Datasheet

AD8310
Rev. F | Page 16 of 24
FREQUENCY (MHz)
14
4
–1
60 15080
DECIBELS
100 110 130
3
2
1
0
70 90 120 140
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.
GAIN
Table 5. Narrow-Band Matching Values
f
C
(MHz)
Z
IN
(Ω)
C1
(pF)
C2
(pF)
L
M
(nH)
Voltage Gain
(dB)
10 45 160 150 3300 13.3
20 44 82 75 1600 13.4
50 46 30 27 680 13.4
100 50 15 13 270 13.4
150 57 10 8.2 220 13.2
200 57 7.5 6.8 150 12.8
250 50 6.2 5.6 100 12.3
500 54 3.9 3.3 39 10.9
10 103 100 91 5600 10.4
20 102 51 43 2700 10.4
50 99 22 18 1000 10.6
100 98 11 9.1 430 10.5
150 101 7.5 6.2 260 10.3
200 95 5.6 4.7 180 10.3
250 92 4.3 3.9 130 9.9
500 114 2.2 2.0 47 6.8
INPUT
9
8
7
6
5
13
12
11
10
01084-032
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
C2
INLO
INHI
AD8310
SIGNAL
INPUT
L
M
8
01084-031
1
Figure 31. Reactive Matching Network
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, 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
IN
IN
C
L
2
1
ω
= (5)
where C
IN
= 1.4 pF. For example, at f
C
= 100 MHz, L
IN
= 1.8 μH.
Step 2: Calculate C
O
and L
O
Now, having a purely resistive input impedance, calculate the
nominal coupling elements, C
O
and L
O
, using
()
C
MIN
O
MIN
C
O
f
RR
L
RRf
C
π
=
π
=
2
;
2
1
(6)
For the AD8310, R
IN
is 1 kΩ. Therefore, if a match to 50 Ω 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
To provide the desired fully balanced form of the network
shown in Figure 31, two capacitors C1 and C2, each of
nominally twice C
O
, 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 C
O
= 6.96 pF.