AD8310_15 Datasheet, PDF (16/24 Pages) Analog Devices – Fast, Voltage-Out, DC to 440 MHz, 95 dB Logarithmic Amplifier

AD8310
NARROW-BAND MATCHING
Transformer coupling is useful in broadband applications.
However, a magnetically coupled transformer might not be
convenient in some situations. Table 5 lists narrow-band
matching values.
Table 5. Narrow-Band Matching Values
fC
ZIN
(MHz) (Ω)
C1
C2
LM
(pF) (pF) (nH)
10
45
160 150 3300
20
44
82
75
1600
50
46
30
27
680
100 50
15
13
270
150 57
10
8.2
220
200 57
7.5
6.8
150
250 50
6.2
5.6
100
500 54
3.9
3.3
39
10
103 100 91
5600
20
102 51
43
2700
50
99
22
18
1000
100 98
11
9.1
430
150 101 7.5
6.2
260
200 95
5.6
4.7
180
250 92
4.3
3.9
130
500 114 2.2
2.0
47
Voltage Gain
(dB)
13.3
13.4
13.4
13.4
13.2
12.8
12.3
10.9
10.4
10.4
10.6
10.5
10.3
10.3
9.9
6.8
At high frequencies, it is often preferable to use a narrow-band
matching network, as shown in Figure 31. This has several advan-
tages. The same voltage gain is achieved, providing increased
sensitivity, but a measure of selectivity is also introduced. The
component count is low: two capacitors and an inexpensive chip
inductor. Additionally, by making these capacitors unequal, the
amplitudes at INP and INM can be equalized when driving from
a single-sided source; that is, the network also serves as a balun.
Figure 32 shows the response for a center frequency of 100 MHz;
note the very high attenuation at low frequencies. The high fre-
quency attenuation is due to the input capacitance of the log amp.
C1
SIGNAL
INPUT
8
INHI
L_M AD8310
INLO
C2
1
Figure 31. Reactive Matching Network
14
13
12
11
GAIN
10
9
8
7
6
5
4
3
INPUT
2
1
0
–1
60 70 80 90 100 110 120 130 140 150
FREQUENCY (MHz)
Figure 32. Response of 100 MHz Matching Network
GENERAL MATCHING PROCEDURE
For other center frequencies and source impedances, the
following steps can be used to calculate the basic matching
parameters.
Step 1: Tune Out C_IN
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
L _IN
=
1
ω² C _IN
(5)
where CIN = 1.4 pF. For example, at fC = 100 MHz, LIN = 1.8 μH.
Step 2: Calculate C_O and L_O
Now, having a purely resistive input impedance, calculate the
nominal coupling elements, CO and LO, using
C_O = _2πfC
1
;
R_IN R_M
L_O =
(R_IN R_M )
2 πf_C
(6)
For the AD8310, RIN is 1 kΩ. Therefore, if a match to 50 Ω is
needed, at fC = 100 MHz, CO must be 7.12 pF and LO must be
356 nH.
Step 3: Split C_O into Two Parts
To provide the desired fully balanced form of the network
shown in Figure 31, two capacitors C1 and C2, each of
nominally twice CO, can be used. This requires a value of
14.24 pF in this example. Under these conditions, the voltage
amplitudes at INHI and INLO are similar. A somewhat better
balance in the two drives can 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, capacitors of C1 = 15 pF and C2 = 13 pF can be
used, making CO = 6.96 pF.
Rev. F | Page 16 of 24