1
\$\begingroup\$

I have this TTL-to-USB-serial device (supposedly from WINGONEER, who doesn't seem to have any documentation on how to use it):

https://www.amazon.com/gp/product/B06XY18XLL

picture of board

I'm using it to hook up serial console on a system that has a 3.3V TTL serial debug connector with 4 pins: ground, tx rx, and a +3.3V supply. Since the USB adapter above produces 5V and 3.3V outputs on those pins, I left them unconnected. My expectation was that the adapter would only power up when the USB port is connected to a USB host, and that it would otherwise remain unpowered.

To my surprise, when I tried powering up the system it's connected to without the USB cable connected, the LEDs on the adapter lit and it seemed to be operational (power LED steady, activity LED flashing). Presumably it's drawing power over the RX line.

Is this potentially damaging?

I found the datasheet for the CP2102 chip and it indicates that the 3.3V can be configured as a supply or an output, but it seems to be configured as an output on this board. Would it be safe to hook it up?

\$\endgroup\$

2 Answers 2

Reset to default
1
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

It is common for pins on a chip to be protected with diodes. These diodes clamp the pin to the rails. This way if a voltage transient occurs such as a static discharge the pin is undamaged. The TX and RX lines are pulled up with resistors somewhere. This may be inside the chip or externally. UART has to have pull-ups. What is most likely happening is that D1 is forward biasing and powering up the rail that it is attached to. You can increase the value of the pull up on the line to alleviate this effect.

\$\endgroup\$
0
\$\begingroup\$

Well if you device is powered, the it will power this adaptor.

You must decide what is powering what.

The +5 is probably connected to USB +5. If you connect it to your (powered) device +5, then your device will try to power the PC through its usb port. Or maybe the other way around. It's confusing. It probably won't end well.

If your device is always powered, don't join 3,5V to your device.

If usb only is powering everything, then connect 5V

If you want either or both ends to be powered, you need to use a schottky diode to connect them (1n5817) so they don't fight

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
6
  • 1
    \$\begingroup\$ I understand it would power the adapter if I connect the +3.3V (but I'm unsure if current might flow the other way back into the system it's connected to if the "3.3V" from the adapter is slightly higher, and if that would be problematic). I don't understand why it's powered with neither the +5V or +3.3V connected, only GND/TX/RX, and that seems bad... \$\endgroup\$
    – R.. GitHub STOP HELPING ICE
    Commented Apr 25, 2018 at 1:38
  • \$\begingroup\$ The intended usage is that the adapter is always connected to the device (the board is easily damaged by repeatedly connecting and removing it), but the USB cable to a USB host is only connected when console access is needed. So I'm concerned about whether having it powered over TX/RX lines most of the time is going to be a problem. \$\endgroup\$
    – R.. GitHub STOP HELPING ICE
    Commented Apr 25, 2018 at 1:40
  • \$\begingroup\$ Put a resistor in series with the TX line going from your device to the adaptor RX pin e.g. 10k or 22k. What powers your device? \$\endgroup\$
    – Henry Crun
    Commented Apr 25, 2018 at 1:48
  • \$\begingroup\$ The device is powered by its own power brick with 5V output. Adding a resistor sounds like it would make the issue go away but there's nowhere really good to put it except cutting out part of the wire and splicing it in the middle. \$\endgroup\$
    – R.. GitHub STOP HELPING ICE
    Commented Apr 25, 2018 at 2:01
  • 1
    \$\begingroup\$ and the problem with that is? The resistor is correct, as you have no control over voltages at each and, so current will flow one way or the other, trying to power one part up through its cmos protection diodes. In practice probably not going to blow anything up from 3V. \$\endgroup\$
    – Henry Crun
    Commented Apr 25, 2018 at 2:06

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.