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K4T51163QI
datasheet
DDR2 SDRAM
Rev. 1.0
K4T51083QI
K4T51043QI
44.
For tAOFD of DDR2-400/533, the 1/2 clock of tCK in the 2.5 x tCK assumes a tCH, input clock HIGH pulse width of 0.5 relative to tCK. tAOF,min and
tAOF,max should each be derated by the same amount as the actual amount of tCH offset present at the DRAM input with respect to 0.5.
For example, if an input clock has a worst case tCH of 0.45, the tAOFmin should be derated by subtracting 0.05 x tCK from it, whereas if an input clock
has a worst case tCH of 0.55, the tAOFmax should be derated by adding 0.05 x tCK to it. Therefore, we have;
tAOFmin(derated) = tAC,min - [0.5 - Min(0.5, tCHmin)] x tCK
tAOFmax(derated) = tAC,max + 0.6 + [Max(0.5, tCHmax) - 0.5] x tCK
or
tAOFmin(derated) = Min(tACmin, tACmin - [0.5 - tCHmin] x tCK)
tAOFmax(derated) = 0.6 + Max(tACmax, [tCHmax - 0.5] x tCK)
where tCHmin and tCHmax are the minimum and maximum of tCH actually measured at the DRAM input balls.
45.
For tAOFD of DDR2-667/800, the 1/2 clock of nCK in the 2.5 x nCK assumes a tCH(avg), average input clock HIGH pulse width of 0.5 relative to
tCK(avg). tAOFmin and tAOFmax should each be derated by the same amount as the actual amount of tCH(avg) offset present at the DRAM input
with respect to 0.5.
For example, if an input clock has a worst case tCH(avg) of 0.48, the tAOFmin should be derated by subtracting 0.02 x tCK(avg) from it, whereas if an
input clock has a worst case tCH(avg) of 0.52, the tAOFmax should be derated by adding 0.02 x tCK(avg) to it. Therefore, we have;
tAOFmin(derated) = tACmin - [0.5 - Min(0.5, tCH(avg)min)] x tCK(avg)
tAOFmax(derated) = 0.6 + [Max(0.5, tCH(avg)max) - 0.5] x tCK(avg)
tAOFmin(derated) = Min(tACmin, tACmin - [0.5 - tCH(avg)min] x tCK(avg))
tAOFmax(derated) = 0.6 + Max(tACmax, [tCH(avg)max - 0.5] x tCK(avg))
where tCH(avg),min and tCH(avg),max are the minimum and maximum of tCH(avg) actually measured at the DRAM input balls.
NOTE
: Tthat these deratings are in addition to the tAOF derating per input clock jitter, i.e. tJIT(duty) and tERR(6-10per). However tAC values used in the equations shown
above are from the timing parameter table and are not derated. Thus the final derated values for tAOF are;
tAOFmin(derated_final) = tAOFmin(derated) + { - tJIT(duty)max - tERR(6-10per)max }
tAOFmax(derated_final) = tAOFmax(derated) + { - tJIT(duty)min - tERR(6-10per)min }
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