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Reset asynchronous assertion synchronous deassertion

Hi Eveyone,

The specifications goes something like this :

The device core asserts reset to the device peripherals asynchronously and releases (deasserts) the reset synchronously after 4 clock periods

there are two possible implementations for the above spec which one is better :

signal reset_reg : std_logic_vector(3 downto 0);

p_reset_reg : process(clk,reset_async) begin if (reset_async = '0') then -- on async reset assertion reset the registers reset_reg '0');

elsif (clk'event and clk = '1') then

reset_reg(0)

Reply to
arant
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Hi -

I don't speak VHDL, so let's see if I've got this right:

You want to create an active-LOW reset signal that is asserted asynchronously (presumably right after the part configures) and deasserts synchronous to some clock.

To do this, you created a 4-bit shift register, all of whose bits are reset LOW by some global asynchronous reset. The D input of the shift register input is HIGH; when the asynch reset is deasserted, each successive clock cycle propagates the HIGH to the next stage.

You've proposed two ways of obtaining your active-LOW, asynch-assert-synch-deassert reset from this shift register:

1) use the Q output of the last shift register FF

or

2) AND the Q outputs of each of the 4 shift register FFs.

You claim that (2) reduces the odds of a metastable propagating to the reset signal.

I disagree: (2) is no better than (1). Here's why.

Immediately after the deassertion of the asynch reset, only one of the four shift register FFs can go metastable, namely the first one. That's the only flip-flop in which the next desired state, HIGH, is different from the current LOW state; the other 3 FFs have a LOW on their D inputs. Metastability arises from trying to change a FF's state with insufficient energy; if there's no possibility of a state change, there's no metastability. You will not see all four FFs, or even three or two, go metastable at once.

If the first FF goes metastable, that metastable state could propagate to the second FF the next clock cycle, but if there's plenty of settling time between the first and second FFs, the odds are low (I'm going to skip the issue of just how low). And the odds of propagating to the third FF are even lower, and so on. Unless you're running this reset circuit at a very high clock speed or haven't guaranteed lots of slack between stages, using the output of the fourth FF as a reset should be fine.

ANDing the four FF outputs buys you nothing. If, against all odds, the metastable manages to propagate to the output of the last FF, ANDing that Q with three HIGH signals does nothing to stop the metastable from propagating through the AND gate. And if the metastable doesn't make it to the last FF, the AND gate is again superfluous.

Bob Perlman Cambrian Design Works

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Reply to
Bob Perlman

Hi,

arant schrieb:

[..]

No, it is invalid. Metastability _may_ affect the output of any gate regardless of the other inputs. On the other side has option 2 a more stable reset release in case of "bouncing" imagine your reset has a low ripple and goes clocked in as

00010110111111 Option 2 waits until all four registers are stable, option 1 may lead to a problem if theres a "weak" '0' which resets only part of the FF but not the shift register properly (e.g. due to metastability). It is very unlikely to have such a problem, but I would prefer a mixture of both: AND at least two FF to release Reset and don't use the first FF or first two FF for reset release to be safe against metastability.

bye Thomas

Reply to
Thomas Stanka

Hi Thomas,

First of all thanks for taking out time and replying to my query...

Seconly what is the meaning of a "weak" '0'' as the flops share a common supply rail ?

If you mean the metastability due to reset removal then I can agree that the shift register will enter an unknown random pattern after the metastability resolution time has expired and the reset to the device will be deasserted only when all the flops reach a '1' state.

In the event that any of the register bits settle to '1' then the specs fail in both the implementations for reset release time of four clock periods.

Is there any way that still meets the specs even if there is metastability ...

Thomas Stanka wrote:

Reply to
arant

That is what I am not so sure of. If that is true, then I agree with your assessment. But I am pretty sure the way gates are designed, if one input is in a state that assures the output condition, the other inputs will not have an effect on the output even if they are not within the proper digital voltage range. However, if this is a LUT in and FPGA, I am not as certain. I believe they use pass transistors to connect the output of memory elements to the output of the LUT. No matter how they decode the inputs to connect the memory element to the output, I can see a single metastable input possibly making the LUT output invalid.

So it depends on whether a LUT is used or a standard gate.

Not a bad idea!

Reply to
rickman

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