Author Options:

Fan needs 12 volts at .44 amps. How can I limit the current to .44 amps from 2 amps, without losing the 12 volts? Answered

I am trying to power a fan that needs 12 volts at 440 milliamps. I have a wall-wart transformer that is 12 volts, 2.0 amps, and I many components. I have thermal compound,  heat sinks, resistors, 12 volt and lm317 voltage regulators. So, how can I limit the current to 440 milliamps from 2.0 amps, without losing the 12 volts? Also, I tried a resistor at 33 ohms and it burnt out. Thank you, Jeremy

(Please note: I am teaching myself electrical engineering, I have not taken physics yet. My knowledge is based on sever books, web resources, and hundreds upon hundreds of youtube videos)



Best Answer 5 years ago

Your wall wart is all you need.

The fan motor description means that at 12VDC the motor will draw
( Use ) 440 ma.

The supply ( wall wort ) needs to be able to deliver 440 ma
and that it can since it could supply way more current .
In fact your wall wart could run up to four fans at once.

0.44 x 4 = 1.76 amps for four fans out of 2 amps which is 2000 ma..


I am doing my camper as a project, and in my idea i want to use:

Car radio
12v 4A TV
and 6 LED spot lights (they will b placed just to work 3 at the same time and 3 separate from each other.

to power this i was thinking to use a 12V 165A, connected to one of the 2 alternators that I have already installed.
My question is, can i create a point connected to the battery and plug everything direct to the batery terminals or I need to do some power controler specially for the tv that is 12v and 4A on a 12v 165A?????

all the best

I have a question for you I have a 12V electric motor that I want to limit to 25 Amps how could I do that?


What kind of motor AC, DC, Stepper etc ?

Does the name plate give a voltage, current and horse power ?

How heavy ?

It is this on on Ebay,

(an outrunner)

OK a 66g 3 phase 4 pole brushless AC motor needs an ESC that is programmable.

No load current is 0.7A, RPM 1050,

Continuous current 25A, Max (means damaging) current 30A

Voltage 6V to 22V And Unbelievable 98% efficiency more like a single high point at 89%. I doubt you will ever get to 24A on a 12V bus.

Lets look at ways we can reduce current.

1] Lowering the voltage will lower the current.

1a] Use a different lower voltage battery pack.

1b] Place a diode in series with the motor plus red wire which will drop the motor voltage by 0.7 Volts.

2] Reducing the load will lower the current.

2a] Use a different propeller with less cut.

2b] Take weight off the vehicle

3] Reducing the top RPM will lower the current.

3a] Set your controller to limit RPM.

3b] Program your ESC for lower RPM.

Thank you i will do this when it arives.

thanks again lewis

Voltage/current = resistance. So 12/.44=27.27273 ohms. So keep the resistance of your circuit to 27 ohms and your circuit will be limited to .44 amps.

However in your case you really don't need to limit the circuit. Your device will only draw the current that it requires.

So what is the point of your comment about a 3 year old question.

BTW did you know,.... when you block the inlet of the fan the current goes down !!

Care to hazard a guess why this happnes ?


"P" is work, "U" is Your mother, Your mother is the same no matter if You work or not, but she will not feed You much "I" food since Your not doing anything usefull like moving air.

Difficult point, mine stopped moving air long ago, like you she was a smoker of killer tobacco.

Not "You" in person, You like in general.
"Work" is wrong word to use also, it should be "P"="muscle". Work is muscle used over time.
It was just silly post, delet it :).

While all of the answers I've seen are informative, I was wondering if there is a way to limit amperage. Instead of a load with a specification, what if you 12v source is a car battery, and a solid core copper wire 3ft long, if left to it's own devices it would catch fire, and would need to limit the amperage.

There is a way using a power TO-220 semiconductor and other components.

Limiting amperage is actually said => Limiting Current...

Bare copper wire cannot "catch fire" it can get red and white hot, melt, ionize oxidize and turn plasmic But not burn with a flame !.

That 3' long wire IS THE LOAD !

You should ask your own Question so you can give your own best answer as members help you acquire electrical knowledge.

While all of the answers I've seen are informative, I was wondering if there is a way to limit amperage. Instead of a load with a specification, what if you 12v source is a car battery, and a solid core copper wire 3ft long, if left to it's own devices it would catch fire, and would need to limit the amperage.

Just to add to iceng excellent answer - A produce draws the current it needs.
As long as the voltage is correct extra current doesn't matter.

Thanks guys, I was wondering the same thing but it makes sense.

So to clarify, the current rating on a DC power source is the maximum current that it can provide. The current in the circuit is based on the resistance (load) of the device.

I Googled this because I am actually doing a very similar project using a 12V A/C adapter to power a 140mm 12V PC case fan mounted on the back of a closed home electronics cabinet (window in front).

First of all, let me thank you all very much for answeringI. It is a bit difficult to excel in electronics beyond the basics without having a source to ask questions. As for the battery example, I thought that they are measured, in terms of current, with amperes per hour.Thanks, Jeremy

(P.S. The reason why I wanted to power was because I built a fume extractor->see picture)

Good for you to study the art of EE..

About batteries :
  1. Fully Charged Voltage --- is the First measure of a battery..
  2. Maximum Current Supply --- is the Second measure of a battery..
  • Consider a 12VDC battery for emergency lighting rated @ 3 Amps
  • You can hold this four pound battery in your hand.
  • Now consider a 12VDC car battery rated @ 220 Amps..
  • You may have trouble lifting this very bigger battery with both hands..
The point being in electronics is a current rating in most parts relates directly
to size and weight...

I'll step-off my soap box now :)


Your fan wants .44 amps. if there's .44 amps available, it will draw .44 amps. There's nothing else to it. 2 amps means the supply will source 2 amps before it melts. Don't worry about it, you're fine.

Or, put it another way. Imagine the watertap. If you open that water tap fully on, an awful lot of water's going to come out, awful fast.

Now, imagine putting a tiny thin pipe on the outlet of the same tap. Put the tap full on. Now what happens ? The tap hasn't lost any of its ABILITY to supply a drenching, but the little pipe limits the flow.

Let me give you this simple example so you understand the concept of AMPERES. If you take 2 small 9volt batteries and connect them in series... you will have 18 volts. That is MORE volts than what comes out of a big car battery..... right? This 18 volt battery can not be used to start your car though... because there are only less than ONE AMP available from the tiny battery. The amps AVAILABLE from the battery are not constantly sent to the fan (in your example). If you connect a very SMALL fan it will only use a fraction of an amp. If you connect a ten foot tall fan with a giant motor, it will use maybe ten amps. Do you see, it is the device that determines amperes used, not the supply. However the supply MUST BE CAPABLE of delivering the amps, or it will be overloaded. If nothing is connected to the wall-wart, then ZERO amps flow, even though 440 milliamps are available.