My company makes a USB-C/USB-PD 2.0 compliant charging device for use in airplanes. It runs off 28VDC, has a built-in DC-DC converter, and uses the TPS25740B for the USB-PD controller. See the picture below for a conceptual block diagram. Given the extreme space constraints in aircraft instrument panels, we want to make a right-angle adapter with these design constraints:

  • Have a USB-C male plug that can plug into an unmodified charging device
  • Light up an indicator LED, with different colors for whether or not an external device is plugged in or not.
  • NOT have a hot VBUS when not in use
  • Fit into an extremely small space. The existing charging device is only about 4cm x 5cm x 1cm, and this adapter needs to be even smaller.
  • Capable of USB data pass-through for possible future applications

Our first idea was to have a pull-down resistor on the CC line, effectively forcing the TPS25740B to provide VBUS at all times to power the indicator LED. We had hoped to find some system to turn the external VBUS connection on and off with a PMOS switch, so as not to have a hot VBUS when not in use.

I discussed the matter at length with an engineer on the TI E2E forums, and he said this idea would never work, or at least never be USB-PD compliant. He recommended instead that I use a USB-C dual-port controller, TPS65988, and a buck-boost converter with the TPS55288. It seems to me that this would effectively be recreating the circuitry of the charging device itself within the right-angle adapter. Given the space constraints, and concerns about cost and development time, I doubt this would be acceptable.

As I was researching this, I had to ask, is there any easier way to do this?

Preferably something that doesn't involve making another DC-DC converter and/or programming a microcontroller. Any help, advice, or resources would be appreciated. Thank you.

Concept Block Diagram

  • 1
    \$\begingroup\$ "NOT have a hot VBUS when not in use" How to determine and transition from VBUS cold to VBUS hot? \$\endgroup\$
    – winny
    Apr 26, 2023 at 14:33
  • \$\begingroup\$ I don't know. When I say "not have a hot VBUS," I mean no voltage on the female receptacle which would be the output of the right-angle adapter. I was hoping there would be some IC that could monitor the CC lines for a connection, and allow VBUS once it sensed one. Or something like that. \$\endgroup\$ Apr 26, 2023 at 14:42
  • \$\begingroup\$ You can't "monitor the CC lines" without some stimulus. USB ports are designed to idle at 5V VBUS. \$\endgroup\$
    – Ben Voigt
    Apr 26, 2023 at 14:43
  • \$\begingroup\$ @BenVoigt no, USB-C ports are designed to idle with no VBUS until a device is detected on CC pin. \$\endgroup\$
    – Justme
    Apr 26, 2023 at 14:44
  • 1
    \$\begingroup\$ Well, overall what is proposed is decoupling the PD negotiation that takes place between the existing supply and the downstream device. So naturally the existing supply will no longer be able to respond to the device's needs, and the interposer will need all the complexity that the TI engineer proposed. Consider keeping the USB connection intact and using an alternative mechanism for powering the LED, such as charging a capacitor when device is present and powered, and using a low duty cycle to make that stored energy last as long as possible. \$\endgroup\$
    – Ben Voigt
    Apr 26, 2023 at 15:30

1 Answer 1


USB C is designed with no power if nothing is plugged in. This means you have to terminate the USB C connection and basically start another connection if you want features like an led that shows up if the cable is disconnected.

Your idea of an extra pulldown on the USB C CC lines doesn't work, if a standard device is plugged it, it would push the values to he undefined region, where the upstream or downstream facing device may not behave correctly. (for example, a passive USB C PD devices sees a lower voltage than it would normally have seen, so charges slower)

If you do not want a DC-DC converter inside your device, one trick you can do is modifying the USB C PD packet to lower the advertised current based on the resistance the extra connection gives. The exact current that you need to lower it depends on extra resistance your adapter provides, as you still have to meet the max 0.25V (eg 0.08333 ohm at 3A) voltage drop on the VBUS cable. This also requires a protection MOSFET so the VBUS is only turned on when a proper connection is detected. Make sure to add a SOP marker to the VCON wire for 5A advertisement, so the original charger can give higher current profiles, you also need to read the SOP marker in the target cable, as having a 5A marker means the cable has lower resistance, so you can switch to the 5A resistance calculation for higher output current

Of course, it would be better to also include a DC-DC converter into the device, as it allows you to boost the voltage back up to the standards, which means you can advertise values closer to the values of the original charger. This also requires a protection MOSFET so the VBUS is only turned on when a proper connection is detected.


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.