I bought an 8 channel USB/UART controlled relay board from banggood.com:


This board features a Nuvoton N76E003 1T 8051-based microcontroller to convert the serial commands (which come from either the UART hook-up or an on-board USB-to-serial converter) to actual relay switching action.

I was able to communicate with it using USB with no problems but I decided that I wanted to use the UART interface instead for my project. I bought an FTDI FT4232HL-based 4-channel UART breakout board to do just that:


I set the break out board outputs to be 5V using the solder-jumper pads on the bottom and hooked it up to the relay board and my computer USB port.

I found that when I plugged the USB into my computer, the N76E003 would boot up and work (read: communicate with computer over serial connection) even when DC power was not applied separately to the relay board.

I made some measurements and there are DC voltages on the TXD and RXD lines coming from the FTDI breakout board. And when the FTDI board is connected (only using the TXD, RXD, and GND lines) to the relay board I'm able to measure around 3.7V on the VDD pin of the microcontroller.

As a further note, the relay board has the UART breakout pins tied to the UART0 of the microcontroller.

Again, this scenario is without any external DC voltage applied to the relay board, and there isn't enough power to drive the relays. I tried toggling the relays with serial commands and I actually thought that making it do this would overload whatever this mystery power delivery mechanism is and kill either or both of the boards, but it didn't, just no relay action.

I poked through the Nuvoton datasheet and couldn't find any mention of a "Power-over-UART" capability. So my question is this: is this supposed to be happening, has anyone else seen this phenomenon of this uC chip, or have I somehow damaged the thing in a really interesting way?


1 Answer 1


You're likely backfeeding the power supply through the ESD-protection diodes integated on the chip. They dissipate static charge that may be applied to a pin, sending it safely to the power rails (which are expected to have bulk capacitance, a power supply, loads, etc which can absorb that charge), protecting the very sensitive thin gate oxides of the chip's transistors from catastrophic damage.


simulate this circuit – Schematic created using CircuitLab

This happens to work, including in interesting projects such as this one, but it's almost certainly not an intended mode of operation designer unless specified explicitly by the datasheet. The diodes have a limited current handling capacity, and because your supply might be coming through a UART pullup resistor, it's possible to experience both diode voltage drop and resistive voltage drop related to the load current.

  • 1
    \$\begingroup\$ Aha! That's it, sort of a "no, duh" now that I know the answer. Thanks! \$\endgroup\$
    – user97497
    Commented Mar 3, 2021 at 17:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.