??? 02/11/07 20:42 Read: times |
#132556 - here's the code again Responding to: ???'s previous message |
Ok, several posts above. It is also below as well.
My code is as follows: jmp main main: mov TCON,5h mov dptr,0 mov IE,85h sjmp $ In the "INT0" interrupt: mov A,P1 movx @dptr,A inc dptr iret In the "INT1" interrupt: mov dptr,0 iret 6 bits is the total number of address bits I need. As I have mentioned before, I want to make a ROM using logic gates, that way, it is already hardcoded, and I don't have to reprogram it. The actual data codes are 8 bits in length. One 8051 command requires all 8 bytes fed to the data line. So basically, I'm taking the 6 address lines (A0 to A5) as logic inputs, and each data bit represents a output. so if I were to convert my code for the first bit that the computer generated into an algebraic expression, it would be something like this: Bit 0: FD'C'BA + FE'D'BA + FED'C'A' + F'EDC'A' + F'ECBA So lets take this equation. What is the easiest way to simplify this so that I use the fewest number of logic gates? No, this is not homework. I'm doing this for myself. |