【发布时间】:2021-12-31 14:50:14
【问题描述】:
我正在做一个 BCD 计数器,它可以根据输入信号向上/向下计数。这是要求:
这是我的 VHDL 代码:
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.numeric_std.ALL;
use IEEE.STD_LOGIC_ARITH.ALL;
use IEEE.STD_LOGIC_UNSIGNED.ALL;
-- main
entity BCDcounter is
port(
D_in: in std_logic_vector(3 downto 0);
enable_in, load_in, up_in, clr_in, clk_50hz: in std_logic;
C_out: out std_logic;
LED0: out std_logic_vector(0 to 6)
);
end BCDcounter;
architecture Behavioral of BCDcounter is
signal Q_temp: std_logic_vector(3 downto 0);
signal clk_1hz: std_logic;
component Clock_Divider is
port ( clk,reset: in std_logic;
clock_out: out std_logic);
end component;
component BCD_counter is
port(
D: in std_logic_vector(3 downto 0);
enable, load, up, clr, clk: in std_logic;
Q: std_logic_vector(3 downto 0);
Cout: out std_logic
);
end component;
component led IS
PORT ( input : IN STD_LOGIC_VECTOR(3 downto 0);
output : OUT STD_LOGIC_VECTOR(6 downto 0));
end component;
begin
stage0: Clock_Divider port map(clk_50hz, clr_in, clk_1hz);
stage1: BCD_counter port map(D_in, enable_in, load_in, up_in, clr_in, clk_1hz, Q_temp, C_out);
stage2: led port map(Q_temp, LED0);
end Behavioral;
-- 1-digit BCD counter
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.numeric_std.ALL;
use IEEE.STD_LOGIC_ARITH.ALL;
use IEEE.STD_LOGIC_UNSIGNED.ALL;
entity BCD_counter is
port(
D: in std_logic_vector(3 downto 0);
enable, load, up, clr, clk: in std_logic;
Q: std_logic_vector(3 downto 0);
Cout: out std_logic
);
end BCD_counter;
architecture bhv of BCDcounter is
signal temp: std_logic_vector(3 downto 0);
begin
process(enable, load, up, clr, clk)
begin
if clr = '0' then
temp <= "0000";
elsif enable = '0' then
temp <= "0000";
elsif load = '1' then -- load = 1, enable = 1
temp <= D;
elsif(rising_edge(clk)) then -- load = 0, enable = 1
if up = '1' then -- count up
if temp = "1001" then
temp <= "0000";
Cout <= '1';
else
temp <= temp + 1;
end if;
else -- count down
if temp = "0000" then
temp <= "1001";
Cout <= '1';
else
temp <= temp - 1;
end if;
end if;
end if;
end process;
Q <= temp;
end bhv;
-- Clock Divider from 50MHz to 1Hz
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.numeric_std.ALL;
entity Clock_Divider is
port ( clk,reset: in std_logic;
clock_out: out std_logic);
end Clock_Divider;
architecture behavioral of Clock_Divider is
signal count: integer:=1;
signal tmp : std_logic := '0';
begin
process(clk,reset)
begin
if(reset='1') then
count <= 1;
tmp <= '0';
elsif(clk'event and clk='1') then
count <= count+1;
if (count = 25000000) then
tmp <= NOT tmp;
count <= 1;
end if;
end if;
clock_out <= tmp;
end process;
end behavioral;
-- LED 7 segments
LIBRARY ieee;
USE ieee.std_logic_1164.all;
ENTITY led IS
PORT ( input : IN STD_LOGIC_VECTOR(3 downto 0);
output : OUT STD_LOGIC_VECTOR(6 downto 0));
END led;
ARCHITECTURE behave OF led IS
BEGIN
PROCESS(input)
BEGIN
CASE input IS -- abcdefg
WHEN "0000" => output <= "0000001";
WHEN "0001" => output <= "1001111";
WHEN "0010" => output <= "0010010";
WHEN "0011" => output <= "0000110";
WHEN "0100" => output <= "1001100";
WHEN "0101" => output <= "0100100";
WHEN "0110" => output <= "0100000";
WHEN "0111" => output <= "0001111";
WHEN "1000" => output <= "0000000";
WHEN "1001" => output <= "0000100";
WHEN OTHERS => output <= "1111111";-- ALL OFF
END CASE;
END PROCESS;
END behave;
【问题讨论】:
-
见How to Ask,关于图像的一点。图像中显示的文本都没有出现以寻找未来的读者。查找错误不需要您分配的详细信息,Tricky 会分析(编译)您的代码。 Clock_Divider 和 led 都不需要演示 clk、Cout、D 和 Q 的图像错误和看不见的错误。您有一些不必要的使用子句使 IEEE 包 numeric_std 和 Synopsys 包 std_logic_arith 声明直接可见,您的代码不依赖这些声明.