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Modifying PSU Fans from 12v to 9v
tntrower
Hi guys!
I have a Diablotek PSDA400 (400Watt) PSU. The fans (2 pin DC 2 x 12v 0.12a 80mm) are loud. I was thinking of soldering a resistor inline on the hotwire (red). I used an online ohms calculator and I am not sure I did it correctly.
My question is, would a 75 Ohm resistor (2 watts) do the job of reducing 12v to 9v?
Thanks!
I have a Diablotek PSDA400 (400Watt) PSU. The fans (2 pin DC 2 x 12v 0.12a 80mm) are loud. I was thinking of soldering a resistor inline on the hotwire (red). I used an online ohms calculator and I am not sure I did it correctly.
My question is, would a 75 Ohm resistor (2 watts) do the job of reducing 12v to 9v?
Thanks!
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I'm guessing you calculated thus:
R = V/I = 9/0.12 = 75 Ohms
Wrong. That answer is close to the effective "resistance" of one fan motor IF it continued to draw 0.12 A when supplied with only 9 V (and it would not). It's actually a good number NOT to use.
First, to reduce 12 VDC down to 9 V, you need the drop across the resistor to be 3 volts, not 9.
Next, the current flowing in the circuit. If I read your post right, EACH fan claims to draw 0.12 A, and there are two of them in parallel. If you are planning on one resistor in the line that is shared by the two fans, then the total current is 0.24 A when they have the full 12 VDC supplied. However, you propose to give them less voltage, so the current will be lower.
So, let's work with the total resistance and see where we get. Apparently each motor behaves as if it were a resistance of R = V/I = 12/0.12 = 100 Ohms, so two in parallel is like 50 Ohms. You want to add a second resistance of R in series with that, so the the voltage split is 9 V across the 50 ohms, and 3 volts across the R. Thus R is 16.67 ohms. Don't know where you can get one of those, but you might get one at 20 ohms. This would make its drop a little over 3 V, and the motors would get 8.57 V, not 9. If you did that, the current in the line would be I = V/R = 12/70 = 0.17 A, and each motor would be running on about 8.57 V with a current of 0.085 A. However, that all assumes that the motors continue to act like pure resistances, which is not true. At lower speeds the probably offer slightly less "resistance" and the real values in operation are harder to predict.
For resistor power rating, P = I^2*R = 0.17*0.17*20 = 0.58 Watts. A 1 Watt will do. However, 2 Watts is a good idea, considering the impact of starting currents.
You should be aware that, for a few seconds at startup, a motor will pull a current of 2 to 3 times its running current because its impedance is very low when nearly stalled. In your proposed circuit, the effective "resistance" of the two motors in parallel during startup is more like 16 to 25 ohms, not 50, and thus the voltage drop across the resistor you plan is greater  perhaps 6 V  leaving only 6 V drop across the motors to start them up. This is getting close to the minimum necessary to start the motors, and they might NOT start up this way. And even if they do when you first put this together, in two years when the motor bearings start to wear and their friction is higher, the motors will need MORE starting voltage. but won't get it. So you may be setting up a failure, now or in future. Consider how the mobo controls case and CPU fans when it is set to do so "automatically". Of course, its control system is more complicated than a simple resistor, but it does accomplish fan speed control by varying the voltage supplied to them (at least, that's how 3pin fans work). If you observe closely, though, you will see that most of these systems work by applying the full 12 VDC to the fan for a few seconds at startup. Then the control system takes over and reduces the voltage AFTER the fan is already running at full speed. A simple series resistor can't do that.
I can understand that you believe your other fans already provide a significant air flow into the case and through, and hence probably also through the PSU, so that the full speed of these two fans is not needed. But you don't know that, and you really have no way of testing whether some components inside your PSU are overheating. So doing what you plan is taking a risk with no way to check it yourself. 
Paperdoc, first off thank you very much for your very thoughtful and thorough response.
The fans each have their own plug on the PCB and I was thinking of a resistor for each fan. I was probably not using the correct terminology when I said inline. I meant to describe one resistor soldered on the hot wire of each fan.
So if they are individual and I need to drop 3v across the resistor I would be looking at 3/0.12=25 Ohms. The wattage would be the same though correct?
Thanks again for the input. 
Best answer
Not quite  the problem is that with only 9 V across the motor, it won't draw the same current of 0.12 A. Do it with the resistances, where you don't need the amperage.
Motor "resistance" = RM = 12/0.12 = 100 ohms
Add a resistor R in series so that voltage drops are 9 V across 100 ohms (I know, at slower speeds the motor's impedance will not be 100 ohms, but it's approximate) and 3 volts across R. Then R must be 100/3 = 33 ohms. Now you can estimate the current as I = 12/133 = 0.090 A, and the resistor power dissipation as P = 3*3/33 = 0.272 Watts.
So you need approximately a 33 ohm, 1/2 Watt (or 1 Watt) resistor for each motor's hot lead.
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