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3.3v circuit from a LiPo cel

Discussion in 'Power Electronics' started by Trikkitt, Feb 23, 2018.

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  1. Trikkitt

    Trikkitt

    2
    0
    Feb 23, 2018
    I'm trying to figure out the best way to run a small circuit from a single LiPo cell. As best I can tell from the research I've done the 3.7v LiPo cells peak at 4.2v, drop to around 3.7v quite rapidly where they stay until almost flat, and are regarded as flat at 3.6v. The circuit components are rated at 3.3v but all appear to have an operating range of 2.5v to 3.6v.

    At the heart of the circuit is a PIC which wakes up every second, does a few checks, flashes an LED and goes back to sleep. The remaining chips get woken up when they need to do something, so spend most of their time powered down. I'm doing my best to keep power usage down to a minimum to get the maximum run-time. I'm also aware that I need to be careful about flattening the LiPo out completely, so I need a low battery circuit, but I also need a way to keep the supply voltage within the tolerated range. Cost per unit is also a factor here, I'm trying to make these as cheap as possible.

    To get the voltage in to the supported range I was thinking that perhaps just putting a simple diode on the positive side of the battery. That means I can't accidentally connect it back to front and damage the circuit, and should result in a 0.7v drop, giving me a 3.5v peak which is just within the supported range. This does seem a little risky - do LiPos float higher than 4.2v? I wanted to avoid the complexity of a buck converter (simple + small is important), and I'm also worried that unlike a PIC that I can turn off, the buck converter will continue to draw power. However I also don't know the efficiency of a buck converter vs a simple diode.

    For the low battery circuit, would I simply need to use the ADC using VDD as the + source and the 2v ref as the input, to then give me a reading of the battery voltage (assuming that VDD isn't regulated). If I ended up using a regulated VDD would I then complete this via a voltage divider? How do you stop that drawing power all the time?
     
    Last edited by a moderator: Feb 23, 2018
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,448
    2,809
    Jan 21, 2010
    Why not pick a microcontroller which operates up to 5V or so?

    A diode is not a great option because at low currents the forward voltage drops.
     
  3. Trikkitt

    Trikkitt

    2
    0
    Feb 23, 2018
    Because while the microprocessor runs at 5v quite happily it has some parts such as a wifi module that are only available at 3.3v. So the circuit as a whole needs to be around the 3.3v mark.
     
  4. Cirkit

    Cirkit

    132
    10
    Oct 28, 2015
    If you have a spare output port on the microcontroller, you could use it to switch a 3.3V LDO when the Wi-Fi module is required. Have you worked out the current profile of your circuit and how long it needs to last on one charge?
     
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