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is the 7V fan mod bad for the PSU?
- Thread starter indokyne
- Start date
If it's on an old PSU, and the fan is the only thing connected to that PSU, then it *might* be bad. Older PSUs didn't react too well to having current return into the 5V line instead of only going out the 5V line. Nowadays, if a fan is hooked up to a PSU with the 7V trick and the fan is the only load, the PSU will just refuse to keep powered.
All of that is moot, however, if you have it in a full system. There's a whole lot more current going out that 5V rail than your fan is bringing in, and the PSU only cares about total current. You could stack a whole case full of fans in there with the 7V trick and the PSU would probably still be fine, given a decent load on the 5V line.
All of that is moot, however, if you have it in a full system. There's a whole lot more current going out that 5V rail than your fan is bringing in, and the PSU only cares about total current. You could stack a whole case full of fans in there with the 7V trick and the PSU would probably still be fine, given a decent load on the 5V line.
It doesn't really matter. You only have to ensure that there is a net current flowing out of both the 12V and 5V lines. Computer fans, even in stall, only draw about .3A, so as long as you satisfy the net current requirement, you'll be fine.
How can such an assurance be made in this arrangement? Current can't flow out of both the 12V and 5V lines when the fan is connected across them; it must flow from the higher potential 12V line to the lower potential 5V line, through the fan.It doesn't really matter. You only have to ensure that there is a net current flowing out of both the 12V and 5V lines.
What limits their current draw?Computer fans, even in stall, only draw about .3A, so as long as you satisfy the net current requirement, you'll be fine.
When the PSU is powering the guts of a computer, there's plenty of load on both the 12V and 5V lines. If all you're hooking up to a power supply is a single fan with the 7V mod, you need to add some sort of load to the 5V line that exceeds the 300mA (or whatever) the fan draws. You could attach a fan running off 5V, or get a 5W 10 Ohm resistor (in which case you might as well just put a resistor in series with the fan and run the two off the 12V line).How can such an assurance be made in this arrangement? Current can't flow out of both the 12V and 5V lines when the fan is connected across them; it must flow from the higher potential 12V line to the lower potential 5V line, through the fan.
Brushless DC motors have an electronic motor controller which controls the current to some extent, and the actual coil resistance limits the current as well.What limits their current draw?
I was wondering along the same lines, as the OP. I'm looking at running 4 120X38 fans and I want to lower their voltage/speed. What resistor would be needed to drop the voltage to say 10v? The equation would be fine, or a link to a calculator. I've seen pre done leads at Sidewindercomputers.com that drop the voltage to 10, but I have bundles of resistors laying around and would rather do it my self.
Remember V=IR. Remarkably, the fans will draw close to the same amount of current even when running off lower voltages. You'll want to drop 2V through the resistor, so R=2/I. You'll want each fan to have its own resistor, and make sure each resistor is big enough to handle the power dissipated (P=4/R in this case).
That is remarkable. What causes R to change with the voltage?Remember V=IR. Remarkably, the fans will draw close to the same amount of current even when running off lower voltages.
It's a motor, and motors don't exactly act like resistors. In fact, for a given voltage, the current flowing through a motor will increase as the speed decreases, all the way to a stall, when you have the maximum current flowing that the coil resistance will allow.That is remarkable. What causes R to change with the voltage?
It's a motor, and motors don't exactly act like resistors. In fact, for a given voltage, the current flowing through a motor will increase as the speed decreases, all the way to a stall, when you have the maximum current flowing that the coil resistance will allow.
Right; I pointed that out before. But you said the speed controller would limit the current.
Remember V=IR. Remarkably, the fans will draw close to the same amount of current even when running off lower voltages. You'll want to drop 2V through the resistor, so R=2/I. You'll want each fan to have its own resistor, and make sure each resistor is big enough to handle the power dissipated (P=4/R in this case).
I understand the Ohms law bit, but the rest has me a bit confused...
The fans I'm using, Panaflo L1's draw 2.16 watts.
watts to amps A=W/V, so 5.5/12=2.75
2/5.5(I)= 2.75
4/2.75=.6875
So...
R=2.75, P=.6875?
so my V=I/R is 5.5/1.81 = 9.95?
So a 1.8 ohm 1 watt resistor would work? Which would be a brown/grey/gold marked resistor?
Sorry for hijacking the thread....
I'm not quite sure where you're getting the 5.5 from. I would suggest that you use a separate resistor for each fan. In this case, since each fan is 2.16W, you'll have I = 2.16/12 = 180mA. To drop it to 10V, you'll need to drop 2V across the resistor, so 2V/180mA = R = about 11Ohms. The power dissipated would be P=IV=.18A * 2V = .36W, so a half-watt resistor would be enough.I understand the Ohms law bit, but the rest has me a bit confused...
The fans I'm using, Panaflo L1's draw 2.16 watts.
watts to amps A=W/V, so 5.5/12=2.75
...
I'm not quite sure where you're getting the 5.5 from. I would suggest that you use a separate resistor for each fan. In this case, since each fan is 2.16W, you'll have I = 2.16/12 = 180mA. To drop it to 10V, you'll need to drop 2V across the resistor, so 2V/180mA = R = about 11Ohms. The power dissipated would be P=IV=.18A * 2V = .36W, so a half-watt resistor would be enough.
5.5 comes from trying to work the numbers while getting ready for work and switching the numbers when dividing.....
Just ordered some 110 Ohm 1/2 watts resistors, I have some but they are 1/4 watt, never fails.
Thanks for the help
Connect two 1/4 watts in parallel. Then connect two more 1/4 watts in parallel. Then connect those two groups in series. Then, you've got a network that can handle 110 ohms at 1/2 watts.Just ordered some 110 Ohm 1/2 watts resistors, I have some but they are 1/4 watt, never fails.
Ummm. I think you misread. It's 11 Ohms, not 110. That O looks a lot like a 0, I know....Just ordered some 110 Ohm 1/2 watts resistors, I have some but they are 1/4 watt, never fails.
Thanks for the help
Yes, that would work fine.Edit: So I could actually run two 20 Ohm 1/4w parallel (drops to 10 Ohm) and then run these in a series with two 2 Ohm 1/4w (drops to 1) to get 11 Ohm at 1/2w?