The USB-PD protocol supports a maximum output of 20V 5A, but directly outputting 12V is limited by the current, resulting in a maximum power of 12V x 5A = 60W. To achieve the full 100W supported by the USB-PD protocol, while also powering a radio, it would require tricking the module into outputting 20V and then converting it to 12V or 13.8V. However, if you have a shortwave device that is compliant with the new B standard, directly outputting 12V should be sufficient, right?
Additionally, could portable power banks and USB-PD deception modules potentially introduce new interference? I recently tested a specific brand of power bank, and the results were quite concerning. After searching through forum posts, I couldn't find any discussions specifically about this issue. Could someone with expertise shed some light on this?