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XRP6670 Datasheet(PDF) 8 Page - Exar Corporation |
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XRP6670 Datasheet(HTML) 8 Page - Exar Corporation |
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8 / 13 page  X XR RP P6 66 67 70 0 3 3A A 3 30 00 0k kH Hz z- -2 2..5 5M MH Hz z S Sy yn nc ch hr ro on no ou us s S Stte ep p D Do ow wn n R Re eg gu ulla atto or r © 2013 Exar Corporation 8/13 Rev. 1.0.1 R2 is the resistor between FB and GND 0.800V is the nominal feedback voltage A resistor selection guide for common values of VOUT is shown in table 1. VOUT R1(kΩ) R2(kΩ) 1.1V 100 267 1.2V 100 200 1.5V 105 120 1.8V 120 95.3 2.5V 100 47 2.8V 75 30 3.3V 75 24 Table 1: Resistor Selection PROGRAMMING THE FREQUENCY Use resistor ROSC between SHDN/RT and GND pins to program the switching frequency. A graph of nominal frequency versus ROSC is shown in figure 16. Fig. 16: Frequency versus ROSC The following equation closely fits the empirical data and can be used to select ROSC for a given frequency. . . DUTY -CYCLE LIMITATION XRP6670 has a “Minimum On-Time” specification of 150ns which imposes a restriction on minimum duty-cycle (see table 2) F (MHz) TTYP(ns) TMIN(ns) Duty-cycleMIN 1.0 1000 800 0.19 2.0 500 400 0.38 Table 2: Minimum duty-cycle arising from “Minimum On-Time) For example if frequency is set at 2MHz then typical switching period is 500ns. Allowing a ±20% uncertainty, minimum period is 400ns and corresponding minimum duty-cycle is 0.38. Recall that for a buck regulator duty- cycle=VOUT/VIN. Therefore when operating at 2MHz with VIN of 5V, a VOUT≤1.9V is not possible (5V x 0.38 = 1.9V). OUTPUT INDUCTOR Select the output inductor for inductance L, DC current rating IDC and saturation current rating ISAT. IDC should be larger than regulator output current. ISAT, as a rule of thumb, should be 50% higher than the regulator output current. Since the regulator is rated at 3A then IDC≥3A and ISAT≥4.5A. Please note that “Peak Switch Current” is rated at 3.2A minimum. Therefore applications that require an output current of 3A should limit the peak-to-peak inductor current ripple to ∆IL≤0.4A. In the following we will use the common practice of ∆IL≤1A. Therefore worst- case maximum output current will be limited to IOUT=3.2A-0.5A=2.7A. Calculate the inductance from: !"# $% & Where: ∆IL is peak-to-peak inductor current ripple nominally set to ≤30% of IOUT fS is nominal switching frequency As an example, inductor values corresponding to 5VIN/1MHz and 3.3VIN/1MHz are shown in tables 3 and 4 for several common output |
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