AD9517-2A/PCBZ Analog Devices Inc, AD9517-2A/PCBZ Datasheet - Page 42
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AD9517-2A/PCBZ
Manufacturer Part Number
AD9517-2A/PCBZ
Description
12-Output Clock Generator With 2.8GHz VC
Manufacturer
Analog Devices Inc
Datasheet
1.AD9517-2ABCPZ-RL7.pdf
(80 pages)
Specifications of AD9517-2A/PCBZ
Main Purpose
Timing, Clock Generator
Embedded
No
Utilized Ic / Part
AD9517-2A
Primary Attributes
2 Inputs, 12 Outputs, 2.2GHz VCO
Secondary Attributes
CMOS, LVPECL & LVDS Compatible
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
Lead Free Status / RoHS Status
Lead free / RoHS Compliant
AD9517-2
Duty Cycle and Duty-Cycle Correction (0, 1)
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 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 DCCOFF 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:
•
•
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 %.
The duty cycle at the output of the channel divider for various
configurations is shown in Table 35 to Table 37.
Table 35. Duty Cycle with VCO Divider; Input Duty Cycle Is 50%
VCO
Divider
Even
Odd = 3
Odd = 5
Even, Odd
Even, Odd
What are the M and N values for the channel?
Is the DCC enabled?
Is the VCO divider used?
What is the CLK input duty cycle? (The internal VCO has
a 50% duty cycle.)
An even division must be set as M = N.
An odd division must be set as M = N + 1.
N + M + 2
1 (divider
bypassed)
1 (divider
bypassed)
1 (divider
bypassed)
Even
Odd
D
X
DCCOFF = 1
50%
33.3%
40%
(N + 1)/
(N + M + 2)
(N + 1)/
(N + M + 2)
Output Duty Cycle
DCCOFF = 0
50%
50%
50%
50%; requires M = N
50%; requires M = N + 1
Rev. B | Page 42 of 80
Table 36. Duty Cycle with VCO Divider; Input Duty Cycle Is X%
VCO
Divider
Even
Odd = 3
Odd = 5
Even
Odd = 3
Odd = 3
Odd = 5
Odd = 5
Table 37. Channel Divider Output Duty Cycle When the
VCO Divider Is Not Used
Input
Clock
Duty
Cycle
Any
Any
50%
X%
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.
N + M + 2
1 (divider
bypassed)
1 (divider
bypassed)
1 (divider
bypassed)
Even
Odd
Even
Odd
Even
Odd
N + M + 2
1
Even
Odd
Odd
D
D
X
X
DCCOFF = 1
50%
33.3%
40%
(N + 1)/
(N + M + 2)
(N + 1)/
(N + M + 2)
(N + 1)/
(N + M + 2)
(N + 1)/
(N + M + 2)
(N + 1)/
(N + M + 2)
(N + 1)/
(N + M + 2)
DCCOFF = 1
1 (divider
bypassed)
(N + 1)/
(M + N + 2)
(N + 1)/
(M + N + 2)
(N + 1)/
(M + N + 2)
Output Duty Cycle
Output Duty Cycle
DCCOFF = 0
50%
(1 + X%)/3
(2 + X%)/5
50%,
requires M = N
50%,
requires M = N + 1
50%,
requires M = N
(3N + 4 + X%)/(6N + 9),
requires M = N + 1
50%,
requires M = N
(5N + 7 + X%)/(10N +
15), requires M = N + 1
DCCOFF = 0
Same as input
duty cycle
50%, requires M = N
50%, requires
M = N + 1
(N + 1 + X%)/(2 × N + 3),
requires M = N + 1