Connect with us

Best way to reducing voltage

Discussion in 'General Electronics Discussion' started by GerCer, Mar 22, 2017.

Scroll to continue with content
  1. GerCer

    GerCer

    8
    0
    Jul 16, 2016
    Hello!

    I'm designing a SMD circuit. Since it's my first time doing a SMD i was wondering which was the best way to reduce the voltage.

    The IC i'm going to use, needs max 3.3V to work. I was gonna use a CR2032 coin cell but is too big for the design. So i was thinking about using a 3.8V Li-Po rechargeable battery and reduce the voltage to no less than 2.7V to power the IC. I know there are some ways to do this, like a voltage divider, diodes or a linear voltage regulator. I must know when the battery needs to be charged so i can´t use a regulator. But i don´t know which one would be the best option for low power consumption and smallest size possible.

    Thank you,
    GerCer
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,397
    2,777
    Jan 21, 2010
    There's nothing magic about SMD. The same components are available.

    A low dropout 3.3V regulator would probably work for you. If the chip can operate down to 2.7V then the whole circuit will probably run until the battery voltage drops to 2.8V (which is considered "flat" for lipo).

    You can use a regulator and still detect the battery voltage, in fact you pretty much MUST have a voltage regulator of some sort to provide a reference voltage to which the battery voltage is compared.

    If your chip can operate at up to 4.1V out higher) then you may not need a regulator at all (for the chip)

    As to how to measure the voltage and annunciate that the battery needs to be charged, that depends what your doing. If that chip is a microcontroller turn you may be able to do everything in software plus a LED on an unused pin (or even dual purpose an existing display element).
     
  3. Audioguru

    Audioguru

    2,869
    644
    Sep 24, 2016
    A 3.8V Li-PO is 4.20V when fully charged and about 3.2V when it should be disconnected from its load, then its half-discharged voltage is 3.7V. If you discharge a lithium battery to 2.8V then it is damaged.

    If you use a simple voltage divider then you should calculate the minimum and maximum voltages and if the IC will operate correctly overt that voltage range.
     
  4. GerCer

    GerCer

    8
    0
    Jul 16, 2016
    Hey! I had to recheck the datasheet.
    The max input is 3.4V and lowest is 2.35V. The manufacturer recommends to use no less than 2.5V.
    I was thinking about using the regulator LXDC2HL33A-055. It has a good efficiency and the max input is 5V.
     
  5. GerCer

    GerCer

    8
    0
    Jul 16, 2016
    Hello!
    First of all, this is the battery i was thinking to buy. So even if it says 3.8V means that the max V is 4.2V?
    If i use a regulator shouldnt be a problem. The problem is that the ADC max input is 1.2V (a 3x attenuator can be used to get a max input of 3.6V).

    What i thought about is that i could use an voltage subtractor. Let's say the battery is 4.2V and the regulator has 3.3V, so it would be 0.9V as input for the adc and i would know it's full and if the battery is 3.3V i would know i need to recharge.

    I'm new designing circuits so those were my ideas. I don't know if is a good way to do it tho...
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,397
    2,777
    Jan 21, 2010
    Yes. That's typically the voltage you charge them to.

    I can't see that a voltage divider will work,. It will just draw power until the battery is flat.

    However, since we don't know what you're doing, we really can't be certain.

    Can't you throw us a bone?
     
  7. GerCer

    GerCer

    8
    0
    Jul 16, 2016
    Oh yeah sorry, i forgot about that.

    I'm designing a small bluetooth circuit using a DA14580 (BLE microcontroller). The main idea is to have the circuit in sleep mode until an interruption happens, it wakes up to read some peripherals and send the info to an app. The sleep mode consumes 600nA and it will consume about 10mA when it's awake. It will only be awake for a few seconds before going to sleep again.
     
  8. GerCer

    GerCer

    8
    0
    Jul 16, 2016
    The problem with the batteries is that dialog semiconductors just offers a few ref. designs using a cr2031 coin cell which is too big for what i'm thinking to do.
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-