有人可以告诉我这个电路如何增加h_count_reg和v_count_reg？我真的没看到它。另外，输出的含义是什么呢？它只是延迟了一个像素？也不是真的看到了。谢谢！
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.NUMERIC_STD.ALL;
entity vgaController is
Port ( clk : in STD_LOGIC;
reset : in STD_LOGIC;
hsync : out STD_LOGIC;
vsync : out STD_LOGIC;
video_on : out STD_LOGIC;
p_tick : out STD_LOGIC;
pixel_x : out STD_LOGIC_VECTOR (9 downto 0);
pixel_y : out STD_LOGIC_VECTOR (9 downto 0));
end vgaController;
architecture Behavioral of vgaController is
-- VGA 640 -by - 480 sync p a r a m e t e r s
constant HD: integer:=640; --horizontal display area
constant HF: integer:=16 ; --h. front porch
constant HB: integer:=48 ; --h. back porch
constant HR: integer:=96 ; --h. retrace "Sync Pulse"
constant VD: integer:=480; -- vertical display area
constant VF: integer:=10 ; -- v. front porch
constant VB: integer:=33 ; -- v. back porch
constant VR: integer:=2 ; -- v. retrace "sync pulse"
-- mod-2 counter
signal mod2_reg, mod2_next : std_logic;--mod-2 counter to generate the 25-MHz enable tick
-- sync counters, two counters for the horizontal and vertical scans
signal v_count_reg, v_count_next : unsigned(9 downto 0);
signal h_count_reg, h_count_next : unsigned(9 downto 0);
--To remove
--potential glitches, output buffers are inserted for the hsync and vsync signals. This leads
--to a one-clock-cycle delay. add a similar buffer for the rgb signal in the pixel
--generation circuit to compensate for the delay.
-- output buffer
signal v_sync_reg, h_sync_reg: std_logic;
signal v_sync_next ,h_sync_next : std_logic;
--status signal
signal h_end , v_end , pixel_tick: std_logic;
begin
--register
process(clk,reset)
begin
if (reset='1') then
mod2_reg <='0';
v_count_reg <=(others=>'0');
h_count_reg <=(others=>'0');
v_sync_reg <='0';
h_sync_reg <='0';
elsif(clk'event and clk='1')then
mod2_reg <=mod2_next;
v_count_reg <=v_count_next;
h_count_reg <=h_count_next;
v_sync_reg <=v_sync_next;
h_sync_reg <=h_sync_next;
end if;
end process;
--mod-2 circuit to generate 25 MHz enable tick
mod2_next <= not mod2_reg;
-- 25 MHz pixel tick
pixel_tick <= '1' when mod2_reg = '1' else '0';
--status
h_end <= --end of horizonal counter
'1' when h_count_reg = (HD+HF+HB+HR-1) else --799
'0';
v_end <= --end of vertial counter
'1' when v_count_reg = (VD+VF+VB+VR-1) else --524
'0';
-- mod-800 horizontal sync counter
process(h_count_reg,h_end,pixel_tick)
begin
if (pixel_tick='1') then --25 MHz tick
if h_end='1' then
h_count_next <= (others=>'0');
else
h_count_next <= h_count_reg+1;
end if;
else
h_count_next <= h_count_reg;
end if;
end process;
-- mode-525 vertical sync counter
process(v_count_reg,h_end,v_end,pixel_tick)
begin
if (pixel_tick='1' and h_end='1') then
if (v_end='1') then
v_count_next <= (others=>'0');
else
v_count_next <= v_count_reg+1;
end if;
else
v_count_next <= v_count_reg;
end if;
end process;
-- horizontal and vertial sync, buffered to avoid glitch
h_sync_next <=
'1' when (h_count_reg >= (HD+HF)) --656
and (h_count_reg <= (HD+HF+HR-1)) else --751
'0';
v_sync_next <=
'1' when (v_count_reg >= (VD+VF)) --490
and (v_count_reg <= (VD+VF+VR-1)) else --491
'0';
--video on/off
video_on <= '1' when (h_count_reg < HD) and (v_count_reg < VD) else '0';
--output signals
hsync <= h_sync_reg;
vsync <= v_sync_reg;
pixel_x <= std_logic_vector(h_count_reg);
pixel_y <= std_logic_vector(v_count_reg);
p_tick <= pixel_tick;
end Behavioral;
答案 0 :(得分：1)
纵向和横向计数器分布在两个过程中：
--register
process(clk,reset)
begin
if (reset='1') then
mod2_reg <='0';
v_count_reg <=(others=>'0');
h_count_reg <=(others=>'0');
v_sync_reg <='0';
h_sync_reg <='0';
elsif(clk'event and clk='1')then
mod2_reg <=mod2_next;
v_count_reg <=v_count_next;
h_count_reg <=h_count_next;
v_sync_reg <=v_sync_next;
h_sync_reg <=h_sync_next;
end if;
end process;
在elsif条件中，计数器从v_count_next和h_count_next加载，这是在两个不同的过程中生成的：
-- mod-800 horizontal sync counter
process(h_count_reg,h_end,pixel_tick)
begin
if (pixel_tick='1') then --25 MHz tick
if h_end='1' then
h_count_next <= (others=>'0');
else
h_count_next <= h_count_reg+1;
end if;
else
h_count_next <= h_count_reg;
end if;
end process;
-- mode-525 vertical sync counter
process(v_count_reg,h_end,v_end,pixel_tick)
begin
if (pixel_tick='1' and h_end='1') then
if (v_end='1') then
v_count_next <= (others=>'0');
else
v_count_next <= v_count_reg+1;
end if;
else
v_count_next <= v_count_reg;
end if;
end process;
（现在你可以想象实际标记流程陈述是个好主意。）
至于＆＃34;缓冲＆＃34;：
--To remove
--potential glitches, output buffers are inserted for the hsync and vsync signals. This leads
--to a one-clock-cycle delay. add a similar buffer for the rgb signal in the pixel
--generation circuit to compensate for the delay.
-- output buffer
signal v_sync_reg, h_sync_reg: std_logic;
signal v_sync_next ,h_sync_next : std_logic;
--status signal
signal h_end , v_end , pixel_tick: std_logic;
这是上述--register流程中的最后两个分配。从评论中，通过触发器的一个时钟延迟是为了消除由关系运算符引起的组合故障：
h_sync_next <=
'1' when (h_count_reg >= (HD+HF)) --656
and (h_count_reg <= (HD+HF+HR-1)) else --751
'0';
v_sync_next <=
'1' when (v_count_reg >= (VD+VF)) --490
and (v_count_reg <= (VD+VF+VR-1)) else --491
'0';
答案 1 :(得分：1)
只是为了兴趣，这里是上面的单流程重构。
它相当短，至少在我看来，更容易理解，修改和正确。简单地在它们之间进行通信的不必要信号的多个琐碎过程使设计变得模糊（正如您所指出的那样！） - 并为臭虫提供了肥沃的滋生地。
一个警告：我还没有确认这是您设计的完全实现;如果您选择使用它，那将是您的责任。
欢迎意见。
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.NUMERIC_STD.ALL;
entity vgaController is
Port ( clk : in STD_LOGIC;
reset : in STD_LOGIC;
hsync : out STD_LOGIC;
vsync : out STD_LOGIC;
video_on : out STD_LOGIC;
p_tick : out STD_LOGIC;
pixel_x : out STD_LOGIC_VECTOR (9 downto 0);
pixel_y : out STD_LOGIC_VECTOR (9 downto 0));
end vgaController;
architecture Behavioral of vgaController is
-- VGA 640 -by - 480 sync p a r a m e t e r s
constant HD: integer:=640; --horizontal display area
constant HF: integer:=16 ; --h. front porch
constant HB: integer:=48 ; --h. back porch
constant HR: integer:=96 ; --h. retrace "Sync Pulse"
constant VD: integer:=480; -- vertical display area
constant VF: integer:=10 ; -- v. front porch
constant VB: integer:=33 ; -- v. back porch
constant VR: integer:=2 ; -- v. retrace "sync pulse"
-- derived constants to make code clearer
constant H_Last : integer := HD+HF+HB+HR-1;
constant H_Sync_First : integer := HD+HF;
constant H_Sync_Last : integer := HD+HF+HR-1;
constant V_Last : integer := VD+VF+VB+VR-1;
constant V_Sync_First : integer := VD+VF;
constant V_Sync_Last : integer := VD+VF+VR-1;
-- sync counters, two counters for the horizontal and vertical scans
signal v_count : unsigned(9 downto 0);
signal h_count : unsigned(9 downto 0);
--status signal
signal pixel_tick: std_logic;
begin
--register
process(clk,reset)
begin
if (reset='1') then
pixel_tick <= '0';
v_count <= (others=>'0');
h_count <= (others=>'0');
vsync <= '0';
hsync <= '0';
elsif(clk'event and clk='1')then
pixel_tick <= not pixel_tick;
-- H and V pixel counters
if pixel_tick = '1' then
if h_count = h_last then
h_count <= (others=>'0');
-- and start the next line
if v_count = v_last then
v_count <= (others=>'0');
else
v_count <= v_count + 1;
end if;
else
h_count <= h_count + 1;
end if;
end if;
-- default assignments for sync
hsync <= '0';
vsync <= '0';
-- H and V sync pulses
if h_count >= H_Sync_First and h_count <= H_Sync_Last then
hsync <= '1';
end if;
if v_count >= V_Sync_First and v_count <= V_Sync_Last then
vsync <= '1';
end if;
end if;
end process;
--video on/off ... was unregistered so I'll leave it that way
video_on <= '1' when (h_count < HD) and (v_count < VD) else '0';
--output signals
pixel_x <= std_logic_vector(h_count);
pixel_y <= std_logic_vector(v_count);
p_tick <= pixel_tick;
end Behavioral;
答案 2 :(得分：0)
需要缓冲区（寄存器是一个更好的名称，因为它不是输出缓冲区） - 不仅要抑制毛刺 - 而且要确保每个引脚的相等输出延迟时序（v_sync，h_sync和颜色） 。如果不在下游输出缓冲区之前使用寄存器，则会在屏幕上产生不良颜色效果。
请注意，您的x，y坐标计数器电路寻址下一个像素而不是当前的像素。因此，计数器值和h / v_sync之间必须存在延迟，具体取决于计数器之间的延迟 - >获取像素信息 - ＆gt;像素颜色输出。