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ACE705151BN+ 数据表(PDF) 6 Page - ACE Technology Co., LTD. |
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ACE705151BN+ 数据表(HTML) 6 Page - ACE Technology Co., LTD. |
6 / 14 page ACE705 Technology High efficiency, multi-funtion step-up DC/DC Controller VER 1.2 6 Selection of the External Components Thus it can be seen, the inductor, schottky diode and external NMOSFET or NPN transistor. affect the conversion efficiency greatly. The inductor and the capacitor also have great influence on the output voltage ripple of the converter. So it is necessary to choose a suitable inductor, a capacitor, an external NMOSFET or NPN transistor and a right schottky diode, to obtain high efficiency and low ripple. Before discussion,we define D≣Vout-Vin / Vout (1)Inductor Selection Above all, we should define the minimum value of the inductor that can ensure the boost DC-DC to operate in the continuous current-mode condition. Lmin≧D(1-D)2RL / 2f The above expression is got under conditions of continuous current mode, neglect Schottky diode’s voltage, ESR of both inductor and capacitor. The actual value is greater that it. If inductor’s value is less than Lmin,the efficiency of DC-DC converter will drop greatly, and the DC-DC circuit will not be stable. Secondly, consider the ripple of the output voltage, Δ I=D‧Vin / Lf Im ax=Vin / (1-D)2RL + DVin / 2Lf If inductor value is too small, the current ripple through it will be great. Then the current through diode and power switch will be great. Because the power switch on chip is not ideal switch, the energy of switch will improve. The efficiency will fall. Thirdly,in general, smaller inductor values supply more output current while larger values start up with lower input voltage and acquire high efficiency. An inductor value of 3uH to 1mH works well in most applications. If DC-DC converter delivers large output current (for example: output current is great than 50mA), large inductor value is recommended in order to improve efficiency. If DC-DC must output very large current at low input supply voltage, small inductor value is recommended. The ESR of inductor will effect efficiency greatly. Suppose ESR value of inductor is rL,Rload is load resistor,then the energy can be calculated by following expression: Δη≈ RL / Rload (1-D) 2 For example: input 1.5V, output is 3.0V, Rload=20Ω, rL=0.5Ω, The energy loss is 10%. Consider all above,inductor value of 47uH、ESR<0.5Ω is recommended in most applications. Large value is recommended in high efficiency applications and smaller value is recommended |
类似零件编号 - ACE705151BN+ |
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类似说明 - ACE705151BN+ |
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