Friday, 25 November 2016

power supply - How to achieve the max efficiency from a Buck converter


I have a question about switching step-down regulators. (As I stated in my previous questions, please consider the fact that that I'm not very expert, so feel free to reply/talk as if I were a student.)


Let's take a practical example of a switching step-down regulator, based on this IC. (I've seen that is largely used and common in various circuits):


We need to feed a device which needs 12V with a power consumption of 200mA. Ok: We'll take a buck converter circuit, and as Vin we'll provide, for example, a voltage of 30V from a batteries pack with a total capacity of 2000mAh, then we will set the Vout of the buck converter to 12V. But If we want to make use of a less number of batteries we can also go with a Vin of 20 or less volt: I have read that for the lm2596 IC, the Vin, should be at least greater of 1,5V than the Vout.


I was thinking: If I reduce 30V (from a batteries pack) to 12V, the difference of 18V could be reason of an higher power consumption from the batteries? Am I right? Eg I know that linear regulators (differently from switching regulators) have a bad efficiency because some of the power will be lost as heat. But what about switching regulators? Some days ago, by a search on Google, I've read of a person which had the needs to get 5V using a Buck converter: someone told him that would be better get the 5V from a Vin of 18V instead of using a Vin of 12V.


So, taking again in consideration my example: when using a switching regulator, is better to start from an higher Vin, for obtaining a same Vout? Why?


I'd also like to see some charts of the switching regulators.




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