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For a project, I am trying to find a way to power a small rotary solenoid.

The solenoid resistance is 27 Ohms and requires 5V to turn, so it will draw 185.2mA. I know that there are 5V across the USB VCC and Ground pins. What I have read online about USB 3.0 current limitations has confused me. I have read that a USB 3.0 port (on a computer) will support up to 900mA to be drawn from it safely, but I have also read a lot of other stuff saying that one port can only support up to 100mA.

Basically, I am confused by the wording of what I have read. My question is: can a computer's USB 3.0 port safely deliver enough power to turn my solenoid? Again, that's 185.2mA.

Thanks for the help.

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  • \$\begingroup\$ Aside from the power requirement, I'd be hesitant to connect a solenoid to a computer's USB 3.0 port. Solenoids can generate massive reverse voltage spikes when disengaged. If this isn't properly suppressed, you may damage your port. \$\endgroup\$ – Daniel Mar 16 '18 at 0:57
  • \$\begingroup\$ @DanielGiesbrecht, while the concern of back-EMF from solenoids is generally true, VBUS on USB ports is usually well protected, usually by a TVS device, and bypassed with a good-size capacitor, minimum 10 uF and usually 47 uF and up to 220-300uF. See electronics.stackexchange.com/a/361875/117785 \$\endgroup\$ – Ale..chenski Mar 16 '18 at 20:05
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There are two NON-CONFLICTING requirements in USB specifications.

First, a USB DEVICE must not draw more than 100/150 mA from USB port before it gets enumerated and finally configured. Then, if its power requirements fit into host budget (and the host knows what it is), the device can use more. Otherwise it won't be configured and used by system.

Second, a USB HOST ports must be able to supply much more current. From Section 7.2.1, "Systems that obtain operating power externally, either AC or DC, must supply at least five unit loads to each port.".

So, if your USB HOST has a AC-DC adapter or is plugged into AC outlet, or has a massive battery, each USB 3.0 port must be capable to feed at least 900 mA.

If it is a skinny smartphone, it depends on how the manufacturer feels about battery life of their product.

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  • \$\begingroup\$ Thank you for your answer! I hand't read that the host will "allow" more current if it fits within the host budget. That answers my question. With your first comment on the original question, would you say that this is safe to do? I hadn't considered back-EMF. \$\endgroup\$ – Taylor Davison Mar 16 '18 at 21:48
  • \$\begingroup\$ @TaylorDavison, the host can't control port's current, it has no means to do this. It only can "allow" a device in, and make note of its maximum needs. Regarding "safe", you might need to do some research on your undisclosed device, and make reasonable decision. I don't recall any USB regulations regarding "quality of power sink", other than to have no more than 10 uF capacitive load and a limit on inrush integrated charge. \$\endgroup\$ – Ale..chenski Mar 16 '18 at 22:24

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