关于VHDL按键去抖如何实现

House · 发表于 2019-5-23 07:30

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关于VHDL按键去抖如何实现

      按键检测需要消抖，一般有硬件和软件两种方式。硬件就是加去抖动电路，这样从根本上解决按键抖动问题。除了用专用电路以外，用可编程FPGA或者cpld设计相应的逻辑和时序电路，对按键信号进行处理，同样可以达到去抖动的目的。
      本例中用状态机实现了消抖电路：
端口描述：clk 输入检测时钟；reset 复位信号；din 原始按键信号输入； dout 去抖动输出信号。

VHDL源码如下：
LIBRARY ieee;
USE ieee.std_logic_1164.all;
USE ieee.std_logic_unsigned.all;
ENTITY xiaod IS
PORT
(
clk    : IN STD_LOGIC ;
reset : IN STD_LOGIC ;
din : IN STD_LOGIC ;
dout : OUT STD_LOGIC
);
END ENTITY;
ARCHITECTURE RTL OF xiaod IS
TYPE state IS( s0,s1,s2,s3);
SIGNAL pre_s, next_s: state;
BEGIN

P0

ROCESS( reset, clk )
BEGIN
if reset = '0' then
pre_s <= s0;
elsif rising_edge( clk ) then
pre_s <= next_s;
else
null;
end if;
END PROCESS P0;

P1

ROCESS( pre_s, next_s, din )
BEGIN
case pre_s is

when s0 =>
   dout <= '1';
   if din = '1' then
   next_s <= s0;
   else
   next_s <= s1;
   end if;

when s1 =>
   dout <= '1';
   if din = '1' then
   next_s <= s0;
   else
   next_s <= s2;
   end if;

when s2 =>
   dout <= '1';
   if din = '1' then
   next_s <= s0;
   else
   next_s <= s3;
   end if;

when s3 =>
   dout <= '0';
   if din = '1' then
   next_s <= s0;
   else
   next_s <= s1;
   end if;
end case;

END PROCESS P1;
END RTL;
多按键去抖动电路VHDL源码，按键个数参数化，每个按键处理调用了上面的模块：

LIBRARY ieee;
USE ieee.std_logic_1164.all;
USE ieee.std_logic_arith.all;
USE ieee.std_logic_unsigned.all;
ENTITY Nxiaod IS
GENERIC( width: positive:= 5 );

PORT
(
clk    : IN STD_LOGIC ;
reset : IN STD_LOGIC ;
din : IN STD_LOGIC_VECTOR( width - 1 DOWNTO 0);
dout : OUT STD_LOGIC_VECTOR( width - 1 DOWNTO 0)
);
END ENTITY;
ARCHITECTURE RTL OF Nxiaod IS

COMPONENT xiaod IS
PORT
(
clk : IN STD_LOGIC ;
reset : IN STD_LOGIC ;
din : IN STD_LOGIC ;
dout : OUT STD_LOGIC
);
END COMPONENT;
BEGIN
g1: FOR i IN 0 to width - 1 GENERATE
ux: xiaod port map( clk => clk, reset => reset, din => din(i), dout => dout(i));
END GENERATE;
END RTL;

helendcany · 发表于 2019-5-23 16:45

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