vhdl按键消抖
时间: 2023-11-27 17:04:47 浏览: 52
在VHDL中,按键消抖可以通过状态机来实现。以下是一个简单的按键消抖状态机的示例代码:
```vhdl
library ieee;
use ieee.std_logic_1164.all;
entity debounce is
port(
clk: in std_logic;
button: in std_logic;
debounced_button: out std_logic
);
end entity debounce;
architecture rtl of debounce is
type debounce_state is (IDLE, PRESS_DETECTED, RELEASE_DETECTED, DEBOUNCING);
signal state: debounce_state := IDLE;
signal counter: integer range 0 to 10 := 0;
begin
process(clk)
begin
if rising_edge(clk) then
case state is
when IDLE =>
if button = '0' then
state <= PRESS_DETECTED;
else
debounced_button <= '1';
end if;
when PRESS_DETECTED =>
if button = '1' then
state <= IDLE;
debounced_button <= '0';
elsif counter = 9 then
state <= DEBOUNCING;
counter <= 0;
else
counter <= counter + 1;
end if;
when RELEASE_DETECTED =>
if button = '0' then
state <= IDLE;
debounced_button <= '1';
elsif counter = 9 then
state <= DEBOUNCING;
counter <= 0;
else
counter <= counter + 1;
end if;
when DEBOUNCING =>
if button = '1' then
state <= RELEASE_DETECTED;
counter <= 0;
elsif button = '0' and counter = 9 then
state <= PRESS_DETECTED;
counter <= 0;
else
counter <= counter + 1;
end if;
end case;
end if;
end process;
end architecture rtl;
```
在这个状态机中,有四个状态:IDLE、PRESS_DETECTED、RELEASE_DETECTED和DEBOUNCING。当按键被按下时,会从IDLE状态转移到PRESS_DETECTED状态,并且计数器开始计数。如果在计数器计数过程中,按键被松开,就会回到IDLE状态;如果计数器计数到10,就会进入DEBOUNCING状态。在DEBOUNCING状态中,如果按键被松开,就会进入RELEASE_DETECTED状态;如果按键被按下,并且计数器计数到10,就会回到PRESS_DETECTED状态。在PRESS_DETECTED和RELEASE_DETECTED状态中,如果按键被按下或松开,就会回到IDLE状态,并且输出稳定的debounced_button信号。