5 V, Rail-to-Rail, High Output Current, xDSL Line Driver Amplifiers
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19 /8 — AD8018. –10. VIN = 2V p-p. –20. G = 2 VS =. 2.5. RL = 5. –30. SIDE A …
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AD8018. –10. VIN = 2V p-p. –20. G = 2 VS =. 2.5. RL = 5. –30. SIDE A DRIVEN. –40. dB –. SIDE B DRIVEN. –50. –60. –70. L = 100. CROSSTALK. –80. –90. RL = 100
AD8018–10VIN = 2V p-p–20G = 2 VS = ⴞ 2.5RL = 5 ⍀ –30SIDE A DRIVEN–40RL = 5 ⍀ dB –SIDE B DRIVEN–50–60R–70L = 100 ⍀ SIDE B DRIVENCROSSTALK–80–90RL = 100 ⍀ SIDE A DRIVEN–100–110100k1M10M100M1GFREQUENCY – Hz TPC 25. Crosstalk vs. Frequency THEORY OF OPERATION The AD8018 is composed of two current feedback amplifiers capable of delivering 400 mA of output current while swinging to within 0.5 V of either power supply, and maintaining low distortion. A differential line driver using the AD8018 can provide CPE performance on a single 5 V supply. This performance is enabled by Analog Device’s XFCB process and a novel, two- VOBIASV stage current feedback architecture featuring a patent-pending PVN rail-to-rail output stage. A simplified schematic is shown in Figure 4. Emitter followers buffer the positive input, VP, to provide low input current and current noise. The low impedance current feedback summing junction is at the negative input, VN. The output stage is another Figure 4. Simplified Schematic high-gain amplifier used as an integrator to provide frequency compensation. The complementary common-emitter output provides the extended output swing. A current feedback amplifier’s dynamic and distortion performance is relatively insensitive to its closed-loop signal gain, which is a distinct advantage over a voltage-feedback architecture. Figure G = 1+ 5 shows a simplified model of a current feedback amplifier. The VO feedback signal is a current into the inverting node. R VIN IN is inversely RIIININT = IINCT RT proportional to the transconductance of the amplifier’s input stage, + gmi. Circuit analysis of the pictured follower with gain yields: –VOUT T V Z S = × OUT V / ( ) IN G – T + + × Z (S ) RF G R IN where: RF G = 1 + RF R / G RG R T T ( ) = Z S 1 + C S Figure 5. Model of Current Feedback Amplifier T (RT ) = 1/ ≅ Ω RIN gmi 125 FEEDBACK RESISTOR SELECTION Recognizing that G ⫻ RIN < RF, and that the –3 dB point is set In current feedback amplifiers, selection of the feedback and gain when TZ(S) = RF, one can see that the amplifier’s bandwidth resistors will impact the MTPR performance, bandwidth, noise, depends primarily on the feedback resistor. There is a value of and gain flatness. Care should be exercised in the selection of these RF below which the amplifier will be unstable, as an actual ampli- resistors so that the optimum performance is achieved. Table I fier will have additional poles that will contribute excess phase shows the recommended resistor values for use in a variety of gain shift. The optimum value for RF depends on the gain and the settings for the test circuit in TPC 1. These values are intended amount of peaking tolerable in the application. to be a starting point when designing for any application. –8– REV. A