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USB standard-A uses a single VBUS, but type-C seems to have two. Why is that?

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    \$\begingroup\$ Type-C has four VBUS pins. And four ground pins. And they are connected together, in group of four. \$\endgroup\$ Feb 17, 2018 at 22:58
  • \$\begingroup\$ Sorry, you're right, I mis-wrote. I meant two in each row. \$\endgroup\$
    – VortixDev
    Feb 17, 2018 at 23:04

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Because the power delivered through the connector is (within reason) limited by the current. Power is current * voltage so more power can be transferred through the connector by increasing the voltage.

The duplicate set of VBus pins allow twice the current (almost, connector heating is still an issue).

On the flip side, having a single 5V power rail on the older USB connectors (and on USB-C unless you negotiate for a higher voltage) means you can power 5V devices directly or use a 3.3V LDO for moderate power applications without generating a bunch of heat.

Part of the desire to allow higher power levels is to allow the USB-C connector to be the DC input for thin ultra-book and tablet devices as well as fast charging phone devices.

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    \$\begingroup\$ I think the question is referring to the fact that the type c connector has duplicated sets of physical vbus pins. \$\endgroup\$
    – Jon
    Feb 17, 2018 at 20:44
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    \$\begingroup\$ Actually, Type-C connector has FOUR VBUS pins. And FOUR grounds. Don't you guys read any specifications or see colorful online pinout pictures? \$\endgroup\$ Feb 17, 2018 at 22:57
  • \$\begingroup\$ Don't forget that type USB-C OTG connectors can request via the data lines 9 volts to 12 volts DC. This way up to 30 watts or so can be used to charge the battery. Very fast charge times. \$\endgroup\$
    – user105652
    Feb 17, 2018 at 23:10
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    \$\begingroup\$ @Sparky256, actually, Type-C connector can only "request" ("provide") 5V over VBUS pins, up to 3.0 A. If one need more, the connector must have the Power Delivery protocol implemented, on both ends of the link, which is a pretty separate specification. And not to 12 V, but to 20 V and 5 A. Just for the sake of accuracy. \$\endgroup\$ Feb 18, 2018 at 19:17

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