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I've got a Full speed USB device which I want to provide 2 ports for - one each on opposite ends of the enclosure. This is so the cable can be connected wherever is most convenient. My MCU (atmega32u4) has only one interface, so the physical ports will shared, but only one should be plugged in at one time.

Of course users can't be trusted to not plug in both sides at once. How to protect against this?

Ideas I've come up with:

• Just wire them up "as-is" and hope the host can deal with joined D+/D- pins


• NAND gate with separate 5V pins as input, output to a MOSFET that disconnects Vcc when both are plugged in.

Does the second option sound reasonable, or is a more complex solution required?

As pointed out, you can't directly connect two hosts to a device - so if you just wired them together and somebody plugged it in on both ends, you'd have a problem.

However you also have a problem if only one end is plugged in. USB, especially high-speed (480Mbps) mode is controlled impedance. If you wire both connectors data lines together you end up with what is known as a stub in high frequency design. The cable going to the unused connector will degrade the performance of the active connector.

To do this properly, you want a multiplexer IC. You can buy dedicated USB2.0 multiplexers designed specifically for this sort of application - something like the TS3USB30. That would allow you to connect the data lines from both ports to the mux inputs, and connect the output to your device internally. The mux will disconnect the unused connector which will disconnect the transmission line stubs.

For power I would probably use a power multiplexer such as a diode OR-ing circuit. The VBUS line from one of the ports (before the power multiplexer) can then be used as the input to the data multiplexer.

Ground would be common (connected) between the two USB ports and your device.

Multiplexer as suggested by Tom Carpenter is a good solution.

But for full speed USB (12 Mbps), the stubs in the signals are not particularly important. If the distance between the stub ends remain below 1/10th of wavelength, i.e. below ~2 meters, the reflections will not distort the waveform much.

Also, the voltage levels on D+ and D- pins will remain within the acceptable range, so there is not much risk of electronic damage to either host.

The main problem that remains is if you connect the +5V pins from both hosts together, there could be large currents involved. You could use a diode from each USB connector's +5V pin to only let current come in, never out.

The lazy electronical design way would just be to use a physical multi connector switch to choose between which one is used.

Note that this would have the added benefit of being able to leave them both plugged in and use the switch to select which input is to be used.

There is no USB compliant way of doing this. USB is not designed to be shared between two hosts and attempting to do so will lead to potentially disastrous situations.

At best, you can use a USB multiplexer or switch IC, with GPIO to determine which connector is connected. You'll have to decide which USB connection has precedence, because both cannot be connected to your USB interface at the same time.

Or look at alternatives. Have one computer communicate to the other. Or use Bluetooth or use Wi-Fi or a different connection type.

Perhaps it would be easier to leave the USB connector at one side of the enclosure and duplicate and / or relocate all other user-facing elements symmetrically. Then the enclosure can be rotated so that the USB port is facing the connector, while all other elements are still conveniently accessible.

Another option is to locate both ports on one side of the enclosure (at the opposite ends) and add a sliding stub inside the enclose which the user can move to open one port or the other. 12 MHz USB has a good chance of working with an extra disconnected port attached to it.