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LM2665: LM2665 as supercapacitor charger

David Dean
Expert 2315 points
Part Number: LM2665

I have a 5V rail and want to charge a 2.7V supercapacitor.  It looks reasonable to use the LM2665 to generate 2.5V (Vin/2) and use that to charge the capacitor.  My question is whether I can rely on the 40mA max current of the LM2665 to operate as constant-current charger, or whether that will potentially harm the device.  The discharged supercap will basically look like a short circuit so I worry that this won't work.  I could put a 62Ω resistor in series with the supercap, thus limiting the max current to 40mA, but as the capacitor voltage rises the charging current will decrease, significantly stretching out charging time.

Any advice on this, either for a way to use the LM2665 or some other circuit I am overlooking, is appreciated!

  • 0
    TI__Mastermind 46905 points
    Hi David,
    the current should be defined by the C1 and the switching frequency. you can change the value of C1 to limit the output current.
    it is basically charge and discharge the C1. if the C1 is not too large, I don't think the device could be damaged.
  • 0 in reply to Jasper Li
    Expert 2315 points

    How does that work with the warning in the Absolute Max section, to limit a short circuit from OUT to GND to less than 1 sec?  

    Are there any particular guidelines for selecting C1 for a given output current limit?

  • 0 in reply to David Dean
    TI__Mastermind 46905 points
    Hi David,
    I think it is related to thermal of the IC. if the output is short to GND, all the power loss occurs in the IC, which would cause high temperature in the IC. what is the capacitance of the super capacitor?

    if the output is zero, the output current should be 2*5v*C1*fs. it is 80mA if the C1 is 0.1uF. we can try 0.01uF to limit the output current at 8mA.

    with a charge pump, how do you guarantee the output voltage may be lower than 2.7V. there may be voltage variation in the 5V rail.
  • 0 in reply to Jasper Li
    Expert 2315 points

    The capacitance would be somewhere between 1.5F and 6F.  That would look like a short circuit for a lot longer than 1 sec...

    The input voltage is a fairly well-regulated 5V, so I'd feel comfortable saying it would never go over 5.4V, and Vin/2 would never go over 2.7V.

    I wouldn't want to limit the current to something as low as 8mA.  Even 40mA is less than I'd really want, but it's the highest Iout I found in a charge pump of this size.  

  • 0 in reply to David Dean
    TI__Mastermind 46905 points
    Hi David,
    it is very old device. it seem the device doesn't have over temperature protection function. I think this is why the device limits its short circuit period. otherwise the thermal could damage the IC. May we can have a >20ohm resistor in series with the C1? the most of the power loss will be at resistor.

    do you also consider the buck converter to charge the super capacitor? as in this application www.ti.com/.../slva678.pdf
  • 0 in reply to Jasper Li
    TI__Mastermind 46905 points
    Hi David,
    I closed the post as I assumed you had solved the problem. if not, just reply below.