Looking over the equations for a boost-converter some more, it seems that if one wants to keep the converter in continuous mode at small currents, one actually has to INCREASE the size of the inductor the lower the current one is providing, to make the ripple current smaller. This is rather counter-intuitive to me, but that's what the equations seem to be saying. It also seems to defeat the purpose of using a switching converter for small currents, as the design goal should be to minimize the size of the inductor used. I therefore assume that in general boost converters supplying small currents at high voltages would be used in discontinuous mode, where the output voltage is inversely proportional to both the output current and the size of the inductor, that is you get a higher voltage the smaller you make the inductor and the lower the output current is. However, using a smaller value inductor means that the ripple current will be larger, and you need to have an inductor sized to handle the peak ripple current! Argh! :yuck: This seems like a catch-22 situation!(adsbygoogle = window.adsbygoogle || []).push({});

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