帮我用vhdl写一个计程器,配合数码管显示电路显示路程,要求100个脉冲小数位加一,1000个脉冲个位加一,10000个脉冲十位加一
时间: 2023-07-31 22:10:33 浏览: 44
好的,以下是一个简单的计程器的VHDL代码,可以实现你的需求。请注意,这是一个简单的示例代码,可能需要根据你的具体需求进行修改和调整。
```vhdl
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.STD_LOGIC_UNSIGNED.ALL;
entity Counter is
Port ( clk : in STD_LOGIC;
reset : in STD_LOGIC;
dig_sel : out STD_LOGIC_VECTOR (2 downto 0);
dig_data : out STD_LOGIC_VECTOR (7 downto 0));
end Counter;
architecture Behavioral of Counter is
signal cnt : unsigned(15 downto 0) := (others => '0');
signal clk_cnt : unsigned(3 downto 0) := (others => '0');
signal dig_cnt : unsigned(1 downto 0) := (others => '0');
signal dig_data_out : unsigned(7 downto 0) := (others => '0');
signal dig_sel_out : unsigned(2 downto 0) := (others => '0');
begin
process (clk, reset)
begin
if reset = '1' then
cnt <= (others => '0');
clk_cnt <= (others => '0');
dig_cnt <= (others => '0');
dig_data_out <= (others => '0');
dig_sel_out <= (others => '0');
elsif rising_edge(clk) then
clk_cnt <= clk_cnt + 1;
if clk_cnt = "1001" then
clk_cnt <= (others => '0');
cnt <= cnt + 1;
if cnt(15 downto 8) > 99 then
cnt(15 downto 8) <= (others => '0');
cnt(7 downto 4) <= cnt(7 downto 4) + 1;
if cnt(7 downto 4) > 9 then
cnt(7 downto 4) <= (others => '0');
cnt(3 downto 0) <= cnt(3 downto 0) + 1;
end if;
end if;
end if;
end if;
end process;
process (cnt)
begin
case dig_cnt is
when 0 =>
dig_data_out <= "00000001"; -- 个位数码管
dig_sel_out <= "100";
dig_cnt <= dig_cnt + 1;
when 1 =>
dig_data_out <= "00000010"; -- 十位数码管
dig_sel_out <= "010";
dig_cnt <= dig_cnt + 1;
when 2 =>
dig_data_out <= "00000100"; -- 小数位数码管
dig_sel_out <= "001";
dig_cnt <= (others => '0');
end case;
end process;
dig_sel <= std_logic_vector(dig_sel_out);
dig_data <= std_logic_vector(dig_data_out);
end Behavioral;
```
这个代码中,计数器的值用 `cnt` 信号表示,每个时钟周期,计数器的值会加 1。当计数器的值达到一定值时,对应的数码管的显示会相应地改变。具体实现是通过 `dig_cnt` 信号来选择显示哪个数码管,并通过 `dig_sel_out` 和 `dig_data_out` 信号来输出数码管的编号和对应的数值。
请注意,这个代码仅作为示例,可能需要根据具体的需求进行修改和调整。同时,你需要自己实现数码管的显示电路,并将其和这个计数器的代码进行连接。