Electronics Using 1n4001 diodes to reduce voltage on fans

I have use 6 1N4001 diodes to drop the voltage on a delta FFB1212EHE to 7.5v but im worried about the heat that is emitted from the diodes when the fan is on. At first I tried to use resistors but that proved to be a bad idea as they ended up very hot and a little brown. I got the idea from cpemma's web site but im not sure about the temp as the FFB1212EHE has a rated current of 2 amps and a rated input of 24 watts. sorry if this sounds noob like any help would be appreciated as this fan is a monster at 12v.

 

That is bad. From what I remember, 1n4001s can only carry 1A max, and you're pushing twice as much through them. Definitely look at replacing them. I'm pretty sure there's something in the 1n range that'll do, but I can't give you a model number.

 

Ok thanks ive did a quick read up about diodes i think i need some 1N5401 they can carry 3 amp so they should be ok.

 

why don't you use a higher wattage resistor? thats what wattage ratings are for.

 

Let's see, bumping down a 2A load from 12 to 7.5v would require at LEAST a 10w resistor, probably more because the wattage dissipated is 9w even, which might be a bit too close to 10w for comfort. Diodes are much cheaper, not to mention much more compact. Not sure which one would be cooler to the touch though, depends on whether the specific heat of silicon is higher than that of concrete or sand.

Edit: A few minutes on Google says that diodes would actually be the hottest. Specific heat of silicon is ~.714 J/g, while sand is about ~.795 J/g, while concrete beats the snot out of them with 3.3 J/g. This is by no means definitive, just food for thought.

 

At that point why not just use the ever popular 7v mod? Use the 12v as your v source and the 5v as the ground. I doubt there is a noticable difference in that half a volt, and you don't have any resistors or heat issues. Unless you're planning a more adjustable circuit or something.

 

It may come down to surface area. And with the diodes there are six sharing out the heat loss.

The 1A diodes are only a bit over-rated, as the fan will be running at around 7V so current will be down from the 2A @12V to around 1.2A. The 1N5401 are still a very good idea. Or just fit a less OTT fan.

The speed wire wouldn't work, if that matters in this case.

 

As cpemma said... it all comes down to surface area. It doesn't matter what type of component you use, if you're dropping 5V in this particular circuit then the same amount of heat will need to be dissipated regardless. You can use a few smaller devices to "split up" the heat dissipation or use 1 component that can handle the heat all on it's own. Usually (up to a point) its cheaper to use a few smaller components rather than one big one.

 

twelve 1-watt resistors in parallel in the right resistances will be equal to a 12W resistor at the same equivalency.

7.5V/1.2A=6.25 Ohms

i tried the math for you but i literally got a nosebleed from concentrating so hard

2 resistors in parallel of same value (X) = 1/2 X
2 resistors of different values in parallel = R1 + R2 / R1*R2

 

An adjustable linear regulator, like LM350 ( http://www.fairchildsemi.com/ds/LM/LM350.pdf ) would do a better job. Perhaps you can find one in an old linear power supply. The adjustment resistors at the output (Figure 13 at above link) can be small, like 1k and 4.7k, 1/16W (check my math). The regulator will shut down if it gets too hot so nothing gets damaged.

 

Given my choices id go for a diode solution over a resistor , much more predictable voltage output. Probalbly an amplified diode circuit would be your best bet, and its adjustable. Use a TO220 package transistor, its easier to heatsink.

 

Yes i still want to mesure the fans speed as im going to get a xp-120 heatsink for the fan to sit on.

 

Nevermind what I said before. I was thinking of something different.

An adjustable voltage regulator seems to be the best solution here, as nick01 suggested. The LM350 will handle 3A so it should handle any fan you decide to put on that heatsink. Just remember the LM350 will get hot so it will need a heatsink of its own.

 

I agree, an LM350 would be a better option for you. Besides, it simplifies the number of components that you'll be using. The regulator will get hot, and as mentioned you should screw a heatsink on. At the very least you can stick it to the chassis for decent thermal dissipation.

 

Well it could...but wiring would be a mess.

 

Good idea, just make sure that its electrically isolated.

 

I think you'd need a dozen 144o resistors... good luck finding those in one watt variety. Only ones I've seen in 1w are really low resistance. Then again I haven't LOOKED.

you need to use better variables.... parallel (all the same) is R/N where R is the resistance in ohms and N is the number of resistors. At least I think it is. Hopefully solving for N in that doesn't give you a nosebleed, or else someone needs to retake prealgebra.

 

shut up ok, i never took prealgebra... but i did take algebra from 6th to 10th...

that was because of bad teachers, one actually said i sucked when i got a question wrong...

 

Can't you use a small schematic with a ne555, to use PWM for the fan, PWM will turn the power supply on and off very fast, regulated from 0% on until 100% on, an example:
at 50% on time, you get: .5 * 12 = 6v average
this also works with leds, but not for regulating power for a chip or something
PWM has the advantage that the used parts will not become hot and you can regulate the speed with a small pot-meter, like this:



The diode's can be 1n4148 or the 1n4001 you already used, the transistor has to be a npn-type and must be able to drive 1,5 amp
ow, and forget the wire from + to -

 

True, mine (installed for another purpose) is on a 3/8" x 6-32 Nylon standoff (came with the case for mobo mounting). The standoff replaces one of the screws that hold a HD in its cage. The regulator is screwed to the standoff and has its own little clip on heatsink (from a dead ATX PS). It sees some air flow from the HD fan, but still shuts down at 5W dissipated power. If you want more power dissipation you may need a bigger sink.