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IRU3048CF Datasheet(PDF) 9 Page - International Rectifier |
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IRU3048CF Datasheet(HTML) 9 Page - International Rectifier |
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9 / 18 page  IRU3048 9 Rev. 1.7 09/12/02 www.irf.com This results to R9=46.4K V; Choose R9=46.4KV 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: Using equations (11),(12) and (13) for Ch2, where: We get: R11 = 38.9K V; Choose R11 = 39.2KV C19 = 1554pF; Choose C19 = 1800pF 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 a general solution for unconditionally stability for any type of output capacitors, in a wide range of ESR values we should implement local feedback with a compensa- tion network. The typically used compensation network for voltage-mode controller is shown in Figure 7. C9 = 1630pF; Choose C9 = 1800pF VIN2 = 5V VOSC = 1.25V FO2 = 30KHz FESR2 = 26.5KHz FLC2 = 3.5KHz R15 = 1K R14 = 442 V gm = 600mhmo FZ ≅ 75%FLC1 FZ ≅ 0.75 3 1 2 p L3 3 CO ---(13) For: L3 = 10.2 mH Co = 300 mF Fz = 2.1KHz R9 = 46.4K V Figure 7 - Compensation network with local feedback and its asymptotic gain plot. In such configuration, the transfer function is given by: The error amplifier gain is independent of the transcon- ductance under the following condition: By replacing ZIN and Zf according to figure 7, the trans- former function can be expressed as: As known, transconductance amplifier has high imped- ance (current source) output, therefore, consider should be taken when loading the E/A output. It may exceed its source/sink output current capability, so that the ampli- fier will not be able to swing its output voltage over the necessary range. The compensation network has three poles and two ze- ros and they are expressed as follows: Ve 1 - gmZf 1 + gmZIN VOUT = CPOLE = ≅ p 3 R9 3 fS - 1 C18 1 1 p 3 R9 3 fS For FP << fS 2 FP = 2 p 3 R9 3 C18 3 CPOLE C18 + CPOLE 1 VOUT VREF R5 R6 R8 C10 C12 C11 R7 Ve FZ1 FZ2 FP2 FP3 E/A Zf ZIN Frequency Gain(dB) H(s) dB Fb Comp gmZf >> 1 and gmZIN >>1 ---(14) For: VIN1 = 12V VOSC = 1.25V FO1 = 30KHz FESR1 = 26.5KHz FLC1 = 2.8KHz R8 = 1K R6 = 1.64K gm = 600mmho H(s)= sR6(C12+C11) 1+sR7 3(1+sR8C10) 1 (1+sR7C11) 3[1+sC10(R6+R8)] 3 C12C11 C12+C11 [ ( )] FP1 = 0 1 2 p3C103(R6 + R8) FZ2 = ≅ 1 2 p3C103R6 FZ1 = 1 2 p3R73C11 FP3 = ≅ 1 2 p3R73 FP2 = 1 2 p3R83C10 1 2 p3R73C12 C12 3C11 C12+C11 ( ) |
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