我是VHDL语言的新手，现在我正在尝试制造一个可控制的步进电机。应该是类似this的东西（我实际上从Tina的代码中得到了这张照片）
这是我的代码：
LIBRARY ieee;
USE ieee.std_logic_1164.ALL;
USE ieee.std_logic_signed.ALL;
ENTITY state_stepper_halfstep IS
PORT(clk, plus, minus, start: IN STD_LOGIC;
q0,q1,q2,q3: OUT STD_LOGIC;
a7,b7,c7,d7,e7,f7,g7,h7: OUT std_logic);
END state_stepper_halfstep;
ARCHITECTURE arc OF state_stepper_halfstep IS
TYPE state_type IS (s0, s1, s2, s3, s4, s5, s6, s7);
SIGNAL state: state_type;
SIGNAL q: STD_LOGIC_VECTOR(3 downto 0);
SIGNAL counter: integer range 0 to 21 :=11;
SIGNAL digit: std_logic_vector (7 downto 0);
BEGIN
PROCESS (plus, minus, start, clk)
BEGIN
IF rising_edge(plus) THEN -- + Rotation value
counter <= counter+1;
IF counter > 20 THEN
counter <= 20;
END IF;
ELSIF rising_edge(minus) THEN -- - Rotation Value
counter <= counter-1;
IF counter < 2 THEN
counter <= 2;
END IF;
END IF;
IF start = '1' THEN
IF counter > 11 THEN -- When value > 0
WHILE counter > 11 LOOP -- Clockwise looping
EXIT WHEN counter = 11;
IF rising_edge(clk) THEN -- Clockwise Steps
CASE state IS
WHEN s0 => state <= s1;
WHEN s1 => state <= s2;
WHEN s2 => state <= s3;
WHEN s3 => state <= s4;
WHEN s4 => state <= s5;
WHEN s5 => state <= s6;
WHEN s6 => state <= s7;
counter <= counter-1;
WHEN s7 => state <= s0;
END CASE;
END IF;
END LOOP;
ELSIF counter < 11 THEN -- When value < 0
WHILE counter < 11 LOOP -- Counter-clockwise looping
EXIT WHEN counter = 11;
IF rising_edge(clk) THEN -- Counter-clockwise Steps
CASE state IS
WHEN s0 => state <= s7;
WHEN s7 => state <= s6;
WHEN s6 => state <= s5;
WHEN s5 => state <= s4;
WHEN s4 => state <= s3;
WHEN s3 => state <= s2;
WHEN s2 => state <= s1;
counter <= counter+1;
WHEN s1 => state <= s0;
END CASE;
END IF;
END LOOP;
ELSIF counter = 11 THEN -- When value = 0
counter <= counter;
END IF;
ELSE state <= state;
END IF;
END PROCESS;
WITH state SELECT
q <= "0001" WHEN s0,
"0011" WHEN s1,
"0010" WHEN s2,
"0110" WHEN s3,
"0100" WHEN s4,
"1100" WHEN s5,
"1000" WHEN s6,
"1001" WHEN s7;
q0 <= q(0);
q1 <= q(1);
q2 <= q(2);
q3 <= q(3);
with counter select
digit<= "11000000" when 11,
"11111001" when 12,
"10100100" when 13,
"10110000" when 14,
"10011001" when 15,
"10010010" when 16,
"10000010" when 17,
"11111000" when 18,
"10000000" when 19,
"10010000" when 20,
"01111001" when 10,
"00100100" when 9,
"00110000" when 8,
"00011001" when 7,
"00010010" when 6,
"00000010" when 5,
"01111000" when 4,
"00000000" when 3,
"00010000" when 2;
a7 <= digit(0);
b7 <= digit(1);
c7 <= digit(2);
d7 <= digit(3);
e7 <= digit(4);
f7 <= digit(5);
g7 <= digit(6);
h7 <= digit(7);
END arc;
这实际上应该如何工作：
1.我可以选择步进电机（用LED指示）与“加”和“减”开关一起旋转的次数（它可以从-9到9计数）
2. 7段的点是显示屏中显示的数字的负号
3.正值使电机顺时针旋转
4.负值（用7段中的点表示）将使电机沿逆时针方向旋转
5.每次电机完成1次旋转时，7段数应减少1位数
6.步进电机只有在启动时才会开始旋转（为“启动”输入提供高逻辑）
每当我试图模拟电路并点击“开始”时就会出现问题。蒂娜只会挂起然后“不回应”。说实话，我不确定我在代码中犯了什么错误，因为每当我尝试用我的VHDL代码“输入新宏”时，Tina都没有显示错误。
我最好的猜测是我在循环命令中犯了错误
IF start = '1' THEN
IF counter > 11 THEN -- When value > 0
WHILE counter > 11 LOOP -- Clockwise looping
EXIT WHEN counter = 11;
IF rising_edge(clk) THEN -- Clockwise Steps
CASE state IS
WHEN s0 => state <= s1;
WHEN s1 => state <= s2;
WHEN s2 => state <= s3;
WHEN s3 => state <= s4;
WHEN s4 => state <= s5;
WHEN s5 => state <= s6;
WHEN s6 => state <= s7;
counter <= counter-1;
WHEN s7 => state <= s0;
END CASE;
END IF;
END LOOP;
ELSIF counter < 11 THEN -- When value < 0
WHILE counter < 11 LOOP -- Counter-clockwise looping
EXIT WHEN counter = 11;
IF rising_edge(clk) THEN -- Counter-clockwise Steps
CASE state IS
WHEN s0 => state <= s7;
WHEN s7 => state <= s6;
WHEN s6 => state <= s5;
WHEN s5 => state <= s4;
WHEN s4 => state <= s3;
WHEN s3 => state <= s2;
WHEN s2 => state <= s1;
counter <= counter+1;
WHEN s1 => state <= s0;
END CASE;
END IF;
END LOOP;
如果有人能指出错误，我真的很高兴。提前谢谢你，祝你有个美好的一天！
答案 0 :(得分：0)
您要实现的是从计算机语言到VHDL的直接端口。 HDL不会以这种方式工作，如果你想实现这个设计，你必须改变你的范例。以下是一些提示：
IF rising_edge(clk) THEN开头，该counter <= counter;可以清楚地显示时钟流程。