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M41T00M6F 参数 Datasheet PDF下载

M41T00M6F图片预览
型号: M41T00M6F
PDF下载: 下载PDF文件 查看货源
内容描述: 串行实时时钟 [Serial real-time clock]
分类和应用: 计时器或实时时钟微控制器和处理器外围集成电路光电二极管PC
文件页数/大小: 25 页 / 206 K
品牌: STMICROELECTRONICS [ ST ]
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M41T00 clock operation  
M41T00  
Therefore, each calibration step has the effect of adding 512 or subtracting 256 oscillator  
cycles for every 125,829,120 actual oscillator cycles, that is +4.068 or –2.034 ppm of  
adjustment per calibration step in the calibration register. Assuming that the oscillator is in  
fact running at exactly 32768 Hz, each of the 31 increments in the calibration byte would  
represent +10.7 or –5.35 seconds per month which corresponds to a total range of +5.5 or  
–2.75 minutes per month.  
Two methods are available for ascertaining how much calibration a given M41T00 may  
require. The first involves simply setting the clock, letting it run for a month and comparing it  
to a known accurate reference (like WWV broadcasts). While that may seem crude, it allows  
the designer to give the end user the ability to calibrate his clock as his environment may  
require, even after the final product is packaged in a non-user serviceable enclosure. All the  
designer has to do is provide a simple utility that accessed the calibration byte.  
The second approach is better suited to a manufacturing environment, and involves the use  
of some test equipment. When the frequency test (FT) bit, the seventh-most significant bit in  
the control register, is set to a '1', and the oscillator is running at 32768 Hz, the FT/OUT pin  
of the device will toggle at 512 Hz. Any deviation from 512 Hz indicates the degree and  
direction of oscillator frequency shift at the test temperature.  
For example, a reading of 512.01024 Hz would indicate a +20 ppm oscillator frequency  
error, requiring a –10(XX00 1010b) to be loaded into the calibration byte for correction. Note  
that setting or changing the calibration byte does not affect the frequency test output  
frequency.  
Figure 12. Crystal accuracy across temperature  
Frequency (ppm)  
20  
0
–20  
–40  
–60  
–80  
ΔF  
F
2
= K x (T –T )  
O
–100  
–120  
–140  
–160  
2
2
K = –0.036 ppm/°C  
0.006 ppm/°C  
T
= 25°C 5°C  
O
–40  
–30  
–20  
–10  
0
10  
20  
30  
40  
50  
60  
70  
80  
Temperature °C  
AI00999b  
16/25  
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