Resistors: Check my math, please!
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Resistors: Check my math, please!
If I use a 56 Ohm resistor like this one on a .30 Amp fan, it would be functionally equivalent to running the fan off of the 5V rail, right? Or did I do something backwards?
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You're creating a voltage divider. Let's double-check, shall we?
Goal: you want 5v to fall across the fan leads, and 7v to fall across the resistor. I am assuming your source is 12v, correct?
Fan load: .3 amps. This same current must flow through the resistor.
V = IR (good enough for this situation, although this is far from complete)
7V = .3A * R
R = 23.3 ohms. I think you missed something.
NOTE: be careful. Make sure you buy a resistor that can drop the maximum power we see here (2.1w), or connect several smaller resistors in-parallel to the the same effect. You might actually be better off finding a way to source 5v direct rather than drop that much power.
Goal: you want 5v to fall across the fan leads, and 7v to fall across the resistor. I am assuming your source is 12v, correct?
Fan load: .3 amps. This same current must flow through the resistor.
V = IR (good enough for this situation, although this is far from complete)
7V = .3A * R
R = 23.3 ohms. I think you missed something.
NOTE: be careful. Make sure you buy a resistor that can drop the maximum power we see here (2.1w), or connect several smaller resistors in-parallel to the the same effect. You might actually be better off finding a way to source 5v direct rather than drop that much power.
As near as I can tell, I replicated the calculations used in the article
I didn't really have a goal, I wanted to see what this old Zalman thing I had would do.
I started by finding the resistance of the fan:
12V=0.3A*R
R=12V/0.3A
R=40Ω
Finding the current of the new series circuit:
I=V/R(total)
I=12V/(56Ω+40Ω)
I=.125A
And then solving for Voltage across the fan:
V(fan)=IR(fan)
V(fan)=.125A*40Ω
V(fan)=5V
This method also produces the figures published in the review...
Anyways, yes, I'd much rather feed 5V to the fan directly by changing pins around on a molex connector.
I was trying to give my Yate Loon D12SL-12s more than 5V (and less than 12) without using the 7-volt trick and using stuff I had on hand, and I wanted to make sure that the 56Ω Zalman thing was pretty useless.
So did they figure wrong back then, too?The exact amount of the voltage drop depends on the fan's impedance. With the 0.2A Zalman fan, the drop is around 6V, which means the fan runs at 6V. The resistor dissipates 0.55 watts, and gets a bit warm.
The voltage drop with the 0.07A Panaflo 80mm "L" fan is only a bit over 3V. The Panaflo gets nearly 9V -- a bit too high for many of us, but still not useless in many apps.
I didn't really have a goal, I wanted to see what this old Zalman thing I had would do.
I started by finding the resistance of the fan:
12V=0.3A*R
R=12V/0.3A
R=40Ω
Finding the current of the new series circuit:
I=V/R(total)
I=12V/(56Ω+40Ω)
I=.125A
And then solving for Voltage across the fan:
V(fan)=IR(fan)
V(fan)=.125A*40Ω
V(fan)=5V
This method also produces the figures published in the review...
Anyways, yes, I'd much rather feed 5V to the fan directly by changing pins around on a molex connector.
I was trying to give my Yate Loon D12SL-12s more than 5V (and less than 12) without using the 7-volt trick and using stuff I had on hand, and I wanted to make sure that the 56Ω Zalman thing was pretty useless.
Several times I have spliced carbon resistors in line with 12 vdc computer fans. The ampere rating on a fan label is a start but calculations seldom are exact. I also want the actual noise level to be the deciding factor, not basic math. Because of this, I use a different approach.
I snip the red wire to the fan and insert a 200 ohm potentiometer in series with the red wires. I connect my voltmeter and experiment. The fan needs a tad more current to start than to run. I play with this to determine how low a voltage I can go before the fan won't start. Of course, I'm listening to the noise of the fan. The general range of play is in the 4 to 7 volt range. Each fan varies.
After I am content with the noise level and confident the fan can start at the selected resistance, I remove and measure the potentiometer resistance. I select two to three 1/2 watt carbon resistors that in parallel provides a resistance close to the final pot setting. More wattage dissipation might be desired by some people, but I never have had a problem.
I don't get fancy thereafter. I twist the resistor leads and wires together, add those really small barrel nuts, then add electrical tape to cover the leads while leaving most of the resistors exposed to dissipate heat.
The process takes about 15 minutes.
You likely will discover the 56 ohm value is close to any final value you select and will satisfy many people. That resistance value is conservative, however. The potentiometer method allows tweaking that extra one half to one volt to further reduce noise.
I did this only a few months ago with a 120 mm PSU fan. I settled for an approximate 6 volt split. With my hand I barely feel air flow from the the PSU, but typically I draw only 55 to 60 watts, occasionally ramping to 81 watts when I run VirtualBox. In other words, the nominal air flow through the PSU case is sufficient.
I snip the red wire to the fan and insert a 200 ohm potentiometer in series with the red wires. I connect my voltmeter and experiment. The fan needs a tad more current to start than to run. I play with this to determine how low a voltage I can go before the fan won't start. Of course, I'm listening to the noise of the fan. The general range of play is in the 4 to 7 volt range. Each fan varies.
After I am content with the noise level and confident the fan can start at the selected resistance, I remove and measure the potentiometer resistance. I select two to three 1/2 watt carbon resistors that in parallel provides a resistance close to the final pot setting. More wattage dissipation might be desired by some people, but I never have had a problem.
I don't get fancy thereafter. I twist the resistor leads and wires together, add those really small barrel nuts, then add electrical tape to cover the leads while leaving most of the resistors exposed to dissipate heat.
The process takes about 15 minutes.
You likely will discover the 56 ohm value is close to any final value you select and will satisfy many people. That resistance value is conservative, however. The potentiometer method allows tweaking that extra one half to one volt to further reduce noise.
I did this only a few months ago with a 120 mm PSU fan. I settled for an approximate 6 volt split. With my hand I barely feel air flow from the the PSU, but typically I draw only 55 to 60 watts, occasionally ramping to 81 watts when I run VirtualBox. In other words, the nominal air flow through the PSU case is sufficient.
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Actually, I think you missed something. The fan load of .3A is only when it's getting 12V, it will be less at a lower voltage.defaultluser wrote:You're creating a voltage divider. Let's double-check, shall we?
Goal: you want 5v to fall across the fan leads, and 7v to fall across the resistor. I am assuming your source is 12v, correct?
Fan load: .3 amps. This same current must flow through the resistor.
V = IR (good enough for this situation, although this is far from complete)
7V = .3A * R
R = 23.3 ohms. I think you missed something.
zoatebix, your math looks right to me. It won't be exact because these fans aren't purely resistive loads, but it should be close enough. Still, a Fanmate is well worth the couple extra dollars to be able to dial in exactly the noise/voltage level you want. I messed around with resistors early in my quieting days and gave it up because of the lack of adjustability.
I might see what they sound like on the motherboard's headers first. I might be able to get my hands on two fanmates for free. They came with the heatsinks my cousin bought for the computer I built her. Later on, I gave her one of these - I couldn't work with it because there was no jumper setting to disable the slow fan alarm (I guess I could have popped the buzzer off). I wonder if she ever got those fanmates un-adhered from the side of her case.Still, a Fanmate is well worth the couple extra dollars to be able to dial in exactly the noise/voltage level you want
Anyways, even though the Yate Loons will start at 5V (well, one of the old dusty ones stalls out or fails to start now), and they sound great at that speed, but I had some blue screen crashes that were probably heat-related. They started after 8 or 9 months of happy overclocking, they didn't go away when I stopped overclocking, but they did once I upped my exhaust fan to 12 volts. I think the fans wore down just enough to push me over some thermal threshold that a component didn't like.
Maybe the motherboard will find a happy medium for me. Maybe I'll find one with the fanmates.
Here are my specs, for the curious:
Antec P180 with 2 Yate Loon D12SL-12 fans (one in the rear exhaust, one wedged in diagonally where the upper HDD cage should be. I don't have a picture uploaded yet.)
Seasonic S12 500w Power Supply
DFI LANPARTY UT nF4 SLI-DR (04/06/2006 BIOS)
Thermalright HR-05 SLI/IFX with no fan (before that a fanless JTS-0006, before that a mangled Zalman)
AMD X2 3800+ dual core
Thermalright XP-90 with Nexus Real Silent Case Fan
2x1024MB Patriot PC3500LLK (PDC2G3500LLK) Infineon CE-5
Asus Radeon HD 4850
Accelero S1 Rev. 2 and Zalman ramsinks with no fan
750GB Samsung F1
Lite-On LH-20A1s SATA DVD+-RW Drive
Creative X-Fi XtremeMusic (I should probably sell it and go back to onboard sound...)
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Yeah, you're right, that's what I get for doing math at night, when I was dead tired. You have to scale the amps based on the voltage split, and use a voltage divider like I mentioned at the top of my post.flyingsherpa wrote:Actually, I think you missed something. The fan load of .3A is only when it's getting 12V, it will be less at a lower voltage.defaultluser wrote:You're creating a voltage divider. Let's double-check, shall we?
Goal: you want 5v to fall across the fan leads, and 7v to fall across the resistor. I am assuming your source is 12v, correct?
Fan load: .3 amps. This same current must flow through the resistor.
V = IR (good enough for this situation, although this is far from complete)
7V = .3A * R
R = 23.3 ohms. I think you missed something.
Impedence of fan Zfan = 12v/0.3A = 40 ohms
Vfan = Zfan/(Zfan+Zresistor) * 12v.
Zresistor = (Zfan/Vfan) * 12v - Zfan
Goal: Vfan = 5v. Zresistor = 56 ohms.
That said, this is just a bit of guesswork, since the fan will not behave as a purely resistive load. You're better off just getting a variable resistor and finding the voltage that works best for you.