# How to power nearly anything off a USB port

6 Steps
Hi , this is my first instructable so go easy on me :)

So i am going to show you how to power nearly anything off a USB port

This is for the USB competition

USB runs at 5v. The max current you can draw is 500ma. Therefore the max load is 5v x 0.5A=2.5. Watts. (W=VxI) If you try and draw more than 500mA, you may overload the port which will cause it to break

UPDATED PICTURES  :)

Just a WARNING to look what your circuit needs if it needs a higher Voltage/current this can break your ports make sure you have got the correct details of the circuit WARNING

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## Step 1: How Many Volts ?

So to power something off a USB port it has to be under 5 volts as thats how much a USB port gives out so you can check this by ;
Looking at what battery is being used
4 x aaa = OK
4 x aa = OK

USB runs at 5v. The max current you can draw is 500ma. Therefore the max load is 5v x 0.5A=2.5. Watts. (W=VxI) If you try and draw more than 500mA, you may overload the port which will cause it to break

Just a warning to look what your circuit needs if it needs higher this can break your ports

Or looking at a the transformer and see how many votls go in to the circuit ( if your 5 or under your ok )

and you can power of the Port :)

Most electronics nowadays are under 10 V and most of them Under 5V which is good for USB

chrispbarlow says: Mar 14, 2013. 11:30 AM
Sorry to resurrect an old discussion, but this is a dangerous thing to say: "So to power something off a USB port it has to be under 5 volts as thats how much a USB port gives out " I think the "How Many Volts ?" step needs to be expanded on.

Voltage is controlled by the supply (the port in this case)
Current is controlled by the device connected to the supply

The port SUPPLIES 5 V
The device has to be rated at 5 V or ABOVE (if it is rated too high, the device won't function as the port won't supply enough voltage. If the device is rated lower than 5 V, you risk damaging the device, unless you limit the voltage before the device)

The device must DRAW no more than 500 mA at the voltage supplied. Remember that Power (voltage x current) is always constant to the device. This is important.

Say you have a device rated at 7 V, 500 mA, 3.5 W
If you supply it with 5 V, assuming it can handle the voltage variation, it will still draw 3.5 W.

So, P = VI, therefore I = P/V
I = 3.5 / 5
I = 0.7 A (700 mA) which is too high for your USB port.

So to check if you can power something:

First check the Voltage rating of the device is 5 V or OVER.
If it is 5 V exactly, check the current rating is 500 mA (0.5 A) or LESS

If the voltage rating is over 5 V, use the power consumption (calculated from the RATED voltage and the current if necessary) to calculated the new current draw at 5 V. If this is under 500 mA, then you can plug in :)

You go some way to explaining this, but saying that the device needs to be 5 V or under is incorrect.
Cobra2001 says: Nov 28, 2012. 7:14 AM
Can you tell me how the voltage and current is increased from a usb port? I tried to power four LED’s off a usb port but could not do to the power coming out of the port, but I have seen night lights with several bright LED lights powered off of one USB port. How do they do it? How are they getting the extra power for the multiple LEDs?
GASSYPOOTS says: Mar 19, 2012. 3:49 PM
dats why u use fuses :D
NativeSs says: Nov 18, 2011. 3:40 PM
my usb wire has red brown yellow blue... red(+) brown(-) and blue and yellow data?
acewardph says: Jun 22, 2011. 7:08 PM
tnx. :)
nomad805 says: May 19, 2011. 2:44 AM
so if i connected a device to 2 usb ports would that be 10volts of power? and if something is 12 volts and im getting only 10 volts power is that safe? considering these speakers are rated a 1.0A but 12 volts...so its getting the 1.0A but not the extra 2 volts of power...im trying to learn so am i way off? lol ! its being powered fine right now and i can bump these JBL speakers havent heard any type of distortion and everything sounds normal. my usb ports have shorted or anything been using it for awhile now lol !
imkwl12345 says: Oct 25, 2010. 1:16 AM
hmm... you said..: Power almost anything off USB Port, well powering my house didnt work. just for future reference.
mrsplooge says: Jun 8, 2010. 9:18 PM
This is wrong. If your circuit requires 3V nominal and a maximum of 4V then you could not use a 5V USB port to power your circuit.. You risk destroying your device. If your circuit requires 7V you could still use 5V to power the circuit. If it works, is dependent upon the specs of the circuit you are powering. ie. can it actually run at 5V minimum? The USB port will not supply a varied voltage, it will only supply 5V. It is the current that changes. You are right by telling people that a maximum of 500mA is available.
eight in reply to mrsploogeJun 10, 2010. 9:39 PM
You could use diodes (in series) to drop the voltage (by 0.6volts, per diode). You could also use a 78xx or 79x regulator.
joshsstuff in reply to eightJul 1, 2010. 4:19 PM
@ eight That is interesting,
I am looking for an instructable that shows how to modify a "mains" to usb wall plug down to 3.7 volts so as to replace the LiPo/Lion batteries with constant power.

Can you go into more detail about how to implement this voltage modification, or point me to an instructable if you know of one?

(i.e. if 2 diodes drops the 5v to 3.8 would that work?, where do the belong in the circuit?)

Thanks
eight in reply to joshsstuffJul 1, 2010. 7:42 PM
Hi Josh ! Ta for writing to ask. I'm not an electronics genius where are other here are. Until they chip in with a concise reply, what I can say (from memory) is that each diode you insert in the circuit will drop output by 0.6 volts. Two in series will drop 1.2 volts. So 5 Volts in, Two diodes in series gives you 3.8 Volts out. As far as where in the circuit... just solder them to the + terminal on the inbound wire from the USB power supply. Diodes are polarised, so you will need to make sure they face the right way. Diodes sometimes have an arrow or a band at one end to show flow direction. Again, Josh, I'm no expert, so if anyone wants to politely correct my assertions and fill some gaps for Josh, please do. One last thought. Am I correct in thinking that different materials give different voltage drops? Germanium, Silicon, etc Josh, lots of things online to read about basic electronics. Good Luck, Dave
joshsstuff in reply to eightJul 1, 2010. 9:51 PM
Thanks Dave,
I've been doing some research since I read your post, I've been looking into your other suggestion of using regulators.
A linear regulator seems to be the best candidate from what I read (for it's size and simplicity, and low voltage)

What would be the difference in using a resister instead of the diodes?

I am open for suggestions of course. In fact I'd be willing to start from scratch if a better suggestions was found.
My current idea is the AC to USB adapter is 95% there.

NOTE: the load will be < 150ma so I'm going for the SMALLEST solution available.
The adapter above is only 300-400ma, so pretty close to my specs.

??? How do those photo LED night lights deal with the mains AC?
Do all of the components operate on full line current? or is it converted somehow?

I tried to read the voltage across the LEDs but didn't have any success.
Here is a picture of what I mean:

eight in reply to joshsstuffJul 2, 2010. 12:16 AM
You are welcome Josh, I have to remind you that I'm no expert, sir ! Regulators are easy to use. Just remember excess voltage is shed as heat. Resistors are for limiting current. ( http://www.doctronics.co.uk/resistor.htm ) USB adaptors supply 500 mA. To use AC on a DC component, one must rectify the Alternating sine wave. A rectifier is 4 diodes in a daisy chain, with in's and out's on opposite corners. ( http://en.wikipedia.org/wiki/Rectifier ) So these night lights, have a rectifier to get DC then current limiting via resistor(s) So in a nutshell, you shouldn't need to convert a USB power suppy as it is DC. You may need to alter voltages and most likely will need to limit the current so you dont burn out your leds. Actually what are you building? Ummm sorry. I'm not sure and may have missed it Josh. I'm at work and my mind is spread thin...
Jayceetoo (author) in reply to eightJun 11, 2010. 8:33 AM
Thanks for this :)
eight in reply to JayceetooJun 11, 2010. 10:54 PM
Hey... You are most welcome ! Thanks for a nice instructable Jaycetoo !
Jayceetoo (author) in reply to mrsploogeJun 8, 2010. 11:40 PM
How is this wrong as i havnt said anything about 3 V
StickStoneBone says: Jun 15, 2010. 1:58 AM
"How to power nearly anything off a USB port" this is unfortunately mislabeled. I of course want to run my vacuum cleaner off my USB. What office geek doesn't want to do this? I am but of the herd... one nameless face among the unwashed masses. I see this instructable and think with big words screaming in my mind like this: "FINALLY!! I CAN POWER MY VACUUM CLEANER OFF OF USB" and sadly... I cannot. I mourn now, never to know the sweet, SWEET victory of... um.. huh... I forgot what point I was making? Oh yeah... you mislabeled your Instructable, it is misleading.
Jayceetoo (author) in reply to StickStoneBoneJun 15, 2010. 8:44 AM
you could actually power a vacuum cleaner if you had 56 ports :)
StickStoneBone in reply to JayceetooJun 15, 2010. 6:43 PM
As fun as that might be, I think I'll stick to my wall outlet.
pocdragon says: Jun 13, 2010. 3:46 PM
im sry jayceetoo but the nominal voltagefor a usb plug is 3V not 5V 5v reaLLY should be a maximum voltage and should not be expectd out of a normal usb port 3-4.5 v would be best and watch your draw! 500ma max
Jayceetoo (author) in reply to pocdragonJun 14, 2010. 4:52 AM
No your completely wrong, USB ports are not 3V at all they will give out 5 at the most , the amount of current you can draw from a given USB port is limited and depends on how many other devices you have on the same bus segment. USB 2.0 Definitely does this and will give out , 5V You could use diodes (in series) to drop the voltage (by 0.6volts, per diode). if you wanted to power anything that is lower than 5V
siedpe13 says: Jun 6, 2010. 3:18 PM
i think "nearly anything" is a large overstatement
Jayceetoo (author) in reply to siedpe13Jun 7, 2010. 8:30 AM
I would disagree as you can power nearly anything , as you could power as much as you want if you had enough USB ports at 5V Each , just add them together
siedpe13 in reply to JayceetooJun 7, 2010. 10:02 PM
to "power nearly anything" is a broad term. when you say "anything" the first thing that come to mind is a car. to power a small car it would require over 100,000 usb ports at 5v and 200ma if i just "add them together". this is simply the point i was trying to make saying this was a large overstatement as really i think that a usb should really be left to powering small electronics and such.
Jayceetoo (author) in reply to siedpe13Jun 8, 2010. 8:31 AM
You are right thank you , i see where you are coming from thats why i put the almost in when i made the instructable , almost is a large overstatement
J-Po says: Jun 6, 2010. 8:06 AM
First, pretty good link for USB tech which might assist you. http://www.hardwarebook.info/Universal_Serial_Bus_(USB) Second, how bout using a powered USB hub instead of a computer, atleast for initial testing. It won't increase your amperage, but will provide a buffer before damaging y our computer.
Jayceetoo (author) in reply to J-PoJun 6, 2010. 9:12 AM
Yeah certainly , that would be good to test the product before plugging into your computer thanks :)
cokecola says: Jun 4, 2010. 10:42 AM
Now I am confused, 500mA is only .166 volts correct?
Jayceetoo (author) in reply to cokecolaJun 4, 2010. 11:40 AM
dear confused, USB runs at 5v. The max current you can draw is 500ma. Therefore the max load is 5v x 0.5A=2.5. Watts. (W=VxI) If you try and draw more than 500mA, you may overload the port which will cause it to break
error32 says: Jun 4, 2010. 9:37 AM
This is not correct, USB root hubs (on the motherboard of your computer) can only supply upto 500mA current. So if you connect a load which needs more you kill your usb ports. There should be a BIG BIG warning about this in the introductory step!
Jayceetoo (author) in reply to error32Jun 4, 2010. 9:41 AM
The port will give out at most 5 V if the circuit requires any higher there is a chance of it killing your ports , thanks for telling me this i will edit the post :)