds of the flop remaining metastable decrease with time in the
absence of a clock. The arrival of the next active clock edge (with
correct setup and hold time for data) will end the metastable state.

State in a flipflop is maintained by a positive feedback loop, except
during a very short time near the clock edge, when the input to the FF
is transferred to the output of the FF. If the input is stable during
this short time, output will be stable after the normal propagation
delay, regardless of the previous state of the FF.


>2. What happens if a metastable flop's output is presented to the
>following (in a chain) flop's input? Will it go metastable too? Or is
>its action undefined?


There is a low probability that the following flipflop will also go
metastable. In buffered CMOS logic (almost anything modern), meta
stability usually shows up as a slower output time. If this slower
output causes the input to the next flipflow to change at just the
wrong time, then that flipflop may go metastable.


--
Phil Hays(Xilinx)
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Old 05-25-2006, 04:46 PM
Peter Alfke
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Default Re: Metastability question (newbie)

Good qustions.
1. On the next clock, with valid and stable data on D, the flip-flop
will react normally.
2. The metastable output of a modern CMOS flip-flop is actually not at
a strange level, but it can change between 0 and 1 at an uncontrolled
time. Therefore, the cascaded flip-flop has no reason to go metastable,
unless (and this is very unlikely) its D input happens to change
exactky at the "moment of truth" where the second flip-flop is being
clocked. That moment of truth is a very tiny window, measured in
femtoseconds ( millionth of a nanosecond). That's why cascading two
flip-flops is the standard method to effectively avoid metastability.
(Avoid means: reducing its probability to a tolerably low level)
Look for the Xilinx app note XAPP094 .(Or google with my name)
Peter Alfke, Xilinx Applications

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