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CS1600 데이터시트(PDF) 11 Page - Cirrus Logic |
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CS1600 데이터시트(HTML) 11 Page - Cirrus Logic |
11 / 18 page CS1600 DS904A7 11 5. FLUORESCENT BALLAST APPLICATION EXAMPLE The following section gives an example for a front-end PFC stage design for an electronic ballast application. The equations that follow may be used as guidelines for any other requirements using the CS1600. 5.1 Component Selection Guidelines The following design example is for a wide-input-voltage fluorescent ballast application using 2 T5 lamps in series for a total nominal power of 108W.The target specifications for the PFC portion of the design, assuming a 94% efficient second stage, are as follows: 5.1.1 IAC and IFB Sense Resistors The rectified line voltage, VAC, and the output voltage of the PFC boost converter, Vlink, are scaled as currents by using sense resistors, whose values are estimated based on the equations below: where RFB = Feedback resistor used to reflect the PFC output voltage RAC = Feedforward resistor used to reflect the rectified line voltage Vlink= PFC Output Voltage VDD = IC Supply Voltage Iref = Target reference current used for feedback 1% or lower tolerance resistors are recommended to maximize the tightly toleranced system behavior provided by the unique digital controller in the CS1600. Resistors may be separated into two or more series elements if voltage breakdown and/or regulatory compliance is of concern. 5.1.2 PFC Input Filter Capacitor For a typical 115 W PFC output stage required to power up a 108 W fluorescent ballast, an input filter capacitance of 0.33 μF is recommended. Capacitor tolerances and the value of the EMI filter capacitor need to be considered when selecting the value of the capacitor to be used in this application. D6 C1 D5 C2 BR1 BR1 BR1 BR1 AC Mains +12V L1 Q1 R3 CS1600 IAC NC FB STBY VDD GND NC GD 2 3 1 7 8 6 4 5 R1a R1b R2b R2a C3a C3b R2c R1c RAC RFB Clink Figure 17. CS1600 Basic Application Circuit Vin(min) 108 VAC Vin(max) 305 VAC Vlink 460 V Po 115 W η 95% R FB V link V dd – I ref ---------------------------- = R FB 460 12 – 130 10 6 – × ---------------------------- = R FB 3.45M Ω = [Eq.4] R AC R FB = R AC 3.45M Ω = [Eq.5] |
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