Designing a circuit to charge 3.7v battery from 12v

Tajiknomi

Joined Mar 18, 2010
34
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).

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

crutschow

Joined Mar 14, 2008
26,781
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.

Last edited:

JUNELER

Joined Jul 13, 2015
183
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..

crutschow

Joined Mar 14, 2008
26,781
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..
That doesn't change the power dissipated.

ian field

Joined Oct 27, 2012
6,539
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).

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
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.

Tajiknomi

Joined Mar 18, 2010
34
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.

crutschow

Joined Mar 14, 2008
26,781
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.
Not at the same time I presume.