AD8310ARMZ-REEL7 Analog Devices Inc, AD8310ARMZ-REEL7 Datasheet - Page 16

• Manufacturer Part Number: AD8310ARMZ-REEL7
• Description: IC,Log/Antilog Amplifier,SINGLE,BIPOLAR,TSSOP,8PIN,PLASTIC
• Manufacturer: Analog Devices Inc
• Type: Logarithmic Amplifierr

Specifications of AD8310ARMZ-REEL7

• Applications: Receiver Signal Strength Indication (RSSI)
• Mounting Type: Surface Mount
• Package / Case: 8-MSOP, Micro8™, 8-uMAX, 8-uSOP,
• Number Of Channels: 1
• Number Of Elements: 6
• Power Supply Requirement: Single
• Input Resistance: 0.0012@5VMohm
• Single Supply Voltage (typ): 3/5V
• Dual Supply Voltage (typ): Not RequiredV
• Power Dissipation: 200mW
• Rail/rail I/o Type: Rail to Rail Input
• Single Supply Voltage (min): 2.7V
• Single Supply Voltage (max): 5.5V
• Dual Supply Voltage (min): Not RequiredV
• Dual Supply Voltage (max): Not RequiredV
• Operating Temp Range: -40C to 85C
• Operating Temperature Classification: Industrial
• Mounting: Surface Mount
• Pin Count: 8
• Package Type: MSOP
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant

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
f
(MHz)
10
20
50
100
150
200
250
500
10
20
50
100
150
200
250
500
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.
C
Z
(Ω)
45
44
46
50
57
57
50
54
103
102
99
98
101
95
92
114
IN
SIGNAL
INPUT
Figure 31. Reactive Matching Network
C1
(pF)
160
82
30
15
10
7.5
6.2
3.9
100
51
22
11
7.5
5.6
4.3
2.2
C1
C2
C2
(pF)
150
75
27
13
8.2
6.8
5.6
3.3
91
43
18
9.1
6.2
4.7
3.9
2.0
L
M
AD8310
INLO
INHI
L
(nH)
3300
1600
680
270
220
150
100
39
5600
2700
1000
430
260
180
130
47
8
1
M
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
Rev. F | Page 16 of 24
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
At a center frequency, f
capacitance, C
a temporary inductor, L
where C
Step 2: Calculate C
Now, having a purely resistive input impedance, calculate the
nominal coupling elements, C
For the AD8310, R
needed, at f
356 nH.
Step 3: Split C
To provide the desired fully balanced form of the network
shown in Figure 31, two capacitors C1 and C2, each of
nominally twice C
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
L
C
14
13
12
11
10
–1
9
8
7
6
5
4
3
2
1
0
IN
O
60
IN
=
=
= 1.4 pF. For example, at f
Figure 32. Response of 100 MHz Matching Network
2
C
ω
π
70
= 100 MHz, C
2
f
IN
C
O
1
O
C
, can be made to disappear by resonating with
into Two Parts
= 6.96 pF.
IN
O
80
1
R
IN
, can be used. This requires a value of
IN
IN
O
is 1 kΩ. Therefore, if a match to 50 Ω is
and L
R
C
90
IN
, the shunt impedance of the input
M
FREQUENCY (MHz)
, whose value is given by
O
;
100
O
must be 7.12 pF and L
O
L
and L
O
110
=
C
GAIN
O
= 100 MHz, L
INPUT
, using
120
(
R
2
IN
π
f
130
R
C
M
)
140
O
IN
must be
= 1.8 μH.
150
(5)
(6)