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AD8310 Datasheet(PDF) 16 Page - Analog Devices |
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AD8310 Datasheet(HTML) 16 Page - Analog Devices |
16 / 24 page AD8310 Rev. D | Page 16 of 24 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 (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 At high frequencies, it is often preferable to use a narrow-band matching network, as shown in Figure 31. This has several advantages. 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 frequency attenuation is due to the input capacitance of the log amp. C1 C2 INHI INLO AD8310 SIGNAL INPUT LM 1 8 Figure 31. Reactive Matching Network FREQUENCY (MHz) 14 4 –1 60 150 80 100 110 130 3 2 1 0 70 90 120 140 INPUT GAIN 9 8 7 6 5 13 12 11 10 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 = 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 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. |
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