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pupillo 09-13-2010 11:13 AM

process with two clks in sensitivity list problem
 
Hi,
in the code below there are two 2 bit free running counters (cnta and
cntb). The o output is generated inside oprc process sampling
alternatively cnta on the rising edge of clk and cntb on the rising
edge of clkn (that is not clk).

The output o1 is generated in the same way, except that it uses a copy
of clk (lck1) and a copy of clkn (clk1n).

The poroblem is: o and o1 are different. WHY?

This code comes from an implementation of a DDR IO pad mux from an
FPGA manufacturer.
In my opinion the problem has something to do with the fact that those
processes use a sensitivity list with two clocks.

Thanks
Pupillo
------------------------------------------------------------------------------------------------------------------------
---------------- ENTITY myff
------------------------------------------------------------------------------------------
------------------------------------------------------------------------------------------------------------------------
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use ieee.std_logic_arith.all;
use ieee.std_logic_unsigned.all;

entity myff is
Port ( clk : in STD_LOGIC;
rst : in STD_LOGIC;
o,o1: out STD_LOGIC_VECTOR (1 downto 0));
end myff;

architecture Behavioral of myff is
signal clkn,clk1,clkn1:std_logic;
signal cnta,cntb:std_logic_vector(1 downto 0);
begin


clkn<=not clk;
clk1<=clk;
clkn1<=clkn;


---- counter cnta ----------
cntaprc:process(clk) begin
if (rising_edge(clk)) then
if (rst='1') then cnta<="00";
else cnta<=cnta+1;
end if;
end if;
end process;

---- counter cntb -------------
cntbprc:process(clk) begin
if (rising_edge(clk)) then
if (rst='1') then cntb<="10";
else cntb<=cntb+1;
end if;
end if;
end process;


----- DDR sampling of cnta / cntb using clk / clkn ---------
oprc:process(clk,clkn) begin
if (rising_edge(clk)) then
o<=cnta;
end if;

if (rising_edge(clkn)) then
o<=cntb;
end if;
end process;


----- DDR sampling of cnta / cntb using clk1 / clkn1 ---------
o1prc:process(clk1,clkn1) begin
if (rising_edge(clk1)) then
o1<=cnta;
end if;

if (rising_edge(clkn1)) then
o1<=cntb;
end if;
end process;

end Behavioral;










------------------------------------------------------------------------------------------------------------------------
----------------------------------- TEST BENCH
-------------------------------------------------------------------------
------------------------------------------------------------------------------------------------------------------------

LIBRARY ieee;
USE ieee.std_logic_1164.ALL;

ENTITY myff_TB IS
END myff_TB;

ARCHITECTURE behavior OF myff_TB IS


COMPONENT myff
PORT(
clk : IN std_logic;
rst : IN std_logic;
o,o1 : OUT std_logic_vector(1 downto 0)
);
END COMPONENT;


--Inputs
signal clk : std_logic := '0';
signal rst : std_logic := '0';

--Outputs
signal o,o1: std_logic_vector(1 downto 0);

-- Clock period definitions
constant clk_period : time := 10 ns;

BEGIN

uut: myff PORT MAP (
clk => clk,
rst => rst,
o => o,
o1 => o1
);

-- Clock process definitions
clk_process :process
begin
clk <= '0';
wait for clk_period/2;
clk <= '1';
wait for clk_period/2;
end process;


-- Stimulus process
stim_proc: process
begin
-- hold reset state for 100 ns.
rst<='1';
wait for 100 ns;
rst<='0';
wait;
end process;

END;




Andy 09-13-2010 03:44 PM

Re: process with two clks in sensitivity list problem
 
Your assignments between clocks consume a delta delay cycle. While
their edges coincide in simulation time, they do not coincide in
execution time, and therefore you get different results. If data is
clocked out by the un-delayed clock, then clocked in by the delayed
clock, the date will ripple through both registers in one clock cycle,
which is not what you expected (or what the synthesized hardware will
do).

If the clock gets delta delayed, the input data needs to be delta
delayed too. But if the input data is delta delayed, the clock does
not need to be delta delayed, since the next clock edge will occur
long after delta delays have passed.

You can also use rising_edge() and falling_edge() to avoid having to
invert the clocks.

Andy

pupillo 09-13-2010 06:36 PM

Re: process with two clks in sensitivity list problem
 
On 13 Set, 17:44, Andy <jonesa...@comcast.net> wrote:
> Your assignments between clocks consume a delta delay cycle. While
> their edges coincide in simulation time, they do not coincide in
> execution time, and therefore you get different results. If data is
> clocked out by the un-delayed clock, then clocked in by the delayed
> clock, the date will ripple through both registers in one clock cycle,
> which is not what you expected (or what the synthesized hardware will
> do).
>
> If the clock gets delta delayed, the input data needs to be delta
> delayed too. But if the input data is delta delayed, the clock does
> not need to be delta delayed, since the next clock edge will occur
> long after delta delays have passed.
>
> You can also use rising_edge() and falling_edge() to avoid having to
> invert the clocks.
>
> Andy


Ok very helpfull!
Thanks


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