
AD9551
Rev. B | Page 22 of 40
The value of K depends on the device configuration, as shown
Table 16. Configuring the Value of K
Mode
fCRYSTAL
K
Description
19.44 MHz
52 MHz
1300/243
Using 2× multiplier
325/243
Using 2× divider
50 MHz
1250/243
Using 2× multiplier
625/486
Using 2× divider
49.86 MHz
277/54
Using 2× multiplier
277/216
Using 2× divider
49.152 MHz
2048/405
Using 2× multiplier
512/405
Using 2× divider
Normal
1
Crystal independent
The user must carefully consider the operating frequency of the
externally connected crystal resonator (assuming that the input
PLL is not bypassed). Because the DCXO is capable of pulling
the crystal over a 50 ppm range only, the output frequency of
the DCXO is essentially identical to the crystal frequency.
The denominator of Equation 1 is the input division factor, which
has an integer part (Nx) due to an integer divider, as well as an
optional fractional part that is associated with the input SDM.
1/2 + FRACx/(2 × MODx)
Note that when bypassing the SDM, the device forces the frac-
tional part to 0 (equivalent to FRACx = MODx).
The numerator of Equation 2 contains the feedback division
factor, which has an integer part (N) due to an integer divider,
as well as an optional fractional part (FRAC/MOD) that is
associated with the feedback SDM.
Equation 1 and Equation 2, along with the constraints placed on
their variables, imply a rational relationship between fREF and fOUT1.
That is, the ratio fOUT1/fREF must be expressible as a ratio of two
integers. For example, it is not possible to configure the device
to satisfy a frequency ratio having a value of 2 because it is
irrational (that is, not expressible as a ratio of two integers).
CALCULATING DIVIDER VALUES
This section provides a 5-step procedure for calculating the divider
values when given a specific fOUT1/fREF ratio. The methodology is
described in general terms, but a specific example is provided
for clarity. The example assumes the use of the frequency control
pins with A[3:0] = 0010 and Y[3:0] = 0100 (see
Table 13 and
Table 14). The example parameters are as follows:
MHz
239
08
.
622
237
=
f
REF
MHz
64
625
=
OUT1
f
MHz
26
=
IF
f
66
Calculation Steps
1.
Ensure that fOUT1 and fREF are rationally related.
As shown in the following equation, fOUT1/fREF is expressible
as a ratio of two integers, so fOUT1 and fREF are rationally related.
568
553
,
951
,
977
)
64
)(
239
(
62208
)
100
)(
237
)(
66
(
625
237
239
08
.
622
64
66
625
=
=
REF
OUT1
f
,
000
,
625
2.
Determine the output divide factor (ODF).
Note that the VCO frequency (fVCO) spans 3350 MHz to
4050 MHz. The ratio, fVCO/fOUT1, indicates the required ODF.
Given the specified value of fOUT1 (~644.53 MHz) and the
range of fVCO, the ODF spans a range of 5.2 to 6.3. The ODF
must be an integer, which means that ODF = 6 (because 6
is the only integer between 5.2 and 6.3).
3.
Determine suitable values for P0 and P1.
The ODF is the product of the two output dividers; that is,
ODF = P0 × P1. It has already been determined that ODF = 6
for the given example. Therefore, we have P0 × P1 = 6, with the
constraints that P0 and P1 are both integers and that 4 ≤ P0 ≤ 11
These constraints lead to the singular solution of P0 = 6, and P1
= 1.
Although this particular example yields a singular solution
for the output divider values with fOUT1 ≈ 644.53 MHz, some
fOUT1 frequencies result in multiple ODFs rather than just
one. For example, if fOUT1 = 100 MHz, the ODF ranges from
34 to 40. This leads to an assortment of possible values for
The P0 and P1 combinations listed in Table 17 are all equally valid. However, note that they yield only three valid ODF
values (35, 36, and 40) from the original range of 34 to 40.
Table 17. Combinations of P0 and P1
P0
P1
ODF
4
9
36
4
10
40
5
7
35
5
8
40
6
36
7
5
35
8
5
40
9
4
36
10
4
40