0

我有一个由这个键盘接口引起的问题。我正在尝试用键盘和放大器制作数码钢琴，但是当我们按下按钮时声音没有发出；有约 1 秒的延迟。你能帮我解决这个问题吗？当我们更改代码部分时

Shift2_next <= Shift1(0) & Shift2(10 downto 1);
到
Shift2_next <= PS2Df & Shift2(10 downto 1);

按键立即发出想要的声音，但现在声音不会停止；在这种情况下，我认为中断代码已损坏。希望你能帮忙。谢谢。

library IEEE;
use IEEE.STD_LOGIC_1164.ALL;

entity keyboard_ctrl is
  port(
    clk25 : in STD_LOGIC;
    PS2C : in STD_LOGIC;
    PS2D : in STD_LOGIC;
    xkey : out STD_LOGIC_VECTOR(16 downto 1)
  );
end keyboard_ctrl;

architecture keyboard of keyboard_ctrl is
  signal PS2Cf, PS2Df: std_logic;
  signal PS2Cf_next, PS2Df_next: std_logic;
  signal ps2c_filter, ps2d_filter: std_logic_vector(7 downto 0);
  signal shift1,shift2: std_logic_vector(10 downto 0);
  signal shift1_next,shift2_next: std_logic_vector(10 downto 0);
begin

  xkey <= shift1(8 downto 1)&shift2(8 downto 1);

  -- filter for PS2 clock and data
  filter: process(clk25)
  begin
    if clk25'event and clk25 = '1' then 
        ps2c_filter(7) <= PS2C;
        ps2c_filter(6 downto 0) <= ps2c_filter(7 downto 1);
        ps2d_filter(7) <= PS2D;
        ps2d_filter(6 downto 0) <= ps2d_filter(7 downto 1);

        PS2Cf <= PS2Cf_next;
        PS2Df <= PS2Df_next;
    end if;
  end process filter;
    PS2Cf_next <= '1' when ps2c_filter = X"FF" else 
                  '0' when ps2c_filter = X"00" else
                  PS2Cf;
    PS2Df_next <= '1' when ps2d_filter = X"FF" else 
                  '0' when ps2d_filter = X"00" else
                  PS2Df;

  --Shift used to clock in scan codes from PS2--
  shift: process(PS2Cf)
  begin
    if (PS2Cf'event and PS2Cf = '0') then
        shift1 <= shift1_next;
        shift2 <= shift2_next;
    end if;
  end process shift;

  Shift1_next <= PS2Df & Shift1(10 downto 1);
  Shift2_next <= Shift1(0) & Shift2(10 downto 1);
end keyboard;
4

1 回答 1

0

使用 PS2 时，您必须专门将设计更改为同步。我建议您检查 PS2 的时钟，确保它连接到 25 MHz 引脚，或者尝试使用更高频率的时钟并将其分频，直到获得正确的时序。附上一个时钟除以3的例子，你可以改变它并使用它

最良好的祝愿，

library ieee;
    use ieee.std_logic_1164.all;
    use ieee.std_logic_unsigned.all;

entity divide_by_3 is
    port (
        cout   :out std_logic;  -- Output clock
        clk    :in  std_logic;  -- Input clock
        reset  :in  std_logic   -- Input reset
    );
end entity;

architecture rtl of divide_by_3 is
    signal pos_cnt :std_logic_vector (1 downto 0);
    signal neg_cnt :std_logic_vector (1 downto 0);
begin
    process (clk, reset) begin
        if (reset = '1') then
            pos_cnt <= (others=>'0');
        elsif (rising_edge(clk)) then
            if (pos_cnt = 2) then
                pos_cnt <= pos_cnt + 1;
            end if;
        end if;
    end process;

    process (clk, reset) begin
        if (reset = '1') then
            neg_cnt <= (others=>'0');
        elsif (falling_edge(clk)) then
            if (neg_cnt = 2) then
                neg_cnt <= neg_cnt + 1;
            end if;
        end if;
    end process;

    cout <= '1' when ((pos_cnt /= 2) and (neg_cnt /= 2)) else
            '0';
end architecture;
-------------------------------------------------------
--  Testbench to check the divide_by_3 logic
-------------------------------------------------------
library ieee;
    use ieee.std_logic_1164.all;
    use ieee.std_logic_textio.all;
    use std.textio.all;

entity div3_tb is
end entity;
architecture test of div3_tb is

    signal cout   :std_logic;
    signal clk    :std_logic := '1';
    signal reset  :std_logic := '1';

    component divide_by_3 is
    port (
        cout   :out std_logic;
        clk    :in  std_logic;
        reset  :in  std_logic
    );
    end component;
begin

    -- Generate clock
    clk   <= not clk after 10 ns;
    reset <= '0' after 20 ns;

    Inst_div3 : divide_by_3


      port map (
            cout   => cout,   -- Output
            clk    => clk,    -- Input
            reset  => reset   -- Iinput
        );
    end architecture

;
于 2014-12-29T14:00:05.757 回答