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recently, for a project, i need to provide an auxiliary power unit, since required power demand is not high, i decided to go with a 1200mAH 18650 li-ion battery with TP4056 charging module and HW-432 Voltage step-up module (to provide 5v needed for the project).

here is schematic i am using (minus the 1n4007 and fuse meant as reverse voltage protection]). enter image description here

If i connect the 5.5v input voltage (since there is 0.5v voltage drop because of the D2 schottky diode [SS34]), everything works fine, when the input voltage is cut, battery kicks in and again everything looks good, until battery is fully discharged! after the when i connect the 5.5v input voltage, it draws around 80mA forever?! after waiting for a good 20H, i connected 4.2V (limited to 1A) directly to the battery for 1 minute and now it charges the battery with the current around 1A as suggested in the datasheet. so could someone please tell me what is wrong? perhaps the battery is faulty?

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  • \$\begingroup\$ Why have you got Q6/D2 etc. when the TP4056 does all that for you anyway? \$\endgroup\$
    – Finbarr
    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
    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
    Jul 24, 2021 at 11:31

1 Answer 1

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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?

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

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