Recently, for a project, I needed to provide an auxiliary power unit. Since required power demand is not high, I decided to go with a 1200 mAh 18650 Li-ion battery with a TP4056 charging module and a HW-432 voltage step-up module (to provide 5 V needed for the project).

Here is the schematic I am using (minus the 1N4007 and fuse meant as reverse voltage protection]):

enter image description here

When I connect the 5.5 V input voltage (since there is a 0.5 V voltage drop because of the D2 Schottky diode [SS34]), everything works fine; when the input voltage is cut, the battery kicks in and again everything looks good, until the battery is fully discharged.

After that, when I connect the 5.5 V input voltage, it draws around 80 mA forever. After waiting for a good 20h, I connected 4.2 V (limited to 1 A) directly to the battery for 1 minute and now it charges the battery with the current around 1 A as suggested in the datasheet.

Could someone please tell me what is wrong? Perhaps the battery is faulty?

  • \$\begingroup\$ Why have you got Q6/D2 etc. when the TP4056 does all that for you anyway? \$\endgroup\$
    – Finbarr
    Commented Jul 24, 2021 at 9:16
  • \$\begingroup\$ @Finbar D2, Q6 are to switch between (Vin to J3) or (U1 to J3) . The TP4056 hadles battery low voltage cutoff. \$\endgroup\$
    – Russell McMahon
    Commented Jul 24, 2021 at 11:17
  • \$\begingroup\$ I had the impression TP4056 modules passed the 5V input straight to the output if present. Maybe not all of them do. \$\endgroup\$
    – Finbarr
    Commented Jul 24, 2021 at 11:31

1 Answer 1


If you are using a typical TP4056 module then it breaks the negative lead between charger proper and battery negative when battery voltage is too low. B+ and Vout+ are joined.
Vin -ve and Vout -ve are joined.

I do not know what the "pin" numbers are for U7 (not the same as on diagrams I have) BUT B+ and Vout+ should be connected (but are not shown connected on your diagram). This seems liable to be a major issue.

Typical TP4056 circuit.

enter image description here

If the charger was charging the battery at 80 mA x 20 hours = 1600 mAh a 1200 mAh LiIon cell shou;ld/would have been very charged +. This suggests things may not be as you think either connection wise or alive wise.

Might the 80 mA be being taken by U1? Why not?

  • \$\begingroup\$ the connection between OUT+ and B+ is made internally in the module, so no problems there. \$\endgroup\$
    – Babak.ds
    Commented Jul 24, 2021 at 18:24

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