Designing a circuit to charge 3.7v battery from 12v

Discussion in 'Power Electronics' started by Tajiknomi, Feb 2, 2017.

  1. Tajiknomi

    Thread Starter Active Member

    Mar 18, 2010
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    0
    Charging 3.6v battery from 12v battery is just a module of the large circuit which i am trying to build. So here is the scenario of the specific module which i want to discuss.

    I am using BD139 NPN as a switch. Input base current to the transistor is given by OP-AMP. What i am trying to do is; when the output current (provided by OP-AMP) reaches the base of transistor(BD139), the 12v battery should somehow reduce its voltage (via voltage divider or whatever) to charge my 3.7v AAA batteries (2800mA). So what I've done is; I have configured the transistor to operate in 'Saturation mode' and used 'Voltage division method' to ensure ~5v/~500mA charge to the battery (shown in the screenshot below).
    [​IMG]
    The problem with this method is; my power dissipation is very high across the resistors (approx 6watts total). I want to somehow save my power (using non-active devices) to ensure conserving energy and i don't want to use Buck converter (costly) or linear regulator (which will result in the same power loss).
    Kindly suggest me any good approach or transistor configuration scenario, I will be grateful.

    Thank you :)
     
  2. crutschow

    Expert

    Mar 14, 2008
    21,731
    6,267
    Sorry, you are at the mercy of Ohm's law. :(
    There's no magic way to efficiently drop the voltage from 12V to 3.6V without using a buck regulator or other type of switching circuit with an inductive device.
    Any form of linear voltage drop (active, non-active) will have to dissipate that same amount of power.
    The minimum dissipation with a linear regulator, such as by placing a transistor in series with the 12V to the battery, will be 3.5W with a 5V output @ 500mA.
    It that's still too much then you need to go to a switching regulator.

    Below is the LTspice simulation of a simple buck regulator you might want to experiment with.
    It uses common, inexpensive parts, and no speciality switching regulator ICs.
    It's a hysteretic (bang-bang) type regulator so doesn't require any feedback compensation.
    Pot U2 adjusts the output voltage.
    M1 can be just about any P-MOSFET with at least a 2A rating.
    It should dissipate less than a watt in converting 12V to 5V @ 500mA.

    upload_2017-2-1_23-47-34.png
     
    Last edited: Feb 2, 2017
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  3. JUNELER

    Member

    Jul 13, 2015
    181
    25
    Hi,
    maybe an easy way is the 12v passes in to 3pin regulator (7805) and the output 5v placed a dropping
    resistor to give a 3.7v. No need a pwm for the gate or base input.

    try it..
     
  4. crutschow

    Expert

    Mar 14, 2008
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    That doesn't change the power dissipated.
     
  5. ian field

    AAC Fanatic!

    Oct 27, 2012
    6,543
    1,195
    Results would be better if you use a constant current charge circuit, but you also need to avoid overcharging, assuming nickel chemistry cells; there's a slight drop in terminal voltage when full charge is reached - most commercial chargers detect that to terminate charge. Cell temperature sensing also works and is simpler to implement.
     
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  6. Tajiknomi

    Thread Starter Active Member

    Mar 18, 2010
    34
    0
    Thank you all for your kind replies. I am already using a buck/boost DC converter to boost 3.7v to 12v in this same project. I am now thinking about "how can I use one boost/buck converter for two purposes i.e. 3.7->12v AND 12--> 3.7v in this project". I hope I will need your helpful suggestions on which buck/boost converter IC will suit "both" purposes.
     
  7. crutschow

    Expert

    Mar 14, 2008
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    Not at the same time I presume. :confused:
     
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