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參數(shù)資料
| 型號(hào)： | SP8852E |
| 廠商： | Mitel Networks Corporation |
| 英文描述： | 2.7GHz Parallel Load Professional Synthesiser(2.7GHz并行負(fù)載專業(yè)合成器) |
| 中文描述： | 2.7GHz的并聯(lián)負(fù)載專業(yè)合成器（2.7GHz的并行負(fù)載專業(yè)合成器） |
| 文件頁數(shù)： | 11/13頁 |
| 文件大小： | 174K |
| 代理商： | SP8852E |
11
SP8852E
+
C1
R2
C2
FROM
CHARGE PUMP
FROM
CHARGE PUMP
REFERENCE
TO VCO
Fig. 8 Third order loop filter circuit diagram
Loop Filter Design
Generally, the third order filter configuration shown in Fig. 8
gives better results than the more commonly used second order
because the reference sidebands are reduced. Three equations
are required to determine values for the three constants, where
t
1
=
C
1
R
1
t
2
=
R
2
(C
1
1
C
2
)
t
3
=
C
2
R
2
The equations are:
…(2)
…(3)
…(1)
t
2
=
1
v
n2
t
3
2
2
tan
F
0
1
t
3
=
v
n
t
1
=
K
f
K
0
v
n2
N
1
1
v
n
1
1
v
n
2
t
2
2
t
3
2
2
1
2
1
cos
F
0
where
K
f
is the phase detector gain factor in mA/radian
K
0
is the VCO gain factor in radians/seconds/V
N
is the division ratio from VCO to reference frequency
v
n
is the natural loop frequency
F
0
is the phase margin, normally set to 45
°
Since the phase detector used is linear over a range of 2
p
radians, the phase detector gain is given by:
These values can now be substituted in equation (1) to obtain
a value for C
1
and in equations (2) and (3) to determine values
for C
2
and R
2
.
Example
Calculate values for a loop with the following parameters:
Frequency to be synthesised
Reference frequency
Division ratio
K
0
VCO gain factor
F
phase margin
Phase comparator current
1000MHz
10MHz
1000MHz/100MHz = 100
2
p3
10MHz/V
45
°
6·3mA
The phase detector gain factor
K
f
= 6·3/2
p
= 1mA/radian
mA/radian
Phase comparator current setting
2
p
K
f
=
From equation (3):
2
tan
45
°
1
t
3
=
100kHz
3
2
p
0·4142
1
cos 45
°
∴
t
3
=
659
3
10
2
9
=
628319
t
2
=
(100kHz
3
2
p
)
2
3
659
3
10
2
9
∴
t
2
=
3·844
3
10
2
6
1
Using these values in equation (1):
t
1
=
100
3
(100kHz
3
2
p
)
2
1
3
10
2
3
3
2
p3
10MHz/V
3
[A]
1
1
v
n
1
1
v
n
1
1
(100kHz
3
2
p
)
2
3
(3·844
3
10
2
6
)
2
1
1
(100kHz
3
2
p
)
2
3
(659
3
10
2
9
)
2
2
t
2
2
t
3
2
2
where A =
=
1
2
t
1
=
∴
t
1
=
3·84
3
10
2
9
From equation (2):
= 1·59
3
10
2
9
3
2·415
39·48
3
10
2
12
62832
6·833
1·1714
1
2
Substituting for C2:
3·844
3
10
2
6
2
659
3
10
2
9
0·0153
3
10
2
6
∴
R
2
=
829·4
t
2
= R
2
C
1
1
t
3
R
2
=
=659
3
10
2
9
829·4
∴
C
2
=
0·794nF
t
3
=
C
2
R
2
=
t
3
R
2
Now,
t
1
= C
1
∴
C
1
=
3·84nF
t
2
= R
2
(C
1
1
C
2
)
t
2
= C
2
R
2
t
2
2
t
3
C
1
or,
R
2
=
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相關(guān)代理商/技術(shù)參數(shù) |
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