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Reducing voltage to a fan

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snobrder27

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I was searching for a place to find an answer to a problem i have concerning a recent project i started. i found google answers to be too impersonal and i stumbled accross this forum. Anyway, i am building a small paint booth for my basement to use in the cold winters here in vermont. In the booth the fumes are obviously too umbearably to stay in without some type of airflow, so, i have a small computer fan that blows air through dryer tubing and into a makeshift mask so i always have fresh air to breath (i also have a fan venting fumes but that is irrelevant) i have previously been running the fan of a 12 volt car battery that happened to be there at the time. I have an old ac/dc converter or what have you for a laptop. on the back it says that it drops the ac (100-240 volts) to dc at 20 volts. after testing the fan on the dc end for 5 seconds at a time between each solder i figured that the fan should be running faster, which it was, and proceded to solder and heat shrink all my connections. then after showing my father my work which i was very happy with i let the fan run for about 30 seconds and found out that what was bound to happen, happend, the fan started to smell because the 20 volts was too much. i was wondering if there is anyway to drop the 20 volts back to the 12 i was using so that i dont have to have the fan sitting on the battery. I was wonder if i would be able to use resistors or if i would be better off finding a different power supply or returning to the battery, thanks in advance to those who took the time to read and thanks for the help.
 
I have an old ac/dc converter or what have you for a laptop. on the back it says that it drops the ac (100-240 volts) to dc at 20 volts. ... i was wondering if there is anyway to drop the 20 volts back to the 12 i was using so that i dont have to have the fan sitting on the battery.
Supply is 20 volts. Fan must be 12 volts. So the resistor must have 8 volts. V=IR. I is the fan current - listed on the fan. 8 volts = fan's current times R - the resistor you need.

Now you have the resistor value. How much power does it consume? P=VI. Power is that voltage (8) times the fans current (I). Now you have resistor ohms and resistor wattage.
 
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Just get a 12V power supply. $10 at a surplus electronics place like Allelectronics

Computer fans use PWM and complex electronics. They cannot be run off a higher voltage power supply and a resistor...
 
If you use Westom's resistor method, put the resistor in front of the fan to cool it. You'd also need a capacitor in parallel with the fan. I, personally, would simply use the proper power supply and avoid the hassle of resistors and extra heat.
 
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I, personally, would simply use the proper power supply and avoid the hassle of resistors and extra heat.
Any 12VDC wall wart supply would work just fine. Resistors are how one learns this stuff. How to convert simple math into a solution. How to really impress Dad.

Or simply buy a 7812 single chip power supply for about $0.60 from most every semiconductor manufacturer - sold in every electronics supply house. Also learn more and impress dad.
 
You could connect two fans in series, that would give each fan approximately 10 volts, increase your air flow too.
 
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Yes, a new 24V fan will work better than a partially burnt out 12V fan. ;)
I don't know if snobrder27 is in the position to buy one though.
 
There are 24 volt fans which should work just fine with your voltage
Which is a solution that costs significantly more money and teaches nothing. If buying a new fan, then select one that meets that voltage AND perform numeric tradeoff between CFM and dBs of noise. How to learn something from the fan purchase.

Buying a new supply or new fan is tens of dollars. Doing simple arithmetic - what is famously called "Work smarter, not harder" - means a solution that costs dimes. Either using the resistor or using a 7812 single chip power supply.

Too many would throw extravagant money at a problem rather than both learn from the experience and spend almost nothing. The OPs problem is classic of what we solve to – first and foremost - learn. Too many instead would spend massive sums rather than “work smarter – not harder”.
 
If he really wanted to "learn something" maybe he should hack the 20V switching power supply so that it puts out 12V. I suspect he just wants the easiest and cheapest solution.
 
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If he really wanted to "learn something" maybe he should hack the 20V switching power supply so that it puts out 12V. I suspect he just wants the easiest and cheapest solution.
A cheapest and easiest solution would be a twenty five cent resistor or a sixty cent single chip power supply. Not tens of dollars for 'more fans' or more power supplies.

Hacking the supply would be interesting and educational. But well beyond what most everyone here could attempt - based upon current replies. Long before he might hack any supply, an OP first must learn the basics: V=IR and P=VI. Which is why one resistor is so educational AND a cheapest and easiest solution.
 
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A cheapest and easiest solution would be a twenty five cent resistor or a sixty cent single chip power supply. Not tens of dollars for 'more fans' or more power supplies.
Considering that the original post included this statement, other ideas besides a dropping resistor are acceptable to snobrder27:
I was wonder if i would be able to use resistors or if i would be better off finding a different power supply or returning to the battery,

Who knows. Maybe the snobrder27 can find a suitable 12V wallwart or other supply from a buddy, dump, E-recycler, etc. Maybe he can find a 120V fan. Maybe snobrder27 has no local source of electronic components and can't find the right resistor?

But well beyond what most everyone here could attempt - based upon current replies.
I think that most of the people who replied to this thread could mod the old 20V brick to 12V with little effort. But, you are right, it is unlikely that snobrder27 is capable of that.
 
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Haha thanks everyone, I think for starters I'll do some quick math and buy a resistor. The only thing about putting in front of the fan to cool it makes perfect sense but would it be giving off anything besides heat that should not be pumped directly into my air supply?
 
but would it be giving off anything besides heat that should not be pumped directly into my air supply?
Not if it has a high enough wattage! ;)
I would recommend also putting a 220-470uF 25V capacitor in parallel with the motor to keep its controller circuitry happy.
 
The only thing about putting in front of the fan to cool it makes perfect sense ...
P=IV formula is about sizing that resistor for sufficient cooling in non-moving 100 degree F air. More than cool enough without a fan.

And that wattage - its heat output. Well, do the numbers. Then learn how much 'heat' requires 'concern'.

A little hint from the numbers. Same current flow through fan and resistor. P=IV. Constant I. V for the fan is 12 volts. V for the resistor is 8 volts. Which one is consuming more power - has the larger P number - generating more heat? Which one would pump too much heat into your air supply? How often does a fan generate too much heat?

Appreciate how much to be learned from the experience. Appreciate why answers without numbers so easily create myths, speculation, or complete lies.
 
When adding the capacitor is there an order I should be aware of when connecting the parts? Positive to resistor to capacitor to fan? Or positive to capacitor to resistor to fan?
 
The resistor can go in either the positive or negative lead. It doesn't really matter, but tradition puts it in the positive lead. The positive lead of the capacitor goes to the positive lead of the fan. The negative lead of the capacitor goes to the negative lead of the fan. Like this:
 

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When adding the capacitor is there an order I should be aware of when connecting the parts?
220 uf capacitor is the size found in power supplies for whole computers. If your 20 volt supply is large, then its capacitor might be that massive. No readon for a cap that large.

I never needed a a capacitor to reduce fan voltage. Keep it simple. Do it without a capacitor. Later get curious. Touch a 1 uf, 0.01 uf or a massive 10 uf cap across the fan to learn - to discover what changes.
 
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I never needed a a capacitor to reduce fan voltage.
It is not there to reduce voltage. It is to keep the ripple on the supply to the fan controller at a manageable level. A computer fan doesn't use a simple DC motor with a commutator, but a complex electronic drive circuitry.

Touch a 1 uf, 0.01 uf or a massive 10 uf cap across the fan
LOL! 10uF = massive???
 
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Which is a solution that costs significantly more money and teaches nothing. If buying a new fan, then select one that meets that voltage AND perform numeric tradeoff between CFM and dBs of noise. How to learn something from the fan purchase.

Buying a new supply or new fan is tens of dollars. Doing simple arithmetic - what is famously called "Work smarter, not harder" - means a solution that costs dimes. Either using the resistor or using a 7812 single chip power supply.

Too many would throw extravagant money at a problem rather than both learn from the experience and spend almost nothing. The OPs problem is classic of what we solve to – first and foremost - learn. Too many instead would spend massive sums rather than “work smarter – not harder”.

I was ONLY pointing out that there are different voltage fans, other than 12 volts.
 
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