AD8306 datasheet(12/16 Pages) AD

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REV. A

AD8306

–12–

Table I.

Match to 50

Match to 100

(Gain = 13 dB)

(Gain = 10 dB)

MHz

140

3500

100.7

4790

10.7

133

3200

94.1

4460

95.0

2250

67.1

3120

71.0

1660

50.3

2290

21.4

66.5

1550

47.0

2120

57.0

1310

40.3

1790

47.5

1070

33.5

1460

40.7

904

28.8

1220

35.6

779

25.2

1047

31.6

682

22.4

912

28.5

604

20.1

804

23.7

489

16.8

644

17.8

346

12.6

448

100

14.2

262

10.1

335

120

11.9

208

8.4

261

150

9.5

155

6.7

191

200

7.1

104

5.03

125

250

5.7

75.3

4.03

89.1

300

4.75

57.4

3.36

66.8

350

4.07

45.3

2.87

52.1

400

3.57

36.7

2.52

41.8

450

3.16

30.4

2.24

34.3

500

2.85

25.6

2.01

28.6

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 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

LIN = 1/{(2

π f

IN} = 10

10/f

(7)

when CIN = 2.5 pF. For example, at fC = 100 MHz, LIN = 1 µH.

Step 2: Calculate CO and LO

Now having a purely resistive input impedance, we can calculate

the nominal coupling elements CO and LO, using

CIN

()

= ()

;

(8)

For the AD8306, RIN is 1 kΩ. Thus, 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

Since we wish to provide the fully-balanced form of network

shown in Figure 28, two capacitors C1 = C2 each of nominally

twice CO, shown as CM in the figure, can be used. This requires

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

CO = 6.96 pF).

Step 4: Calculate LM

The matching inductor required to provide both LIN and LO is

just the parallel combination of these:

LM = LINLO/(LIN + LO)

(9)

With LIN = 1 µH and LO = 356 nH, the value of LM to complete

this example of a match of 50

Ω at 100 MHz is 262.5 nH. The

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

GAIN











 =













log

(10)

Altering the Logarithmic Slope

Simple schemes can be used to increase and decrease the loga-

rithmic slope as shown in Figure 30. For the AD8306, only

power, ground and logarithmic output connections are shown;

refer to Figure 24 for complete circuitry. In Figure 30(a), the op

amp’s gain of +2 increases the slope to 40 mV/dB. In Figure

30(b), the AD8031 buffers a resistive divider to give a slope of

Figure 30. Altering the Logarithmic Slope

VPS1

VPS2

PADL, COM1, COM2

AD8306

0.1 F

AD8031

0.1 F

+5V

40mV/dB

VLOG

(a)

VPS1

VPS2

PADL, COM1, COM2

AD8306

0.1 F

AD8031

0.1 F

10mV/dB

+5V

VLOG

(b)

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