Does reducing the voltage to fans hurt the PSU?

[xringx]

I am planning to change the molex connectors around to get my fans to run at 7v, but I read one little something that said it could hurt the PSU.

Is that correct? Is it likely?

My PSU is an Antec True Power Trio 650w

Thanks

 

[niz]

I'm presuming you're planning to connect the fan between +5 and +12 (so get the 7v) rather than +12 and ground.
There are many good reasons not to do this, including breaking the isolation between different rails, introducing noise onto your +5 line, and doing god-knows-what to current limits.
You mention changing your molex connectors around... I REALLY hope you're not stupid enough to move the pins around on the connector attached to the power-supply. Imagine what happens much later when you forget what you've done and try to use that connector on a hard drive or something.

A much more proper way to reduce voltage to your fans is to put a resistor (or potentiometer rated at suitable wattage) in series with the fan.

 

[cnumartyr]

You could just buy a resistor and put it inline with the 12V line to change it to 7V. Can't remember off hand what size it is.

 

[xringx]

I was going to make the change on the fan side, not the psu side.

As seen here http://www.bleedinedge.com/guides/7volt_mod/7volt_fan_mod.html

 

[cnumartyr]

Still going to give feed back to 5V line.

 

[xringx]

OK. How do I wire in a resistor? I tried looking on google, but maybe doing a search for the wrong thing.

 

[joefriday]

if any of you have taken physics you would understand that there is really nothing to worry about when doing a 7 volt mod. You have 12 volt potential driving one way, and a 5 volt potential the other way, making the total voltage potential across the circuit as 7 volts. All 7 volts are used by the rotation of the fan motor, which provides a fixed resistance to the circuit. Nothing to worry about here folks.

As for using a resistor inline of the 12volt supply, the size of resistor depends entirely upon the resistance inherent in the fan, following the equation V=IR.

The fan should tell you what it's typical amperage is under 12v right on the label. Since amperage is dependent upon voltage when the resistance is fixed, the amperage will be less at 7 volts than at 12. So, using the equation above, you should be able to calculate the fixed resistance of the fan at 12 volts using the amperage given on the label, and then use that value along with your desired voltage potential (7 volts in this case) to calculate the amperage when ran at the new voltage, like this example where the amperage on the fan label was .12:

12v= .12amps * x ohms

x ohms=12v/.12amps=100 ohms

7v= y amps * 100 ohms
y amps= 7v/100 ohms= 0.07 amps

Finally, take the calculated amperage and the calculated fixed resistance, and plug it back into the equation, this time with a 12 volt potential, making an algebraic equation to solve for value of the desired resistor, like so:

12v= (100 + z ohms) * 0.07 amps
z ohms= 12v/0.07 amps - 100 ohms= 71.4 ohms


since you can't just order any resistor value under the sun, we'll pick a resistor close to this number. Looks like a 75 ohm resistor will work great.

Not so fast though. We know the value of resistor we need, but what about its ability to handle all that energy flowing through it? If you don't account for this, you could have a melted resistor, which will cause an open in the circuit and fan failure. We must be sure that the resistor can adequately handle the amount of current flowing through the circuit. To calculate that, we need to find the wattage of the circuit, using the equation P=IV. Using the previous example values we get this:

P=.07 * 12=0.84 watt

The maximum load of most small resistors are measured in fractions of a watt. In this case, we should be looking for a 75 ohm resistor that can dissipate up to 1 watt of energy, like this one:

http://cgi.ebay.com/Allen-Bradley-Carbon-Comp-75-ohm-1-watt-5-1_W0QQitemZ150162236896QQihZ005QQcategoryZ4664QQcmdZViewItem

Now you should be able to calculate your own resistor, or you could just stick the fan in between the 12 v and 5 v wires!

 

[xringx]

Kudos to Joe.... Excellent info with facts, numbers and a weblink to solve the problem...

My math is weak and I didnt quite follow what you did, so here are the specs of the fan DC 12v 0.7A.


Side notes. I had written the original posting prior to recieveing the fans. They have just now come via FedEx.

Per http://www.jab-tech.com/Yate-Loon-140mm-Case-Fan-D14SM-12-pr-3594.html the dBA of the fan is 29.6 and the dBA is ISO certified

I have an some Apevia fans that according to Newegg, the dBA is 29.6 and I find that to be very loud. Hence the reason I was worried about the the noise and interested in reducing the fan speed to help the situation.

I plug in the fan and it whispers....... I would say even more quite than my favored Thermaltake Silent wheels....

 

[icepop456]

if any of you have taken physics you would understand that there is really nothing to worry about when doing a 7 volt mod. You have 12 volt potential driving one way, and a 5 volt potential the other way, making the total voltage potential across the circuit as 7 volts. All 7 volts are used by the rotation of the fan motor, which provides a fixed resistance to the circuit. Nothing to worry about here folks.

My only concern is that I do not believe the 5V rails in an ATX box are rated for sinking current. However, there's probably enough capacitors in the silver box and everything else on that rail to prevent significant noise problems. Not sure I would do it though. Zalman Fanmates are easy enough to do and aren't expensive.

 

[Zorg]

Not so fast though. We know the value of resistor we need, but what about its ability to handle all that energy flowing through it? If you don't account for this, you could have a melted resistor, which will cause an open in the circuit and fan failure. We must be sure that the resistor can adequately handle the amount of current flowing through the circuit. To calculate that, we need to find the wattage of the circuit, using the equation P=IV. Using the previous example values we get this:

P=.07 * 12=0.84 watt

You calculated the wattage of the entire circuit.

The wattage consumed by the resistor would be P=I²R or P=.07²*71.4 ohms = .35w
The wattage consumed by the fan would be P=.07²*100 ohms = .49w
Total circuit wattage = .84w

So a ½ watt resistor would do the job.

 

[joefriday]

Ah, nice catch.

Yes, I didn't think the last part through well enough. Of course with two "resistors" in the circuit load would be shared among them. DOH!

Thanks for the correction.

 

[Zorg]

12v= .12amps * x ohms

x ohms=12v/.12amps=100 ohms

This is what I would do from here, it's quick and dirty but it works.

You want 7 volts but have 12 so you know that you want roughly 7/12ths of the voltage.

So the motor has to have a value approx 7/12 of the total resistance. 7/12 = 100/X so X = 171.43 ohms
171.43 ohms - 100 ohms = 71.43 ohm resistor.

To check for accuracy
The current is 12/171.43 = .0699A
Voltage dropped across resistor "ER1" = 71.43 ohms*.0699A = 4.99V
Voltage dropped across the fan 12 - 4.99 = 7.1V

Wattage used by the resistor is .0699² * 71.43 ohms = .349W or just .0699*4.99V =.3488W