0
\$\begingroup\$

Lets say i powered a led strip rated at 5V and 3.6A by my laptop usb port via usb cable, what will happen?

-will it just be limited to 900mA since thats max current for usb 3.0

-or will it try draw out 3.6A from the usb port and break it

extra: what if i wired the led strip to an arduino that has a max current of 1A and powered it with usb port, what will happen then?

\$\endgroup\$

marked as duplicate by Eugene Sh., Bimpelrekkie, AndrejaKo, brhans, Harry Svensson Jul 10 '18 at 15:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ So many similar questions on different sites and SE... \$\endgroup\$ – Eugene Sh. Jul 10 '18 at 14:21
  • \$\begingroup\$ Can you think how to do Ohm's Law from USB port view to decide what it can supply???? If you cannot learn Ohm's Law, what are you doing? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 10 '18 at 14:47
0
\$\begingroup\$

If you try to draw 3.6 A from a laptop USB port, what happens depends on the design of the electronics powering the USB port.

The USB port: - might just support 3.6 A (not very likely) - does not support 3.6 A and will deliver a certain maximum current and might shut down immediately or after a while (most likely) - it might break if it is poorly designed (not that likely)

What happens when powering from an Arduino is similar, it depends on the design of the Arduino but also how you power the Arduino.

If you power the Arduino from 5 V then that power supply might limit the current. If there is a (resettable) fuse in the 5 V line of the Arduino it might blow (or go open if it is resettable).

If you feed the Arduino via Vraw an internal regulator is used to regulate this voltage down to 5 V. This regulator cannot support that much current so you will not get 5 V out of the Arduino for a long time. When the regulator heats up too much it might shut down.

In general: you simply should not power a device which needs 3.6 A from anything that cannot supply at least 3.6 A. Especially "dumb" loads like LED strips which aren't "intelligent" in the sense that they will draw less current when a high current cannot be supplied. Some smartphones are "intelligent" and adapt their power consumption to the source used.

\$\endgroup\$
  • \$\begingroup\$ Ahh i see, thanks. uh recently i saw a guy power de led strip (5V 3.6A) through an arduino connected to a 900mA usb port and it was working fine, is it just cuz the led is "dumb" and drew out less current then it needs and thus lights up not as bright? \$\endgroup\$ – user193234 Jul 10 '18 at 14:47
  • \$\begingroup\$ Yes it works "fine" but that is stressing the components (in the 900 mA port most likely). So if you run this for a long time at some point the USB port might break. \$\endgroup\$ – Bimpelrekkie Jul 10 '18 at 15:08
  • \$\begingroup\$ One more constraint: most normal USB Type-A receptacles have contact rating of 1.5 A. Higer-rated connectors (designed for old-style charging ports) are rare and expensive. If the port can supply more than a bare limit (which it usually does), drawing 3.6A will likely overheat the contacts and possibly melt the housing plastics. Normally designed devices, however, will not allows this, and use some overcurrent protection. \$\endgroup\$ – Ale..chenski Jul 10 '18 at 16:29

Not the answer you're looking for? Browse other questions tagged or ask your own question.