参数资料
| 型号： | AD8306AR |
| 厂商： | Analog Devices Inc |
| 文件页数： | 4/16页 |
| 文件大小： | 0K |
| 描述： | IC LOGARITHM AMP 5-400MHZ 16SOIC |
| 标准包装： | 1 |
| 类型： | 限制-对数放大器 |
| 应用： | 接收器信号强度指示（RSSI） |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 16-SOIC（0.154"，3.90mm 宽） |
| 供应商设备封装： | 16-SO |
| 包装： | 管件 |
REV. A
AD8306
–12–
Table I.
Match to 50
Match to 100
(Gain = 13 dB)
(Gain = 10 dB)
fC
CM
LM
CM
LM
MHz
pF
nH
pF
nH
10
140
3500
100.7
4790
10.7
133
3200
94.1
4460
15
95.0
2250
67.1
3120
20
71.0
1660
50.3
2290
21.4
66.5
1550
47.0
2120
25
57.0
1310
40.3
1790
30
47.5
1070
33.5
1460
35
40.7
904
28.8
1220
40
35.6
779
25.2
1047
45
31.6
682
22.4
912
50
28.5
604
20.1
804
60
23.7
489
16.8
644
80
17.8
346
12.6
448
100
14.2
262
10.1
335
120
11.9
208
8.4
261
150
9.5
155
6.7
191
200
7.1
104
5.03
125
250
5.7
75.3
4.03
89.1
300
4.75
57.4
3.36
66.8
350
4.07
45.3
2.87
52.1
400
3.57
36.7
2.52
41.8
450
3.16
30.4
2.24
34.3
500
2.85
25.6
2.01
28.6
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 CIN
At a center frequency fC, the shunt impedance of the input
capacitance CIN can be made to disappear by resonating with a
temporary inductor LIN, whose value is given by
LIN = 1/{(2
π f
C)
2C
IN} = 10
10/f
C
2
(7)
when CIN = 2.5 pF. For example, at fC = 100 MHz, LIN = 1 H.
Step 2: Calculate CO and LO
Now having a purely resistive input impedance, we can calculate
the nominal coupling elements CO and LO, using
C
fR
R
L
RR
f
O
CIN
M
O
IN
M
C
=
()
= ()
1
2
π
;
(8)
For the AD8306, RIN is 1 k. Thus, if a match to 50 is
needed, at fC = 100 MHz, CO must be 7.12 pF and LO must be
356 nH.
Step 3: Split CO Into Two Parts
Since we wish to provide the fully-balanced form of network
shown in Figure 28, two capacitors C1 = C2 each of nominally
twice CO, shown as CM 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
CO = 6.96 pF).
Step 4: Calculate LM
The matching inductor required to provide both LIN and LO is
just the parallel combination of these:
LM = LINLO/(LIN + LO)
(9)
With LIN = 1 H and LO = 356 nH, the value of LM 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
GAIN
R
IN
S
IN
S
=
=
20
10
log
(10)
Altering the Logarithmic Slope
Simple schemes can be used to increase and decrease the loga-
rithmic slope as shown in Figure 30. For the AD8306, only
power, ground and logarithmic output connections are shown;
refer to Figure 24 for complete circuitry. In Figure 30(a), the op
amp’s gain of +2 increases the slope to 40 mV/dB. In Figure
30(b), the AD8031 buffers a resistive divider to give a slope of
Figure 30. Altering the Logarithmic Slope
VPS1
VPS2
PADL, COM1, COM2
AD8306
10
0.1 F
5k
AD8031
0.1 F
10
+5V
40mV/dB
VLOG
(a)
VPS1
VPS2
PADL, COM1, COM2
AD8306
10
0.1 F
AD8031
0.1 F
10
10mV/dB
5k
+5V
VLOG
(b)
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