# Li-Po charger booster circuit

I'm trying to design a simple circuit that has the ability to charge a single cell Li-Po battery and can be used as +5v power source. I chose the circuit from here.

It seems that when VUSB is active, the +5V pin would select the VUSB instead of the step-up regulator because of two Schottky diodes, and the charger chip would charge the battery at the same time. When the VUSB pin disconnected, the charger would go off obviously and it seems that the EN pin of step-up would go high (Why?) and enables the regulator, then the +5v pin selects the output of the step-up regulator. My first question is what is happening to FET switch?

My second question is, does this circuit protect the LiPo battery from under-voltage discharge? For example, when VBAT goes below 3.2V. I'm using a different step-up regulator (AP3031) and I think I should calculate a different value for R15.

EDIT: here is the second schematic to make sure step-up regulator become active only when VUSB is off.

My first question is what is happening to FET switch?

As far as I can tell, the idea is that VUSB applied to MOSFET gate through D1 should open it, pulling PROG to the ground and enabling charging (2k resistor sets charging current to 500 uA), while at the same time disabling step-up via EN pin.

When VUSB removed MOSFET closes and this supposed to enable step-up and disable charging.

I don't think it would happen, though. AP3031 requires logic "High" to enable, and there is no pull-up on the input. On the other hand PROG input should be left floating to disable, so you cannot just add pull-up there.

My second question is, are this circuit protects the LiPo battery for under voltage discharge?

There is no discharge protection in the circuit whatsoever, not under-voltage, nor over-current/overheating. MCP73831 only manages charging. You can use battery with built-in protection PCB.

What I also don't like about this circuit is that step-up regulator is connected to battery directly. This can affect charging profile with unknown consequences.

UPDATE: Have you considered using more sophisticated BMS like MCP73871 ? It has all those MOSFETs and diodes inside the chip itself. All you have to do is connect your step-up (including EN pin) to "system load" output.

• Thanks, for your last tip, what is your suggestion? Is there any other option I need to get 5v from the battery? What do you mean by "On the other hand PROG input should be left floating to disable". I think the charger would disable if the VUSB becomes low. Can I use another FET switch and connect its gate to VUSB, when the VUSB becomes high it would connect the EN to the ground and remove the wire from R15? – M. R. Jul 21 '18 at 20:41
• I don't see a simple way to isolate battery from output while charging. Maybe MOSFET can be used. Check out AN1149 for ideas. Yes, the charger should be off when VUSB is low. And if you use another MOSFET to control EN input then you can add pull-up there to connect EN to VBAT (this is another reason to isolate battery from step-up, you don't want pull-up connected to charging circuit). And you don't need R15 at all. – Maple Jul 21 '18 at 22:44
• Here are couple links on battery disconnect. I am sure you can find more. one, two – Maple Jul 21 '18 at 23:23
• Thanks. I read the AN1149 and it seems that it uses the same idea like here it delivers current from the battery to load by a FET switch and In the above schematic the power delivers by the step-up regulator which has an enable pin like the gate of FET. I added another FET as I said before but I think it is not good because when the VUSB is on it draw current with EN pin of step-up from the battery. I edited the first post here – M. R. Jul 22 '18 at 15:49
• I think you can simplify circuit by connecting MOSFET gates together. As for drawing current from the battery when VUSB is up, see my previous comment: "this is another reason to isolate battery from step-up, you don't want pull-up connected to charging circuit". Note that I mentioned AN1149 because in this case step-up IS the load. EN pin disables it's internal circuitry, but VBAT remains connected to pins 1, 5, 6. I have no idea what is inside that chip or how it will affect the charging. But placing switch between battery and VBAT will cut it all off, including pull-up. – Maple Jul 22 '18 at 17:38