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The new USB Type C connector doesn't have a physical reverse polarity protection anymore. You can plug it in any way you want on both ends, there is also no A and B end anymore, it's all the same.

So how does this new USB type handle that the polarity doesn't end up being reversed? Do the devices have to agree on something in hardware and route the connection appropriately?

Or is there some sort of routing magic going on in the connector and the devices don't have to handle anything and can be sure the polarity is always correct?

Type C connector and receptor

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    \$\begingroup\$ geometrical symmetry. \$\endgroup\$ – Vladimir Cravero Aug 13 '14 at 15:27
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    \$\begingroup\$ It's obviously more than just that. \$\endgroup\$ – PTS Mar 12 '15 at 15:31
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Below is the pinout for the receptacle:

GND  TX1+ TX1- Vbus CC1   D+   D-  SBU1 Vbus RX2- RX2+ GND
 |    |    |    |    |    |    |    |    |    |    |    |
=+====+====+====+====+====+====+====+====+====+====+====+=
 |    |    |    |    |    |    |    |    |    |    |    |
GND  RX1+ RX1- Vbus SBU2  D-   D+  CC2  Vbus TX2- TX2+ GND

You will note that all the pins are rotationally symmetric, so if you flip the connector, TX1+ connects to TX2+, TX1- connects to TX2-, etc. and most importantly, Vbus and GND always match up.

The trick lies in the controller and cable -- the CC pins are used to detect orientation, at which point the controller routes appropriately:

2.3.2 Plug Orientation/Cable Twist Detection

The USB Type-C plug can be inserted into a receptacle in either one of two orientations, therefore the CC pins enable a method for detecting plug orientation in order to determine which SuperSpeed USB data signal pairs are functionally connected through the cable. This allows for signal routing, if needed, within a DFP or UFP to be established for a successful connection.

Source: blogspot link Source: blogspot link

As you might imagine, the cables are going to be a fair bit heftier due to the extra wires.

  • A minimum of 15 wires plus braid required for full-featured Type-C (i.e. USB 3.1 -- recommended 4-6mm outer diameter)
  • 10 wires plus braid for legacy Type-C USB 3.0/3.1 cables (intended to connect to Type-A or Type-B on the other end -- recommended 3-5mm outer diameter)
  • For USB 2.0 or earlier, whether connecting to Type-C or a legacy type on the other end, the usual four wire configuration is permitted (recommended 2-4mm outer diameter)

Source: USB 3.1 Specification @ usb.org -- specifically, the Universal Serial Bus Revision 3.1 Specification PDF available for download at the top of the page)

Also a great blog post explaining all the details about the Configuration Channel pin:

http://kevinzhengwork.blogspot.de/2014/09/usb-type-c-configuration-channel-cc-pin.html

Archive.org (in case it goes offline)

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    \$\begingroup\$ Why not have it exactly rotationally symmetric and not have to worry about orientation whatsoever and cut down on pin count?? \$\endgroup\$ – ACD Aug 13 '14 at 18:25
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    \$\begingroup\$ @ACD to do that, you'd have to add four more wires after removing the two CC wires, which is two more than the wiring that detects orientation. \$\endgroup\$ – Funkyguy Aug 13 '14 at 18:36
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    \$\begingroup\$ @Funky I meant why care about orientation at all. If you made the connector like this instead: imgur.com/VKqyvJg it's the same number of pins and no need to have a controller change routing if it's plugged in one way or the other. \$\endgroup\$ – ACD Aug 13 '14 at 18:57
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    \$\begingroup\$ @ACD In the image you linked, half of the superspeed signals are omitted. You've accounted for full rotational symmetry but forgotten to add the other half of the signals. The D+/D- signals are proper, but that is USB 2.0, in 3.0, you have two more differential pairs. en.wikipedia.org/wiki/USB_3.0#Pinouts \$\endgroup\$ – Funkyguy Aug 13 '14 at 19:06
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    \$\begingroup\$ <s>The pins are rotationally symmetrical, so why does either end care which orientation it is? The CC pins aren't needed?</s> Ohh, because there are 2 transmit pairs and 2 receive pairs. \$\endgroup\$ – endolith Nov 18 '14 at 21:45
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Since the cables are passive and are meant to be backwards compatible, the signals are duplicated top and bottom. This has the advantage of doubling the power pins and thus increasing current capacity.

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    \$\begingroup\$ So you also have every cable twice? Doesn't that make the cables pretty thick? Is that also the reason why they simply doubled the data rate for 3.1? They just have the double amount of everything? \$\endgroup\$ – PTS Aug 13 '14 at 14:35
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    \$\begingroup\$ @ProfessorSparkles (more for anyone else who happens to be reading this now) all the pairs are actually used, which allows for increased bandwidth and power transmission. The "CC" pins are where the magic happens, that allows the devices to determine which TX/RX pairs are which. \$\endgroup\$ – Doktor J Aug 11 '16 at 17:19
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The 2×12 (i.e. 24) pins are arranged in a way that inserting them in both ways will direct the electrical energy to the same path. As Vladimir says geometrical symmetry. Each of the pins has a clone pin on the other row of 12 pins.

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  • \$\begingroup\$ Originally written as comment, but I decided to post it as answer. There are already answers, but I just wanted to add my wording. \$\endgroup\$ – neverMind9 Apr 14 at 16:40
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    \$\begingroup\$ You might want to double-check this. My reading is that both TX/RX quads are used all the time but that rotating the plug swaps them. The controller needs to route them correctly and does so using CC1 and CC2. Read Doctor J's answer again. It looks good to me (but I don't know much about the topic). \$\endgroup\$ – Transistor Apr 14 at 17:17

protected by Nick Alexeev Aug 13 '14 at 20:26

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