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HMC832LP6GETR Datasheet(PDF) 28 Page - Analog Devices

Part # HMC832LP6GETR
Description  Cellular infrastructure
PDF  49 Pages
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Manufacturer  AD [Analog Devices]
Direct Link  http://www.analog.com
Logo AD - Analog Devices

HMC832LP6GETR Datasheet(HTML) 28 Page - Analog Devices

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Data Sheet
HMC832
Using Exact Frequency Mode
If the constraint in Equation 16 is satisfied, the HMC832 is able
to generate signals with zero frequency error at the desired
VCO frequency. Exact frequency mode can be reconfigured for
each target frequency, or be setup for a fixed fGCD that applies to
all channels.
Configuring Exact Frequency Mode for a Particular
Frequency
1. Calculate and program the integer register setting
Register 0x03 = NINT = floor(fVCO/fPD)
where the floor function is the rounding down to the
nearest integer.
2. Then calculate the integer boundary frequency
fN = NINT × fPD.
3. Calculate and program the exact frequency register value
Register 0x0C = fPD/fGCD
where fGCD = GCD(fVCO,fPD).
4. Calculate and program the fractional register setting
Register 0x04


−
=
PD
N
VCOk
FRAC
f
f
f
ceil
N
)
(
2
24
where ceil is the ceiling function meaning round up to the
nearest integer.
Example: to configure the HMC832 for exact frequency mode
at fVCO = 2800.2 MHz, where the PD rate (fPD) = 61.44 MHz,
proceed as follows:
1. Check Equation 16 to confirm that the exact frequency
mode for this fVCO is possible.
≥
=
14
2
)
,
(
PD
GCD
PD
VCO
GCD
f
f
and
f
f
GCD
f
fGCD = GCD(2800.2 × 106, 61.44 × 106) =
120 × 103 >
14
6
2
10
44
.
61
×
= 3750
Because Equation 16 is satisfied, the HMC832 can be
configured for exact frequency mode at fVCO = 2800.2 MHz by
continuing with the remaining steps.
2. Calculate NINT
NINT = Register 0x03 =
D
2
x
0
d
45
10
44
.
61
10
2
.
2800
6
6
1
=
=


×
×
=


floor
f
f
floor
PD
VCO
3. Calculate the value for Register 0x0C
Register 0x0C =
00
xC
0
d
3072
20000
10
44
.
61
)
10
44
.
61
,
10
100
(
10
44
.
61
)
),
((
6
6
3
6
1
=
=
×
=
×
×
×
=
−
+
GCD
f
f
f
GCD
f
PD
VCOk
VCOk
PD
4. To program Register 0x04, the closest integer-N boundary
frequency (fN) that is less than the desired VCO frequency
(fVCO) must be calculated: fN = fPD × NINT. Using the current
example
fN = fPD × NINT = 45 × 61.44 × 106 = 2764.8 MHz, then
Register 0x04 =
938000
x
0
d
9666560
10
44
.
61
)
10
8
.
2764
10
2
.
2800
(
2
)
(
2
6
6
6
24
24
=
=


×
×
−
×
=


−
ceil
f
f
f
ceil
PD
N
VCO
Exact Frequency Channel Mode
When multiple, equally spaced, exact frequency channels are
needed that fall within the same interval (that is, fN ≤ fVCOk <
fN + 1) where fVCOk is shown in Figure 46 and 1 ≤ k ≤ 214, it is
possible to maintain the same integer-N (Register 0x03) and
exact frequency register (Register 0x0C) settings and only
update the fractional register (Register 0x04) setting. The exact
frequency channel mode is possible when Equation 16 is
satisfied for at least two equally spaced adjacent frequency
channels, that is, the channel step size.
To configure the HMC832 for exact frequency channel mode,
initially and only at the beginning, the integer (Register 0x03)
and exact frequency (Register 0x0C) registers need to be
programmed for the smallest fVCO frequency (fVCO1 in Figure 46),
as follows:
1. Calculate and program the integer register setting Regis-
ter 0x03 = NINT = floor(fVCO1/fPD), where fVCO1 is shown in
Figure 46 and corresponds to the minimum channel VCO
frequency. Then, the lower integer boundary frequency is
given by fN = NINT × fPD.
2. Calculate and program the exact frequency register value
Register 0x0C = fPD/fGCD, where fGCD = GCD((fVCOk+1 − fVCOk),
fPD) = greatest common divisor of the desired equidistant
channel spacing and the PD frequency ((fVCOk + 1 − fVCOk)
and fPD).
Rev. A | Page 27 of 48



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