I’m charging a Li-ion 3.7 V, 3500 mAh using a TP4056 charge controller IC and feeding a 6 V, 9 W monocrystalline solar panel to charge it.

When I connect the solar panel at good sun light at 2 PM in the noon, though what ever the programmable resistor I connect Rprog, 1k, 2k, 3k or 4.7k, the 4056 IC is heating very badly, at times my skin got peel off. What is the mistake we are doing?enter image description here

  • 1
    \$\begingroup\$ Post schematic or it didn’t happen. \$\endgroup\$
    – winny
    Jun 19, 2018 at 9:29
  • \$\begingroup\$ Schematic with Problem at the point is uploaded \$\endgroup\$
    – Shiv
    Jun 19, 2018 at 10:08
  • 1
    \$\begingroup\$ Better, but what's behind your "speaking boubble"? \$\endgroup\$
    – winny
    Jun 19, 2018 at 10:27
  • 2
    \$\begingroup\$ It is a linear regulator so power loss is in the nature of this thing. Datasheet is lacking to say the least, but your average SO8 capsule is in the 100 K/W range so it will heat up badly if you don't have a PCB heatsink for it. \$\endgroup\$
    – winny
    Jun 19, 2018 at 10:31
  • \$\begingroup\$ What is the voltage at the BAT pin when no battery is connected? \$\endgroup\$
    – User323693
    Jun 19, 2018 at 10:35

1 Answer 1


As @winny says, there is no place for the heat to go without a heat sink, so the temperature will rise. This device detects its own junction temperature, and will automatically reduce its output current to maintain the junction temperature at 145 Celsius. So with no heat sink, the chip will reach this temperature regardless of Rprog.

  • \$\begingroup\$ Thank you for this explanation. Could it get dangerours for the charged LiPo battery if they share the same case? If yes, what could one do to prevent it? \$\endgroup\$
    – Helyon
    Jan 14, 2021 at 18:01
  • 1
    \$\begingroup\$ If you are not in contact with the battery, I wouldn't worry too much. The thermal resistance of junction-to-case is not listed on the data sheet, but if you aren't attached to a heat sink you are probably going to shut down when you get a few watts dissipated in the part. The part also has a thermistor on it that you can use to monitor temperature and reduce power if you are worried about the battery overheating. \$\endgroup\$ Jan 14, 2021 at 21:53
  • \$\begingroup\$ Okay nice, thank you again for this good explanation! I think having some air between the battery should do the trick. \$\endgroup\$
    – Helyon
    Jan 15, 2021 at 0:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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