Continue to Site

Welcome to

Welcome to our site! is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Re-wiring a 3v hand-fan to run off a USB cable, need help reducing voltage.

Not open for further replies.


Newbie level 4
Aug 10, 2015
Reaction score
Trophy points
Activity points
Hi All,

I have bought a 3v (2 X 1.5 v AA batteries) hand fan and also bought a USB phone charger, cut off the phone charger bit, and am trying to get the USB cable to power the hand fan without burning out the motor. Initially I just hooked the red/black cables to the corresponding points on the fan - I call that "Fan No.1" now. Fan No.2, I got a potentiometer from my local Maplins (electronics shop in the UK) and when I wired it up it worked, but had a very, very short "adjustment" phase on turning the knob, that, and it started to smoke and glow red. I did tell the guy at Maplins what it was for so I don't know why he gave me the pot that wasn't the correct voltage or whatever. Got a refund on the pot and bought 3 resistors, a 100ohm, 150ohm, and 300ohm (based on some googling of people attempting similar projects). I think those are the levels, I asked for them but the 100 one looks like it might actually be a different ohm level the colors on it are brown-brown-black-black and what looks like either brown or a darker shade of brown. Which confused me, but anyway. I tried hooking fan no.2 up with the resistors in between the positive and negative wires but the fan wouldn't turn at all, tried jiggling the connections etc no joy. It did work when I briefly touched the pos/neg usb cables directly to the motors points so I know the fans motor is operative, however I only ran it for a second so as not to move onto fan no.3.

This is just a fun project for me, I know I can buy a USB fan for £3 or whatever, I just like tinkering - any ideas on what level of ohms would work if I was to use a resistor or is there some easier way to reduce the voltage from 5 to 3 - keeping in mind I don't own a soldering iron, don't really want to buy a breadboard or make a proper circuit, although if it is really simple/cheap I'd be ok with that... it's just a very simple project and I want to keep it cheap-ish.

I have also done a search of your forums for "reduce 5v to 3v" but couldn't find anything that would help.

Any ideas?


if you want to reduce 5v to 3v for the fan ,
my suggestion is to connect 2 diodes in series with 5v powersupply.

(you can coonect three diodes also).
this is the simplest solution for you.
Have you tried connecting two 3V fans in series and powered from the 5V USB? They will probably work because then each fan gets 2.5V. Two AA batteries produce 3V only when they are brand new. Their voltage runs down to less than 2V when they are dead.
I have burned up several potentiometers. It's very easy to do accidentally, by dialing it down to a low ohm value, causing current flow to increase.

A rheostat would be a better choice to reduce current to a load such as your fan. It's a potentiometer made with thicker wire, or, nichrome wire. Rheostats are not so readily available as they used to be.

Therefore the above replies give satisfactory solutions.
You can also add a series fixed resistor that is calculated by the maximum amount of current you want running through the fan. That way you at least dont have to worry about dialing the pot down to zero.

I did a bit of research and decided on the silicon diode solution, I put 3 in series and it seems to work although the diodes are getting quite warm/hot when it's running... I assume they will eventually burn out? Is it dangerous?

Regarding the resistors I tried all 3 but the fan would not turn at all, even slowly, it was just dead.

I'm considering putting a pot in with the diodes so I can dial it down, however ideally I'd like to remove one of the diodes as the fan seems a bit slower and provides less air than when it was running on batteries. However I have achieved my goal, I do not need to buy batteries for my fan anymore! Hooray! However it is going into winter anyway so it was more just to have a bit of a fiddle around.

Thank you for all the responses, this has been a great re-learning experience (I say re-learning because I used to do electronics as a kid).

You simply use Ohm's Law to calculate resistor values. Since you used resistors but the fan did nothing then the resistor values were much too high for the motor current which is high enough to make diodes hot. The motor current was also high enough to destroy the pot you tried. The high motor current probably used a lot of batteries.
You simply use Ohm's Law to calculate resistor values. Since you used resistors but the fan did nothing then the resistor values were much too high for the motor current which is high enough to make diodes hot. The motor current was also high enough to destroy the pot you tried. The high motor current probably used a lot of batteries.

Ok, thanks for that info. I went to and entered 5v for the voltage, and 1.5 amps for the current (based on an Ebay 3v fan motor that looks exactly the same) and it gave me 3.33333etc ohms. So I need a 3ohm resistor?? It also output the Watts as 7.5. This is assuming the motor in the hand fan I bought uses the same amps as the one on Ebay. It looks exactly like this: **broken link removed**

Hope this post doesn't get wiped as I'm posting ebay and website links.

- - - Updated - - -

PS - If I want to add a potentiometer, they have 1k, 10k and 22k - which one would I use? I have the 3 diodes in series could I remove 1 and add a 1k pot or should I use a 10k? Should I remove 1 of the diodes, will the pot take up some of the excess voltage?

Hope this post doesn't get wiped as I'm posting ebay and website links.

For future reference, it's preferable to post the stated specs:


Flat Model Motor

The two flat sides enable easy mounting onto a panel or PCB.
Rotation is counter-clockwise (as viewed from shaft end).
Dimensions: L: 25 (exc. shaft) W:20.1 H:15.1mm. Shaft length 8.1mm x 2.0mm dia. Weight 19g approx.
Nominal voltage: 3.0V
Operating range: 1.5-4.5V
No load speed: 16000 RPM
No load current: 0.4A
Speed at max. efficiency: 12400 RPM
Current at max. efficiency:1.5A
Torque: 11.9g-cm
RoHS compliant

And possibly upload an image:


Rather than post an eBay listing which in a few months will have expired and vanished. Please keep in mind those members whom maybe searching for answers to similar questions in the future.


Ok Big Dog, will do in future.

Ultimately, what is the intended power source, a wall socket or your PC's USB port?

If the latter, I would strongly suggest purchasing an inexpensive powered USB hub and power your design from it, accidents do happen, it's typically preferable to damaging or destroying an external USB hub, rather than your PC motherboard. At the very least an intermittent USB fault may trigger your PC/OS to disable the USB port/hub, requiring it be re-enabled, which can be quite inconvenient if you are using a USB keyboard and mouse.


You are just guessing about the current required by your fan's motor. Motors are completely different but look exactly the same. MAYBE your motor uses 1.5A at 3V.
Use Ohm's Law to calculate the resistance of the motor. 3V/1.5A= 2 ohms.

If you use a 1k ohms pot then when it is set to 1k the motor gets 5V/(1k + 2 ohms) x 2 ohms= 0.0998V which is nothing.
If the 1k pot is turned to 1/10th (100 ohms) then it will burn while the motor gets 5V/(100 + 2 ohms) x 2 ohms= 0.098V which is almost nothing.
If the 1k pot is turned to 100th (impossible) then it will burn while the motor gets 5V/(10 + 2 ohms) x 2 ohms= 0.833V and it will not turn.
Instead of a 1k pot you need a 10 ohms or 20 ohms rheostat, or a DC motor speed control circuit.

A DC motor draws a high current when it starts because then it is "stalled". If you reduce the voltage to it then it might not start running.

When Pulse-Width-Modulation (PWM) is used to control the speed then it starts fine because each pulse has full power.

Not open for further replies.

Part and Inventory Search

Welcome to