AD9516-2BCPZ Analog Devices Inc, AD9516-2BCPZ Datasheet - Page 44

• Manufacturer Part Number: AD9516-2BCPZ
• Description: IC,Ten Distributed-Output Clock Driver,LLCC,64PIN,PLASTIC
• Manufacturer: Analog Devices Inc
• Type: Clock Generator, Fanout Distributionr

Specifications of AD9516-2BCPZ

• Pll: Yes
• Input: Clock
• Output: CMOS, LVDS, LVPECL
• Number Of Circuits: 1
• Ratio - Input:output: 1:14
• Differential - Input:output: Yes/Yes
• Frequency - Max: 2.33GHz
• Divider/multiplier: Yes/No
• Voltage - Supply: 3.135 V ~ 3.465 V
• Operating Temperature: -40°C ~ 85°C
• Mounting Type: Surface Mount
• Package / Case: 64-LFCSP
• Frequency-max: 2.33GHz
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• For Use With: AD9516-2/PCBZ - BOARD EVAL FOR AD9516-2 2.2GHZ
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant

AD9516-2
Duty Cycle and Duty-Cycle Correction (Divider 3 and
Divider 4)
The same duty cycle and DCC considerations apply to Divider 3
and Divider 4 as to Divider 0, Divider 1, and Divider 2 (see the
Duty Cycle and Duty-Cycle Correction (0, 1, and 2) section;
however, with these channel dividers, the number of possible
configurations is even more complex.
Duty-cycle correction on Divider 3 and Divider 4 requires the
following channel divider conditions:
•
•
•
•
The possibilities for the duty cycle of the output clock from
Divider 3 and Divider 4 are shown in Table 40 through Table 44.
Table 40. Divider 3, Divider 4 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 3, Divider 4 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
Even, odd
Even, odd
Even, odd
Even, odd
N
1
1
Even, odd
Even, odd
1
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
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
Output Duty
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
Cycle
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. A | Page 44 of 80
Table 42. Divider 3, Divider 4 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 3, Divider 4 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
N
1
1
1
1
1
1
Even (N
Even (N
Even (N
Even (N
Odd (M
Odd (M
+ 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
+ 2
+ 1)
+ 1)
+ 1)
)
)
)
)
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
+ 13 + X%)/
N
X.1
+ 4 + X%)/
+ 9)
+ 7 + X%)/
+ 3))
X.1
N
+ 3))
+ 22 + X%)/
X.2
+ 15)
+ 3)
X.2
Output
Duty Cycle
50%
50%
50%
50%
50%
50%
50%
50%
50%
50%
50%
50%
+ 3)
+ 9N
+ 15N
X.1
X.1
+
+