Using a boost-buck IC for charging batteries

Discussion in 'General Electronics Chat' started by Supernerd Sven, Mar 11, 2016.

  1. Supernerd Sven

    Thread Starter New Member

    Feb 16, 2015
    6
    0
    Hello AAC, I've been thinking about building a simple system to charge batteries for a while. While looking up DC-to-DC converters I came across this:
    http://www.mouser.com/ds/2/405/mc34063a-443483.pdf
    By the look of the datasheet, it can take in a good range of input voltages, which is great for me since I'm planning to start by using a repurposed BLDC computer fan and would probably have somewhat inconsistent input, and I can just set it to whatever voltage is appropriate for the battery. As long as I then find a way to monitor the state of charge and stop it when the battery is fully charged, could I use this? It seems so to me, but then I wonder why I haven't heard about anyone using them for small-scale battery charging before. (Also, I would have thought they'd list that as a possible application in the datasheet.)

    On a side note, what happens with voltage and current in a typical voltage regulator - say, the LM317? Suppose I put in 10V at 100mA and set it to put out 5V; my intuition is that I would just get 5V at 100mA rather than 5V at 200mA since the latter generally requires more machinery. Is that correct? If so, what happens to the other half a watt? Is it dissipated as heat, or is it in some way "blocked" so that the extra voltage and hence power isn't used in the first place?

    Thank you for your help!
     
  2. gerty

    AAC Fanatic!

    Aug 30, 2007
    1,153
    304
    You'll have to give a little more info !! What is the power source, type and voltage of batteries?

    On a side note, what happens with voltage and current in a typical voltage regulator - say, the LM317? Suppose I put in 10V at 100mA and set it to put out 5V; my intuition is that I would just get 5V at 100mA rather than 5V at 200mA since the latter generally requires more machinery. Is that correct? If so, what happens to the other half a watt? Is it dissipated as heat, or is it in some way "blocked" so that the extra voltage and hence power isn't used in the first place?

    On your LM317 question, the regulator will only pass the current that the load requires.If you have a 100ma load, that's all that will pass through the regulator.
     
  3. Dodgydave

    Distinguished Member

    Jun 22, 2012
    4,969
    744
    What type and voltage of battery is it,?

    You're better using the deadicated charger ic for the battery you are using.
     
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