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Quote:
Originally Posted by Daredevil 720

So if I were to undervolt a Gentle Typhoon AP-15 to 9V using a resistor:

One of them draws 0.083A at 12V (confirmed), so its resistance should be ~145Ω. Using a DMM I measured 0.033A at 5V (molex cables swapped), which shows a resistance of ~150Ω. So I'll assume the fan's resistance is between 145-150Ω.

Using a simple voltage divider, I found out that to get 9V to the AP-15 through the 12V rail I would have to put a ~50Ω resistor in series to the fan.

Through measuring with the DMM I also found out that the current draw of the fan also goes down linearly with the voltage, so at 9V the current draw should be around 0.062A, giving us ~0.56W power consumption on the fan.

The power consumption of the resistor would be different though. If I'm correct it should be 3V (the voltage drop) * 0.062A ~= 0.19W. So the ideal resistor should be a 50Ω resistor rated at 1/4W, right?
__________________

If I were to connect multiple AP-15s to a single molex connector, I would actually be connecting them in parallel, right? In this case should I use a 50Ω 1/4W resistor in series with each fan, or could I use a single resistor on the 12V node? If the latter is possible/advisable how would I go about calculating what resistor is needed?

There is a way better approach to this, use a voltage divider circuit:

You would put this between your power supply and whatever you want powered at 9V, effectively creating a 9V rail.

Example:

Say use 1k and 3k resistors

Vout = Vs * R2/(R1+R2) = 12v * 3k/(1k+3k) = 12 * 3/4 = 9v

Again, with this approach, adding more fans to that 9V rail will not effect the voltage divider.

Whats more is, you can use add a potentiometer to give you a variable controller. All yo do here is use a constant for R2 and a pot for R1 to vary the output voltage

Example

Vs = 12v
R1 [10k Potentiometer] = 0->10k
R2 = 10k

Vmax (R1 = 0) = 12 * 10k/(10k + 0) = 12 * 1 = 12v
Vmin (R1 = 10k) = 12 * 10k/(10k + 10k) = 12 * 1/2 = 6v

Now you can turn the potentiometer and have a linear ramp from 6v-12v. Oh look, we just built a simple fan speed controller

EDIT:

If you'd like an example, I have all the parts lying around and could put together an example video
Edited by cyphon - 8/31/13 at 10:04am
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Quote:
Originally Posted by cyphon

Unfortunately the gigabyte amd cards don't use reference design so the blocks don't fit right

Also, the EK amd blocks I think mostly all come in clean acrylic anyway so polishing wouldn't be necessary

I believe gigabyte 7950/7970 windforces are compatible though
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Quote:
Originally Posted by kyismaster

I believe gigabyte 7950/7970 windforces are compatible though

The first rev was, but the 2nd rev (which are all I could find for the 7950s now) are not
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CoolingCoolingCoolingOS
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PowerCase
COOLER MASTER Silent Pro M2 RSA00-SPM2D3-US 1000W Fractal Arc Midi R2
Quote:
Originally Posted by cyphon

Quote:
Originally Posted by Daredevil 720

So if I were to undervolt a Gentle Typhoon AP-15 to 9V using a resistor:

One of them draws 0.083A at 12V (confirmed), so its resistance should be ~145Ω. Using a DMM I measured 0.033A at 5V (molex cables swapped), which shows a resistance of ~150Ω. So I'll assume the fan's resistance is between 145-150Ω.

Using a simple voltage divider, I found out that to get 9V to the AP-15 through the 12V rail I would have to put a ~50Ω resistor in series to the fan.

Through measuring with the DMM I also found out that the current draw of the fan also goes down linearly with the voltage, so at 9V the current draw should be around 0.062A, giving us ~0.56W power consumption on the fan.

The power consumption of the resistor would be different though. If I'm correct it should be 3V (the voltage drop) * 0.062A ~= 0.19W. So the ideal resistor should be a 50Ω resistor rated at 1/4W, right?
__________________

If I were to connect multiple AP-15s to a single molex connector, I would actually be connecting them in parallel, right? In this case should I use a 50Ω 1/4W resistor in series with each fan, or could I use a single resistor on the 12V node? If the latter is possible/advisable how would I go about calculating what resistor is needed?

There is a way better approach to this, use a voltage divider circuit:

You would put this between your power supply and whatever you want powered at 9V, effectively creating a 9V rail.

Example:

Say use 1k and 3k resistors

Vout = Vs * R2/(R1+R2) = 12v * 3k/(1k+3k) = 12 * 3/4 = 9v

It wouldn't quite work that way . . . .

Reason being that the effective resistance of the fan motor would then be a parallel resistance with R2, and being much, much lower, would be the only effective resistance to figure with, so you'd be looking at 150 ohms out of 1150 ohms which would give the fan ~1.5 volts.

In a series circuit, the OP was correct with the 50 Ohm resistor, and also correct with his 3V drop across the resistor times the current thru the loop to get 0.19W as the power dissipation of the resistor

Pretty much all the resistors I've seen in the low voltage adapters that come with a lot of fans use 1W resistors, as they have more surface area to dissipate heat, especially where they are all covered in a layer or two of heat shrink.

47, 51, & 56 Ohms are the standard values in that range for common resistor types.

I'd go with a 1W for the same reason that most fan makers that include low voltage adapters do.

Darlene
Edited by IT Diva - 8/31/13 at 10:13am
Quote:
Originally Posted by cyphon

There is a way better approach to this, use a voltage divider circuit:

You would put this between your power supply and whatever you want powered at 9V, effectively creating a 9V rail.

Example:

Say use 1k and 3k resistors

Vout = Vs * R2/(R1+R2) = 12v * 3k/(1k+3k) = 12 * 3/4 = 9v

Again, with this approach, adding more fans to that 9V rail will not effect the voltage divider.

Whats more is, you can use add a potentiometer to give you a variable controller. All yo do here is use a constant for R2 and a pot for R1 to vary the output voltage

Example

Vs = 12v
R1 [10k Potentiometer] = 0->10k
R2 = 10k

Vmax (R1 = 0) = 12 * 10k/(10k + 0) = 12 * 1 = 12v
Vmin (R1 = 10k) = 12 * 10k/(10k + 10k) = 12 * 1/2 = 6v

Now you can turn the potentiometer and have a linear ramp from 6v-12v. Oh look, we just built a simple fan speed controller

EDIT:

If you'd like an example, I have all the parts lying around and could put together an example video

I used a voltage divider too, putting the fan in place of R2 (150Ω) and a resistor with 1/3rd of the resistance as R1 (50Ω).

What you're suggesting is actually adding the fans in parallel to R2, which in my original plan is adding the fans in parallel to the first fan, using just one 50Ω resistor. I would have to use a much bigger resistor with a bigger wattage rating though as the current draw will be multiplied by the number of fans effectively multiplying the wattage by the same amount.

Am I correct?
Quote:
Originally Posted by IT Diva

It wouldn't quite work that way . . . .

Reason being that the effective resistance of the fan motor would then be a parallel resistance with R2, and being much, much lower, would be the only effective resistance to figure with, so you'd be looking at 150 ohms out of 1150 ohms which would give the fan ~1.5 volts.

In a series circuit, the OP was correct with the 50 Ohm resistor, and also correct with his 3V drop across the resistor times the current thru the loop to get 0.19W as the power dissipation of the resistor

Pretty much all the resistors I've seen in the low voltage adapters that come with a lot of fans use 1W resistors, as they have more surface area to dissipate heat, especially where they are all covered in a layer or two of heat shrink.

47, 51, & 56 Ohms are the standard values in that range for common resistor types.

I'd go with a 1W for the same reason that most fan makers that include low voltage adapters do.

Darlene

I knew it wouldn't be that simple..

So what are you suggesting, usual daisy-chaining and one resistor per fan?
Edited by Daredevil 720 - 8/31/13 at 10:16am
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Quote:
Originally Posted by Daredevil 720

Quote:
Originally Posted by cyphon

There is a way better approach to this, use a voltage divider circuit:

You would put this between your power supply and whatever you want powered at 9V, effectively creating a 9V rail.

Example:

Say use 1k and 3k resistors

Vout = Vs * R2/(R1+R2) = 12v * 3k/(1k+3k) = 12 * 3/4 = 9v

Again, with this approach, adding more fans to that 9V rail will not effect the voltage divider.

Whats more is, you can use add a potentiometer to give you a variable controller. All yo do here is use a constant for R2 and a pot for R1 to vary the output voltage

Example

Vs = 12v
R1 [10k Potentiometer] = 0->10k
R2 = 10k

Vmax (R1 = 0) = 12 * 10k/(10k + 0) = 12 * 1 = 12v
Vmin (R1 = 10k) = 12 * 10k/(10k + 10k) = 12 * 1/2 = 6v

Now you can turn the potentiometer and have a linear ramp from 6v-12v. Oh look, we just built a simple fan speed controller

EDIT:

If you'd like an example, I have all the parts lying around and could put together an example video

I used a voltage divider too, putting the fan in place of R2 (150Ω) and a resistor with 1/3rd of the resistance as R1 (50Ω).

What you're suggesting is actually adding the fans in parallel to R2, which in my original plan is adding the fans in parallel to the first fan, using just one 50Ω resistor. I would have to use a much bigger resistor with a bigger wattage rating though as the current draw will be multiplied by the number of fans effectively multiplying the wattage by the same amount.

Am I correct?
Quote:
Originally Posted by IT Diva

It wouldn't quite work that way . . . .

Reason being that the effective resistance of the fan motor would then be a parallel resistance with R2, and being much, much lower, would be the only effective resistance to figure with, so you'd be looking at 150 ohms out of 1150 ohms which would give the fan ~1.5 volts.

In a series circuit, the OP was correct with the 50 Ohm resistor, and also correct with his 3V drop across the resistor times the current thru the loop to get 0.19W as the power dissipation of the resistor

Pretty much all the resistors I've seen in the low voltage adapters that come with a lot of fans use 1W resistors, as they have more surface area to dissipate heat, especially where they are all covered in a layer or two of heat shrink.

47, 51, & 56 Ohms are the standard values in that range for common resistor types.

I'd go with a 1W for the same reason that most fan makers that include low voltage adapters do.

Darlene

I knew it wouldn't be that simple..

So what are you suggesting, usual daisy-chaining and one resistor per fan?

Yep,

That's what's been proven to work the most reliably if you're not using a controller.

The low voltage adapters that comes with some fans use two 1W resistors in series to spread the heat dissipation across more surface area., but the AP15's current draw is very low compared to some fans, so a single 1W per fan would be fine.

Darlene
Quote:
Originally Posted by IT Diva

It wouldn't quite work that way . . . .

Reason being that the effective resistance of the fan motor would then be a parallel resistance with R2, and being much, much lower, would be the only effective resistance to figure with, so you'd be looking at 150 ohms out of 1150 ohms which would give the fan ~1.5 volts.

In a series circuit, the OP was correct with the 50 Ohm resistor, and also correct with his 3V drop across the resistor times the current thru the loop to get 0.19W as the power dissipation of the resistor

Pretty much all the resistors I've seen in the low voltage adapters that come with a lot of fans use 1W resistors, as they have more surface area to dissipate heat, especially where they are all covered in a layer or two of heat shrink.

47, 51, & 56 Ohms are the standard values in that range for common resistor types.

I'd go with a 1W for the same reason that most fan makers that include low voltage adapters do.

Darlene

You're right, just woke up and wasn't thinking it all through, lol.

I overlooked the output would be in parallel with the R2...Ignore what I said previously.

To accomplish what I was saying, you'd need some op amps in there and control the gain level with a pot
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Quote:
Originally Posted by B NEGATIVE

BNEG's jealous at our mad science skillz..
Quote:
Originally Posted by IT Diva

Yep,

That's what's been proven to work the most reliably if you're not using a controller.

The low voltage adapters that comes with some fans use two 1W resistors in series to spread the heat dissipation across more surface area., but the AP15's current draw is very low compared to some fans, so a single 1W per fan would be fine.

Darlene

Thanks for the info. I'll proceed with one 50Ω 1W per fan, and if they're getting hot I'll make them double 25s.
Edited by Daredevil 720 - 8/31/13 at 10:36am
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CoolerMaster CM690 II Advanced Logitech MX518 CoolerMaster Stacker Bundle
GraphicsCoolingCoolingCooling
3x MSI R9-290X 3x EK-FC R9-290X - Acetal 3x EK-FC R9-290X Backplate 2x Alphacool NexXxos UT60 480
Quote:
Originally Posted by cyphon

The first rev was, but the 2nd rev (which are all I could find for the 7950s now) are not

and to think I sold the rev 1.
 [ WOMD ] Rainuke (17 items)
CPUMotherboardGraphicsRAM
i7-2600k Asrock Z77E-ITX R9 290X Crucial Ballistix Sport
Hard DriveHard DriveCoolingOS
Seagate 7200 RPM Samsung 830 128GB XSPC AX240 kit - With custom Components windows 8 professional
MonitorKeyboardPowerCase
2x Asus VG248QE Coolermaster TK Coolermaster Hybrid Define R4
Razer Mamba Rainbow dash custom Schitt Modi Bose Companion 3
Audio
Objective 2
 [ WOMD ] Rainuke (17 items)
CPUMotherboardGraphicsRAM
i7-2600k Asrock Z77E-ITX R9 290X Crucial Ballistix Sport
Hard DriveHard DriveCoolingOS
Seagate 7200 RPM Samsung 830 128GB XSPC AX240 kit - With custom Components windows 8 professional
MonitorKeyboardPowerCase
2x Asus VG248QE Coolermaster TK Coolermaster Hybrid Define R4