LiIon tp4056 with protection, avoid drop out voltage at output
u will get a dropout voltaje of 0.4 - 0.7V in the output

I doubt that. Actually I'm quite sure it is simply not true.

The DW01A datasheet (page 8, section 11.1) confirms this:

Normal condition ... M1 and M2 are both turned on.

Also: Most phone batteries have a protection circuit similar to what you show. On phones with removable battery the protection circuit is inside the battery. Accepting a 0.4 V drop due to this protection would severely limit the battery life.

So I think you're forgetting that both MOSFETs must be on when the battery is in use. Perhaps you're thinking that only one MOSFET is on at any time. This is not the case!

Why use to MOSFETs in series then?

Well, both MOSFETs can only block the current in one direction due do the drain-source diode. So two MOSFETs in anti-series have to be used so that current can be blocked in both directions.

I think you should simply do what everyone else does and that is simply this:







simulate this circuit &ndash; Schematic created using CircuitLab

This is how it is supposed to be used!

The overcharge protection of the DW01A is not the same as what the TP4056 has! You want both. The TP4056 charges until the battery is full and then it stops charging. The DW01A's protection is there to prevent fire/smoke when the TP4056 does not stop charging (for whatever reason). Some cheap Chinese gadgets rely on the DW01A to stop the charging but actually that's a really bad idea. The DW01A stops the charging at a much higher voltage than any decent charger. So the battery will be overcharged all the time and wear out more quickly.