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BD9104FVM 데이터시트(PDF) 21 Page - Rohm |
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BD9104FVM 데이터시트(HTML) 21 Page - Rohm |
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21 / 29 page ![]() 21/27 BD9102FVM BD9104FVM BD9106FVM Datasheet 02.MAR.2012 Rev.001 www.rohm.com © 2011 ROHM Co., Ltd. All rights reserved. TSZ22111・15・001 TSZ02201-0J3J0AJ00080-1-2 Switching Regulator Efficiency Efficiency ŋ may be expressed by the equation shown below: Efficiency may be improved by reducing the switching regulator power dissipation factors PDα as follows: Dissipation factors: 1) ON resistance dissipation of inductor and FET:PD(I 2R) 2) Gate charge/discharge dissipation:PD(Gate) 3) Switching dissipation:PD(SW) 4) ESR dissipation of capacitor:PD(ESR) 5) Operating current dissipation of IC:PD(IC) 1)PD(I 2R)=IOUT2×(RCOIL×RON) (RCOIL[Ω]:DC resistance of inductor, RON[Ω]:ON resistance of FET IOUT[A]:Output current.) 2)PD(Gate)=Cgs×f×V (Cgs[F]:Gate capacitance of FET,f[H]:Switching frequency,V[V]:Gate driving voltage of FET) 4)PD(ESR)=IRMS 2×ESR (IRMS[A]:Ripple current of capacitor,ESR[Ω]:Equivalent series resistance.) 5)PD(IC)=Vin×ICC (ICC[A]:Circuit current.) Consideration on Permissible Dissipation and Heat Generation As this IC functions with high efficiency without significant heat generation in most applications, no special consideration is needed on permissible dissipation or heat generation. In case of extreme conditions, however, including lower input voltage, higher output voltage, heavier load, and/or higher temperature, the permissible dissipation and/or heat generation must be carefully considered. For dissipation, only conduction losses due to DC resistance of inductor and ON resistance of FET are considered. Because the conduction losses are considered to play the leading role among other dissipation mentioned above including gate charge/discharge dissipation and switching dissipation. P=IOUT 2×(RCOIL+RON) RON=D×RONP+(1-D)RONN D:ON duty (=VOUT/VCC) RCOIL:DC resistance of coil RONP:ON resistance of P-channel MOS FET RONN:ON resistance of N-channel MOS FET IOUT:Output current If VCC=5V, VOUT=3.3V, RCOIL =0.15Ω, RONP=0.35Ω, RONN=0.25Ω IOUT=0.8A, for example, D=VOUT/VCC=3.3/5=0.66 RON=0.66×0.35+(1-0.66)×0.25 =0.231+0.085 =0.316[Ω] P=0.8 2×(0.15+0.316) ≒298[mV] As RONP is greater than RONN in this IC, the dissipation increases as the ON duty becomes greater. With the consideration on the dissipation as above, thermal design must be carried out with sufficient margin allowed. η= VOUT×IOUT Vin×Iin ×100[%]= POUT Pin ×100[%]= POUT POUT+PD α ×100[%] Vin 2×CRSS×IOUT×f IDRIVE 3)PD(SW)= (CRSS[F]:Reverse transfer capacitance of FET,IDRIVE[A]:Peak current of gate.) Ambient temperature:Ta [℃] Fig.50 Thermal derating curves 0 200 400 600 800 1000 ②387.5mW ①587.4mW ①using an IC alone θj-a=322.6℃/W ②mounted on glass epoxy PCB θj-a=212.8℃/W 0 25 50 75 100 125 150 85 |
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