OT: Fan mod help needed


queeg

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Sorry about the OT but I know there is an electrical engineer on the list somewhere.

 

I'm converting a 12 volt case fan to run off USB 5 volts.  I bought a  2 drive sata dock.  I love it.  I have experimented glueing a 50mm fan to it for cooling the hard drive.  It's simple and works so well it keeps the drive at or below 31c even when using it heavily.  

Now I want to use a bigger fan so it will cool both drive slots but I'm not sure how to do the math to make sure my USB can handle it.  I found a very cool 100mm x 12mm fan.

 

1000rpm Version 14.5dBA 15.23CFM 12V 0.09A

http://www.newegg.com/Product/Product.aspx?Item=N82E16835185071

 

 

USB is supposed to provide 100ma of current at 5v.  So that is the limit I'd like to stay in.  

 

The math:

The fan: 12v at .09A makes the power output of 1.08w and 133ohms of impedance.

 

Running that fan at 5v:

 

A = V/R :   5/133 = .038A

P = VI :     5*.038 = .19w

 

Is this right?  The watts seems really low.

 

 

EDIT:  In the end the solution is to use an external 12v psu like this http://www.coolerguys.com/840556029977.html

SataDockFan.jpg.c160a69ae4242d8ec5834f85f7004845.jpg

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Your calculations are correct. Realize that if you run the fan at 5 volts it may not start. If you intend to boost the 5 volts to 12 volts you will consume the 1.08 watts plus the losses in your DC-DC converter. Since I bet you really didn't mean you wanted to wire up a DC-DC converter, I think you should find a fan that will start and run at 5 volts. See Silent PC Review for a list:

 

http://www.silentpcreview.com/article63-page1.html

 

Why not pull the 12 volts from your PC power supply? Are you using a laptop?

 

Scott

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The dock sits in a cabinet and currently the fan is plugged into a usb port on my server.  I cut a usb cable open and wired up the red and black leads to just get the power.  I also have a usb fan that I use to exhaust warm air from the top of the cabinet.  I have a usb hub ordered that has an AC plug and I'll plug them into it when it gets here.

 

I also don't really want the fans to run at 12v because they would be noisier and I don't need that much air flow.

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I wish I could get the power for the fan from the dock itself.  It's got at least 2 12v lines in there.  There is a usb connection and esata connection on the back.  I tried just plugging the fan into that usb socket but it doesn't get any power when the esata is being used.  If the 100mm fan runs on 5v without problems I'll probably open the dock up to see if I can tap some power from there.

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Though your calculations are correct for a "resistive" load, a DC brushless fan is anything but... you're driving a 3 phase motor controller and a 3 phase brushless motor.  It will behave anything but linearly as you change voltage.  That being said, if it draws 90 milliamps at 12 volts , you can rest assured it WILL draw less than that at 5 volts  :D.

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Though your calculations are correct for a "resistive" load, a DC brushless fan is anything but... you're driving a 3 phase motor controller and a 3 phase brushless motor.  It will behave anything but linearly as you change voltage.  That being said, if it draws 90 milliamps at 12 volts , you can rest assured it WILL draw less than that at 5 volts  :D.

 

Is there a general rule or some kind of estimate about how the resistance changes with voltage?  A 12v to 5v change is a 59% drop.  Can I assume resistance would also drop somewhat?  It would be really handy to have some rough estimate (like "resistance drops with voltage at a 10% rate"  which would be 5.9% * 133 = 78.5 ohms) or anything that's fairly close.  

 

 

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I thought USB can supply 500ma of power.

I know they have the two headed USB cables for the external drives, so that it can supply up to 1a of power.

 

Further reading here may help.

http://www.girr.org/mac_stuff/usb_stuff.html#devices

 

Thanks for the link.  I did some more reading.  Here is a excerpt from the Wikipedia USB page which I thought explained it in more english terms for me - even these are somewhat confusing.  After reading that I believe the issue of whether each USB connector on a motherboard constitutes a hub or not.  If not, I'd still say they supply 100ma.  I purchased a powered hub because it said it supplied the higher current on each port and the text below seems to confirm that it is typical of powered hubs so it was reassuring to see that.  Going back to the non-hub ports, the text seems to indicate that if a device wants to draw more than the 100ma which is the standard 1 unit then it has to request it via the data lines and my fan can't do that.

I see warnings that some fans can overdraw current which makes sense because that would smoke the USB power source.  But there seems to be high and low power sources.  So the warnings should probably be don't try to draw more than the USB port can supply.  The first warning just scares people away from it.  The second is more informational and that's where I am.

 

So for connecting a fan, which is a non-communicating device and can't request more power, the trustable limit is 100ma on a non-hub USB port, 500ma (or more if specified) on a powered hub port.  Of couse the next test is that if you use a high rpm fan it's possible it won't have enough starting current to get running.  But I've read that lower rpm fans typically referred to as quiet fans will likely be able to start. 

 

What started as a "this probably won't work" has turned into a "this is kind of fun project" for me.

 

 

Excerpt of Wikipedia:  http://en.wikipedia.org/wiki/Universal_Serial_Bus

A unit load is defined as 100 mA in USB 2.0, and was raised to 150 mA in USB 3.0. A maximum of 5 unit loads (500 mA) can be drawn from a port in USB 2.0, which was raised to 6 (900 mA) in USB 3.0. There are two types of devices: low-power and high-power. Low-power devices draw at most 1 unit load, with minimum operating voltage of 4.4 V in USB 2.0, and 4 V in USB 3.0. High-power devices draw the maximum number of unit loads supported by the standard. All devices default as low-power but the device's software may request high-power as long as the power is available on the providing bus.[34]

 

A bus-powered hub is initialized at 1 unit load and transitions to maximum unit loads after hub configuration is obtained. Any device connected to the hub will draw 1 unit load regardless of the current draw of devices connected to other ports of the hub (i.e. one device connected on a four-port hub will only draw 1 unit load despite the fact that all unit loads are being supplied to the hub).[34]

 

A self-powered hub will supply maximum supported unit loads to any device connected to it. A battery-powered hub may supply maximum unit loads to ports. In addition, the VBUS will supply 1 unit load upstream for communication if parts of the Hub are powered down.[34]

 

In Battery Charging Specification[35], new powering modes are added to the USB specification. A host or hub Charging Downstream Port can supply a maximum of 1.5 A when communicating at low-bandwidth or full-bandwidth, a maximum of 900 mA when communicating at high-bandwidth, and as much current as the connector will safely handle when no communication is taking place; USB 2.0 standard-A connectors are rated at 1500 mA by default. A Dedicated Charging Port can supply a maximum of 1.8 A of current at 5.25 V. A portable device can draw up to 1.8 A from a Dedicated Charging Port. The Dedicated Charging Port shorts the D+ and D- pins with a resistance of at most 200?. The short disables data transfer, but allows devices to detect the Dedicated Charging Port and allows very simple, high current chargers to be manufactured. The increased current (faster, 9W charging) will occur once both the host/hub and devices support the new charging specification.

 

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A fan is not a resistor so your calculations are not true.

 

The power the fan draws is used to run the losses in the motor and to move air. The motor losses remain fairly constant but the power required to move air goes down very quickly as the fan speed is reduced. It's probably somewhere along the lines of being at 1/2 speed the air movement uses 1/4 the power compared to full speed.

 

At 5V or 42% voltage the fan will be turning much slower. So, the power to move the air is much less too.

 

You fan is rated for 0.09A at 12V. I would put the current draw at less than 1/2 that at 5V or <0.045A.

 

The USB port is rated for a minimum of 0.1A and with a powered hub you have 0.5A that you can use. So, you will be fine trying to draw 0.045A from any USB port.

 

You would still be fine even if you assumed the fan required the full 0.09A at 5V.

 

Peter

 

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Wait a second.  Power is power. If the fan needs N watts of power to run then it needs N watts of power to run no matter what the voltage is.  If you give it a specific voltage, then the fan will draw current so it gets its N watts of power.  If you halve the voltage to the fan then the fan will try and draw twice as much current to satisfy its power needs.  That is why larger motors (e.g. HVAC) are often run at 240V as opposed to 120V to halve the current draw and reduce the I^2 R losses in the wire. That is why Power transmission lines run at thousands of volts and are stepped down to deliver 240/120 to each home.

 

Motor power rattings are always given for their steady state operation at their nominal speed. 

 

I would use a 12 volt motor on 12 volts, and a 5 volt motor on 5 volts.

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Wait a second.  Power is power. If the fan needs N watts of power to run then it needs N watts of power to run no matter what the voltage is.  If you give it a specific voltage, then the fan will draw current so it gets its N watts of power.  If you halve the voltage to the fan then the fan will try and draw twice as much current to satisfy its power needs. 

 

No, this is not true when applied to a typical case fan. The fan turns slower when the voltage is lowered. The slower turning fan moves less air. Moving less air requires less power.

 

Peter

 

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A fan is not a resistor so your calculations are not true.

 

The power the fan draws is used to run the losses in the motor and to move air. The motor losses remain fairly constant but the power required to move air goes down very quickly as the fan speed is reduced. It's probably somewhere along the lines of being at 1/2 speed the air movement uses 1/4 the power compared to full speed.

 

At 5V or 42% voltage the fan will be turning much slower. So, the power to move the air is much less too.

 

You fan is rated for 0.09A at 12V. I would put the current draw at less than 1/2 that at 5V or <0.045A.

 

The USB port is rated for a minimum of 0.1A and with a powered hub you have 0.5A that you can use. So, you will be fine trying to draw 0.045A from any USB port.

 

You would still be fine even if you assumed the fan required the full 0.09A at 5V.

 

Peter

 

 

I don''t think it's easy to estimate the reduction in rpm in order to come up with the power rating.  But it seems like you are saying the current drop might be sort of linear with the voltage drop - did I read that right?  If so, then a fan with .2A at 12v reduced to 5v would pull current at around or just below .1A?

 

 

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I don''t think it's easy to estimate the reduction in rpm in order to come up with the power rating.  But it seems like you are saying the current drop might be sort of linear with the voltage drop - did I read that right?  If so, then a fan with .2A at 12v reduced to 5v would pull current at around or just below .1A?

Actually, it is probably anything but linear. (that would be pure resistive)

 

In fact, if the voltage is not sufficient to always get the fan spinning, its stall current might be much higher than its spinning current.

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In fact, if the voltage is not sufficient to always get the fan spinning, its stall current might be much higher than its spinning current.

 

From what I read so far it seems to only be a problem with high speed (performance) fans, not the ones classified as quiet or low power requirement. 

 

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Power is power.  Power (watts) is always volts*amps*cosine(powerangle). for DC resistive circuits powerangle is always 0 (current and voltage are in phase) so power reduces/simplifies to volts*amps.

 

If a motor runs slower it should draw less power, and at a fixed voltage less current.  If a DC motor draws 12 watts then on a 12 volt circuit it will draw 1 amp. If you now put it on a 6 volt circuit and the voltage were high enough to get it started, and the motor was to draw 12 watts the current draw would be 2 amps.

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Yeah, the fans run slower and are silent.  That's practically the whole purpose of going from 12v to 5v.  The other part is it's super handy and cheap to use USB as a power source.  Only a gentle puff of air is required to keep the drives cool.  The 50mm I'm using couldn't blow out a birthday candle but it dropped the drive temp from 45c to 31c.

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I don''t think it's easy to estimate the reduction in rpm in order to come up with the power rating.  But it seems like you are saying the current drop might be sort of linear with the voltage drop - did I read that right?  If so, then a fan with .2A at 12v reduced to 5v would pull current at around or just below .1A?

 

No, I didn't say that. The power will not be proportional to the voltage reduction.

 

Thinking about it more, the power is used for 2 things. Power is used to actually move air and power is "wasted" as losses in the motor.

 

The airflow - the air flow (cfm) is fairly proportional to the current and the air flow (cfm) is also fairly proportional to the rpm. This means the power used to move the air is about proportional to the speed reduction squared. In other words, at 50% speed the air movement would use about 25% of full speed power.

 

The motor losses - there are certain motor losses that will likely fairly linearly track the voltage reduction.

 

I would then further guess and say the fan speed reduces fairly close to the voltage reduction.

 

Without testing you can't know how much power goes to the airflow and how much goes to the motor losses. However, I'd bet the motor is fairly efficient so most of the power goes to moving air.

 

So, my best guess would be that the power will easily be less than 1/2 of the 12V rated value when run at 5V. At the low end it would be 17% of rated and at the high end it would be 41% of rated.

 

Peter

 

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Power is power.  Power (watts) is always volts*amps*cosine(powerangle). for DC resistive circuits powerangle is always 0 (current and voltage are in phase) so power reduces/simplifies to volts*amps.

 

If a motor runs slower it should draw less power, and at a fixed voltage less current.  If a DC motor draws 12 watts then on a 12 volt circuit it will draw 1 amp. If you now put it on a 6 volt circuit and the voltage were high enough to get it started, and the motor was to draw 12 watts the current draw would be 2 amps.

 

The motor would draw 12W on a 6V circuit only IF the motor still ran at the same speed and moved the same airflow. The fan will not run the same speed or move the same air when the voltage is 50% of rated.

 

The power does not remain constant when you reduce the voltage on a DC computer case fan.

 

Peter

 

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The new fan and USB hub came today.  The new 100mm fan does spin up at 5v but it spins so slowly even when plugged into the hub that it's not effective.  Disappointing.  This particular fan even spins pretty slowly at 12v so maybe it wasn't a good candidate for 5v.  

 

I'm going to comtemplate whether I have enough nerve to open up the sata dock and try to tap into a 12v source from there.  It does have two sata drives that pull 12v but I'm not sure if doing so might make a drive unhappy.  I think I'll try a different fan first.  I've tried some others but they are mostly 25mm wide and I want a skinny one that can sit on the edge of the dock.

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I'm also wondering if I add a resistor in parallel with the fan if it would pull up the voltage to 10v or so?

No, it would just use power.  If it does anything at all to the voltage it would cause it to go lower

(and only then if the power supply was unable to supply the additional current)

 

if you have the physical space you could just power it with a wall-wart.  I know it is not the most elegant, but it would work.  Or, you could use a boot regulator to convert 5 volts to 12, or easier yet

 

Get a 5volt fan.

http://www.coolerguys.com/840556069188.html

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if you have the physical space you could just power it with a wall-wart.  I know it is not the most elegant, but it would work.  Or, you could use a boot regulator to convert 5 volts to 12

What wall-wart?  I did find an AC to usb plug that I had sitting around - never had a use for it.  It will supply 700ma and 5v.

What is a boot regulator?

 

 

That's only a 60mm fan.  I have a 50mm and it's not wide enough to cool both drives.  I could use two 50mm fans but I'd like to use a single fan if possible.

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if you have the physical space you could just power it with a wall-wart.  I know it is not the most elegant, but it would work.  Or, you could use a boot regulator to convert 5 volts to 12

What wall-wart?  What is a boot regulator?

 

 

That's only a 60mm fan.  I have a 50mm and it's not wide enough to cool both drives.  I could use two 50mm fans but I'd like to use a single fan if possible.

So you think I linked to you the specific for your specific hardware installation???  I linked to an example...  You can so your own google search for a 5 volt fan that fits your enclosure.  They are out there.  Almost every laptop has a tiny one.  There are bigger.

 

If you had taken a few more seconds, even on that same site you would have found a whole collection of 5 volt fans of various sizes.  http://www.coolerguys.com/5vf.html

(and I just picked coolerguys.com at random from google, I have no idea if they are good or bad to order from, but they have a web-site with 5 volt fans.  )

 

Here is another site: http://www.buyextras.com/dc5vfaforlan.html

and yet site another with all sorts of sizes, many only 15mm thick :

http://www.frozencpu.com/cat/l2/g36/c231/list/p1/Fans-5_Volt_Fans.html

 

Joe L.

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if you have the physical space you could just power it with a wall-wart.  I know it is not the most elegant, but it would work.  Or, you could use a boot regulator to convert 5 volts to 12

What wall-wart?  What is a boot regulator?

 

 

That's only a 60mm fan.  I have a 50mm and it's not wide enough to cool both drives.  I could use two 50mm fans but I'd like to use a single fan if possible.

So you think I linked to you the specific for your specific hardware installation???  I linked to an example...  You can so your own google search for a 5 volt fan that fits your enclosure.   They are out there.   Almost every laptop has a tiny one.  There are bigger.

 

If you had taken a few more seconds, even on that same site you would have found a whole collection of 5 volt fans of various sizes.   http://www.coolerguys.com/5vf.html

(and I just picked coolerguys.com at random from google, I have no idea if they are good or bad to order from, but they have a web-site with 5 volt fans.  )

 

Here is another site: http://www.buyextras.com/dc5vfaforlan.html

and yet site another with all sorts of sizes, many only 15mm thick :

http://www.frozencpu.com/cat/l2/g36/c231/list/p1/Fans-5_Volt_Fans.html

 

Joe L.

 

Relax Joe, I spent a while yesterday searching for a 5v fan.  80mm is the largest I could find that was also thin.  Sorry if it looked like I gave you the "talk to the hand" treatment.  Not intended.

 

The part about the wall wart hit me right in the forehead.  :)  I didn't think about all the useless powersupplies sitting around with old unused devices.  Dang, thanks for opening my eyes.  I have the powersupply listed below.  Now I just have to figure out if I have an adapter or if I need to cut it open.

http://www.newpowersupply.com/leadman_ky_05036s_12_5_pin_ac_to_dc_power_adapter_bulk_pack-pr-224.html

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