AD8310

Rev. F | Page 16 of 24

FREQUENCY (MHz)

14

4

–1

60

150

80

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

fC

(MHz)

ZIN

(Ω)

C1

(pF)

C2

(pF)

LM

(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

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

LM

8

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 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

IN

IN

C

L

2

1

ω

=

(5)

where CIN = 1.4 pF. For example, at fC = 100 MHz, LIN = 1.8 μH.

Step 2: Calculate CO and LO

Now, having a purely resistive input impedance, calculate the

nominal coupling elements, CO and LO, using

()

C

M

IN

O

M

IN

C

O

f

R

R

L

R

R

f

C

π

=

π

=

2

;

2

1

(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 CO 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.