7使用VHDL显示段

Question

我目前正在使用vhdl并且遇到7段显示问题.我在网上找到了这个代码,但我很难理解它的确切含义.有些人可以通过以下代码帮助我了解正在发生的事情:

ARCHITECTURE Structure OF multi IS
 SIGNAL C : STD_LOGIC_VECTOR(2 DOWNTO 0);
BEGIN
 LEDR <= SW;
 C(2 DOWNTO 0) <= SW(2 DOWNTO 0);

 HEX0(0) <= NOT( (NOT(C(2)) AND NOT(C(1)) AND C(0)) OR 
 (NOT(C(2)) AND C(1) AND C(0)) ); 
 HEX0(1) <= NOT( (NOT(C(2)) AND NOT(C(1)) AND NOT(C(0))) OR
 (NOT(C(2)) AND C(1) AND C(0)) ); 
 HEX0(2) <= NOT( (NOT(C(2)) AND NOT(C(1)) AND NOT(C(0))) OR 
 (NOT(C(2)) AND C(1) AND C(0)) ); 
 HEX0(3) <= NOT( (NOT(C(2)) AND NOT(C(1)) AND C(0)) OR 
 (NOT(C(2)) AND C(1) AND NOT(C(0))) OR
 (NOT(C(2)) AND C(1) AND C(0)) ); 
 HEX0(4) <= NOT( (NOT(C(2)) AND NOT(C(1)) AND NOT(C(0))) OR 
 (NOT(C(2)) AND NOT(C(1)) AND C(0)) OR 
 (NOT(C(2)) AND C(1) AND NOT(C(0))) OR (NOT(C(2)) AND C(1) AND C(0)) );
 HEX0(5) <= NOT( (NOT(C(2)) AND NOT(C(1)) AND NOT(C(0))) OR 
 (NOT(C(2)) AND NOT(C(1)) AND C(0)) OR
 (NOT(C(2)) AND C(1) AND NOT(C(0))) OR (NOT(C(2)) AND C(1) AND C(0)) ); 
 HEX0(6) <= NOT( (NOT(C(2)) AND NOT(C(1)) AND NOT(C(0))) OR
 (NOT(C(2)) AND NOT(C(1)) AND C(0)) );
END Structure;

我不理解所有NOT和OR语句中的逻辑.

非常感谢!

Answer 1

当代码从混淆的VHDL竞赛中在线逃脱时会发生这种情况.

或者可能更慈善,有人在20世纪70年代早期将一个八进制到七段解码器芯片(或电路板!)的原理图交给了他,并要求用VHDL重写它,因为原始组件不再可用.他/她用经典的"产品总和"形式写出来,而不是试图将其最小化......

我相信你能做的最好的事情就是把整个事情写成一个查找表,一点一滴,而不用担心逻辑的细节.

从每个表达开始......

 HEX0(6) <= NOT( (NOT(C(2)) AND NOT(C(1)) AND NOT(C(0))) OR
 (NOT(C(2)) AND NOT(C(1)) AND C(0)) );

并尽量减少它

 HEX0(6) <= NOT((NOT(C(2)) AND NOT(C(1)));

并为每个值写出来

C  C2  C1  C0  H6  H5  H4  H3  H2  H1  H0   HEX0
0   0   0   0   0   0   0   1   0   0   1   0001001 = 09
1   0   0   1   0
2   0   1   0   1
3   0   1   1   1
4   1   0   0   1
5   1   0   1   1
6   1   1   0   1
7   1   1   1   1

(不完整,也不保证正确......)

然后沿着这些方向重写事物:

subtype Seven_Seg is std_logic_vector(6 downto 0);

constant Lookup : array(0 to 7) of Seven_Seg := ( 0 => "0001001",
                                                  1 => ...
                                                  ...
                                                  7 => ... );

Hex0 <= Lookup(to_integer(unsigned(C)));

并完成它.