Forum Discussion
Altera_Forum
Honored Contributor
14 years agoHmm, that's even more complicated...
I'm not even sure how it would work with a dual-ported version? 4096 x 64 = 262,144 bits of data, which is not divisible by 80. I see problems happening at the boundaries... Worst case I could put some garbage data in the last word. It would be written/read and then ignored, the cost seems small compared to the size of the RAM. Thanks for the response! Edit: So for a dual-port version, I'd still use a width of 12 for the address? since I would need 3276.8 words of 80-bit each. Which is between 2^11 and 2^12. Hmm... I hope it's doable :).