quartus 18.0 六路抢答器 vhdl
时间: 2023-12-16 11:04:46 浏览: 77
很抱歉,根据提供的引用内容,我无法回答您的问题或提供相关介绍或演示。引用中提到了一个使用Quartus 5.0建立的抢答器,而引用中提到了一个破解过的Quartus软件。然而,这些引用都没有提供关于Quartus 18.0下六路抢答器的VHDL代码或相关信息。如果您能够提供更多的信息或引用,我将非常乐意为您提供帮助。
相关问题
quartus ii 18.0六路抢答器VHDL
抢答器是一种常见的电子设备,可以用于各种竞赛或者考试等场合。Quartus II是一款常用的数字电路设计软件,可以用于抢答器的设计。下面是一个六路抢答器的VHDL代码示例:
```vhdl
library ieee;
use ieee.std_logic_1164.all;
use ieee.numeric_std.all;
entity quiz_buzzer is
port (
clk: in std_logic;
reset: in std_logic;
buzzer: out std_logic_vector(5 downto 0);
button: in std_logic_vector(5 downto 0)
);
end entity quiz_buzzer;
architecture rtl of quiz_buzzer is
signal counter: unsigned(15 downto 0);
signal winner: std_logic_vector(5 downto 0);
begin
process(clk, reset)
begin
if reset = '1' then
counter <= (others => '0');
winner <= (others => '0');
buzzer <= (others => '0');
elsif rising_edge(clk) then
if counter = 0 then
buzzer <= (others => '0');
winner <= button;
counter <= 10000;
else
counter <= counter - 1;
if button /= winner then
buzzer <= winner;
end if;
end if;
end if;
end process;
end architecture rtl;
```
这个抢答器的设计中,有一个时钟信号clk和一个复位信号reset。当reset信号为高电平时,计数器counter、胜者winner和蜂鸣器buzzer都会被清零。当clk信号上升沿到来时,计数器counter会递减,如果计数器counter的值为0,则会将当前按下的按钮的编号存储到胜者winner中,并将计数器counter的值设为10000。在接下来的10000个时钟周期内,如果有其他的按钮按下,则会触发蜂鸣器buzzer响起,并且胜者winner的编号会一直显示在LED上,直到下一次比赛开始。
quartus实现六路抢答器的vhdl语言代码
以下是一个基于VHDL语言实现六路抢答器的代码示例:
```vhdl
-- 六路抢答器的计数器模块
library ieee;
use ieee.std_logic_1164.all;
entity counter is
port (
clk : in std_logic;
reset : in std_logic;
count : out integer range 0 to 63
);
end counter;
architecture rtl of counter is
signal cnt : integer range 0 to 63 := 0;
begin
process (clk, reset)
begin
if reset = '1' then -- 复位信号
cnt <= 0;
elsif rising_edge(clk) then -- 时钟信号
if cnt = 63 then -- 计数器达到最大值
cnt <= 0;
else
cnt <= cnt + 1;
end if;
end if;
end process;
count <= cnt; -- 输出计数器的值
end rtl;
-- 六路抢答器的主模块
library ieee;
use ieee.std_logic_1164.all;
entity sixway_arbiter is
port (
input1 : in std_logic;
input2 : in std_logic;
input3 : in std_logic;
input4 : in std_logic;
input5 : in std_logic;
input6 : in std_logic;
output : out integer range 1 to 6
);
end sixway_arbiter;
architecture rtl of sixway_arbiter is
signal cnt1 : integer range 0 to 63 := 0;
signal cnt2 : integer range 0 to 63 := 0;
signal cnt3 : integer range 0 to 63 := 0;
signal cnt4 : integer range 0 to 63 := 0;
signal cnt5 : integer range 0 to 63 := 0;
signal cnt6 : integer range 0 to 63 := 0;
signal min_count : integer range 0 to 63 := 0;
begin
-- 连接计数器模块
counter1 : entity work.counter port map (clk => clk, reset => reset, count => cnt1);
counter2 : entity work.counter port map (clk => clk, reset => reset, count => cnt2);
counter3 : entity work.counter port map (clk => clk, reset => reset, count => cnt3);
counter4 : entity work.counter port map (clk => clk, reset => reset, count => cnt4);
counter5 : entity work.counter port map (clk => clk, reset => reset, count => cnt5);
counter6 : entity work.counter port map (clk => clk, reset => reset, count => cnt6);
process (cnt1, cnt2, cnt3, cnt4, cnt5, cnt6)
begin
min_count <= cnt1;
if cnt2 < min_count then min_count <= cnt2; end if;
if cnt3 < min_count then min_count <= cnt3; end if;
if cnt4 < min_count then min_count <= cnt4; end if;
if cnt5 < min_count then min_count <= cnt5; end if;
if cnt6 < min_count then min_count <= cnt6; end if;
end process;
process (min_count)
begin
if min_count = cnt1 then output <= 1;
elsif min_count = cnt2 then output <= 2;
elsif min_count = cnt3 then output <= 3;
elsif min_count = cnt4 then output <= 4;
elsif min_count = cnt5 then output <= 5;
elsif min_count = cnt6 then output <= 6;
end if;
end process;
end rtl;
```
在这个代码示例中,首先定义了一个计数器模块,用于计算每个输入信号的响应时间。然后,定义了一个主模块,用于控制六个输入信号和一个输出信号。在主模块中,使用了六个计数器实例来计算每个输入信号的响应时间,并使用一个比较器来确定哪个信号先到达。最后,将最先到达的信号编号写入输出信号。