M41T00AUD
Clock calibration
25/44
Example 2:
Sign is 0 and N is 3 (00010b). The 16-minute interval will be:
3 * 513 + (60-3) * 512 + 15 * 60 * 512 = 491523 cycles long out of a possible
512 * 60 * 16 = 491520 cycles of the 512Hz clock in an 16-minute span.
This gives a net correction of (491520-491523) / 491520 = +6.104ppm
Therefore, each calibration step has an effect on clock accuracy of either -4.068 or +2.034
ppm. Assuming that the oscillator is running at exactly 32,768Hz, each of the 31 steps in the
calibration byte would represent subtracting 10.7 or adding 5.35 seconds per month, which
corresponds to a total range of -5.5 or +2.75 minutes per month.
Note:
The modified pulses are not observable on the frequency test (FT) output, nor will the effect
of the calibration be measurable real-time, due to the periodic nature of the error
compensation.
