AD8309 Analog Devices, AD8309 Datasheet - Page 15

• Manufacturer Part Number: AD8309
• Description: 5 - 500 Mhz, 100 DB Demodulating Logarithmic Amplifier With Limiter Output
• Manufacturer: Analog Devices

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
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
impedances Z
shown, and some judgment is needed in utilizing the nearest
standard values.
f
MHz
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
REV. B
C
Z
IN
Figure 33. High Frequency Input Matching Network
C1 = C
C2 = C
M
M
IN
0.1 F
Match to 50
(Gain = 13 dB)
C
pF
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
L
M
10
M
of nominally 50
1
2
3
4
5
6
7
8
NC = NO CONNECT
COM2
VPS1
PADL
INHI
INLO
PADL
COM1
ENBL
Table I.
L
nH
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
AD8309
M
and 100 . Exact values are
LMDR
VLOG
LMLO
PADL
PADL
VPS2
FLTR
LMHI
C
16
15
14
13
12
11
10
9
10
and matching from
NC
(Gain = 10 dB)
C
pF
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
Match to 100
R
M
LIM
0.1 F
to 1000 ).
LIMITER
OUTPUT
V
RSSI
S
L
nH
4790
4460
3120
2290
2120
1790
1460
912
804
644
448
335
191
125
89.1
66.8
52.1
41.8
34.3
28.6
1220
1047
261
M
–15–
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
At a center frequency f
capacitance C
temporary inductor L
when C
Step 2: Calculate C
Now having a purely resistive input impedance, we can calculate
the nominal coupling elements C
For the AD8309, R
needed, at f
356 nH.
Step 3: Split C
Since we wish to provide the fully-balanced form of network
shown in Figure 33, two capacitors C1 = C2 each of nominally
twice C
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
Step 4: Calculate L
The matching inductor required to provide both L
just the parallel combination of these:
With L
this example of a match of 50
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
O
= 6.96 pF).
Figure 34. Response of 100 MHz Matching Network
L
L
GAIN
C
IN
M
O
IN
IN
O
= L
, shown as C
= 1/{(2
= 1 H and L
= 2.5 pF. For example, at f
2
14
13
12
11
10
–1
9
8
7
6
5
4
3
2
1
0
C
IN
60
f
= 100 MHz, C
L
IN
C
20
O
O
/(L
can be made to disappear by resonating with a
70
Into Two Parts
log
1
R R
IN
f
IN
O
IN
C
IN
M
80
)
IN
M
and L
+ L
2
C
is 1 k . Thus, if a match to 50
C
O
, whose value is given by
M
R
in the figure, can be used. This requires
, the shunt impedance of the input
R
IN
= 356 nH, the value of L
O
IN
90
TERMINATION
S
;
} = 10
)
FREQUENCY – MHz
INPUT AT
O
O
L
100
must be 7.12 pF and L
O
GAIN
10
10
O
at 100 MHz is 262.5 nH. The
110
/f
and L
log
C
R R
2
C
2
IN
120
= 100 MHz, L
R
f
R
O
C
IN
S
, using
M
130
140
AD8309
M
IN
150
to complete
and L
O
IN
must be
= 1 H.
is
O
(10)
(11)
is
(8)
(9)