AD7441BRTZ-R2 datasheet(17/24 Pages) AD

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AD7441/AD7451

Rev. D | Page 17 of 24

Timing Example 2

Timing Example 1

Having fSCLK = 5 MHz and a throughput rate of 315 kSPS gives a

cycle time of

Having fSCLK = 18 MHz and a throughput rate of 1 MSPS gives a

cycle time of

1/Throughput = 1/315,000 = 3.174 μs

1/Throughput = 1/1,000,000 = 1 μs

A cycle consists of

t2 + 12.5 (1/fSCLK) + tACQUISITION = 3.174 μs

t2 + 12.5 (1/fSCLK) + tACQUISITION = 1 μs

Therefore, if t2 is 10 ns, then

Therefore, if t2 = 10 ns, then

10 ns + 12.5 (1/5 MHz) + tACQUISITION = 3.174 μs

tACQUISITION = 664 ns

10 ns + 12.5 (1/18 MHz) + tACQUISITION = 1 μs

tACQUISITION = 296 ns

This 664 ns satisfies the requirement of 290 ns for tACQUISITION.

This 296 ns satisfies the requirement of 290 ns for tACQUISITION.

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.

where t8 = 35 ns. This allows a value of 122 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.

tCONVERT

SCLK

12.5(1/

fSCLK)

tACQUISITION

1/THROUGHPUT

tQUIET

10ns

Figure 28. Serial Interface Timing Example

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