I have a very small TP4056 based li-ion battery charging circuit board. It has integral LED charging status lights, but I would like to add a remote LED to indicate when a battery is inserted and actively being charged.

The TP4050 has an input of +-5VDC and output two connectionsfor the battery under charge.

How could I hook up a simple LED using only the +5V input connections and/or the output connections to the battery so that when the battery is connected and charging is active, the external LED lights ups? I thought about just piggy backing on the internal LEDs, but they are very tiny surface mount devices and I cannot solder to them without most likely damaging the board.

Any ideas?

  • \$\begingroup\$ Doesn't seem very simple if you can't access the CHRG and STDBY pins. You could use a shunt resistor between +5 connections and measure voltage (will depend on current drawn by battery). Then use the voltage to determine state of device. \$\endgroup\$
    – Ernesto
    Commented Mar 31, 2021 at 6:23
  • \$\begingroup\$ Could you use a light pipe? vcclite.com/product-category/lightpipes/flexiblelightpipes \$\endgroup\$
    – user57037
    Commented Mar 31, 2021 at 7:18
  • 1
    \$\begingroup\$ Can you place a few other LEDs infront of the internal LED to generate a photovoltage that could be enough to open a small mosfet, which drives your indicator LED off the 5V? \$\endgroup\$
    – tobalt
    Commented Mar 31, 2021 at 8:32
  • \$\begingroup\$ I will run measure the voltage change for the first option (shunt). the light pipe is an intriguing idea, but might be difficult as the project ends up in a solid piece of wood about 3" square. The last idea (photo diode) is also interesting, although the smd led's are very tiny, and would they be able to generate enough signal? \$\endgroup\$
    – handyguy
    Commented Mar 31, 2021 at 20:50
  • 1
    \$\begingroup\$ It would really be much better to access the signals directly from the IC. Are you sure there is nowhere you can solder some wires? \$\endgroup\$
    – user57037
    Commented Apr 1, 2021 at 4:56

1 Answer 1


I will just post this comment from above as an answer as handyguy confirmed that it could be useful. Still, somehow drawing out a wire wth the control signal will be simpler of course.

enter image description here

Things to pay attention to:

  • The color of LEDs D2-D5 has to be "redder" than that of D1. It wont work if those are e.g. blue LEDs and D1 is green. A safe bet would be photodetector diodes or red LEDs.. The area is not so important because the necessary current is small. Still bring them as close as possible to D1. There are 4 Diodes drawn because the total voltage has to be enough to turn M1 on.
  • The transistor M1 is an N Mosfet. Choose the smallest type you can find with the lowest gate charge possible. R2 is to turn off the Mosfet eventually, but might not be needed if the leakage current turns it off fast enough. If the MOSFET has enough ON-resistance (at least a few 100 Ohm), you could even skip R1.
  • Shield the assembly from external light, or the transistor will never turn off.


As mentioned in the comments, it is also possible with fewer parts:

enter image description here

Here, D2 is a single photodiode or LED in photoconductive mode. R3 has to be tuned a bit to work. It has to be small enough that it does not turn on the MOSFET due to D2's dark current. And it has to be big enough, that it does turn on the MOSFET with light on.

  • \$\begingroup\$ Is there any specific reason for using LEDs rather than actual photodiodes or phototransistors here, other than that the LEDs may be cheaper? \$\endgroup\$
    – nekomatic
    Commented Apr 1, 2021 at 13:22
  • \$\begingroup\$ No specific reason. I think one could also use a single photodiode in photoconductive Mode to 5V. But would need to experiment a bit with the pulldown vs dark current \$\endgroup\$
    – tobalt
    Commented Apr 1, 2021 at 16:14
  • \$\begingroup\$ Thanks for the circuit! D1 is a blue LED, right next to a red LED... I may still attempt the piggy back. If I do wreck the board, I can always order a new one. \$\endgroup\$
    – handyguy
    Commented Apr 1, 2021 at 16:19
  • \$\begingroup\$ Then maybe choosing green LEDs for the illuminated side will make your circuit only sensitive to the blue light and (mostly) ignore the red LED \$\endgroup\$
    – tobalt
    Commented Apr 1, 2021 at 16:47

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