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AD9522-5
Rev. 0 | Page 39 of 76
The channel dividers feeding the output drivers contain one
2-to-32 frequency divider. This divider provides for division-by-1
to division-by-32. Division-by-1 is accomplished by bypassing
the divider. The dividers also provide for a programmable duty
cycle, with optional duty-cycle correction when the divide ratio
is odd. A phase offset or delay in increments of the input clock
cycle is selectable. The channel dividers operate with a signal at
their inputs up to 1600 MHz. The features and settings of the
dividers are selected by programming the appropriate setup
VCO Divider
The VCO divider provides frequency division between the CLK
input and the clock distribution channel dividers. The VCO
divider can be set to divide by 1, 2, 3, 4, 5, or 6 (see
Table 50,
0x1E0[2:0]). However, when the VCO divider is set to 1, none
of the channel output dividers can be bypassed.
The VCO divider can also be set to static, which is useful for
applications where the only desired output frequency is the
CLK input frequency. Making the VCO divider static increases
the wideband spurious-free dynamic range (SFDR). The same
improvement in SFDR performance can also be achieved by
setting the VCO divider to 1.
Channel Dividers
A channel divider drives each group of three LVDS outputs.
There are four channel dividers (0, 1, 2, and 3) driving 12 LVDS
outputs (OUT0 to OUT11).
Table 28 gives the register locations
used for setting the division and other functions of these dividers.
The division is set by the values of M and N. The divider can be
bypassed (equivalent to divide-by-1, divider circuit is powered
down) by setting the bypass bit. The duty-cycle correction can
be enabled or disabled according to the setting of the disable
divider DCC bits.
Table 28. Setting DX for the Output Dividers
Divider
Low Cycles M
High Cycles N
Bypass
Disable
Div DCC
0
0x190[7:4]
0x190[3:0]
0x191[7]
0x192[0]
1
0x193[7:4]
0x193[3:0]
0x194[7]
0x195[0]
2
0x196[7:4]
0x196[3:0]
0x197[7]
0x198[0]
3
0x199[7:4]
0x199[3:0]
0x19A[7]
0x19B[0]
Channel Frequency Division (0, 1, 2, and 3)
For each channel (where the channel number x is 0, 1, 2, or 3),
the frequency division, DX, is set by the values of M and N
(four bits each, representing Decimal 0 to Decimal 15), where
Number of Low Cycles = M + 1
Number of High Cycles = N + 1
The high and low cycles are cycles of the clock signal currently routed
to the input of the channel dividers (VCO divider out or CLK).
When a divider is bypassed, DX = 1.
Otherwise, DX = (N + 1) + (M + 1) = N + M + 2. This allows
each channel divider to divide by any integer from 1 to 32.
Duty Cycle and Duty-Cycle Correction
The duty cycle of the clock signal at the output of a channel is a
result of some or all of the following conditions:
The M and N values for the channel
DCC enabled/disabled
VCO divider enabled/bypassed
The CLK input duty cycle
The DCC function is enabled by default for each channel divider.
However, the DCC function can be disabled individually for
each channel divider by setting the disable divider DCC bit for
that channel.
Certain M and N values for a channel divider result in a non-
50% duty cycle. A non-50% duty cycle can also result with an
even division, if M ≠ N. The duty-cycle correction function
automatically corrects non-50% duty cycles at the channel
divider output to 50% duty cycle.
Duty-cycle correction requires the following channel divider
conditions:
An even division must be set as M = N.
An odd division must be set as M = N + 1.
When not bypassed or corrected by the DCC function, the duty
cycle of each channel divider output is the numerical value of
(N + 1)/(N + M + 2) expressed as a percent.
The duty cycle at the output of the channel divider for various
Table 29. Channel Divider Output Duty Cycle with VCO
Divider ≠ 1, Input Duty Cycle Is 50%
VCO
Divider
DX
Output Duty Cycle
N + M + 2
Disable Div
DCC = 1
Disable Div
DCC = 0
Even
Channel
divider
bypassed
50%
Odd = 3
Channel
divider
bypassed
33.3%
50%
Odd = 5
Channel
divider
bypassed
40%
50%
Even, odd
Even
(N + 1)/(N + M + 2)
50%, requires
M = N
Even, odd
Odd
(N + 1)/(N + M + 2)
50%, requires
M = N + 1