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HYB18TC1G800BF 参数 Datasheet PDF下载

HYB18TC1G800BF图片预览
型号: HYB18TC1G800BF
PDF下载: 下载PDF文件 查看货源
内容描述: 1千兆位双数据速率- SDRAM双 [1-Gbit Double-Data-Rate-Two SDRAM]
分类和应用: 动态存储器
文件页数/大小: 65 页 / 3555 K
品牌: QIMONDA [ QIMONDA AG ]
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Internet Data Sheet  
HYB18TC1G[80/16]0BF  
1-Gbit Double-Data-Rate-Two SDRAM  
7.3  
Jitter Definition and Clock Jitter Specification  
Generally, jitter is defined as “the short-term variation of a signal with respect to its ideal position in time”. The following table  
provides an overview of the terminology.  
TABLE 52  
Average Clock and Jitter Symbols and Definition  
Symbol  
Parameter  
Description  
Units  
tCK.AVG  
Average clock period tCK.AVG is calculated as the average clock period within any consecutive ps  
200-cycle window:  
N
1
N
.
---  
tCK.AVG =  
tCKj  
(1)  
j = 1  
N = 200  
tJIT.PER  
Clock-period jitter  
t
t
JIT.PER is defined as the largest deviation of any single tCK from tCK.AVG  
JIT.PER = Min/Max of {tCKi tCK.AVG} where i = 1 to 200  
:
ps  
t
t
JIT.PER defines the single-period jitter when the DLL is already locked.  
JIT.PER is not guaranteed through final production testing.  
t
JIT(PER, LCK)  
Clock-period jitter  
during DLL-locking  
period  
t
JIT(PER,LCK) uses the same definition as tJIT.PER, during the DLL-locking ps  
period only.  
t
JIT(PER,LCK) is not guaranteed through final production testing.  
tJIT.CC  
Cycle-to-cycle clock  
period jitter  
t
JIT.CC is defined as the absolute difference in clock period between two  
ps  
consecutive clock cycles:  
t
JIT.CC = Max of ABS{tCKi+1 tCKi}  
t
t
JIT.CC defines the cycle- to- cycle jitter when the DLL is already locked.  
JIT.CC is not guaranteed through final production testing.  
t
JIT(CC, LCK)  
Cycle-to-cycle clock  
period jitter during  
DLL-locking period  
t
JIT(CC,LCK) uses the same definition as tJIT.CC during the DLL-locking  
ps  
period only.  
t
JIT(CC,LCK) is not guaranteed through final production testing.  
tERR.2PER  
Cumulative error  
across 2 cycles  
t
ERR.2PER is defined as the cumulative error across 2 consecutive cycles ps  
from tCK.AVG  
:
i + n 1  
tERR(2per) =  
tCKj n × tCK(avg)  
(2)  
j = i  
n = 2 for tERR(2per)  
where i = 1 to 200  
Rev. 1.21, 2007-07  
53  
02282007-F8UP-4HSU  
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