H5PS5162FFR-S5 HYNIX [Hynix Semiconductor], H5PS5162FFR-S5 Datasheet - Page 38

• Manufacturer Part Number: H5PS5162FFR-S5
• Description: 512Mb DDR2 SDRAM
• Manufacturer: HYNIX [Hynix Semiconductor]

Release
H5PS5162FFR series
43. When the device is operated with input clock jitter, this parameter needs to be derated by { -
tJIT(duty),max - tERR(6-10per),max } and { - tJIT(duty),min - tERR(6-10per),min } of the actual input
clock.(output deratings are relative to the SDRAM input clock.)
For example, if the measured jitter into a DDR2-667 SDRAM has tERR(6-10per),min = - 272 ps, tERR(6-
10per), max = + 293 ps, tJIT(duty),min = - 106 ps and tJIT(duty),max = + 94 ps, then tAOF,min(derated)
= tAOF,min + { - tJIT(duty),max - tERR(6-10per),max } = - 450 ps + { - 94 ps - 293 ps} = - 837 ps and
tAOF,max(derated) = tAOF,max + { - tJIT(duty),min - tERR(6-10per),min } = 1050 ps + { 106 ps + 272 ps
} = + 1428 ps. (Caution on the min/max usage!)
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 tAOF,min 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 tAOF,max should be derated by adding 0.05 x
tCK to it. Therefore, we have;
tAOF,min(derated) = tAC,min - [0.5 - Min(0.5, tCH,min)] x tCK
tAOF,max(derated) = tAC,max + 0.6 + [Max(0.5, tCH,max) - 0.5] x tCK
or
tAOF,min(derated) = Min(tAC,min, tAC,min - [0.5 - tCH,min] x tCK)
tAOF,max(derated) = 0.6 + Max(tAC,max, tAC,max + [tCH,max - 0.5] x tCK)
where tCH,min and tCH,max 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). tAOF,min and tAOF,max 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 tAOF,min should be derated by subtract-
ing 0.02 x tCK(avg) from it, whereas if an input clock has a worst case tCH(avg) of 0.52, the tAOF,max
should be derated by adding 0.02 x tCK(avg) to it. Therefore, we have;
tAOF,min(derated) = tAC,min - [0.5 - Min(0.5, tCH(avg),min)] x tCK(avg)
tAOF,max(derated) = tAC,max + 0.6 + [Max(0.5, tCH(avg),max) - 0.5] x tCK(avg)
or
tAOF,min(derated) = Min(tAC,min, tAC,min - [0.5 - tCH(avg),min] x tCK(avg))
tAOF,max(derated) = 0.6 + Max(tAC,max, tAC,max + [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 that 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;
tAOF,min(derated_final) = tAOF,min(derated) + { - tJIT(duty),max - tERR(6-10per),max }
tAOF,max(derated_final) = tAOF,max(derated) + { - tJIT(duty),min - tERR(6-10per),min }
Rev. 1.0 / July. 2008
38