SI514 Datasheet, PDF (16/36 Pages) Silicon Laboratories – ANY-FREQUENCY IC PROGRAMMABLE XO

Si514
6. Determine values for LP1 and LP2 according to Table 13:
Table 13. LP1, LP2 Values
Fvco_max
Fvco_min
M_max
M_min
LP1
LP2
2500000000.00000 2425467616.18572
78.173858662
75.843265046
4
4
2425467616.18572 2332545246.89005
75.843265046
72.937624981
3
4
2332545246.89005 2170155235.53450
72.937624981
67.859763463
3
3
2170155235.53450 2087014168.27005
67.859763463
65.259980246
2
3
2087014168.27005 2080000000.00000
65.259980246
65.040650407
2
2
7. Write new LP1, LP2, M_Frac, M_Int, HS_DIV and LS_DIV register values (be sure to write M_Int[8:3] (Register
9) after writing to the M_Frac registers (Registers 5-8)
8. Write FCAL (Register 132, bit 0) to a 1 (this bit auto-resets, so it will always read as 0).
9. Enable the output: Write OE register bit to a 1.
The Si514 does not automatically detect large frequency changes. The user needs to assert the FCAL register bit
to initiate the calibration cycle required to re-center the VCO around the new frequency. Large frequency changes
are discontinuous and output may skip to intermediate frequencies or generate glitches. Resetting the OE bit
before FCAL will prevent intermediate frequencies from appearing on the output while Si514 completes a
calibration cycle and settles to F'_CENTER. Settling time for large frequency changes is 10 msec maximum.
Example 2.2:
The user has a part that is programmed with SPEED_GRADE_MIN = 20 and SPEED_GRADE_MAX = 250 that is
programmed from the factory for F_OUT = 50 MHz and wants to change to an STS-1 rate of 51.84 MHz. This
represents a change of +36,800 ppm which exceeds ±1000 ppm and therefore requires a large frequency change
process.
1. Write Reg 132, bit 2 to a 0 to disable the output.
2. Since 51.84 MHz is not in Table 2.1, the divider parameters must be calculated.
3. Calculate LS_DIV by using Eq 2.7:
a. LS_DIV = 2080/(51.84 x 1022) = 0.039
b. Since 0.039 < 1, use a divide-by-one (bypass), therefore LS_DIV = 0
4. Calculate HS_DIV(MIN) by using Eq 2.9:
a. HS_DIV(MIN) = 2080/(51.84 x 1) = 40.123
b. Since 40.123 > 40, use HS_DIV(MIN) = 42 = 0x2A
5. From Eq 2.11:
a. M = 1 x 42 x 51.84/31.98 = 68.08255159474
b. M_Int = 68 = 0x44
c. M_Frac = 0.08255259474 x 2²⁹ = 44,320,087 = 0x2A44557
6. From Table 2.2:
a. LP1 = 3
b. LP2 = 3
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Rev. 1.0