0
\$\begingroup\$

I am learning the USB powered modes and am a little confused about the self-powered mode.

So, I understand, a USB mouse is a bus-powered device, and an OTG is a self-powered device.

So, in the case of a self-powered device, an OTG, the host's VBUS line has 5 V. At the same time, does the OTG's VBUS pin have either 3.3 V or 5 V? Am I correct?

In that case, won't there be a problem? Won't current flow from the 3.3 V to the 5 V on the USB VBUS line?

Question 1: In the self-powered mode, it is mentioned that the OTG has its own supply. What is the supply used for? What is the current from the self-powered USB used for? Like, sending data on D+ and D-? Am I correct?

\$\endgroup\$
3
  • \$\begingroup\$ An OTG device is not necessarily self powered. A phone for example is an OTG device (OTG really is just the possibility to switch between being a host and a device) but is not necessarily self powered as it can be charged from a USB host. Something like a printer is typically self powered. The VBUS line is always 5V, never 3.3V. \$\endgroup\$
    – Klas-Kenny
    Commented Aug 16, 2022 at 9:47
  • \$\begingroup\$ Thank you for your comment. So, irrespective of whether its a host or a peripheral, the VBUS line is always 5V, right? \$\endgroup\$
    – user220456
    Commented Aug 16, 2022 at 9:50
  • \$\begingroup\$ @Klas-Kenny, could you also provide an answer with simple explanation to my question \$\endgroup\$
    – user220456
    Commented Aug 16, 2022 at 9:51

2 Answers 2

Reset to default
1
\$\begingroup\$

In a self-powered device, all the current needed for operation of the self powered device is drawn from its own power supply. The USB VBUS line (always 5V) connection from the host to the self-powered device is purely for informing the self-powered device that the USB cable has been plugged in or plugged out. The self-powered device will not draw any current from this for its operation.

If you are worried that the self-powered device may have 3.3V supply and since USB VBUS will be 5V, there will be a current path from USB VBUS to the chip's power supply, yes, to avoid this, a 5V tolerant IO will be used for USB VBUS pin on the chip so that even if chip power supply is only 3.3V, there will be no such current path from USB VBUS to the chip's power supply

\$\endgroup\$
0
\$\begingroup\$

An OTG device is built so that it can either be a host that can connect to a device or be the device which can connect to host. Obviously, the mode between device and host depends on which kind of cable you plug in so the OTG device knows in which mode it must be so it knows if it must give out power or accept power in.

So there can be no short circuits as it must not be allowed.

OTG device like a mobile phone or a host device like a laptop must obviouly have it's own power source to work (batteries) to be able to act as a host to give power to a connected device which requires power from USB (like a mouse).

\$\endgroup\$
2
  • \$\begingroup\$ Thank you for the answer. So, what is the power from the VBUS line (either in case of host or device) used for? To send data in D+ and D- lines? \$\endgroup\$
    – user220456
    Commented Aug 16, 2022 at 10:00
  • 1
    \$\begingroup\$ How would bus powered devices work without bus power? It also depends which USB you are talking about. USB-C will not output VBUS unless a cable with device is detected. And for charging it supports up to 20V. \$\endgroup\$
    – Justme
    Commented Aug 16, 2022 at 11:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.