The process you describe is a method to put the ESP32 into flash mode (bootloader mode) to load new code into the microcontroller. Here's how the steps you take work: Connect GND from USB to UART to IO0: When you connect IO0 to GND and then power the ESP32, this puts the ESP32 into bootloader mode. In this mode, the ESP32 is ready to receive new codes from the computer via the serial interface. The green LED on the ESP32 lights up to indicate that the device is in this mode. Connect EN to GND: When you connect EN to GND, you are disabling ESP32. All microcontroller operations will stop and the green LED will turn off, indicating that the ESP32 has been turned off. Disconnect EN and GND: When you disconnect the EN connection from GND, the ESP32 will be enabled again. If IO0 remains connected to GND during this time, the ESP32 will reboot in bootloader mode, ready to receive codes from the computer via the serial interface. Flash code: Once the ESP32 is in bootloader mode (IO0 is connected to GND and EN is connected again to VCC), you can start loading new code into the ESP32 from the computer via a flash tool (e.g. esptool) . Process summary: Put the ESP32 into bootloader mode: Connect IO0 to GND. Connect EN to GND to turn off ESP32. Disconnect EN from GND to turn the ESP32 back on in bootloader mode (with IO0 still connected to GND). Load code: Use a code loading tool (like esptool) to flash new code to the ESP32. Note: After loading the code, you need to unplug IO0 from GND and reset the ESP32 so that it boots into normal operating mode and runs the newly loaded program. This process is a manual way to put the ESP32 into bootloader mode, usually used when there is no push button to enter this mode.

Yes it can, It has ethernet port built in. You can find out some libraries for it. Maybe I will make a video about it.

@@YaserAliHusen thankyou, that will be great

It is also not good to supply power to the USB plug if GND is not connected. It is also not good to remove the GND when the board is still connected to the power supply. How should the voltage regulator generate the 3.3V if you interrupt GND?

Hi, Thank you for information, That is my experiment, I just found that way to upload the code. Hope your comment is useful for others Thank You

Hi, Thank you for your feedback It may help us. Regards

This is my new video: kzfaq.info/get/bejne/ZsemotZz2LfUmac.html

Thank you, I will find out about this

Thanks for your feedback

Thank You for your correction. While it's true that LEDs can work without a resistor in some scenarios, it's worth mentioning that using a resistor is generally recommended. It helps safeguard the LED and provides stable performance over time. So, for simplicity, I demonstrate without a resistor, but in real-world applications, using a resistor is advisable

The resistor is not needed in all situations. The role of the resistors in the circuit to create the desired voltage drop, however if you dont need any drop you can avoid it at all. Example: Power 3,3V -> transistor (drop 0,7V) -> LED (Uf = 2,6V). In this case you dont need any resistor as voltage drop is realized by tranzistor. You can easily verify it by creating test circuit with A-meter in series.