Dual Edge

R

R Quijano

Guest
Hi

You can use my code for a Dual Edge D flipflop which synthetize fine
with ISE form Xilinx

library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.STD_LOGIC_ARITH.ALL;
use IEEE.STD_LOGIC_UNSIGNED.ALL;

entity LatchDD is
Port (D,CLK,R : in STD_LOGIC;
Q,QN : out STD_LOGIC);
end LatchDD;

architecture Behavioral of LatchDD is
signal Q1,Q2 : std_logic;
begin
-- Positive Edge ---
process(D,clk,R)
begin
if R='1' then Q2<='0';
elsif(rising_edge(clk)) then Q2<=D;
else null; end if;
end process;
-- Negative Edge ---
process(D,clk,R)
begin
if R='1' then Q1<='0';
elsif(falling_edge(clk)) then Q1<=D;
else null; end if;
end process;

Q <= (Q1 or Q2);
QN <= not(Q1 or Q2);

end Behavioral;

This one checks, both edges. In case you can use the clk'event the
code will be

library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.STD_LOGIC_ARITH.ALL;
use IEEE.STD_LOGIC_UNSIGNED.ALL;

entity LatchDD is
Port (D,CLK,R : in STD_LOGIC;
Q,QN : out STD_LOGIC);
end LatchDD;

architecture Behavioral of LatchDD is
signal Q1,Q2 : std_logic;
begin
-- Positive Edge ---
process(D,clk,R)
begin
if R='1' then Q2<='0';
elsif(clk'event and clk='1') then Q2<=D;
else null; end if;
end process;
-- Negative Edge ---
process(D,clk,R)
begin
if R='1' then Q1<='0';
elsif(clk'event and clk='0') then Q1<=D;
else null; end if;
end process;

Q <= (Q1 or Q2);
QN <= not(Q1 or Q2);

end Behavioral;


Have a nice day and hope it help you.
 
R Quijano schrieb:

You can use my code for a Dual Edge D flipflop which synthetize fine
with ISE form Xilinx
Click to expand...
....

use IEEE.STD_LOGIC_ARITH.ALL;
use IEEE.STD_LOGIC_UNSIGNED.ALL;
Click to expand...
Don't use these libraries. They are not standard! Furthermore you don't
need them for this task.


entity LatchDD is
Port (D,CLK,R : in STD_LOGIC;
Q,QN : out STD_LOGIC);
end LatchDD;

architecture Behavioral of LatchDD is
signal Q1,Q2 : std_logic;
begin
-- Positive Edge ---
process(D,clk,R)
begin
if R='1' then Q2<='0';
elsif(rising_edge(clk)) then Q2<=D;
else null; end if;
end process;
-- Negative Edge ---
process(D,clk,R)
begin
if R='1' then Q1<='0';
elsif(falling_edge(clk)) then Q1<=D;
else null; end if;
end process;

Q <= (Q1 or Q2);
QN <= not(Q1 or Q2);

end Behavioral;
Click to expand...
Uhmm ... that is not a dual-edge D-Flipflop.

If D='1' and the rising_edge(clk) arrives, the output will go '1'. Fine.
If then D='0' and the falling_edge(clk) arrives, then Q1='0' but the
output stays at '1' till the next rising_edge(clk). I would expect, that
the output should fall to '0'.

If you need a working pseudo dual-edge D-flipflop, have a look at:
<http://www.ralf-hildebrandt.de/publication/pdf_dff/pde_dff.pdf>.
The outputs have to be XORed and the flipflops cross-coupled.

Ralf
 
On Feb 20, 9:34 am, Ralf Hildebrandt <Ralf-Hildebra...@gmx.de> wrote:
R Quijano schrieb:

You can use my code for a Dual Edge D flipflop which synthetize fine
with ISE form Xilinx

...

use IEEE.STD_LOGIC_ARITH.ALL;
use IEEE.STD_LOGIC_UNSIGNED.ALL;

Don't use these libraries. They are not standard! Furthermore you don't
need them for this task.



entity LatchDD is
Port (D,CLK,R : in STD_LOGIC;
Q,QN : out STD_LOGIC);
end LatchDD;

architecture Behavioral of LatchDD is
signal Q1,Q2 : std_logic;
begin
-- Positive Edge ---
process(D,clk,R)
begin
if R='1' then Q2<='0';
elsif(rising_edge(clk)) then Q2<=D;
else null; end if;
end process;
-- Negative Edge ---
process(D,clk,R)
begin
if R='1' then Q1<='0';
elsif(falling_edge(clk)) then Q1<=D;
else null; end if;
end process;

Q <= (Q1 or Q2);
QN <= not(Q1 or Q2);

end Behavioral;

Uhmm ... that is not a dual-edge D-Flipflop.

If D='1' and the rising_edge(clk) arrives, the output will go '1'. Fine.
If then D='0' and the falling_edge(clk) arrives, then Q1='0' but the
output stays at '1' till the next rising_edge(clk). I would expect, that
the output should fall to '0'.

If you need a working pseudo dual-edge D-flipflop, have a look at:
http://www.ralf-hildebrandt.de/publication/pdf_dff/pde_dff.pdf>.
The outputs have to be XORed and the flipflops cross-coupled.

Ralf
Click to expand...
Ralf is correct, you need XOR (or XNOR) encoding and decoding to get
true dual edge flip flop behavior. These functions have the unique
feature that any input can control the state of the output, if the
other inputs are known. It is also expandable to more than two clocks/
edges (think parity).

The folowing code is synthesizable in quartus and synplify, but not
XST (two clock edges), and Precision has/had a bug that does not
implement one of the registers correctly from the variable.

process (rst, clk) is
variable qr, qf : std_logic;
begin
if rst = '1' then
qr := '0';
qf := '0';
elsif rising_edge(clk) then
qr := d xor qf;
elsif falling_edge(clk) then
qf := d xor qr;
end if;
q <= qr xor qf; -- combo xor of reg'd qr & qf
end process;

Andy
 
Andy schrieb:


The folowing code is synthesizable in quartus and synplify, but not
XST (two clock edges), and Precision has/had a bug that does not
implement one of the registers correctly from the variable.

process (rst, clk) is
variable qr, qf : std_logic;
begin
if rst = '1' then
qr := '0';
qf := '0';
elsif rising_edge(clk) then
qr := d xor qf;
elsif falling_edge(clk) then
qf := d xor qr;
end if;
q <= qr xor qf; -- combo xor of reg'd qr & qf
end process;
Click to expand...
Very few synthesis tool support this style. Just have a look at my link
and code it as it can be seen there in the schematic. The approach of R
Quijan using 2 processes (one for each edge) will lead you to the right way.

Ralf
 
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