si514 Silicon Laboratories, si514 Datasheet - Page 11

• Manufacturer Part Number: si514
• Description: Any-frequency I 2c Programmable Xo 100 Khz To 250 Mhz
• Manufacturer: Silicon Laboratories

2. Calculate Mnew:
3. Write Mnew to Registers 5-7:
4. Write Mnew to Register 8:
5. Write Mnew to Register 9:
2.3. Programming a Large Frequency Change (> ±1000 ppm)
Large frequency changes are those that vary the F
operating F
F
operating center frequency. The below steps are recommended when performing large frequency changes:
1. Disable the output: Write OE register bit to a 0 (Register 132, bit2)
2. If using one of the standard frequencies listed in Table 9, then write the new LP1, LP2, M_Frac, M_Int, HS_DIV
3. Determine the minimum value of LS_DIV (minimizing LS_DIV minimizes the number of dividers on the output
4. Determine the minimum value for HS_DIV (this optimizes timing margins)
Note: SPEED_GRADE_MIN (Reg 48) ≤ LS_DIV x HS_DIV ≤ SPEED_GRADE_MAX (Reg 49); If outside this range, the output
5. Determine a value for M according to the following formula (all values are in decimal format):
CENTER
and LS_DIV register values according to the table (be sure to write M_Int[8:3] (Register 9) after writing to the
M_Frac registers (Registers 5-8)). Skip to Step 9. If the desired frequency is not in the table, then follow steps
4-8 below.
stage, thus minimizing jitter) according to the following formula:
e. Register 9 = 0x09 (M_Int[8:3])
f.
g. M_Frac = 0x097C65D3 = 159,147,475
h. M= M_Int + M_Frac/2
a. Mnew = 74.296435272321105 x 148.352/148.5 = 74.2223889933965
b. M_Intnew = 74 = 0x4A
c. M_Fracnew = 0.2223889933965 x 2
a. Register 5 = 0x85
b. Register 6 = 0xCF
c. Register 7 = 0x1D
a. Register 8 = 0x47
a. Register 9 = 0x09
a. LS_DIV = F
b. LS_DIV = 2080/(F
a. HS_DIV(MIN) = F
b. HS_DIV(MIN) = 2080/(F
will be forced to the disabled state.
a. M = LS_DIV x HS_DIV x F
b. M = LS_DIV x HS_DIV x F
c. M_Int = INT[M] (Eq 2.12)
d. M_Frac = (M – INT[M]) x 2
to F'
M_Int = 0b001001010 = 0x4A = 0d74
CENTER
i. Since LS_DIV is restricted to: dividing by 1,2,4,8,16,32, choose the next largest value over the
CENTER
result derived in Eq 2.7 (e.g., if result is 4.135, choose LS_DIV = 8)
. Figure 2 illustrates the difference between large and small frequency changes. Changing from
VCO
i.HS_DIV(MIN) will be the next even number greater than or equal to the result derived in Eq 2.9
requires a calibration cycle that resets internal circuitry to establish F'
(keeping in the range of 10-1022)
(MIN)/(F
VCO
OUT
29
(MIN)/(F
(MHz) x 1022) (Eq 2.7)
OUT
= 74 + 159,147,475/2
OUT
OUT
OUT
29
(MHz) x LS_DIV) (Eq 2.9)
x HS_DIV(MAX)) (Eq 2.6)
(Eq 2.13)
OUT
/F
(MHz)/31.98 (Eq 2.11)
XO
x LS_DIV) (Eq 2.8)
(Eq 2.10)
29
Preliminary Rev. 0.9
= 119,394,181 = 0x071DCF85
VCO
29
frequency by an amount greater than ±1000 ppm from an
= 74.296435272321105
CENTER
Si514
as the new
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