键盘编码器为什么有8个状态?

时间:2014-05-06 12:48:13

标签: vhdl state-machine keypad

我需要在vhdl中编码键盘编码器。我想用4个州来做。当在columnn1扫描row1,row2,row2,row4时。根据我的说法,这样的4个州应该足够了,但我从互联网上找到的一个例子就是8个州:

key_scanner_sm : process (clk)
  begin  -- process key_scanner
    if clk'event and clk = '1' then

      if state_inc = '1' then

        -- reset scan_complete
        scan_complete <= '0';

        case key_state is

          when pulse_row_1 =>
            key_read  <= (others => '0');
            key_row   <= "0001";
            key_state <= read_row_1;

          when read_row_1 =>
            case key_col is
              when "0001" => key_read <= X"31";  -- 1
              when "0010" => key_read <= X"32";  -- 2
              when "0100" => key_read <= X"33";  -- 3
              when "1000" => key_read <= X"41";  -- A
              when others => null;
            end case;
            key_state <= pulse_row_2;

          when pulse_row_2 =>
            key_row   <= "0010";
            key_state <= read_row_2;

          when read_row_2 =>
            case key_col is
              when "0001" => key_read <= X"34";  -- 4
              when "0010" => key_read <= X"35";  -- 5
              when "0100" => key_read <= X"36";  -- 6
              when "1000" => key_read <= X"42";  -- B
              when others => null;
            end case;
            key_state <= pulse_row_3;

          when pulse_row_3 =>
            key_row   <= "0100";
            key_state <= read_row_3;

          when read_row_3 =>
            case key_col is
              when "0001" => key_read <= X"37";  -- 7
              when "0010" => key_read <= X"38";  -- 8
              when "0100" => key_read <= X"39";  -- 9
              when "1000" => key_read <= X"43";  -- C
              when others => null;
            end case;
            key_state <= pulse_row_4;

          when pulse_row_4 =>
            key_row   <= "1000";
            key_state <= read_row_4;

          when read_row_4 =>
            case key_col is
              when "0001" => key_read <= X"2A";  -- *
              when "0010" => key_read <= X"30";  -- 0
              when "0100" => key_read <= X"23";  -- #
              when "1000" => key_read <= X"44";  -- D
              when others => null;
            end case;
            key_state     <= pulse_row_1;
            scan_complete <= '1';

          when others => null;
        end case;

      end if;
    end if;
  end process key_scanner_sm;

这背后是否有充分的理由,有人知道吗?

1 个答案:

答案 0 :(得分:1)

您提供的示例需要额外的状态,因为它是作为单个case语句实现的。对key_row的分配需要一个额外的周期才能在key_col被读取之前生效。由于这是一个简单的循环扫描,因此可以通过从前一个pulse_row_n状态分配key_row的下一个值来消除read_row_n个状态