BU9873 Datasheet, PDF (18/40 Pages) Rohm – I2C BUS Serial Interface RTC with High-precision Oscillation Adjustment

BU9873
Datasheet
4-2. Configuration Method of Time Adjustment
Time adjustment amount can be calculated following the rules below.
Case 1:
When oscillation frequency ^(Note1) >target frequency ^(Note2) (clock gains)
(Oscillation frequency − Target frequency + 0.1)
Adjustment amount ^(Note3) = --------------------------------------------------------------
Oscillation frequency × 3.051 × 10−6
≈ (Oscillation frequency − Target frequency) × 10 + 1
(Note1) Oscillation frequency : Clock frequency output from the 32KOUT pin at room temperature.
(Note2) Target frequency : A frequency to be adjusted to.
Since temperature characteristics of a 32.768 kHz crystal oscillator generally generates the highest frequency at a room temperature, we recommend to set
the target frequency to approx. 32768.00Hz to 32768.10Hz (+3.05ppm to 32768Hz).
Note that this value may differ based on the environment or place where the device will be used.
(Note3) Adjustment amount: A value to be set finally to F6 to F0 bits. This value is expressed in 7bit binary digits with sign bit.
Example of Calculations
When oscillation frequency=32768.85 kHz; target frequency=32768.05 kHz
Oscillation adjustment value = (32768.85 - 32768.05 + 0.1) / (32768.85 × 3.051 × 10^-6)
≈ (32768.85 - 32768.05) × 10 + 1
= 9.001 ≈ 9
In this instance, write the settings (DEV, F6, F5, F4, F3, F2, F1, F0) = (0, 0, 0, 0, 1, 0, 0, 1) in the oscillation adjustment
register. Thus, an appropriate oscillation adjustment value in the presence of any time count gain represents a distance
from 01h.
Case 2:
When oscillation frequency=target frequency (no clock gain or loss)
(F6, F5, F4, F3, F2, F1, F0) = (*, 0, 0, 0, 0, 0, *).
In this case, the correction is not performed.
Case 3:
When oscillation frequency<target frequency (clock loses)
(Oscillation frequency − Target frequency)
Adjustment amount = ---------------------------------------------------------
Oscillation frequency × 3.051 × 10−6
≈(Oscillation frequency - Target frequency) × 10
Example of Calculations
When actual oscillation frequency=32763.95 kHz; target frequency=32768.05 kHz
Oscillation adjustment value = (32763.95 - 32768.05) / (32768.05 × 3.051 × 10^-6)
≈ (32763.95 - 32768.05) × 10 + 1
= -41.015 ≈ -41
To express -41 in 7bit binary digits with sign bit, subtract 41 (29h) from 128 (80h) in the above case, 80h-29h=57h. Thus,
set (F6, F5, F4, F3, F2, F1, F0) to (1, 0, 1, 0, 1, 1, 1). As this example shows, adjustments to be used when the clock loses
shall be distance from 80h.
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13.Sep.2013 REV.001