Originally Posted by astaris
To find the correct strap values let's put what tony said in formula:
Ram_Freq=2*FSB_freq*north_multi/mem_div
where:
Ram_Freq is DDR freq (DDR clock*2, this explains the "2" term in the right)
FSB_freq is the base FSB freq (so for 1066 FSB_freq=1066/4=266)
north_multi give us the Northbridge frequency. We have:
533_strap -> north_multi = 3
800 strap -> north_multi = 2
1066 strap -> north_multi = 3/2
1333 strap -> north_multi = 6/5
Note for 533_strap tony speaked of 6 multi, but it referred to FSB_freq/2, so i'm saying the same thing.
mem_div can be set to the following values:
2
3/2
6/5
1
So for a 266 FSB we can have the following "mem ratios":
533_strap-> 800, 1066, 1333, 1600
800_strap-> 533, 711, 889, 1066
1066_strap-> 400, 533, 667, 800
1333_strap-> 320, 427, 533, 640
Because P5W uses only 800_strap and 1066_strap for 1066 FSB we have the following:
Auto -> it depends on RAM spd
400 -> 1066_strap
533 (1:1) -> can be either 800_strap or 1066_strap (but i think p5W uses 1066 strap)
667 -> 1066_strap
711 -> 800_strap
800 -> 1066_strap
889 -> 800_strap
1067 -> 800_strap
Now based on the strap table above, as tony said, we can have 1:1 with 1333 strap too, in this case the FSB at which the north would be overclocked is:
FSB=400/(6/5)=333, so if the north has its wall at 375 FSB with 1066 strap in this case the wall would be shifted at 469...