參數(shù)資料
型號(hào)： PRISM1KIT-EVAL
廠商： Intersil Corporation
英文描述： ()
中文描述： （）
文件頁(yè)數(shù)： 16/21頁(yè)
文件大?。?/td> 140K
代理商： PRISM1KIT-EVAL
16
Appendix D
A linear control system model of the phase feedback for a
PLL in the locked state is shown in Figure 8. The open loop
gain is the product of the phase detector gain (K
PD
), the
VCO gain (KVCO/s), and the loop filter gain Z(s) divided by
the gain of the feedback counter modulus (N). The passive
loop filter configuration used is displayed in Figure 9.
Synthesizer Loop Analysis
This analysis calculates the loop filter components for the
PRISM HFA3524A Dual PLL.
A. Schaldenbrand and R.D. Schultz, 5/8/96
Where:
LO1, 2132MHz to 2204MHz,
f
loop
is the desired bandwidth of the PLL, in Hz,
f
ref
is the reference frequency of the loop, in Hz,
φ
is the desired phase margin of the PLL, in degrees,
Kvco is the VCO Tuning Voltage Constant in Hz/V,
Kpd is the Phase Detector Pump Constant in A/rad,
N is the divider ratio,
1.424 is a constant to account for impact of R3/C3 pole.
Loop Filter Component Values
T1 = 4.068
10
-6
T2 = 4.472
10
-5
T3 = 1.592
10
-6
f
c
= 1
10
4
C1 = 6.47
10
-9
C2 = 6.47
10
-8
C1 : = 0.0068
10
-6
C2 : = 0.068
10
-6
R2 = 691.064
R2 : = 680
Rules of thumb for selecting the values of the reference
frequency suppression filter are:
1. Choose C3 < C1/10
2. Choose R3 < 2
R2
K
VCO
s
Z(s)
K
PD
1/N
∑
-
Θ
E
Θ
R
Θ
I
+
Θ
O
FIGURE 8. PLL LINEAR MODEL
C2
R2
C1
VCO
D
O
FIGURE 9. PASSIVE LOOP FILTER
R3
C3
f
loop
:
10
10
3
=
ω
p
:
2
π
f
loop
=
ω
p
6.283
10
4
=
f
ref
:
1
10
6
=
ω
r
:
2
π
f
ref
=
ω
r
6.283
10
6
=
φ
:
50
°
=
φ
r
:
π
°
------------
=
Kvco :
53
10
6
=
Kvco :
2
π
Kvco
=
Kvco
3.33
10
8
=
Kpd :
0.004
=
Kpd :
π
-----------
=
Kpd
6.366
10
4
–
=
N :
2168
=
T3 :
10
p
sec
---------------------------------------------
-------------------
=
ω
p
:
1.424
ω
p
=
T1 :
φ
(
)
p
φ
r
(
)
tan
–
=
T1
4.068
10
6
–
=
ω
c
:
φ
(
)
tan
T1
T3
+
(
)
)
T1
T3
+
(
T1
T3
+
[
]
------------------------------------------------------------
1
φ
r
)
2
(
T1
T3
+
(
)
tan
T1
T3
+
)
(
)
2
------------+
+
1
–
=
T2 :
ω
c
T1
T3
+
(
)
-----------------------------------------
=
f
c
:
ω
c
π
-----------
=
C1 :
--------------------------------------------
T2
ω
c
N
1
ω
)
T2
(
)
2
+
1
ω
c
T1
(
+
[
]
1
ω
c
T3
(
)
+
[
]
[
]
------------------------------------------------------------------------------------------------
=
C2 :
C1
------
1
–
=
R2 :
-------
=
C3 :
-------
=
R3 :
-------
=
R3
234.051
=
C3
6.8
10
9
–
=
R3 :
2
R2
=
R3
1.36
10
3
=
R3
1300
=
C3 :
-------
=
C3
1.224
10
9
–
=
C3 :
1200
10
12
–
=
k
10..70
=
kp k
:
10
k
10
------
=
ω
(k) :
2j
π
kp k
=
GH(k) :
Kpd
Kvco
k
ω
k
( )
(
)
C1
N
ω
k
( )
(
1
ω
k
( )
T1
+
)
----------------------------------------+
=
------
k
)
-------+
MAGdb_GH(k) :
20 Log GH k
GH k
)
=
PHI_GH(k) :
)
)
------------------------------
atan
=
f_GH(k) :
--------------------------------------------
180
–
=
MAG_GH(k) :
GH k
GH k
(
)
=
Phim(k) :
180
f_GH(k)
+
=
Application Note 9624
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