0
\$\begingroup\$

I have replicated the li-po battery charger with the famous TP4056 , the protector DW01A and the mosfets FS8205A... exactly like the comertial board.

I design a board that contains this circuit (and MCU, sensors linear regulators etc... all conected to "OUT+/OUT-"), so when the user charges the battery, he can still use the product.

In the comertial board they solder the thermal pad of the TP4056 to the VIN+ ( ??????? not like the datasheet recommends) What happens to me is that i have solder it to GND (like the datasheet recomends)...I don't know if is for this or not but my battery charger circuit is salowing 1.6mA from the batteries all the time (not in charging periods)...what is super high...

Also if i disconect the R5 resistor, the current consumption decreases .... so i supose that the DW01A is consuming a lot ...and i dont know why

Anyone of you experiment so much current consumption on this devices???

THANKS A LOT

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ Can you link the datasheet? The one I found via Google doesn't show the thermal pad connected to ground. \$\endgroup\$ Oct 23, 2019 at 15:39
  • \$\begingroup\$ TP4056 DW01A \$\endgroup\$ Oct 23, 2019 at 17:57

2 Answers 2

2
\$\begingroup\$

Also if i disconect the R5 resistor, the current consumption decreases .... so i supose that the DW01A is consuming a lot ...and i dont know why

This is likely a red herring. When you disconnect R5 you remove power from the DW01A and thus shut off the battery charge/discharge FETs. So with R5 removed, nothing should be able to discharge from the battery.

If you wish measure the current of the DW01, measure the voltage across R5 in circuit. Or add ammeter in series with R5.

\$\endgroup\$
5
  • \$\begingroup\$ I measure 0,156V so it is consumming 1.56mA... exactly the value of that strange consumption.... so.. is just the DW01A who is salowing all this current???? Why this happens? It is suposed that this IC is an ultra low consumption device... \$\endgroup\$
    – Nasib
    Oct 23, 2019 at 16:47
  • \$\begingroup\$ Well you are getting closer to the root cause. It is downstream of R5. Could be damaged or counterfeit parts. DW01 may be placed incorrectly, i.e. rotated 180 deg. Bad solder joints etc... \$\endgroup\$
    – sstobbe
    Oct 23, 2019 at 17:22
  • \$\begingroup\$ I have checked all the layout.. and everything seems to be ok... i am going to repeat the same readings with all the prototypes... lets see what hapens \$\endgroup\$
    – Nasib
    Oct 24, 2019 at 13:17
  • \$\begingroup\$ With the load connected and the battery too... this is the state of the DW01A-G IC: VCC 2.94V TD: 1.75V OC: 2.13V CS: 1.55V OD: 1.7V If i change the 100ohm res in the vcc with a 1000ohm VCC: 2.2V TD: 1V OC: 1.4V CS: 0.89V OD: 1V \$\endgroup\$
    – Nasib
    Oct 25, 2019 at 8:27
  • \$\begingroup\$ Well, when measuring voltages of the DW01 make sure you reference its ground which is the negative terminal of the battery, not the system ground after the protection FETs. Unless the OC/OD pins are oscillating they are totem pole outputs and should be either ~0V or ~Vcc, or the protection FETs are damaged and dragging the outputs down. Can you try using a known good working DW01 IC in your circuit? \$\endgroup\$
    – sstobbe
    Oct 25, 2019 at 15:46
1
\$\begingroup\$

This statement from the datasheet leads me to believe that the thermal pad should be connected to ground.

enter image description here

Source: https://dlnmh9ip6v2uc.cloudfront.net/datasheets/Prototyping/TP4056.pdf

The datasheet itself is badly translated, contacting the manufacturer directly would be the best way to clarify their intention of the datasheet. Datasheets from chinese suppliers are often incomplete, and do not have the same level of testing as other companies, I would avoid them if possible.

As far as the current consumption goes, this is a complex circuit, the best thing to do would be to find spice models if available and simulate in a spice simulator.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.