Battery Switching Circuit using Transistor and PMOS

Question

I am designing a circuit for charging Lipo battery with control and using the same circuit for switching to Lipo battery supply when there is no power source. I am using the following circuit:

(You can also see the simulation at http://tinyurl.com/htwynv2)

Control Input is from micro-controller (3.3V or GND). 4.18V Power Supply is used which is converted from 12V using SMPS.

Working:

3.3V/GND Control Input is given to transistor (NPN) for turning charging ON/OFF.

When Power Source is available,

• At 3.3V control input, the transistor is ON and voltage at point P is low, so PMOS is ON and battery is On charging.
• At 0V (GND) control input, the transistor is OFF and voltage at point P is 4.18V (High) and PMOS is OFF and battery is isolated.

When Power Source is not available,

• Due to the intrinsic diode of PMOS (which is in the forward direction from drain to source), Battery voltage will be reflected to the power source and current will flow through this diode.

4.18V supply is used for driving GSM module, which operated between 3.4V to 4.4V and while posting to server, the module takes 2A burst current (~1ms).

I have some basic doubts:

1. Is this circuit is suitable for switching to Lipo battery supply?
2. Is working of this circuit is same as I described above?
3. When there is no power source, circuit switches to battery supply with a drop of around 0.4V between drain and source of the PMOS. Can I reduce it by using any other PMOS?
4. When power is from the battery, then GSM module is facing power reset while posting to the server. I believe this is because of surge current of 2A. Maybe due to the current limitation of a diode or anything else? Should I switch to other PMOS with the low dropout of diode?

Also, it will be great if you people can suggest any better circuit for this purpose.

Answer 1

Something along the lines of this will eliminate the body diode drop.

^{simulate this circuit – Schematic created using CircuitLab}

When Control is low M4 is On. The gates of M1/M2 are high and the battery is isolated.

When The control is high M4 turns off. The gates of M1/M2 are pulled low connecting the battery to external power.

When external power is off R1 pulls the the gates low no matter what the status of the charge control line and the battery is connected to the load.

D1 is required to prevent the battery turning itself off. R1 should be small enough that the gates of M1/M2 are discharged sufficiently quickly to avoid an interruption to the loads power.

You could potentially eliminate M2 if you can be 100% certain that external power will always be over the battery voltage when it is applied. M2 ensures that if external power is present and the control line is idle the battery is always isolated even if it's voltage is above that of external power, this is needed if you want to be able to run off a voltage below 4.5V without partially discharging the battery.

-- Added detail --
This is all assuming that when high the control input is sufficiently close to the external power voltage to turn M4 off fully. (3.3V and 4.5V should be ok assuming a threshold of around -2V which is fairly typical). If however that isn't guaranteed then instead pull the gate of M4 to external power with a 1k and connect the control input to the gate of an n channel. The n channel then connects the gate of M4 to ground when the control line goes high. This inverts the operation of the control input. It's extra parts and complexity but eliminates the voltage dependency.

This is all assuming you have a smart battery pack with a charger built in and not just a basic protection circuit. Since you refuse to give part details don't blame me if your battery bursts into flames. The external power needs to be around 4.5V so that you get at least 4.2 reaching the battery, I have no idea what headroom the charger needs but it's going to need something over the raw cell voltage in order to fully charge them.