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M41T00S Fiches technique(PDF) 16 Page - STMicroelectronics |
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M41T00S Fiches technique(HTML) 16 Page - STMicroelectronics |
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16 / 25 page ![]() M41T00 clock operation M41T00 16/25 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 AI00999b –160 0 10203040506070 Frequency (ppm) Temperature °C 80 –10 –20 –30 –40 –100 –120 –140 –40 –60 –80 20 0 –20 ΔF = K x (T –TO) 2 K = –0.036 ppm/ °C2 ± 0.006 ppm/°C2 TO = 25°C ± 5°C F |
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