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IRF7457 bảng dữ liệu(PDF) 8 Page - International Rectifier |
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IRF7457 bảng dữ liệu(HTML) 8 Page - International Rectifier |
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8 / 21 page ![]() 8 Rev. 2.8 05/10/04 IRU3037 / IRU3037A www.irf.com Note that this method requires that the output capacitor should have enough ESR to satisfy stability requirements. In general the output capacitor’s ESR generates a zero typically at 5KHz to 50KHz which is essential for an acceptable phase margin. The ESR zero of the output capacitor expressed as fol- lows: Figure 6 - Compensation network without local feedback and its asymptotic gain plot. The transfer function (Ve / VOUT) is given by: The (s) indicates that the transfer function varies as a function of frequency. This configuration introduces a gain and zero, expressed by: The gain is determined by the voltage divider and E/A's transconductance gain. First select the desired zero-crossover frequency (Fo): Use the following equation to calculate R4: Where: VIN = Maximum Input Voltage VOSC = Oscillator Ramp Voltage Fo = Crossover Frequency FESR = Zero Frequency of the Output Capacitor FLC = Resonant Frequency of the Output Filter R5 and R6 = Resistor Dividers for Output Voltage Programming gm = Error Amplifier Transconductance This results to R4=104.4K V. Choose R4=105KV To cancel one of the LC filter poles, place the zero be- fore the LC filter resonant frequency pole: Using equations (11) and (13) to calculate C9, we get: One more capacitor is sometimes added in parallel with C9 and R4. This introduces one more pole which is mainly used to supress the switching noise. The additional pole is given by: The pole sets to one half of switching frequency which results in the capacitor CPOLE: For: VIN = 5V VOSC = 1.25V Fo = 30KHz FESR = 26.52KHz FLC = 2.9KHz R5 = 1K R6 = 1.65K gm = 600 mmho C9 = 698pF Choose C9 = 680pF FP = 2 p 3 R4 3 C9 3 CPOLE C9 + CPOLE 1 VOUT VREF R5 R6 R4 C9 Ve E/A FZ H(s) dB Frequency Gain(dB) Fb Comp FESR = ---(8) 1 2 p 3 ESR 3 Co FZ ≅ 75%FLC FZ ≅ 0.75 3 ---(13) 1 2 p LO 3 CO For: Lo = 10 mH Co = 300 mF FZ = 2.17KHz R4 = 86.6K V Fo > FESR and FO [ (1/5 ~ 1/10)3 fS H(s) = gm 3 3 ---(9) ( ) R5 R6 + R5 1 + sR4C9 sC9 FZ = ---(11) 1 2 p3R43C9 |H(s)| = gm 3 3 R4 ---(10) R5 R6 3R5 R4 = ---(12) VOSC VIN Fo 3FESR FLC2 R5 + R6 R5 1 gm 3 3 3 CPOLE = p3R43fS - 1 for FP << fS 2 1 C9 ≅ 1 p3R43fS |
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