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Li -ion CC/CV circuit

  1. Mar 8, 2009 #1
    Hi All

    i am thinking of using an LM317 to deliver a constant current (CC) of aprox 1A to li-ion batteries for charging. but as you may know it is then recommended to then apply a constant voltage (CV) once the constant current section has made the battery voltage 4.2V per cell. the constant voltage should be 4.2V and shoud terminate when the charge current has dropped to approx 0.1C.

    my question is how do i switch from constant current to constant voltage at the required time. is it just a matter of switching of the CC circuit and turning on a CV circuit? I.E can i have two seperate circuits one for CC and the other for CV and just turn them on and off as required. Or do they have to be incorporated within the one circuit and if so how?

  2. jcsd
  3. Mar 8, 2009 #2
    I'm sure someone else will give you a more informed answer.

    Just if your interested for off the shelf stuffs,, have you looked on digikey (or other supplier?) for li-ion chargers?
    For example searching "lithium charger" on digikey got tons of small and dedicated chargers.
    like this one http://ww1.microchip.com/downloads/en/DeviceDoc/22036b.pdf
  4. Mar 8, 2009 #3
    ya no thanks, but i'm looking for a circuit that i could build myself.
  5. Mar 8, 2009 #4


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    As long as you don't mind the current "set point" being somewhat approximate then what I normally do is just use a low dropout regulator to handle the float voltage (eg 4.2 volts) and then just a series resistor on the input side of the regulator to handle the constant current part. You need the overall supply voltage to be reasonably constant which may or maynot require a second regulator depending on what supply you've got. The main advantage of this method is that it's simple and the "switching" is automatic.

    As an example, imagine that you had a somewhat constant supply voltage of say 8 volts and you connected that through a 3.3 ohm series resistor to the Vin terminal of a low dropout (or even better an ultra low dropout) adjustable regulator that is set for an output of 4.2 volts.

    What happens is that when the battery is not fully charged, say for example the terminal voltage is 3.8 volts, then the regulator starts to dropout when the input side is approx 3.8 + 0.7 = 4.4 volts, at which point the current cannot exceed (8-4.4)/3.3 which is about 1.06 amps. As the battery voltage rises during charging the current tapers slightly and once the voltage reaches the regulator set point it wont rise any further and the circuit is said to be in "float" mode.
    Last edited: Mar 8, 2009
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