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參數(shù)資料
| 型號： | GS9035CCTJ |
| 廠商： | Gennum Corporation |
| 英文描述： | GENLINX II -TM GS9035C Serial Digital Reclocker |
| 中文描述： | GENLINX二，商標GS9035C串行數(shù)字時鐘恢復(fù)器 |
| 文件頁數(shù)： | 10/14頁 |
| 文件大小： | 490K |
| 代理商： | GS9035CCTJ |
GENNUM CORPORATION
20582 - 3
10 of 14
G
Fig. 14 Bode Plot for PLL Transfer Function
The 3dB bandwidth of the transfer function is approximately
9.2 Transfer Function Peaking
There are two causes of peaking in the PLL transfer function
given by Equation 1.
The first is the quadratic
which has
and
This response is critically damped for Q = 0.5.
Thus, to avoid peaking:
or
Therefore,
w
P2
> 4 w
BW
However, it is desirable to keep w
P2
as low as possible to
reduce the high frequency content on the loop filter.
The second is the zero-pole combination:
This causes lift in the transfer function given by
To keep peaking to less than 0.05dB,
w
Z
< 0.0057 w
BW
9.3 Selection of Loop Filter Components
Based on the above analysis, select the loop filter
components for a given PLL bandwidth,
3dB
, as follows:
1. Calculate
where
Ι
CP
is the charge pump current and is a function of the
R
VCO
resistor and is obtained from Figure 15.
K
= 90MHz/V for VCO frequencies corresponding to
the
L
curve.
K
= 140MHz/V for VCO frequencies corresponding to
the
H
curve.
N is the divider modulus.
(
L
,
H
and N can be obtained from Table 2 or Table 3).
2. Choose R
LF
= 2(3.14)
3dB
(0.78)L
3. Choose C
LF1
= 174 L / (R
LF
)
2
4. Choose C
LF2
= L/4(R
LF
)
2
W
Z
0
W
P1
W
BW
W
P2
FREQUENCY
A
w
3dB
w
1
2w
w
P2
-----------
–
w
w
2w
–
---------------------------------
)
2
1
P2
-----------
+
---------------------------------------------------------------------
w
0.78
-----------
≈
=
s
2
C
LF2
L
sL
R
LF
1
+
+
w
O
C
LF2
L
--------------------
=
Q
R
LF
C
L
------------
=
R
LF
C
L
------------
1
2
--
<
R
LF
C
LF2
----------------------
R
LF
---------
4
>
sC
R
----------------------------------------------------------
1
+
s C
LF1
R
LF
LF
----------
–
1
+
w
w
P1
------
-------------------
1
+
---------
1
+
=
20 LOGw
w
Z
---------
20 LOG
1
BW
-----------
–
---------------------
=
2N
Ι
CP
K
L =
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