running a 12v .5A fan from 24v 6.5A source

[ulysses]

I'm trying to figure out how to wire a cheap computer fan off the second output from my 24v6.5a CNC stepper power supply. I bought a voltage regulator IC that had input range 3-30v input with 12v output but it was only rated for 20watts P (I think I fried it which probably wasn't safe lol) my best guess is I need to put some kind of resistance chain together but I don't know where to start with that, watt rating, series or parallel, how big, how many. Any help would be appreciated thank you.

 

[Gryd3]

I'm trying to figure out how to wire a cheap computer fan off the second output from my 24v6.5a CNC stepper power supply. I bought a voltage regulator IC that had input range 3-30v input with 12v output but it was only rated for 20watts P (I think I fried it which probably wasn't safe lol) my best guess is I need to put some kind of resistance chain together but I don't know where to start with that, watt rating, series or parallel, how big, how many. Any help would be appreciated thank you.

Do you have any ratings on the cheap computer fan you want to try? (Amps, Watts, ...)

If this is to keep the stepper motor cooler, have you looked into passive alternatives? Some stepper controllers can reduce the power being sent to a stepper while it is motionless or staying still.

 

[KrisBlueNZ]

Your power supply should have worked. The fan you have needs only 6W (power is equal to voltage multiplied by current; 12V × 0.5A = 6W).

Assuming you don't have another 12V 0.5A fan to connect in series with the one you already have, you can try dropping the voltage using a resistor. The resistor must drop 12V with 0.5A of current flowing through it. Actually, the current may not be this high; the 0.5A rating is probably the maximum current drain for the fan, but let's assume that's how much it draws.

Using Ohm's Law, R = V / I:

R = V / I
= 12 / 0.5
= 24Ω

You probably won't find a 24Ω resistor; use a 27Ω one instead.

Now calculate the power dissipation in the resistor. The resistor will have 12V across it, and about 0.5A of current runing through it. The Power Law says:

P = V × I
= 12 × 0.5
= 6W (same as the fan power calculated above).

So you should use a resistor rated for at least 10W dissipation. It will get fairly hot, so mount it properly, away from anything that could be damaged. Here's a suitable resistor from Digi-Key: http://www.digikey.com/product-detail/en/SQP10AJB-27R/27W-10-ND/18745 (USD 0.64)

Here's a 15W resistor that can be mounted onto a heatsink: http://www.digikey.com/product-detail/en/MP915-27.0-1%/MP915-27.0F-ND/3013433 (USD 3.75)

Radio Shack have a range of 10W resistors: http://www.radioshack.com/family/in...filterName=Type&filterValue=10-watt+resistors. They don't have 27Ω but they do have 33Ω.

Also you can make values by connecting resistors in series (resistances add) and/or in parallel (the formula is R_TOTAL = 1 / ( 1/R₁ + 1/R₂ + 1/R₃ + ... + 1/R_n).

You might want to get several values and find the one that gives the right voltage across the fan (because we don't know exactly how much current it will draw).

A regulator is the proper way to go, but a resistor is cheaper.

 

[debe]

This is what ive used for the same job, of Ebay LM2596 adjustable reg module, input voltage max 37V. Output adjustable 1.5-35V, max 3Amp. can be purchased one for $3.00 ea free freight or 5 for $4.99 + $1.25 freght. I buy them in lots of 5 as they are so easy to use.

 

[ulysses]

Thank you for the replies, I picked up some 10w27ohm resistors today af ter work wired it up (kinda i think one of my connections coming out of the resistor is shoddy but long story short it works. I still have a few questions however:

Since the 27ohm resistor is on a 24v source I thought only the amperage would be limited to .88A using ohms law, but the circuit should still be 24V I thought (also giving a wattage of 21.33 across the resistor, which could mean i should add a second resistor to share the load possibly?). I feel like i still need the voltage reg IC to change the circuit to correct V for the fan, but at least im closer to a reasonable wattage for the IC i initially selected. I know there are cheap prebuilt options, but im trying to learn mostly as a hobby not just fix it, thanks all.

 

[Gryd3]

How were you calculating the limited Amperage using Ohm's Law?
Keep in mind that everything has a resistance (or equivalent resistance). So to be able to properly calculate how much current you are limited to, you need to first figure out the apparent resistance of the Fan you are using.
12V / 0.5A = 24Ω
(Keep in mind that an electric motor's circuit resistance will vary depending on it's load. It's resistance will be lower when it is starting)

So you total circuit should have approximately 24Ω + 27Ω = 51Ω of resistance
Take your 24V source, and run it across your circuit and you end up with:
24V / 51Ω = 0.47A through the entire circuit.
0.47A x 27Ω = 12.7V will drop across the Resistor.
0.47A x 24Ω = 11.3V will drop across the Fan.

When using Ohm's Law, you can't ignore the resistance of any parts you may be using in your circuit. Even if it does not show a resistance. (This gets tricky with Capacitors, and Inductors, but there are equations to figure out the resistance of these items to use in Ohm's Law)

*If I have missed anything, let me know!

 

[ulysses]

Upon closer inspection I've found the ratings of the fan and its lower power than I thought (12v.2A) I recauclated for this and ran it thru a circuit Sim using 2 2W 33ohm resistors in parallel. I have one last question about current. My power supply is ratted at 6.5 amps, are the amps only pulled out when the circuit has load demanding it?

 

[Gryd3]

Your circuit appears to have all the resisters connected in 'series', as in a chain... It is only parallel if there is an alternative path parallel to the other part.
For example.
R1 R2 and R3 are all connected in series.
R1 and V1 are connected in parallel to each other, so is R2 and V2, and finally R3 and V3.

You are catching on. By using the two 33Ω resisters in series like that, you get the equivalent resistance of 66Ω which is damn close to providing the required 12V to your fan.

You are also correct about the amps. Your fan will only pull the required amps when it is running. When you switch the fan off it will not draw power.
(Double edged sword, because of how a motor works, it will draw a little more power if you stick your finger on the fan and force it to stop while it is running. Only use a switch in-line with your fan circuit to turn it off to ensure things will run as planned )

 

[ulysses]

Say I wanted to run a led as a power indicator on this circuit. How do I get amperage to vary within the circuit?

 

[Gryd3]

Say I wanted to run a led as a power indicator on this circuit. How do I get amperage to vary within the circuit?

Amperage will change if the Supply voltage changes, or if the resistance of the circuit changes.

 

[Rleo6965]

How about placing a 12V zener diode in series with the 12V dc fan.