Question

我有应该执行以下操作的VHDL代码：当按下KEY（0）（RESET）时，下一个上升时钟将启动该过程（50MHz）。它设置status_flag，并有一个新进程查看status_flag，并且每1000000（+ const）clk个周期，将更新一个称为DAC的值。我想随着clk-cntr的更新，它需要几个时钟周期，因此是常数。（我使用了数据记录器，可以看到〜20.02ms）在第二个过程的底部，将clk-cntr重置为零。目标是在按下KEY（0）之后经历第二个过程，并等待下一次按下KEY。可以看到，我把status_flag注释掉了，因为编译器的响应是“无法解析多个常量驱动程序”。如何重置status_flag或类似代码以使代码等待KEY（0）？我正在使用实时响应，而不是模拟。

  -- ---------------------------------------------------------------------
    -- Global signals ------------------------------------------------------
    -- ---------------------------------------------------------------------
    CLK   : in std_logic;
    RESET : in std_logic;


);

结束实体test_top;

test_top的架构rtl是

 shared variable status_flag  : std_logic;
 signal clk_cntr        : unsigned(31 downto 0);
 signal DAC             : std_logic_vector(11 downto 0);

开始

DAC_Out_Rising_Edge: process(CLK)
begin 

if rising_edge(CLK) then
    if RESET = '1' then -- KEY(0) switch
            status_flag := '1'; -- The encoder is triggered on the rising edge of the clock
    end if;
end if;

end process;

    Servo_routine: process(CLK)
    begin

    if rising_edge(CLK) then -- 
        if (status_flag = '1') then
            clk_cntr <= clk_cntr + 1;

            if clk_cntr = 4 then 
                DAC <= "000000000000"; -- initialize value
            end if;
            if clk_cntr = 1000000 then
                DAC <= "000000000010";
            end if;
            if clk_cntr = 2000000 then
                DAC <= "000000000100";
            end if;
            if clk_cntr = 3000000 then
                DAC <= "000000001000";
            end if;
            if clk_cntr = 4000000 then
                DAC <= "000000010000";
            end if;
            if clk_cntr = 5000000 then
                DAC <= "000000100000";
            end if;
            if clk_cntr = 6000000 then
                DAC <= "000001000000";
            end if;
            if clk_cntr = 7000000 then
                DAC <= "000010000000";
            end if;
            if clk_cntr = 8000000 then
                DAC <= "000100000000";
            end if;
            if clk_cntr = 9000000 then
                DAC <= "001000000000";
            end if;
            if clk_cntr = 1000000 then
                DAC <= "000100000000";
            end if;
            if clk_cntr = 1100000 then
                DAC <= "000010000000";
            end if;
            if clk_cntr = 1200000 then
                DAC <= "000001000000";
            end if;
            if clk_cntr = 1300000 then
                DAC <= "000000100000";
            end if;
            if clk_cntr = 14000000 then
                DAC <= "000000010000";
            end if;
            if clk_cntr = 15000000 then
                DAC <= "000000001000";
            end if;
            if clk_cntr = 16000000 then
                DAC <= "000000000100";
            end if;
            if clk_cntr = 17000000 then
                DAC <= "000000000010";
            end if;
            if clk_cntr = 18000000 then
                DAC <= "000000000000";
            end if;

          if clk_cntr > 18000000 then
                DAC <= "000000000000";                  -- resets flags/data
                clk_cntr <= (others => '0');            -- resets flags/data
                if RESET = '0' then
                    --status_flag := '0'; -- The encoder is reset
                end if;
            end if;

        end if;  

    end if;

    end process;

Answer 1

“多个常量驱动程序” 错误是SE中经常出现的问题。如果将其放在搜索框中，则会得到175个答案！

它们全部都用相同的解决方案完成：将所有分配移到一个过程中。

if RESET = '1' then 
    status_flag := '1'; // start 
else
    if clk_cntr > 18000000 then
        status_flag := '0'; // stop 
    end if;

如果结束，

我的VHDL标志需要终止

1 个答案: