Datasheet BL8530 (Belling) - 5

BL8530
Vout − Vin Before discussion，we define D ≡ . Vout （
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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. D 1 ( D)2 − R L L min ≥ 2 f 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, D • Vin I ∆ = Lf Vin DVin Im ax = + 1 ( − D)2 R 2Lf L 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 ： rL ∆η ≈ 2 R 1 ( − D) load 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 （
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Capacitor Selection
Ignore ESR of capacitor，the ripple of output voltage is: Vout ∆ D r = = Vout R Cf load So large value capacitor is needed to reduce ripple. But too large capacitor value will slow down system reaction and cost will improve. So 100uF capacitor is recommended. Larger capacitor value will be used in large output current system. If output current is small (<10mA), small value is needed. Consider ESR of capacitor，ripple will increase： ax • R r = Im ' r ESR + Vout When current is large, ripple caused by ESR will be main factor. It may be greater than 100mV。 The ESR will affects efficiency and increase energy loss. So low-ESR capacitor (for example: tantalum capacitor) is recommend or connect two or more filter capacitors in parallel. www.belling.com.cn Ver1.0 5/ 14