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MAX8545 Datasheet(PDF) 13 Page - Maxim Integrated Products |
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MAX8545 Datasheet(HTML) 13 Page - Maxim Integrated Products |
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13 / 18 page ![]() Low-Cost, Wide Input Range, Step-Down Controllers with Foldback Current Limit ______________________________________________________________________________________ 13 has a gain drop of 40dB per decade, and a phase shift of 180°. The error amplifier must compensate for this gain drop and phase shift to achieve a stable high- bandwidth, closed-loop system. The basic regulator loop consists of a power modulator (Figure 3), an output feedback divider, and an error amplifier. The power modulator has DC gain set by VIN/VRAMP, with a double pole set by the inductor and output capacitor, and a single zero set by the output capacitor (COUT) and its equivalent series resistance (ESR). Below are equations that define the power mod- ulator: The DC gain of the power modulator is: where VRAMP = 1V. The pole frequency due to the inductor and output capacitor is: The zero frequency due to the output capacitor’s ESR is: The output capacitor is usually comprised of several same capacitors connected in parallel. With n capaci- tors in parallel, the output capacitance is: COUT = n X CEACH The total ESR is: The ESR zero (fZESR) for a parallel combination of capacitors is the same as for an individual capacitor. The feedback divider has a gain of GFB = VFB/VOUT, where VFB is 0.8V. The transconductance error amplifier has DC gain GEA(dc) of 72dB. A dominant pole (fDPEA) is set by the compensation capacitor (CC), the amplifier output resistance (RO) equals 37MΩ, and the compensation resistor (RC): The compensation resistor and the compensation capacitor set a zero: The total closed-loop gain must equal unity at the crossover frequency. The crossover frequency should be higher than fZESR, so that the -1 slope is used to cross over at unity gain. Also, the crossover frequency should be less than or equal to 1/5 the switching fre- quency (fSW) of the controller: The loop-gain equation at the crossover frequency is: VFB/VOUT x GEA(fC) x GMOD(fC) = 1 where GEA(fC) = gmEA × RC, and GMOD(fC) = GMOD(DC) × (fPMOD)2 / (fZESR × fC). The compensation resistor, RC, is calculated from: RC = VOUT/ gmEA x VFB x GMOD(fC) where gmEA = 108µS. Due to the underdamped (Q > 1) nature of the output LC double pole, the error-amplifier compensation zero should be approximately 0.2 fPMOD to provide good phase boost. CC is calculated from: A small capacitor, CF, can also be added from COMP to GND to provide high-frequency decoupling. CF adds another high-frequency pole, fPHF, to the error-amplifier response. This pole should be greater than 100 times the error-amplifier zero frequency to have negligible impact on the phase margin. This pole should also be less than 1/2 the switching frequency for effective decoupling: 100 fZEA < fPHF < 0.5 fsw Select a value for fPHF in the range given above, then solve for CF using the following equation: PCB Layout Guidelines Careful PCB layout is critical to achieve low switching losses and stable operation. If possible, mount all the power components on the top side of the board with their C Rf F C PHF = ×× 1 2π C Rf C C PMOD = ×× 5 2π ff f ZESR C SW < ≤ 5 f CR ZEA CC = ×× 1 2π f CR R DPEA CO C = ×× + () 1 2π ESR ESR n EACH = f ESR C ZESR OUT = ×× 1 2π f LC PMOD OUT = 1 2π G V V MOD DC IN RAMP () = |
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