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MP1567 Datasheet(PDF) 8 Page - Monolithic Power Systems |
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MP1567 Datasheet(HTML) 8 Page - Monolithic Power Systems |
8 / 11 page MP1567 – 1.2A SYNCHRONOUS RECTIFIED STEP-DOWN CONVERTER MP1567 Rev. 2.3 www.MonolithicPower.com 8 1/3/2006 MPS Proprietary Information. Unauthorized Photocopy and Duplication Prohibited. © 2006 MPS. All Rights Reserved. TM 75KHz or lower to insure stable operation. Lower crossover frequencies result in slower response and worse transient load recovery. Higher crossover frequencies degrade the phase and/or gain margins and can result in instability. Choosing the Compensation Components The values of the compensation components given in Table 1 yield a stable control loop for the output voltage and capacitor given. Table 1—Compensation Values for Typical Output Voltage/Capacitor Combinations VOUT C2 R3 C3 C4 1.8V 4.7µF Ceramic 3.3kΩ 2.2nF None 2.5V 4.7µF Ceramic 5.1kΩ 1.5nF None 3.3V 4.7µF Ceramic 6.8kΩ 1.2nF None 1.8V 10µF Ceramic 7.5kΩ 1nF None 2.5V 10µF Ceramic 10kΩ 820pF None 3.3V 10µF Ceramic 10kΩ 820pF None 1.8V 47µF Tantalum (300mΩ) 10kΩ 2.2nF 1.5nF 2.5V 47µF Tantalum (300mΩ) 10kΩ 3.3nF 1.5nF 3.3V 47µF Tantalum (300mΩ) 10kΩ 4.7nF 1.5nF To optimize the compensation components for conditions not listed in Table 1, use the following procedure. Choose the compensation resistor to set the desired crossover frequency. Determine the value by the following equation: FB CS EA C OUT V G G f V 2 C 2 3 R × × × × × π = Putting in the known constants and setting the crossover frequency to the desired 75KHz: OUT 8 V 2 C 10 36 . 4 3 R × × × ≈ In this case, the actual crossover frequency is less than the desired 75KHz, and it is calculated by: OUT FB CS EA C V 2 C 2 V G G 3 R f × × π × × × = Choose the compensation capacitor to set the zero to one fourth of the crossover frequency. Determine the value by the following equation: FB CS EA 2 OUT V G G 3 R V 2 C 4 3 C × × × × × = Determine if the second compensation capacitor, C4, is required. It is required if the ESR zero of the output capacitor occurs at less than four times the crossover frequency, or: 1 f R 2 C 8 C ESR ≥ × × × π Where RESR is the equivalent series resistance of the output capacitor. If this is the case, then add the second compensation capacitor. Determine the value by the equation: 3 R R 2 C 4 C ) MAX ( ESR × = Where RESR(MAX) is the maximum ESR of the output capacitor. For Example: Given: VOUT = 1.8V C2 = 10µF Ceramic (ESR = 10mΩ max.) Calculate: Ω = × µ × ≈ k 85 . 7 ) V 8 . 1 ( ) F 10 ( 10 36 . 4 3 R 8 (Use the nearest standard value of 7.5kΩ.) nF 05 . 1 V 8 . 1 F 10 10 9 . 1 3 C 14 = × µ × = − (Use 1nF since it is a standard value.) 19 . 0 f R 2 C 8 C ESR = × × × π which is less than 1, therefore the second compensation capacitor (C4) is not required. |
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