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參數(shù)資料
| 型號： | LMX1501MDA |
| 廠商： | NATIONAL SEMICONDUCTOR CORP |
| 元件分類： | PLL合成/DDS/VCOs |
| 英文描述： | PLL FREQUENCY SYNTHESIZER, 1100 MHz, UUC |
| 封裝： | DIE |
| 文件頁數(shù)： | 7/25頁 |
| 文件大?。?/td> | 410K |
| 代理商： | LMX1501MDA |
Application Information
LOOP FILTER DESIGN
A block diagram of the basic phase locked loop is shown
TLW12340 – 30
FIGURE 1 Basic Charge Pump Phase Locked Loop
An example of a passive loop filter configuration including
the transfer function of the loop filter is shown in
Figure 2
TLW12340 – 31
Z(s) e
s (C2
R2) a 1
s2 (C1
C2 R2) a sC1 a sC2
FIGURE 2 2nd Order Passive Filter
Define the time constants which determine the pole and
zero frequencies of the filter transfer function by letting
T2 e R2
C2
(1a)
and
T1 e R2
C1
C2
C1 a C2
(1b)
The PLL linear model control circuit is shown along with the
open loop transfer function in
Figure 3 Using the phase
detector and VCO gain constants Kw and KVCO and the
loop filter transfer function Z(s)
the open loop Bode plot
can be calculated The loop bandwidth is shown on the
Bode plot (0p) as the point of unity gain The phase margin
is shown to be the difference between the phase at the unity
gain point and b180
TLW12340 – 33
Open Loop Gain e ii ie e H(s) G(s)
e
Kw Z(s) KVCO Ns
Closed Loop Gain e io ii e G(s) 1 a H(s) G(s)
TLW12340 – 32
FIGURE 3 Open Loop Transfer Function
Thus we can calculate the 3rd order PLL Open Loop Gain in
terms of frequency
G(s)
H(s)
ls e j 0 e
b
Kw
KVCO (1 a j0 T2)
02C1
N(1 a j0 T1)
T1
T2
(2)
From equation 2 we can see that the phase term will be
dependent on the single pole and zero such that
w
(0) e tanb1 (0
T2) b tanb1 (0 T1) a 180
(3)
By setting
dw
d0
e
T2
1 a (0
T2)2
b
T1
1 a (0
T1)2
e
0
(4)
we find the frequency point corresponding to the phase in-
flection point in terms of the filter time constants T1 and T2
This relationship is given in equation 5
0p e 1
0T2 T1
(5)
For the loop to be stable the unity gain point must occur
before the phase reaches b180 degrees We therefore
want the phase margin to be at a maximum when the magni-
tude of the open loop gain equals 1 Equation 2 then gives
C1 e
Kw
KVCO T1
0p2 N T2
(1 a j0p T2)
(1 a j0p T1)
(6)
15
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