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I recently came across a USB charger (Xiaomi MDY-11-EZ) with the following specs:

5.0V=3A 15.0W
9.0V=3A 27.0W
12.0V=2.25A 27.0W
20.0V =1.35A Max 27.0W
11.0V=3.0A Max 33.0W Max

What confuses me about this is the fact that the charger has only a USB-A socket and comes with a USB-A -> USB-C adapter cable. Since the USB-A connector does not provide the necessary signals to negotiate any power delivery settings; how does the charger decide the output voltage?

As a side note: I also used this charger to power (/charge) a custom Li-ion-powered board which led to the destruction of the device after plugging in the USB cable. On the device I used the 5.1k resistors on the CC lines to get 5V with a PD-capable charger, which worked with no problems with other USB-C chargers...

Any ideas?

Best, Josh

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    \$\begingroup\$ Please contact the supplier of the device. \$\endgroup\$
    – Andy aka
    Commented Jan 13, 2023 at 12:40

1 Answer 1

Reset to default
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The power supply uses Quick Charge protocol, that is communicated via the USB data pins.

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  • \$\begingroup\$ Thanks for the reply. When I look into the datasheet of a QC2.0 charger IC (CHY100), I find the same voltage options as stated above: mouser.com/datasheet/2/328/chiphy_family_datasheet-269468.pdf Unfortunately, I cannot see what happens when the D+/D- voltage does not match the 4 cases in the datasheet (Table 1). I would assume that it uses the "default" (5V) but I am not sure how it would react to floating D+/D- pins. I guess that, depending on the input impedance of their measurement circuit, also measuring a voltage > 0V would be possible on floating pins!? \$\endgroup\$
    – MoTex_42
    Commented Jan 13, 2023 at 10:50
  • \$\begingroup\$ @MoTex_42 note it says it looks at the voltage after a power negotiation. That is just one step of the power negotation. Seems that first both devices agree to use Quick Charge and then the data pin voltage is read to determine the power voltage. \$\endgroup\$
    – user20574
    Commented Jan 13, 2023 at 12:23
  • \$\begingroup\$ The voltage output must of course be 5V like with any other standard USB device, if the other device is not even connected and the pins float. \$\endgroup\$
    – Justme
    Commented Jan 13, 2023 at 12:42
  • \$\begingroup\$ @user253751: With that it seems to be very unlikely that the charger "accidentally" enters the quick charge mode and sets the output voltage to a level > 5V... In this case something else must have caused the problem on my device. I'll try to investigate that further. Thanks for clarifying that it is a QC device! \$\endgroup\$
    – MoTex_42
    Commented Jan 13, 2023 at 16:52

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