參數(shù)資料
型號： AD9518-4A/PCBZ
廠商： Analog Devices Inc
文件頁數(shù)： 30/64頁
文件大小： 0K
描述： BOARD EVALUATION FOR AD9518-4A
設(shè)計資源： AD9518 Schematics
AD9518 Gerber Files
AD9518-4 BOM
標準包裝： 1
主要目的： 計時，時鐘發(fā)生器
嵌入式： 否
已用 IC / 零件： AD9518-4A
主要屬性： 2 輸入，14 輸出，1.6GHz VCO
次要屬性： LVPECL 輸出邏輯
已供物品： 板
AD9518-4
Data Sheet
Rev. B | Page 36 of 64
The duty cycle at the output of the channel divider for various
configurations is shown in Table 33 to Table 35.
Table 33. Duty Cycle with VCO Divider, Input Duty Cycle Is 50%
VCO
Divider
DX
Output Duty Cycle
N + M + 2
DCCOFF = 1
DCCOFF = 0
Even
1 (divider
bypassed)
50%
Odd = 3
1 (divider
bypassed)
33.3%
50%
Odd = 5
1 (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
Table 34. Duty Cycle with VCO Divider, Input Duty Cycle Is X%
VCO
Divider
DX
Output Duty Cycle
N + M + 2
DCCOFF = 1
DCCOFF = 0
Even
1 (divider
bypassed)
50%
Odd = 3
1 (divider
bypassed)
33.3%
(1 + X%)/3
Odd = 5
1 (divider
bypassed)
40%
(2 + X%)/5
Even
(N + 1)/
(N + M + 2)
50%,
requires M = N
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 35. Channel Divider Output Duty Cycle When the
VCO Divider Is Not Used
Input Clock
Duty Cycle
DX
Output Duty Cycle
N + M + 2
DCCOFF = 1
DCCOFF = 0
Any
1
1 (divider
bypassed)
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
The internal VCO has a duty cycle of 50%. Therefore, when the
VCO is connected directly to the output, the duty cycle is 50%.
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 (0, 1, and 2)
Each channel divider allows for a phase offset, or a coarse time
delay, to be programmed by setting register bits (see Table 36).
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) that are
programmed for the divider.
The sync function must be used to make phase offsets effective
Table 36. Setting Phase Offset and Division for Divider 0,
Divider 1, and Divider 2
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]
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
By giving each divider a different phase offset, output-to-output
delays can be set in increments of the channel divider input
clock cycle. Figure 40 shows the results of setting such a coarse
offset between outputs.
CHANNEL DIVIDER OUTPUTS
DIV = 4, DUTY = 50%
0
123
45
678
9
10 11 12 13 14 15
Tx
DIVIDER 0
DIVIDER 1
DIVIDER 2
CHANNEL
DIVIDER INPUT
SH = 0
PO = 0
SH = 0
PO = 1
SH = 0
PO = 2
1 × Tx
2 × Tx
06433-
071
Figure 40. Effect of Coarse Phase Offset (or Delay)
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