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MAX8745 Datasheet(PDF) 33 Page - Maxim Integrated Products |
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MAX8745 Datasheet(HTML) 33 Page - Maxim Integrated Products |
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33 / 36 page ![]() High-Efficiency, Quad Output, Main Power- Supply Controllers for Notebook Computers ______________________________________________________________________________________ 33 The output capacitor and the load resistance create the dominant pole in the system. However, the internal ampli- fier delay, the pass transistor’s input capacitance, and the stray capacitance at the feedback node create additional poles in the system, and the output capacitor’s ESR gen- erates a zero. For proper operation, use the following steps to ensure the linear-regulator stability: 1) First, calculate the dominant pole set by the linear regulator’s output capacitor and the load resistor: where COUTA is the output capacitance of the aux- iliary LDO and RLOAD is the load resistance corre- sponding to the maximum load current. The unity- gain crossover of the linear regulator is: fCROSSOVER = AV(LDO)fPOLE(LDO) 2) The pole caused by the internal amplifier delay is at approximately 1MHz: fPOLE(AMP) ≈ 1MHz 3) Next, calculate the pole set by the transistor’s input capacitance, the transistor’s input resistance, and the base-to-emitter pullup resistor. Since the tran- sistor’s input resistance (hFE/gm) is typically much greater than the base-to-emitter pullup resistance, the pole can be determined from the simplified equation: where gm is the transconductance of the pass tran- sistor, and fT is the transition frequency. Both para- meters can be found in the transistor’s data sheet. Therefore, the equation can be further reduced to: 4) Next, calculate the pole set by the linear regulator’s feedback resistance and the capacitance between FBA and ground (approximately 5pF including stray capacitance): 5) Next, calculate the zero caused by the output capacitor’s ESR: where RESR is the equivalent series resistance of COUTA. 6) To ensure stability, choose COUTA large enough so that the crossover occurs well before the poles and zero calculated in steps 2 through 5. The poles in steps 3 and 4 generally occur at several MHz, and using ceramic output capacitors ensures the ESR zero occurs at several MHz as well. Placing the crossover frequency below 500kHz is typically suf- ficient to avoid the amplifier delay pole and gener- ally works well, unless unusual component selection or extra capacitance moves the other poles or zero below 1MHz. A capacitor connected between the linear regula- tor’s output and the feedback node can improve the transient response and reduce the noise cou- pled into the feedback loop. If a low-dropout solution is required, an external p- channel MOSFET pass transistor could be used. However, a pMOS-based linear regulator requires higher output capacitance to stabilize the loop. The high gate capacitance of the p-channel MOSFET lowers the fPOLE(CIN) and can cause instability. A large output capacitance must be used to reduce the unity-gain bandwidth and ensure that the pole is well above the unity-gain crossover frequency. Applications Information Duty-Cycle Limits Minimum Input Voltage The minimum input operating voltage (dropout voltage) is restricted by the maximum duty-cycle specification (see the Electrical Characteristics table). For the best dropout performance, use the slowest switching fre- quency setting (200kHz, FSEL = GND). However, keep in mind that the transient performance gets worse as the step-down regulators approach the dropout volt- age, so bulk output capacitance must be added (see the voltage sag and soar equations in the SMPS Design Procedure and Transient Response sections). The absolute point of dropout occurs when the inductor cur- rent ramps down during the off-time ( ΔIDOWN) as much as it ramps up during the on-time ( ΔIUP). This results in a minimum operating voltage defined by the following equation: f CR ZERO ESR OUTA ESR () = 1 2 π f CR R POLE FBA FBA () (|| ) = 1 25 6 π f f h POLE CIN T FE () ≈ f CR C g f POLE CIN IN IN IN m T () ≈ = 1 2 2 π π f CR POLE LDO OUTA LOAD () = 1 2 π |
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