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There's this Rp/Rd switch schematic mentioned in TI's primer on USB-C that I don't quite understand. The original text on page 6 goes:

A DRP/DRD can present itself as either a UFP or DFP. As a result, this design must have a method to pull the CC lines up with Rp or pull the lines down with Rd (default on a dead battery in order to charge), as shown in Figure 6. Notice how the switch can toggle between pulling the CC line up (in this case, with a current source to create a specific voltage across Rd), or pulling the CC line down to GND.

AFAIK a DFP pulls CC1 and CC2 pin high with pull-up resistor Rp, and a UFP pulls CC2 and CC1 low with pull-down resistor Rd. The two resistors, Rp and Rd, form a voltage divider and controller on either side of DFP or UFP can sense a specific voltages values and regulate the Vbus voltage or current. So a DRP/DPD device should be able to switch in between being pulled up by Rp, or pulled down to ground by Rd.

  1. But the only resistors shown in the schematic are two Rd resistors pulling both CC1 and CC2 down to ground (UFP). If the device is switched to DFP, there are no pull-up resistors, but instead CC1 and CC2 are driven by an unspecified current source. What does the current source do? Where are the pull-up resistors Rp, and where is the voltage source Vbus that Rp should be pulling up to?

  2. I don't get where it says "... a current source to create a specific voltage across Rd ...". When CC1 or CC2 is pulled up to the current source, Rd will become an open circuit tied to ground - how could there be any voltage across Rd?

z

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  • \$\begingroup\$ one thing to understand about usb c is that they went to some length to make the connector reversible and can charge both ways. In this case it appears to involve some amount of duplicate circuitry to activate things one way or the other depending on the connection. \$\endgroup\$
    – Abel
    May 29, 2021 at 4:06
  • \$\begingroup\$ the Rp and Rd definitions are on page 4, figure 3. Each CC is expected to be one of the setups like on that figure 3. \$\endgroup\$
    – Abel
    May 29, 2021 at 4:23

2 Answers 2

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The current sources do the same as pullup resistors. They source a current into Rd on the other end of the cable. A current source is used instead of a pullup resistor probably because accurate current sources are easier to make on a chip.

When this device is a power source, the other end of the cable will be a power sink. It has Rd's.

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  • \$\begingroup\$ As shown in the circuit, when CC1/CC2 is switched to connect to the current source, the \$ R_d \$ node is left floating (an open circuit to ground). I see no current through \$ R_d \$ but I may have misinterpret the circuit diagram hence the question. \$\endgroup\$
    – KMC
    May 29, 2021 at 4:07
  • \$\begingroup\$ Suppose the diagram shows the inside of you laptop's USB port. If you were to charge your cell phone using this port, your laptop's Rd's will be disconnected and the current sources in your laptop will source current into your cell phone's Rd's. \$\endgroup\$
    – jy3u4ocy
    May 29, 2021 at 4:20
  • \$\begingroup\$ ohh... so the current source in the diagram connects Rd at the opposite end (not shown) when the device acts as DFP. And when the device switches to UFP, the Rd in the diagram sinks current from current source at the opposite end (not shown)? \$\endgroup\$
    – KMC
    May 30, 2021 at 1:35
  • \$\begingroup\$ You get the idea. However, the setup at the opposite end does not have to be the same. \$\endgroup\$
    – jy3u4ocy
    May 30, 2021 at 23:00
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The schematic representing a single device naturally doesn't have a complete circuit. The complete circuit is made once you connect two devices with a cable, so that Rd and Rp are located in different devices:

enter image description here

The DFP (the source) will then detect what's connected to it according to a table:

enter image description here

Whether the source uses actual Rp resistors or current sources doesn't matter, as long as the current corresponds to standard Rp values connected to +5V (56k for default USB power, 22k for 1.5A and 10k for 3A) In both cases, connecting to Ra (~1k) will result in a voltage drop that is lower than Rd (5.1k), which in turn is lower than the open-circuit voltage, allowing the DFP to tell which one is connected.

The UFP (sink) detects connections using voltages on CC lines according to the following table (voltage levels above 0.2V correspond to different DFP current capabilities, as encoded by Rp resistors):

enter image description here

As you can deduce, connecting two DFPs will result in both of them detecting nothing (Open/Open) or just the active cable (Ra/Open), and connecting two UFPs will result in detecting nothing at all (both CC lines at 0V).

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