I asked a while ago, how to build an automatic light switch and finally got around to actually building it.

My board is an ESP8266 mini D, and ignoring all the sensor parts, my problem right now is powering the actual light.

It’s just a small LED array and I connected it directly to the 5V and GND pins (controlled via a transistor).

Measuring from the wall (so including the PSU), this whole setup pulls about 3W (so far expected), however, one small component close to the USB connector gets uncomfortably warm, and I’m not sure, whether that’s ok.

The hot component is one of the two small thingies circled in the picture. I thought the 5V get pulled directly from the USB plug, so I’m not sure, why there is any circuitry involved.

  • leisesprecher@feddit.orgOPEnglish
    2·
    3 months ago

    Two stupid questions:

    1.) why is there a 5V regular for a USB plug that’s supposed to be 5V only?

    2.) will this regulator output more than 2W and burn itself out or is there a self limiting thingy? The 3W from the wall include power supply and board, so 2W don’t seem too far off.

    • zorro@lemmy.worldEnglish
      5·
      3 months ago

      1. It helps ensure that levels between 2.5v and 5.5v can be fed to the controller without breaking anything. For some cheaper power supplies you might get a voltage drop when starting to pull load, this will clean that up and prevent the voltage from dropping too low for the microcontroller.

      2. If I’m reading the correct datasheet I can see it is current limiting so it should shutdown when overdrawn.

      • 5ymm3trYEnglish
        2·
        3 months ago

        The voltage regulator is only used to generate the 3.3V supply for the ESP. But OP is using the 5V from the input of the regulator to power the LEDs.

        At max it is wasting about 0.7W if the 400mA max current is right. But it will still get hot due to the current draw of the ESP. Even at half of the max. rated current this device is probably 30-40 degrees above your ambient temperature.