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參數資料
| 型號： | AD8309-EVAL |
| 廠商： | Analog Devices, Inc. |
| 元件分類： | 運動控制電子 |
| 英文描述： | 5 MHz.500 MHz 100 dB Demodulating Logarithmic Amplifier with Limiter Output |
| 中文描述： | 5 MHz.500兆赫和100分貝解調對數放大器輸出限幅 |
| 文件頁數： | 15/20頁 |
| 文件大小： | 311K |
| 代理商： | AD8309-EVAL |
REV. B
AD8309
–15–
sensitivity
, but now a measure of
selectively
is simultaneously
introduced. Second, the component count is low: two capacitors
and an inexpensive chip inductor are needed. Third, the net-
work also serves as a balun. Analysis of this network shows that
the amplitude of the voltages at INHI and INLO are quite simi-
lar when the impedance ratio is fairly high (say, 50
to 1000
).
1
2
3
4
5
6
7
8
VLOG
VPS2
PADL
LMHI
LMLO
PADL
FLTR
LMDR
COM2
VPS1
PADL
INHI
INLO
PADL
COM1
ENBL
AD8309
9
10
11
14
15
16
0.1
m
F
10
V
NC
R
LIM
RSSI
LIMITER
OUTPUT
0.1
m
F
10
V
C2 = C
M
Z
IN
NC = NO CONNECT
12
13
V
S
C1 = C
M
L
M
Figure 33. High Frequency Input Matching Network
Figure 34 shows the response for a center frequency of 100 MHz.
The response is down by 50 dB at one-tenth the center frequency,
falling by 40 dB per decade below this. The very high frequency
attenuation is relatively small, however, since in the limiting
case it is determined simply by the ratio of the AD8309’s input
capacitance to the coupling capacitors. Table I provides solu-
tions for a variety of center frequencies f
C
and matching from
impedances Z
IN
of nominally 50
and 100
. Exact values are
shown, and some judgment is needed in utilizing the nearest
standard values.
Table I.
Match to 50
V
(Gain = 13 dB)
C
M
pF
Match to 100
V
(Gain = 10 dB)
C
M
pF
f
C
MHz
L
M
nH
L
M
nH
10
10.7
15
20
21.4
25
30
35
40
45
50
60
80
100
120
150
200
250
300
350
400
450
500
140
133
95.0
71.0
66.5
57.0
47.5
40.7
35.6
31.6
28.5
23.7
17.8
14.2
11.9
9.5
7.1
5.7
4.75
4.07
3.57
3.16
2.85
3500
3200
2250
1660
1550
1310
1070
904
779
682
604
489
346
262
208
155
104
75.3
57.4
45.3
36.7
30.4
25.6
100.7
94.1
67.1
50.3
47.0
40.3
33.5
28.8
25.2
22.4
20.1
16.8
12.6
10.1
8.4
6.7
5.03
4.03
3.36
2.87
2.52
2.24
2.01
4790
4460
3120
2290
2120
1790
1460
1220
1047
912
804
644
448
335
261
191
125
89.1
66.8
52.1
41.8
34.3
28.6
FREQUENCY – MHz
14
13
12
11
10
60
D
9
8
7
6
5
4
3
2
1
0
70
80
90
100
110
120
130
–1
140
150
GAIN
INPUT AT
TERMINATION
Figure 34. Response of 100 MHz Matching Network
General Matching Procedure
For other center frequencies and source impedances, the following
method can be used to calculate the basic matching parameters.
Step 1: Tune Out C
IN
At a center frequency f
C
, the shunt impedance of the input
capacitance C
IN
can be made to disappear by resonating with a
temporary inductor L
IN
, whose value is given by
L
IN
=
1/{(2
π
f
C
)
2
C
IN
}
=
10
10
/f
C
2
when
C
IN
= 2.5 pF. For example, at
f
C
= 100 MHz,
L
IN
= 1
μ
H.
Step 2: Calculate C
O
and L
O
Now having a purely resistive input impedance, we can calculate
the nominal coupling elements C
O
and L
O
, using
(8)
C
f
R R
L
R R
2
π
f
O
C
M
O
M
C
=
(
)
=
(
)
1
2
π
;
(9)
For the AD8309, R
IN
is 1 k
. Thus, if a match to 50
is
needed, at f
C
= 100 MHz, C
O
must be 7.12 pF and L
O
must be
356 nH.
Step 3: Split C
O
Into Two Parts
Since we wish to provide the fully-balanced form of network
shown in Figure 33, two capacitors C1 = C2
each of nominally
twice C
O
, shown as C
M
in the figure, can be used. This requires
a value of 14.24 pF in this example. Under these conditions, the
voltage amplitudes at INHI and INLO will be similar. A some-
what better balance in the two drives may be achieved when C1
is made slightly larger than C2, which also allows a wider range
of choices in selecting from standard values. For example, ca-
pacitors of C1 = 15 pF and C2 = 13 pF may be used (making
C
O
= 6.96 pF).
Step 4: Calculate L
M
The matching inductor required to provide both L
IN
and L
O
is
just the parallel combination of these:
L
M
= L
IN
L
O
/(
L
IN
+
L
O
)
With
L
IN
= 1
μ
H and
L
O
= 356 nH, the value of
L
M
to complete
this example of a match of 50
at 100 MHz is 262.5 nH. The
nearest standard value of 270 nH may be used with only a slight
loss of matching accuracy. The voltage gain at resonance de-
pends only on the ratio of impedances, as is given by
(10)
GAIN
R
R
R
R
IN
S
IN
S
=
=
20
10
log
log
(11)
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