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There's a wall outlet with a USB-A fast charging port

It has a USB-A charging port that supports 9V 1.67A fast charging.

It uses FP6601Q and supports QC2.0/QC3.0.

enter image description here I'm trying to change the USB-A charging port to a USB-C charging port, but it doesn't work well.

Can we simply switch A to C port to support 9V 1.67A charging speed without USB-PD negotiations?

Below is a circuit that I made simple. I simply changed the port and connected only pull-up resistance to the cc pin of the usb-c port. enter image description here

However, it initially charges 9V 1.67A, but after a few minutes, the charging speed drops to 5V 1.67A

I tried all 56k/22k/10k resistance with cc pull up resistance, but the result was the same.

What should I do?

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  • \$\begingroup\$ Without negotiation, how would the PD know what your desired voltage level is? USB D-/D+ are important for this reason. You saw the results on what happens without proper negotiations, the voltage dropped to a "default" state. Use the USB spec as guidance on making a proper solution. \$\endgroup\$
    – Colin
    Commented Feb 19 at 2:28
  • \$\begingroup\$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. \$\endgroup\$
    – Community Bot
    Commented Feb 19 at 2:29
  • \$\begingroup\$ Even if I connect pull-up resistance to cc pin and charge 9v using fp6601q ic, does it drop to 5v if there is no additional pd negotiation? \$\endgroup\$
    –  8 minikhh
    Commented Feb 19 at 2:35
  • \$\begingroup\$ What kind of device are you trying to charge with your modification? \$\endgroup\$
    – Ale..chenski
    Commented Feb 19 at 4:51
  • \$\begingroup\$ Devices with 15w charging support are targeted. It's a general-purpose 15w fast-charger supported quick charge\ \$\endgroup\$
    –  8 minikhh
    Commented Feb 19 at 5:01

1 Answer 1

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Yes, but not the way you are approaching it

You'll need QC-to-USB PD translation. You can do this by interfacing a QC downstream (powered device) chip with a PD upstream chip (power source) using a small microcontroller. Translator "widgets" may also exist on the market, I didn't check.

Some USB power testers offer this translation. I know that the Fnirsi FNB58 I got does it: you can plug it into a QC charger, select the translation option, and it will appear to other devices like it was a USB-C PD charger with higher voltages available.

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  • \$\begingroup\$ The reason I tried to do this is because of the USB Type-c conversion gender. Simply plug in a c-type conversion gender in the USB-a charging port and charge my phone using the C to C cable, I confirmed that 9V 1.67A fast charging is supported stably. So I thought it would be good to take out the USB-A and insert the USB-C reatacle. \$\endgroup\$
    –  8 minikhh
    Commented Feb 19 at 2:45

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