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AD9522-5/PCBZ датащи(PDF) 40 Page - Analog Devices

номер детали AD9522-5/PCBZ
подробное описание детали  12 LVDS/24 CMOS Output Clock Generator
PDF  76 Pages
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производитель  AD [Analog Devices]
домашняя страница  http://www.analog.com
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AD9522-5/PCBZ датащи(HTML) 40 Page - Analog Devices

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AD9522-5
Rev. 0 | Page 40 of 76
Table 30. Channel Divider Output Duty Cycle with VCO
Divider ≠ 1, Input Duty Cycle Is X%
VCO
Divider
DX
Output Duty Cycle
N + M + 2
Disable Div
DCC = 1
Disable Div DCC = 0
Even
Channel
divider
bypassed
50%
50%
Odd = 3
Channel
divider
bypassed
33.3%
(1 + X%)/3
Odd = 5
Channel
divider
bypassed
40%
(2 + X%)/5
Even
Even
(N + 1)/
(N + M + 2)
50%, requires M = N
Even
Odd
(N + 1)/
(N + M + 2)
50%, requires M = N + 1
Odd = 3
Even
(N + 1)/
(N + M + 2)
50%, requires M = N
Odd = 3
Odd
(N + 1)/
(N + M + 2)
(3N + 4 + X%)/(6N + 9),
requires M = N + 1
Odd = 5
Even
(N + 1)/
(N + M + 2)
50%, requires M = N
Odd = 5
Odd
(N + 1)/
(N + M + 2)
(5N + 7 + X%)/(10N + 15),
requires M = N + 1
Table 31. Channel Divider Output Duty Cycle When the
VCO Divider Is Enabled and Set to 1
Input
Clock
Duty Cycle
DX
Output Duty Cycle
N + M + 2
Disable Div
DCC = 1
Disable Div DCC = 0
Any
Even
(N + 1)/
(M + N + 2)
50%, requires M = N
50%
Odd
(N + 1)/
(M + N + 2)
50%, requires M = N + 1
X%
Odd
(N + 1)/
(M + N + 2)
(N + 1 + X%)/(2 × N + 3),
requires M = N + 1
Note that the channel divider must be enabled when the VCO
divider = 1.
Table 32. Channel Divider Output Duty Cycle When the
VCO Divider Is Bypassed
Input
Clock
Duty Cycle
DX
Output Duty Cycle
N + M + 2
Disable Div
DCC = 1
Disable Div DCC = 0
Any
Channel
divider
bypassed
Same as input
duty cycle
Same as input duty
cycle
Any
Even
(N + 1)/
(M + N + 2)
50%, requires M = N
50%
Odd
(N + 1)/
(M + N + 2)
50%, requires M = N + 1
X%
Odd
(N + 1)/
(M + N + 2)
(N + 1 + X%)/(2 × N + 3),
requires M = N + 1
If the CLK input is routed directly to the output, the duty cycle of
the output is the same as the CLK input.
Phase Offset or Coarse Time Delay
Each channel divider allows for a phase offset, or a coarse time
delay, to be programmed by setting register bits (see Table 33).
These settings determine the number of cycles (successive rising
edges) of the channel divider input frequency by which to offset, or
delay, the rising edge of the output of the divider. This delay is
with respect to a nondelayed output (that is, with a phase offset
of zero). The amount of the delay is set by five bits loaded into
the phase offset (PO) register plus the start high (SH) bit for
each channel divider. When the start high bit is set, the delay is
also affected by the number of low cycles (M) programmed for
the divider.
It is necessary to use the SYNC function to make phase offsets
effective (see the Synchronizing the Outputs—SYNC Function
section).
Table 33. Setting Phase Offset and Division
Divider
Start
High (SH)
Phase
Offset (PO)
Low Cycles
M
High Cycles
N
0
0x191[4]
0x191[3:0]
0x190[7:4]
0x190[3:0]
1
0x194[4]
0x194[3:0]
0x193[7:4]
0x193[3:0]
2
0x197[4]
0x197[3:0]
0x196[7:4]
0x196[3:0]
3
0x19A[4]
0x19A[3:0]
0x199[7:4]
0x199[3:0]
Let
Δt = delay (in seconds).
Δc = delay (in cycles of clock signal at input to DX).
TX = period of the clock signal at the input of the divider, DX (in
seconds).
Φ =
16 × SH[4] + 8 × PO[3] + 4 × PO[2] + 2 × PO[1] + 1 × PO[0]
The channel divide-by is set as N = high cycles and M = low
cycles.
Case 1
For Φ ≤ 15,
Δt = Φ × TX
Δc = Δt/TX = Φ
Case 2
For Φ ≥ 16,
Δt = (Φ − 16 + M + 1) × TX
Δc = Δt/TX



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