Quote:
Originally Posted by srwatters
This is only conjecture, but if you look at it in binary and mask off the least significant two bits (11) that aren't changing and ignore the top three bits, you see this:
0x27 = xxx0 01xx = 1
0x2f = xxx0 11xx = 3
0xbb = xxx1 10xx = 6
Did you try 0x3b or 0x1b? If that also gives you 6 blinks, then my hypothesis would hold. If it doesn't, then the top three bits might have other usage. I would also try 0x17 or 0x37 to see if that results in 5 blinks or 0x13 or 0x33 to get 4.
(I write embedded controller software for a living so this is sort of like a second language...)
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You are 100% correct 0x17 = 5 blinks and 0x13 = 4 blinks.
Now please explain the logic?