Wait - Less speed takes more power? DC electrical question

[greens]

Alright i have a bit of an LED light project going on. The LEDs are high power and require cooling.

I have these perfect little turbine style fans from old DELL server racks i've been too sentimental towards to throw away.

The fans have two... fans built in, one spins clockwise the other counter, with blade designs to make that work nicely. They are super focused and have a massive static pressure (i'm guessing) for their use in 1U servers.

So i hooked the wires from each individual fan motor into one, so i have them in series. But they are probably too noisy and fast for my needs.

The fans are marked 12v .68A. Holy crap! So i hook them up to a 12v power supply and like i thought - way too loud and more powerful than i need.

So i put a resistor on it. They are only 2W but they are rated for high voltage and are a nice square ceramic variety.

Still not quite enough so I put another resistor on it.

Now it is perfect... but my question is... I am making my fans slower by drawing MORE power?

See my logic is that i have
12v*.68a=8.16W
Then i add the resistors to it.
8.16W+2W+2W=12.16W
So wait - did i really just make my fans slower by sucking in MORE power?

Or did I rob the fan of current some other way? Does it work more like this?
12v*(.68a(5/6)*(5/6))=(12v*.47a)=5.64W But the resistors generate heat so this math doesn't make sense, some energy must be lost.

This is the thing i don't get.

What using a voltage buckbooster be a better plan? I could drop a 12v power supply down to maybe 8v.

TLDR: Does using resistors to slow down fans increase over all power demand? Would using a buckbooster to alter voltage and leave current alone be a better option?

I don't care how i make the fans slower - i just want to make sure i'm not wasting energy for a lesser result which seems silly.

I've been working with computers forever but this is the first project i've done that is very involved with DC electronics.

Please only respond if you have a very clear and concise answer.

 

[aldan]

resistors are measured in ohms,and not amps.this is where your math is flawed,so to speak.in terms of power draw,measured in amps,these resistors will make literally no difference.in terms of voltage draw,you can not draw more than the rated output of the power supply.in this case 12v.if you wanted to figure out the draw (amps) of a circuit simply divide watts by volts.if you want to figure out the draw in amps of your particular circuit use ohms law.to figure draw divide your voltage by the resistance including fans.

 

[scout_03]

did you look at these on web https://www.amazon.ca/dp/B007PPHKO2/ref=pd_lpo_sbs_dp_ss_1?pf_rd_p=1977604502&pf_rd_s=lpo-top-stripe&pf_rd_t=201&pf_rd_i=B00880S6KA&pf_rd_m=A3DWYIK6Y9EEQB&pf_rd_r=MNYN9D6TFPQAAZ8A40NW

 

[greens]

resistors are measured in ohms,and not amps.this is where your math is flawed,so to speak.in terms of power draw,measured in amps,these resistors will make literally no difference.in terms of voltage draw,you can not draw more than the rated output of the power supply.in this case 12v.if you wanted to figure out the draw (amps) of a circuit simply divide watts by volts.if you want to figure out the draw in amps of your particular circuit use ohms law.to figure draw divide your voltage by the resistance including fans.

Sure. That is fine but doesn't answer the question.

If my resistor is being given 12v, and the resistor is adding 2W. Then we know the resistor is eating .1666 amps at 12v.

It still doesn't answer the question - does adding a resistor to slow down the fan increase current overall. 2W might be negligible, but 2W is also 25% of the overall load, so in terms of % it is huge.

 

[scout_03]

what about using a fan controller .

 

[greens]

what about using a fan controller .

Well that is sort of what I'm asking. A buckbooater can step up or step down voltage. I believe that is what fan controllers do - alter voltage.

The question here guys is this-

Is it more effexient to slow down a fan by adding a resistor, or stepping down voltage?

 

[scout_03]

i wouls say safer by stepping down voltage like a regulator .

 

[aldan]

resistors do not ad watts,which is a measurement of power and not resistance.you cant use that formula to measure draw.use the formula i gave you.google ohms law and look for the triangle.

 

[greens]

A resistor generates heat. It HAS to increase load. They do too add watts. Ohm's law. Current=volts/resistance.

Increasing resistance decreases current. This is known.

The question is whether or not a step down in voltage is more efficient than a resistor.

 

[molletts]

You're probably better off using an off-the-shelf fan controller - it will be more efficient apart from anything else - but I'll try to explain how you might be able to use resistors anyway.

The resistors will reduce the overall power but they will dissipate some power themselves.

If we model the fans as a resistor (this is something of an over-simplification as the fans probably don't draw a current proportional to the voltage but it might be good enough), we can say:

R(fans) = 12 / 0.68 = about 18Ω [R=V/I - Ohm's Law]

(I'm assuming that the 0.68A is for both fans together, not 0.68A each.)

So, we want to achieve 8V across the fans by putting resistors in series with them. That's ⅔ of 12V so the 18Ω of the fans needs to be ⅔ of the total resistance. Therefore we need about 9Ω for the series resistance. If your resistors are 4.7Ω each, that might be close enough with them both in series with the fan (resistors in series - add up the resistances). If you've got 18Ω ones (or thereabouts), stick them in parallel with each other (several identical resistors in parallel - divide the resistance by the number of resistors) then put the pair in series with the fan. Otherwise, you'll need to find a way of making up a resistance of about 9Ω. (This is all based on a pretty poor approximation of the fans' electrical characteristics so it probably doesn't matter if it's not that close!)

Assuming that gives us a total resistance of somewhere around 27Ω and ignoring once again the probable lousiness of our modelling, we can calculate how much power will be drawn and dissipated in each part of the circuit.

The current through the circuit will be:

I = 12 / 27 = 0.44...A [Ohm's law again]

The total power dissipated will be:

P(total) = 0.44... × 12 = 5.33W [P=IV]

The power dissipated in the "series resistance" (i.e. whatever arrangement of resistors you use in series with the fans) will be:

P(resistors) = 0.44...² × 9 = 1.77...W [P=IV, V=IR so P=I²R]

So it's within the 2W limit of the resistors you have. If you make up the resistance out of two identical resistors, the power dissipation will be shared equally between them.

Hope you followed that!

Stephen

 

[greens]

You're probably better off using an off-the-shelf fan controller - it will be more efficient apart from anything else - but I'll try to explain how you might be able to use resistors anyway.

The resistors will reduce the overall power but they will dissipate some power themselves.

If we model the fans as a resistor (this is something of an over-simplification as the fans probably don't draw a current proportional to the voltage but it might be good enough), we can say:

R(fans) = 12 / 0.68 = about 18Ω [R=V/I - Ohm's Law]

(I'm assuming that the 0.68A is for both fans together, not 0.68A each.)

So, we want to achieve 8V across the fans by putting resistors in series with them. That's ⅔ of 12V so the 18Ω of the fans needs to be ⅔ of the total resistance. Therefore we need about 9Ω for the series resistance. If your resistors are 4.7Ω each, that might be close enough with them both in series with the fan (resistors in series - add up the resistances). If you've got 18Ω ones (or thereabouts), stick them in parallel with each other (several identical resistors in parallel - divide the resistance by the number of resistors) then put the pair in series with the fan. Otherwise, you'll need to find a way of making up a resistance of about 9Ω. (This is all based on a pretty poor approximation of the fans' electrical characteristics so it probably doesn't matter if it's not that close!)

Assuming that gives us a total resistance of somewhere around 27Ω and ignoring once again the probable lousiness of our modelling, we can calculate how much power will be drawn and dissipated in each part of the circuit.

The current through the circuit will be:

I = 12 / 27 = 0.44...A [Ohm's law again]

The total power dissipated will be:

P(total) = 0.44... × 12 = 5.33W [P=IV]

The power dissipated in the "series resistance" (i.e. whatever arrangement of resistors you use in series with the fans) will be:

P(resistors) = 0.44...² × 9 = 1.77...W [P=IV, V=IR so P=I²R]

So it's within the 2W limit of the resistors you have. If you make up the resistance out of two identical resistors, the power dissipation will be shared equally between them.

Hope you followed that!

Stephen

This is perfect!
I see now where my W=V*A was incorrect. I was using P(total) instead of Ohm's law in my initial set up.
OK so if we say my whole setup uses 5.33W, that means that my resistors burn up 1.77W.

Now all i need to do is see how efficient the step down unit i'd need would be, correct? I realize there are differences depending on what unit i get. But is there any way to calculate, roughly, the power lost when stepping from 12v to 7v using a constant current step down?
https://www.amazon.com/DROK-Converter-Step-down-Transformer-Regulator/dp/B0179OMIR2/ref=sr_1_2?ie=UTF8&qid=1488831270&sr=8-2&keywords=12v+step+down
A unit like that could be used to bring my 12v down to a nice sweet spot where noise and airflow is acceptable.
But obviously it must lose some current as well - as VRMs do get hot too.

How can we figure out which is more efficient?

 

[TJ Hooker]

You'd need to look at the data sheet/product spec for that converter to see what kind of efficiency you'd get, but a DC-DC converter will be more efficient than just using series resistors.

Edit: And when looking at the "2W" rating of your resistors, that doesn't mean they're always going to be consuming 2 W. It just means they are capable of dissipating up to 2 W of heat. Actual power consumption is dictated by the voltage across the resistor and the resistance value (P=V^2/R).

 

[greens]

Oh wow, most of them advertise being 96% efficient.
Looking at our math, the DC-DC would have to beat... 66% efficiency to win... Which is horrible and i can't imagine any even exist that perform that poorly.

A DC-DC step down it is! Thank you.

The resistor is 5ohm i believe - 2w is significantly lower than its rating. I don't have them in front of me so i can't look up their specs. 2W is the number I saw in use, not their rating.