AD9517-1BCPZ Analog Devices Inc, AD9517-1BCPZ Datasheet - Page 44

• Manufacturer Part Number: AD9517-1BCPZ
• Description: 12-Output Clock Generator With 2.5GHz VC
• Manufacturer: Analog Devices Inc
• Type: Clock Generator, Fanout Distributionr

Specifications of AD9517-1BCPZ

• Pll: Yes
• Input: CMOS, LVDS, LVPECL
• Output: CMOS, LVDS, LVPECL
• Number Of Circuits: 1
• Ratio - Input:output: 1:12
• Differential - Input:output: Yes/Yes
• Frequency - Max: 2.65GHz
• Divider/multiplier: Yes/No
• Voltage - Supply: 3.135 V ~ 3.465 V
• Operating Temperature: -40°C ~ 85°C
• Mounting Type: Surface Mount
• Package / Case: 48-LFCSP
• Frequency-max: 2.65GHz
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• For Use With: AD9517-1/PCBZ - BOARD EVAL FOR AD9517-1
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant

AD9517-1
Duty Cycle and Duty-Cycle Correction (Divider 2 and
Divider 3)
The same duty cycle and DCC considerations apply to Divider 2
and Divider 3 as to Divider 0 and Divider 1 (see the Duty Cycle
and Duty-Cycle Correction (0, 1) section); however, with these
channel dividers, the number of possible configurations is even
more complex.
Duty-cycle correction on Divider 2 and Divider 3 requires the
following channel divider conditions:
•
•
•
•
The possibilities for the duty cycle of the output clock from
Divider 2 and Divider 3 are shown in Table 40 through
Table 44.
Table 40. Divider 2 and Divider 3 Duty Cycle; VCO Divider
Used; Duty Cycle Correction Off (DCCOFF = 1)
VCO
Divider
Even
Odd = 3
Odd = 5
Even
Odd
Even
Odd
Table 41. Divider 2 and Divider 3 Duty Cycle; VCO Divider
Not Used; Duty Cycle Correction Off (DCCOFF = 1)
Input Clock
Duty Cycle
50%
X%
50%
X%
50%
X%
An even D
cycles).
An odd D
low cycles must be one greater than the number of high
cycles).
If only one divider is bypassed, it must be the second
divider, X.2.
If only one divider has an even divide-by, it must be the
second divider, X.2.
X.Y
X.Y
N
1
1
1
Even, odd
Even, odd
Even, odd
Even, odd
N
1
1
Even, odd
Even, odd
Even, odd
Even, odd
X.1
X.1
must be set as M
must be set as M
+ M
+ M
D
D
X.1
X.1
X.1
X.1
+ 2
+ 2
N
1
1
1
1
1
Even, odd
Even, odd
N
1
1
1
1
Even, odd
Even, odd
X.2
X.2
X.Y
X.Y
+ M
+ M
= N
D
D
= N
X.2
X.2
X.2
X.2
X.Y
X.Y
+ 2
+ 2
+ 1 (the number of
(low cycles = high
Output
Duty Cycle
50%
33.3%
40%
(N
(N
(N
(N
(N
(N
(N
(N
Output
Duty Cycle
50%
X%
(N
(N
(N
(N
(N
(N
(N
(N
X.1
X.1
X.1
X.1
X.2
X.2
X.2
X.2
X.1
X.1
X.1
X.1
X.2
X.2
X.2
X.2
+ 1)/
+ M
+ 1)/
+ M
+ 1)/
+ M
+ 1)/
+ M
+ 1)/
+ M
+ 1)/
+ M
+ 1)/
+ M
+ 1)/
+ M
X.1
X.1
X.2
X.2
X.1
X.1
X.2
X.2
+ 2)
+ 2)
+ 2)
+ 2)
+ 2)
+ 2)
+ 2)
+ 2)
Rev. B | Page 44 of 80
Table 42. Divider 2 and Divider 3 Duty Cycle; VCO Divider
Used; Duty Cycle Correction Is On (DCCOFF = 0); VCO
Divider Input Duty Cycle = 50%
VCO
Divider
Even
Odd
Even
Odd
Even
Odd
Even
Odd
Even
Odd
Even
Odd
Table 43. Divider 2 and Divider 3 Duty Cycle; VCO Divider
Used; Duty Cycle Correction On (DCCOFF = 0); VCO
Divider Input Duty Cycle = X%
VCO
Divider
Even
Odd = 3
Odd = 5
Even
Odd
Even
Odd = 3
Odd = 5
Even
Odd
Even
Odd
Even
Odd = 3
Odd = 5
N
1
1
Even (N
Even (N
Odd (M
Odd (M
Even (N
Even (N
Odd (M
Odd (M
Odd (M
Odd (M
N
1
1
1
Even
(N
Even
(N
Odd
(M
Odd
(M
Odd
(M
Even
(N
Even
(N
Odd
(M
Odd
(M
Odd
(M
Odd
(M
Odd
(M
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
+ M
+ M
= M
= M
= N
= N
= N
= M
= M
= N
= N
= N
= N
= N
D
X.1
X.1
X.1
X.1
X.1
X.1
X.1
D
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
= N
= N
= N
= N
= N
= N
= M
= M
= M
= M
+ 1)
+ 1)
+ 1)
+ 1)
+ 1)
+ 1)
+ 1)
+ 1)
)
)
)
)
+ 2
+ 2
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
X.1
+ 1)
+ 1)
+ 1)
+ 1)
+ 1)
+ 1)
)
)
)
)
N
1
1
1
1
1
1
1
1
Even
(N
Even
(N
Even
(N
Even
(N
Odd
(M
Odd
(M
Odd
(M
X.2
X.2
X.2
X.2
X.2
X.2
X.2
X.2
+ M
N
1
1
1
1
1
1
Even (N
Even (N
Even (N
Even (N
Odd (M
Odd (M
= M
= M
= M
= M
= N
= N
= N
D
X.2
X.2
X.2
X.2
X.2
X.2
+ M
X.2
X.2
X.2
X.2
+ 1)
+ 1)
+ 1)
)
)
)
)
+ 2
X.2
X.2
D
X.2
X.2
X.2
X.2
X.2
X.2
= N
= N
= M
= M
= M
= M
+ 2
X.2
X.2
Output
Duty Cycle
50%
(1 + X%)/3
(2 + X%)/5
50%
50%
50%
(3N
(6N
(5N
(10N
50%
50%
50%
50%
50%
(6N
9N
(3(2N
(2N
(10N
15N
(5(2 N
(2 N
X.2
X.2
X.2
X.2
+ 1)
+ 1)
)
)
)
)
X.2
X.1
X.1
X.1
X.1
X.2
X.2
X.2
X.1
X.1
N
+ 13 + X%)/
X.1
+ 4 + X%)/
+ 9)
+ 7 + X%)/
+ 3))
X.1
N
+ 22 + X%)/
+ 3))
X.2
+ 15)
X.2
+ 3)
Output
Duty Cycle
50%
50%
50%
50%
50%
50%
50%
50%
50%
50%
50%
50%
+ 3)
+ 9N
+ 15N
X.1
X.1
+
+