Ok this is concerning a DC circuit Questions: 1. If I hook up a fan rated at 12V .42amps to a power source of 12V and up to .5 amps, will the fan still draw just .42amps? 2. If I hook up a fan rated at 12V .42amps to a power source of 9V and up to .5 amps, will the fan still draw .42amps or will it draw more to compensate for the voltage loss? 3. If the answer to question 2 is yes, what can I do to make the current stay the same but be able to lower the voltage?

1) Yes. Think of a fan as a resistor. It cannot draw more power than it can dissipate (assuming it is rotating and not stalled). 2) No. The smaller voltage will actually draw less amperage. This will also mean that the fan will not rotate as quickly. 3) Here's what you need: V=voltage R=Resistance A=Amperage V/R=A V/A=R A*R=V If you want to look at power P=power in watts P=A*V P/A=V P/V=A The fan has a constant R, but variable V. Assuming that you know the A (0.43) and V(12) you can get the constant R (12/.43=28 Ohms). Note: The fan will draw more than 0.43 amps at start-up. This is a property of the electromechanical system, but is disregarded in simple calculations.

There are such things as constant-current regulators, but finding a plug in "box" for a low voltage system won't be easy. Here's some more light reading http://en.wikipedia.org/wiki/Current_source

Sorry for being a stickler, you dropped the 0 in the amps (.042) This leads to 285 Ohms and current draw of .032 amps at 9v

Fixed. My bad. It really is .42 amps. Now, is there a way to reduce voltage without affecting the current?

Yes there is, add a resistor in parallel with the fan: 1) Assume 12 volts: V=12 A=0.42 R=28 2) Assume the following voltages: V=10 A=0.42 Rfan=28 Rsup=143 Ohms Download the linked spreadsheet (in a rar) for an idiots calculation on what you will need. It can be found here:http://www.megaupload.com/?d=HWCPYMCP

Much thanks for the link! Some questions again: 1. Does the wattage of the resistor matter? 2. Why parallel not in series? 3. Regarding with the spreadsheet you gave me, is the voltage under "Input values" the one that the fan will be running on?

1) Not really. You're dealing with a small voltage and amperage, so the resistors you find at a Radioshack (or whatever the equivalent in your area is) should be plenty. You'll likely find 1/4 watt or 1/2 watt resistors; either value should do you just fine. 2) Basic electronics theory. When you hook up parallel resistors the effective resistance is lower than that of the lowest single value; the spreadsheet shows this under the proof heading, where the Rcalculated is the reciprocol of the sum of the reciprocol of the resistances (wow that was a convoluted answer). In a series of Resistors the resistance values are additive. To put this simply: Rseries = R1+R2+R3+......Rn Rparallel = 1/(1/R1+1/R2+1/R3+...Rn) Because you're trying to create less resistance (keep amperage constant), the resistor has to be in parallel with the fan. 3) Change the value in yellow. This voltage will be what you are putting into the system, rather than 12 volts. The value highlighted in green will be the resistor value you need. Note that if you put 12 into the voltage you get a division by 0 error, because the resistor basically cannot exist. Other notes: There is a 99% likelihood that whatever resistance you need will not be a commonly available value. You can either do a series/parallel resistor combination (using the equation from 1)), or you can have something close to but above the recomended value (ie 22 Ohms rather than the recomended 205 Ohms). The former solution will require calculation of effective resistances, while the latter slightly decrease amperage draw. You're obviously not aware of basic electronics theory. This throws up some red flags as to your ability to complete this project. For your own safety, I have to ask why you are doing this? A little bit of advice on my part, combined with knowledge of what you are doing, may help you quite a bit. My real concern though is you hurting yourself or others, because you do not know what you are doing.

Thanks again! I am a student(not from the US/UK) and I really did not take my lessons in basic electronic theories(which I now regret). I am gonna be using this fan - an old intel heatsink fan which we all know is as loud as fuck - to cool the lighting fixture in my aquarium(yes I am an aquarist ). This is the only fan i could find lying around that could move enough air to do the job even at low speeds.

Ok, I think I understand your thrust now. If you have a 12 volt supply, you can run the fan directly from it, without causing any issues. You can also decrease the 12 volt supply,so the fan only receives a fraction of it, using a resistor in series with the fan. If you have a resistor in series you'll drive up the resistance of the circuit, decreasing the amperage drawn. My guess is that you want to decrease voltage to the fan, thereby decreasing fan speed and noise. If you want to do this your calculations will be entirely different. Amperage of the circuit does not matter, as increased resistance will bring the drawn amperage lower (which the PSU should have no problem with). If you want to calculate the resistor you need to drop the voltage you need to know intended voltage for the fan: Vresdrop = A*Rres A=12/Rtotal (12 is 12 volts, Rtotal is sum of series resistance or Rres+Rfan) Vresdrop = (12*Rres)/(Rres+Rfan) You'll have to set a value for the resistor voltage drop (Vresdrop), you know the fan resistance is 28 Ohms. The rest is some dirty math. Is this more of what you were thinking?

Hm...why not just use a fan controler or a low noise adaptor(basicaly pre-built thing you are trying to make, way easier! And you could probably get one for free, from someone who doesnt need them.