The second is the zero-pole combination:
0
s
------ + 1
sCLF1RLF + 1
wZ
---------------------------------------------------------- = -------------------
s
L
--------- + 1
s CLF1RLF – ---------- + 1
wP1
RLF
This causes lift in the transfer function given by
wP1
---------
wZ
1
---------------------
20 LOG
= 20 LOG
wZ
1 – -----------
wBW
W
W
W
W
P2
Z
P1
BW
To keep peaking to less than 0.05dB,
wZ < 0.0057 wBW
FREQUENCY
Fig. 14 Bode Plot for PLL Transfer Function
9.3 Selection of Loop Filter Components
The 3dB bandwidth of the transfer function is approximately
wBW wBW
Based on the above analysis, select the loop filter
components for a given PLL bandwidth, ƒ3dB, as follows:
--------------------------------------------------------------------- -----------
w3dB
=
≈
0.78
2
wBW
-----------
wP2
(wBW ⁄ wP2
+ ---------------------------------
wBW
-----------
)
1. Calculate
where
1 – 2
1 – 2
wP2
Ι
CP is the charge pump current and is a function of the
RVCO resistor and is obtained from Figure 15.
9.2 Transfer Function Peaking
There are two causes of peaking in the PLL transfer function
given by Equation 1.
Kƒ = 90MHz/V for VCO frequencies corresponding to
the ƒL curve.
The first is the quadratic
Kƒ = 140MHz/V for VCO frequencies corresponding to
the ƒH curve.
s2CLF2L + s
+ 1
L
RLF
---------
N is the divider modulus.
which has
1
(ƒL, ƒH and N can be obtained from Table 2 or Table 3).
2. Choose RLF = 2(3.14) ƒ3dB (0.78)L
CLF2
Q = RLF ------------
L
wO = --------------------
CLF2
L
and
2
3. Choose CLF1 = 174 L / (RLF)
This response is critically damped for Q = 0.5.
Thus, to avoid peaking:
2
4. Choose CLF2 = L/4(RLF)
2N
CPKƒ
L =
CLF2
RLF ------------ <
L
1
2
Ι
--
or
1
L
-------------------------------
> 4
RLFCLF2RLF
Therefore,
wP2 > 4 wBW
However, it is desirable to keep wP2 as low as possible to
reduce the high frequency content on the loop filter.
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