combine resistors for wattage

[shaiss]

I'm following this tool: http://www.blackfiveservices.co.uk/fanspeedcalc.php?Voltage=12&Current=0.15&Target=6

I need to reduce my 12v .14a case fan from 12v to 7v.  The tool results in "Power dissipated by resistor:  5V * 0.088A  =  0.438W Target resistance:  57.143 Ohms".  The way I read this is that I could use a 68 ohm 1/2watt resistor and I should be good. 

However, I'm curious if I could/should use two 1/2watt resistors in series or parallel to reduce the load on the one resistor.  My thinking is two 120ohm resistors in parallel would result in 60ohms and 1/4watt each.

Is my thinking correct here?

(I've attached a picture of the fan.  I know it says 14V, but the mesured dc voltage going to it from the PSU is 12v)

I know I can go to radioshack and just buy a 1 watt resistor, but I'm trying to avoid another trip to the store since its quiet out of the way.  Thank you,

[PA0PBZ]

Your thinking is correct, I didn't check your math

[Ian.M]

Fans are motors, not resistors.  The current they draw is determined by the difference between the back EMF (which is proportional to the speed) and the supply voltage.  The power required to spin the blades is proportional to the cube of the fan speed.   That means that you cannot simply use Ohm's law to predict the effects of adding a resistor in series with the fan with reasonable accuracy.

You need to 'suck it and see' e.g. power up the fan from a variable PSU and measure the voltage and current at your desired fan speed then calculate the resistor to drop the difference between the supply voltage and your desired voltage at the actual measured current, or simply use a string of diodes as a dropper.  Its better to use a series string of resistors or diodes to get the drop you want because it distributes the waste heat over a larger area (unless you space parallel resistors well apart) and is easier to insulate properly with heatshrink sleeving.   Before sleeving, the resistors shouldn't run hotter than warm to the touch.   You dont have to worry much about diodes heating because you typically need three for a drop of about 2V so the dissipation is spread over far more parts.

[shaiss]

Thanks Ian!  this makes sense.  I'll use my bench PSU set to 7v, measure the current as you stated.  Once I have the voltage and current would you suggest using a tool like: http://www.gtsparkplugs.com/Dropping_Resistor_Calc.html to figure it out the proper resitor to use?

If that calculator comes back with 40ohm .7w, I could use 2 resistors in series, each @ 20ohm and 1/2w.  Or even better as you suggest, 3 13ohm resistors in series 1/2w each to distribute the heat even further.  The latter option as you suggest only seems needed if the 2 resistor option is too hot to hold in my hand.

Thank you for the details!

[Ian.M]

There is no need to use an online calculator for something as simple as Ohm's law.  Simply rearrange it to R=V/I and put the desired voltage drop in as V.  Power dissipation is easy: P=I²R, which is the form you want the equation in if you are splitting the dissipation across multiple series resistors.

Resistors come in preferred values (e.g. the 10% tolerance E12 series 1.0, 1.2, 1.5, 1.8, 2.2, 2.7, 3.3, 3.9, 4.7, 5.6, 6.8, 8.2, with a power of ten multiplier).   Although you can get other values they tend to be more expensive or even in some cases need to be special ordered.  As such, if you wanted a 40R resistor, you'd make it up from 18R + 22R,  or, near enough, 2x 12R + 15R, or 4x 10R.  Of course the dissipation wont split equally when the resistors aren't equal, but if you use adjacent values, it will usually be close enough.   

N.B. There is no point whatsoever in trying to 'tweak' the total theoretical resistance closer than say +/-5% if you are using 10% tolerance resistors.

[dacman]

I would suggest looking at the complete specifications of the resistors.  The full power temperature rating, for example, may be up to 70 degrees Celsius, and they may be linearly derated to 0 watts at 125 degrees Celsius.  For continuous usage, I would not recommend running them at more than about 25% of their rated power.  I've seen lots of carbon resistors short out (after several years of use) that were run at about half power.

[Audioguru]

I agree that you should not use a resistor near its power rating because then it will be very hot and burn you if you touch it or melt a plastic cased capacitor that touches it. Also, I did not look at the guarantee time for a very hot resistor.

[shaiss]

Thanks for the tips everyone.  It sounds pretty clear that I should go buy an appropriate watt resistor.  For info I did some measurements and the starting voltage is 14v and 120ma.  St 7v it's 62ma.   This is ideally where I'd like the fan. 

I took a "speed control" from an existing PC on the bench.  The resistor in that "controller" is ~150ohm but I can't see the wattage. 

When using this speed controller at 14v the fan drew 52ma.   This is a bit lower than like but not bad.   So my thinking is I need a 130 or so resistor and have to figure out how many watts to get and double or x4 whatever that is to be on the safe side. 

I'll find a calculator and post my results.  Ty everyone!

[Ian.M]

Assuming the 12V rail is actually 12.0V, and you want to drop that to 7V, with a load current of 62mA,  the voltage drop across the resistor will be 5V.  5V/0.062A=80.6R.   Nearest preferred value 82R.   82*0.062²=0.315W.  A 1W resistor shouldn't  run too hot, or use two 39R 1/2W resistors in series or three 27R 1/2W or 1/4W though the latter may run a little warmer than you'd prefer.

[shaiss]

awesome, thx for the help.  I have the two 39R conected in series as suggested and going to leave it for a bit to see how hot it's going to get.  This is going into a PSU thats for a 3dprinter.  I have an additional 80mm fan blowing on all the electronics, so this 60mm is just to remove some of the heat from the PSU.  at most, it'll be on for a day or two. 

Is there a way to calculate how long it'll take the resistor to reach it's max heat?

And I verified the 12v rail is @ 12.14

[shaiss]

it's only been a few min and the resistors are hot.  not burning your had, just "oh, that's hot".  To be on the safe side, I think I'll stop by radioshack and look for an 82R 2W resistor.  this should at least give me plenty of head room.

[Ian.M]

Sounds like you are on the right track.  Its theoretically possible to calculate the maximum temperature - if you have a full datasheet for the resistor with all the thermal data.  Most of us building a one-off would just estimate a suitable derating factor then test to see if the max temperature rise is acceptable.

Another approach would be to pick up a dirt cheap buck converter module off EBAY, and dial it down to 7V to run the fan with minimum waste heat.

[shaiss]

Ian, thank you so much.   I actually happen to have a buck w/ an LM2596S on it laying around.  I bought it a while back to play with and never had a use for it.  Now I do and that saves me the trip to RadioShack!

I plugged it in to my bench power supply, set it to 12v.  Then set the pot on the buck to 7V and then connected the fan, works like a charm! 

by your suggestion I take it the buck converter is probably the best way to go so I'll go ahead and set it up tomorrow and order another buck as a spare.  Thanks again!