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I would like to design a PCB in order to charge a smartphone (1.5 Amps) and in parallel pass through data to a smartphone from an external PCB. This is applied to CDP specifications. I think that the best way it's to provide the appropriate power in the USB port through the Vbus and the smartphone will take the power that it needs. Do I have to use the data buses also or can I charge the phone only through the Vbus in order not to interrupt the pass through mode?

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A direct answer to this question is yes, you must use data lines" upon first connect to the port. A "a smartphone" won't "take the power that it needs" unless it sees a "port charging signature" that it can (is designed to) recognize. Since we have no idea which signatures your phone does recognize, no practical answer can be given; some signatures are incompatible with USB signaling protocol. If no recognizable signature is detected by the phone, it will take only 500mA, or 900 mA if both ports are USB 3.0 capable. For more details, see references in the answer by Passerby.

As a side note, the concept of "CDP" is/was a part of so-called USB Battery Charging specifications, now BC1.2. These specifications were a flop, and only few manufacturer's made it to the market. Although the backward compatibility can be still traced across USB 3.0, 3.1, and USB Type-C specifications, you will be hard pressed to find a BC1.2 compliant device or host, maybe some DELL docking stations. Current mainstream is USB Power Delivery, and Type-C basic identification protocol, but most chargers have a simple "Chinese" signature with D+ and D- shorted.

ADDITION: If you want to convert a SDP (standard port on an ordinary PC) to CDP (charging data port), you can look at Maxim offerings (as well as Cypress, Microchip, Texas Instruments, NXP, etc), and add the necessary electronics to the port, and put all this stuff into an adapter. However, proper functionality might not be achievable without interaction of the host software with new charging electronics over some GPIO/I2C channel.

ADDITION2: No, USB 3.0 Rx and Tx pairs are not a part of any signature or BC1.2/QuickCharge/etc. negotiations. The USB 3.0 signals should be left alone.

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CDP requires active switching on the data pins. First you initiate high current charging with the right signal on the data pins, then you switch the data pins to the usb transceiver that's handling the data. You cannot initiate it simply by providing the right voltage and current on Vusb.

See Maxim's explanation of the usb charging standards and their single IC solution for switching of CDP standards https://www.maximintegrated.com/en/app-notes/index.mvp/id/5801

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You are assuming that a USB Host will pull current beyond the USB specified 500mA limit with out negotiating for it over the data lines. This is not supposed to occur.

That said, there are various schemes used by major companies which allow for high current charging. These schemes usually assume the data lines are not being used and place a "weakly driven" voltage on the D+ and D- lines.

However, in your project, you are using the data lines for communications by a 3rd device. So neither of the above two ways to draw more than 500mA from the USB Host should work.

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  • \$\begingroup\$ Interesting. So a simple 5V source is not a proper USB source. \$\endgroup\$ – tuskiomi Jul 24 '17 at 14:50
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    \$\begingroup\$ The voltage on a normal USB port should be 5 volts. The USB specification limits the current a USB Device can draw from a USB Host to 500mA. People design USB Hosts to supply at least this amount of current. People may not have designed their USB Host to supply any more current. And People may have not designed protection into the USB Host from USB Devices which (incorrectly) draw more then 500mA (with out asking for it first). So drawing more then 500mA with out asking for it may damage the USB Host, USB Device or both. \$\endgroup\$ – st2000 Jul 24 '17 at 15:26
  • \$\begingroup\$ Even if I incorporate usb 3.0 should I use these data buses? \$\endgroup\$ – Ange Mechanic Jul 24 '17 at 16:20
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    \$\begingroup\$ And also then why there is the CDP, which incorporates both charging and data transfer? \$\endgroup\$ – Ange Mechanic Jul 24 '17 at 17:15
  • \$\begingroup\$ @tuskiomi, a simple 5V source is a proper USB source. However, in legacy USB (Type-A) ports, the charge current is controlled by device, not "proper host". The device will be looking for "charger signature", and take max current (1.5A in this case) only if it finds a proper signature on D+/D- lines. \$\endgroup\$ – Ale..chenski Jul 24 '17 at 20:38

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