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BD8960NV Datasheet(PDF) 9 Page - Rohm |
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BD8960NV Datasheet(HTML) 9 Page - Rohm |
9 / 15 page Technical Note BD8960NV 9/14 www.rohm.com 2010.04 - Rev.B © 2010 ROHM Co., Ltd. All rights reserved. ● Selection of components externally connected 1. Selection of inductor (L) * Current exceeding the current rating of the inductor results in magnetic saturation of the inductor, which decreases efficiency. The inductor must be selected allowing sufficient margin with which the peak current may not exceed its current rating. If VCC=3.3V, VOUT=1.8V, f=1MHz, ΔIL=0.2×2A=0.4A, for example,(BD8960NV) *Select the inductor of low resistance component (such as DCR and ACR) to minimize dissipation in the inductor for better efficiency. 2. Selection of output capacitor (CO) 3. Selection of input capacitor (Cin) A low ESR 22μF/10V ceramic capacitor is recommended to reduce ESR dissipation of input capacitor for better efficiency. The inductance significantly depends on output ripple current. As seen in the equation (1), the ripple current decreases as the inductor and/or switching frequency increases. ΔIL= (VCC-VOUT)×VOUT L×VCC×f [A]・・・(1) Appropriate ripple current at output should be 20% more or less of the maximum output current. ΔIL=0.2×IOUTmax. [A]・・・(2) L= (VCC-VOUT)×VOUT ΔIL×VCC×f [H]・・・(3) (ΔIL: Output ripple current, and f: Switching frequency) Output capacitor should be selected with the consideration on the stability region and the equivalent series resistance required to smooth ripple voltage. Output ripple voltage is determined by the equation (4): ΔVOUT=ΔIL×ESR [V]・・・(4) (ΔIL: Output ripple current, ESR: Equivalent series resistance of output capacitor) *Rating of the capacitor should be determined allowing sufficient margin against output voltage. A 22µF to 100µF ceramic capacitor is recommended. Less ESR allows reduction in output ripple voltage. Input capacitor to select must be a low ESR capacitor of the capacitance sufficient to cope with high ripple current to prevent high transient voltage. The ripple current IRMS is given by the equation (5): IRMS=IOUT× VOUT(VCC-VOUT) VCC [A]・・・(5) √ When Vcc is twice the VOUT, IRMS= IOUT 2 Fig.26 Output capacitor (3.3-1.8)×1.8 0.4×3.3×1M L= =2.05μ → 2.2[μH] < Worst case > IRMS(max.) IRMS=2× 1.8(3.3-1.8) 3.3 =0.99[ARMS] √ Fig.27 Input capacitor ΔIL VCC IL L Co VOUT Fig.25 Output ripple current IL VCC L Co VOUT ESR VCC L Co VOUT Cin If VCC=3.3V, VOUT=1.8V, and IOUTmax.=2A, (BD8960NV) |
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