Forum Discussion
Altera_Forum
Honored Contributor
18 years agoAhh...
The age old question of "gate' to FPGA resources. First he easy answer - No, it is not correct to say "1 ALUT equals 4 logic gates and 1 logic register = 2 logic gates". (an ALUT is upto an 8 input Boolean function) (A Register is made up of much more then 2 logic gate, maybe as many as 10 -12) But the real answer is much more challenging then that . And every one you talk to will most likely give a different answer as well. One itimized list I have heard used is to multiply the FPGA stated resources by 9 to 12 to get to ASIC gates. On average this is not a bad in the ball park guess. Other info I have read on the subject might also say; 4 transistors = 1 gate (2 input NAND) A memory bit (SRAM) consumes 6 transistors. (You may have heard of a 6-T memory cell) IBM counts all of the transistors (logic and memory) and divides by 4. So….A 2S180 would look like this. Logic gates 180k *12 (gates we claim in a LUT) = 2.16M DSP gates = ~ 1M (8mb memory *6) /4 = 12M Clocks, test, length base buffer insertion, slew rate fixing = 10% overhead of logic = .216M So a 2S180 = 2.16 + 1 + 12 + .216 = 15.376 million gates. Of course, in an ASIC they are 100% utilized and in an FPGA or Structured AASIC you only use what you use. So, if someone uses an Altera memory to gate ratio, it looks like a big part. If the design is logic intensive, it looks pretty small. I hope this helps.