我要在Altera GX入门套件上设置adv7513。设置后,我从寄存器中读取的数据与我发送的数据不同。
我试图将写入划分为单个寄存器而不是一个长事务,并更改写入和读取的顺序,但是我总是从其余寄存器中读取0x98中的0x03和其余寄存器中的0x00。
我使用了以下i2c控制器:https://www.digikey.com/eewiki/pages/viewpage.action?pageId=10125324
我已经在vhdl中制作了状态机来控制它,并写入rom中存储的数据
if rising_edge(clk) then
case state is
when idle =>
if (beg='1') then
state <= start;
else
state <= idle;
end if;
when start =>
rom_addr <= count;
state <= lut_addr_i2c;
when lut_addr_i2c =>
ena <= '1';
rw <= '0';
addr <= "0111001";
data_wr <= lut_data(15 downto 8);
state <= lut_data_i2c;
when lut_data_i2c =>
if (busy='1' and busy_prev='0') then
data_wr <= lut_data(7 downto 0);
state <= cleanup;
count <= count+1;
end if;
when cleanup =>
if (busy='1' and busy_prev='0') then
state <= next_lut;
end if;
when next_lut =>
if (count = 31) then
state <= rd;
ena <= '0';
else
state <= start;
end if;
when rd =>
ready <= '1';
count <= 0;
if (rd_delay = 10000) then
state <= start_rd;
else
rd_delay <= rd_delay+1;
state <= rd;
end if;
when start_rd =>
rom_addr <= count;
if (next_rd = 20000) then
state <= rd_lut_addr;
next_rd <= 0;
else
next_rd <= next_rd+1;
state <= start_rd;
end if;
when rd_lut_addr =>
ena <= '1';
rw <= '0';
addr <= "0111001";
data_wr <= lut_data(15 downto 8);
state <= rd_lut;
when rd_lut =>
if (busy='1' and busy_prev='0') then
rw <= '1';
count <= count+1;
state <= rd_cleanup;
end if;
when rd_cleanup =>
if (busy='1' and busy_prev='0') then
state <= rd_next_lut;
end if;
when rd_next_lut =>
if (count = 31) then
state <= fin;
ena <= '0';
else
state <= start_rd;
ena <= '0';
end if;
when fin =>
state <= fin;
end case;
end if;
答案 0 :(得分:0)
我认为您的问题是您只是超载了i2c从站。提供第二个数据字节之后,您应该最晚等待not busy
处于清理状态。
您的代码似乎根本没有等待not busy
吗?