I have been experimenting with Wemos D1 mini for deep sleep and came across the video on this topic ( Youtube Link: Wemos D1 Mini Low Power) to reduce the power consumption of the board by disabling the UART during battery operation.

As I am supplying 3.3V directly to the Wemos, after removing the copper trace as shown in the video (so that CH340 does not get 3.3V during normal operation via battery, but it gets its 5V supply when connected to the USB for programming), everything should be normal and wemos should have a better power consumption.

However, the build in blue LED on the ESP12-F is lit dimly during the deep sleep mode and I am afraid that it might eat up quite a lot of current. I do not have the possibility to measure the current at the moment, but I assume the GPIO2 has some leakage currents and/or it is pulled down during the deep sleep, which cause the LED to turn off (which does not fit totally, as the GPIO 0 and 2 of ESP8266 shall be pulled up internally for booting up).

Or the suspicion is that, the removed trace causes the CH340 UART chip behave weirdly and having some floating outputs, causing the LED lit dimly, which is supported by the interesting observation, when USB is connected to supply power to the CH340, the LED is turned off.

Does anyone have an idea, what the reason could be and how the dimly lit LED can be turned off during the deep sleep? What am I missing here?

Thank you

Wemos D1 Mini Schematic

  • \$\begingroup\$ Videos really shouldn't be relied on to provide key parts of a question. Please edit the question with something which can definitively identify the cut trace in the schematic and indicate the reference designator of the "blue" LED. Generally speaking, allowing any I/O voltages to an unpowered chip is a recipe for trouble, as you may power the supply rail of the chip (and even board) through an ESD protection diode. This is out of spec for the diode and may lead to damage, typically it raises the power net to the point where something non-linear just begins to draw real current. \$\endgroup\$ Sep 7, 2020 at 22:05

1 Answer 1


A quick look at the schematic makes me think that 3.3V current is flowing from the led to ground across the GPIO2 pin when the UART is not powered. I would advise to increase the impedance of the LED resistor to at least 10K or to remove the led.

Remember that leds can shine dimly with extremely small current. If the UART was intentionally turning on the led it would shine brightly.

  • \$\begingroup\$ That's rather heavy on suspicious and rather short on fact compared to the expectations on a Stack Exchange site - answers here really shouldn't be based on vague guesses, but rather on things which can be stated with certainty, which sometimes means waiting until the question has more details. Also, even quite small currents will quickly deplete the battery of a device design to spend most of its time asleep. \$\endgroup\$ Sep 7, 2020 at 23:24
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    \$\begingroup\$ @ChrisStratton Stratton This was a strong suspicion and I completed my answer after looking at the schematic. \$\endgroup\$
    – Fredled
    Sep 7, 2020 at 23:25
  • \$\begingroup\$ Which LED in the schematic are you seeing connected to any UART? \$\endgroup\$ Sep 7, 2020 at 23:26
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    \$\begingroup\$ But LED1 is connected to GPIO2, not any UART pin of the ESP32 or the CH340. GPIO2 is used for mode strapping on boot, and has a pullup. If GPIO2 is ending up low, that's doubly bad, as even without the LED asserting a signal against a pulling resistor wastes battery in sleep. \$\endgroup\$ Sep 7, 2020 at 23:30
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    \$\begingroup\$ Update after 2 months, I think it was some small uA level leakage current that was causing the dimly lit LED. With a single 18650 battery cell, by taking measurements once and sleeping for 10min, it took almost 2 months to deplete the battery to 3.6V, which is more than sufficient for my purposes. It might slightly improve the battery life, if the LED is removed. \$\endgroup\$ Oct 27, 2020 at 12:35

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