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AD7451ARTZ-REEL データシートの表示(PDF) - Analog Devices

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AD7451ARTZ-REEL Datasheet PDF : 24 Pages
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AD7441/AD7451
Timing Example 1
Having fSCLK = 18 MHz and a throughput rate of 1 MSPS gives a
cycle time of
1/Throughput = 1/1,000,000 = 1 μs
A cycle consists of
t2 + 12.5 (1/fSCLK) + tACQUISITION = 1 μs
Therefore, if t2 = 10 ns, then
10 ns + 12.5 (1/18 MHz) + tACQUISITION = 1 μs
tACQUISITION = 296 ns
This 296 ns satisfies the requirement of 290 ns for t . ACQUISITION
From Figure 28, tACQUISITION comprises
2.5 (1/fSCLK) + t8 = tQUIET
where t8 = 35 ns. This allows a value of 122 ns for tQUIET,
satisfying the minimum requirement of 60 ns.
Timing Example 2
Having fSCLK = 5 MHz and a throughput rate of 315 kSPS gives a
cycle time of
1/Throughput = 1/315,000 = 3.174 μs
A cycle consists of
t2 + 12.5 (1/fSCLK) + tACQUISITION = 3.174 μs
Therefore, if t2 is 10 ns, then
10 ns + 12.5 (1/5 MHz) + tACQUISITION = 3.174 μs
tACQUISITION = 664 ns
This 664 ns satisfies the requirement of 290 ns for t . ACQUISITION
From Figure 28, tACQUISITION comprises
2.5 (1/fSCLK) + t8 = tQUIET
where t8 = 35 ns. This allows a value of 129 ns for tQUIET,
satisfying the minimum requirement of 60 ns.
As in this example and with other slower clock values, the signal
can already be acquired before the conversion is complete, but it
is still necessary to leave 60 ns minimum tQUIET between conver-
sions. In Example 2, the signal is fully acquired at approximately
Point C in Figure 28.
CS
10ns
t2
t5
tCONVERT
B
C
SCLK
1
2
3
4
5
13
14
15
16
t6
t8
12.5(1/fSCLK)
1/THROUGHPUT
tACQUISITION
tQUIET
Figure 28. Serial Interface Timing Example
Rev. D | Page 17 of 24

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