Voltage divider not working for fan

[davidwinth]

I have a 24 volt power supply to power a 12 volt fan. Since I don't want to burn the motor out I used a voltage divider with two 1kΩ resisters to cut the voltage in half. I measured that the voltage on the divider is 12 V, so this is good. However, it won't drive the fan. When I hook the fan to the voltage divider nothing happens and the voltage between terminals goes to zero.

What could be going wrong? Thanks

 

[phinds]

I have a 24 volt power supply to power a 12 volt fan. Since I don't want to burn the motor out I used a voltage divider with two 1kΩ resisters to cut the voltage in half. I measured that the voltage on the divider is 12 V, so this is good. However, it won't drive the fan. When I hook the fan to the voltage divider nothing happens and the voltage between terminals goes to zero.

What could be going wrong? Thanks

What is undoubtedly going wrong is that you are completely ignoring the resistance of the fan, which is messing up what you think you should be getting from your "divider".

 

[davidwinth]

What is undoubtedly going wrong is that you are completely ignoring the resistance of the fan, which is messing up what you think you should be getting from your "divider".

O.k. So where does that leave me? Is there some practical response you could give? I checked that the fan does indeed work on a 12 Volt supply, but I am stuck using the 24 Volt supply. Also, I did verify that the divider results in 12 Volts, not some other value.

 

[phinds]

O.k. So where does that leave me? Is there some practical response you could give? I checked that the fan does indeed work on a 12 Volt supply, but I am stuck using the 24 Volt supply.

What is the power requirement of the fan? Say, for example, that it is 12 watts. That would mean it needs a 1 amp source and thus has a resistance of one ohm. You have put a 1K ohm in series with it, making the bottom half of your "divider" utterly negligible and reducing the maximum current to 12V/1K ohm = 12 milliamps, which your fan hardly even notices.

If it is, again just as an example, such a fan, you would need a 1 ohm resistor in series with it to get the actual division you want. That means your fan will be drawing 12 watts and you will be wasting 12 watts is the series resistor (which, by the way, should be something like a 25Watt resistor, not some dinky 1/4Watt kind of thing, else it will just evaporate when the power goes on).

Do you understand Ohm's Law?

 

[davidwinth]

What is the power requirement of the fan

The fan is marked as 12 V and 4.3 Amps. It is a little duct fan.

Do you understand Ohm's Law?

V = I*R if I remember correctly. But it is confusing to me because I took apart the voltage divider since it isn't working anyway, and just put the two resisters inline with the positive terminal to see how much voltage drop that would give me. It still reads 24 Volts but I thought it would read less with more resistance.

 

[phinds]

The fan is marked as 12 V and 4.3 Amps. It is a little duct fan.
V = I*R if I remember correctly.

OK, so 4.3amps and 12 volts requires what amount of current? How much power is used? The same amount of power will be used in the series resistor. What size resistor do you need?

 

[davidwinth]

OK, so 4.3amps and 12 volts requires what amount of current? How much power is used? The same amount of power will be used in the series resistor. What size resistor do you need?

The current is 4.3 Amps, yes? The power is I*V = 52 W. The resistance will be 12/4.3 = 2.8 Ohms. Is that the resistance of the fan, then?

 

[davidwinth]

[This post and the ones below have been merged into this existing thread]

I have a 12V DC fan but a 24V DC power supply. I thought I could use a voltage divider to get to around 12V to prevent the fan from burning out. So I think I have found a voltage divider calculator online. The given data is:

Fan Voltage = 12V DC
Fan Current = 4.3 A
Fan Resistance Calculated = 12/4.3 = 2.79 Ohms
Fan Power Calculated = 12*4.3 = 51.6 W

I need to do the calculation with the fan resistance in there and I want to minimize the current in the circuit. Here are two sample calculations found by trial and error from the same site. Both look O.K. as far as voltage supplied to the fan. Am I doing this correctly? In the first calculation, it says the fan will be pulling 53 Watts and the second 55 Watts.

• Is it a concern that these values are higher than the 51.6 Watts required?

The largest difference appears to be on the total supplied power, which is huge.

• Is this value the value I need to use when specifying the wattage capacity of both R1 and R2?

Clearly the first circuit is better if so!

Thanks

 

[Tom.G]

The 1% to 3% increase in fan power shouldn't be a problem, after all it is 'forced air cooled'.

To cut down the total power just put a 2.7 Ohm resistor in series with the fan. No divider (and its losses) needed. The drawback to using resistors in any of these designs is they are going to dissipate a LOT of power.

Watts= (Volts x Volts)/Ohms or W=E²/R

In any of the above designs there is 12V across each resistor; 12*12 = 144.

1st ckt. 144/2 =72W, 144/8=18W
2nd ckt. 144/1=144W, 144/1.75=82W
Suggested ckt. 144/2.7 = 53W

For resistors you need a safety factor of two in power rating, so double the above wattages to get the resistor design wattages. When you are talking about 100W to 300W resistors you are talking serious money; and they may require mounting on a heatsink... all that heat has to go somewhere! You could put them in the fan air stream and maybe get away without the factor-of-two safety factor. That would work until the fan fails or the airflow gets blocked.

If at all possible, it is cheaper to get a 24V fan, a 12V power supply, or a 24V to 12V convertor. 24V to 12V convertors are available, at least in the U.S., for use in large transport trucks so the truck drivers can use common 12V accessories with the truck 24V electrical system.

If a wall outlet from the local power company is available you could even use a computer power supply. They all have a 12V output, some as high as 40Amps. You just have to jumper a couple pins on their big output connector to get them to turn on.

Have fun! And let us know what works for you.
P.S. What is the fan being used for?

 

[davidwinth]

1st ckt. 144/2 =72W, 144/8=18W
2nd ckt. 144/1=144W, 144/1.75=82W
Suggested ckt. 144/2.7 = 53W

Thanks for your response, but I am confused about what all this means. Does "ckt" stand for circuit? If so:

• How come you get 18 W when the Hyperphysics guy gets 141 Watts for the first circuit?

I do not know who is correct between you guys, so can you explain what his numbers are if you are sure your calculations are correct?

Also, perhaps I am showing my ignorance but I took the positive lead from the power supply and put one of the 1k Ohm resistors on it then measured the voltage from there to ground. It was still at 24V. The resister had no effect. I thought voltage dropped across resistors like water pressure drops across obstacles in the flow. So:

• Will putting such a small resister as 2.7 Ohms really drop the voltage to 12 if a 1k Ohm resistor did nothing?

it is cheaper to get a 24V fan, a 12V power supply, or a 24V to 12V convertor.

I had no idea it would be so involved to simply step down the voltage! Wow. I do have a spare computer power supply somewhere, maybe I can try that. I have no idea where the "big output connector" (or even what that looks like) on a PC power supply is located, but I will try to find it.

What is the fan being used for?

I am using it to force air from a small duct to an even smaller tube. 4" down to 1/2". I am a mechanical engineer, and electronics seems like voodoo to me. I am never right about it - ever. Even for such a simple circuit, I feel like all I can do is guess.

 

[Tom.G]

OK, sorry I took so many shortcuts in the analysis!

Yes, 'ckt.' is abbreviation for 'circuit'.

How come you get 18 W when...

1st ckt. 144/2 =72W, 144/8=18W

I get 18W dissipated by the 8 Ohm resistor (From the two previous lines in Post #2.)
You have 12V across the 8 Ohm resistor, so square the 12 (12*12=144) and divide that by 8 (which is the resistance of R₂), yielding 18W.
I did the same thing with R₁, 144/2 Ohms = 72W.
If you add the power dissipated by of each of the elements, 18W_(R2) + 72W_(R1) + 52W_(motor) = 142Watts total. Which is within the roundoff error of the Hyperphysics answer. (Note that I used 12V across each element rather than the 12.202277V, etc that they used.)

I took the positive lead from the power supply and put one of the 1k Ohm resistors on it then measured the voltage from there to ground. It was still at 24V. The resister had no effect. I thought voltage dropped across resistors. So:

• Will putting such a small resister as 2.7 Ohms really drop the voltage to 12 if a 1k Ohm resistor did nothing?

Try it with TWO 1k Ohm resistors.
Connect one lead of a 1k resistor (call it R3) to +24V.
Connect one lead of another 1k resistor (call it R4) to supply ground.
Connect the free lead of R3 to the free lead of R4.

Turn on the supply and measure its output voltage, it should be 24V.
Now measure the voltage across ONE resistor, it doesn't matter which one, it should read 12V.
What you have here is a voltage divider. The voltage divides across the resistors in proportion to their resistances. Since the resistors are the same value there is the same voltage across each of them. (Well, the voltage may be slightly different, no two resistors are exactly identical.)

This is the way a 2.7 Ohm resistor would work in the fan ckt. You already determined the motor resistance to be 2.79 Ohms. I chose a standard value resistor closest to the motor resistance to end up with about half the supply voltage at the motor.

 

[davidwinth]

I get 18W dissipated by the 8 Ohm resistor (From the two previous lines in Post #2.)
You have 12V across the 8 Ohm resistor, so square the 12 (12*12=144) and divide that by 8 (which is the resistance of R2), yielding 18W.

Oh! Now I see. Hyperphysics guy was giving the total wattage summed over all components and you were calculating the wattage for each component separately. Now this makes sense. Thanks for the clarification.

Try it with TWO 1k Ohm resistors.
Connect one lead of a 1k resistor (call it R3) to +24V.
Connect one lead of another 1k resistor (call it R4) to supply ground.
Connect the free lead of R3 to the free lead of R4.

That is exactly what I did for my first attempt at reducing the voltage to 12V for this project.
I built a voltage divider just how you describe and hooked the fan up to it and got nothing from that 12V spot. Zilch. Nada. That is why I started to look for a way to figure this out more correctly.

But the reason I asked if a 2.7 Ohm resistor would really drop the voltage from 24V to 12V is because I know the 1k Ohm didn't drop the voltage whatsoever. You suggested a circuit like this:

(24+) ------ (2.7Ω) ------ (Fan+) ------ (Fan-) ------- (ground)

But when I did this with the 1k resistor (without the fan) I found the voltage between the end of the resistor and ground to be 24V as if the resistor was not even there. So this makes me wonder how a 2.7Ω resistor could do any different.

Is the reason you chose a value close to the fan resistance that you want half the voltage drop from 24+ to ground to occur across the resistor and the other half of the voltage drop to occur across the fan? I think that might make sense to me! But the problem is I have seen no voltage drop across even the 1k resistor even though I thought there would be one like you say there should.

Thanks again, Tom G.

 

[Tom.G]

By chance did your ckt with a single 1k resistor look like this?

(24+) ------ (1kΩ) ------ (Meter+) ------ (Meter-) ------- (ground)

If so, I would expect the meter to read 24V. What you have is again a voltage divider with the Meter being a very high resistance, probably 500k if it's an analog meter or several megohms if it's digital.

Is the reason you chose a value close to the fan resistance that you want half the voltage drop from 24+ to ground to occur across the resistor and the other half of the voltage drop to occur across the fan?

Exactly!

 

[davidwinth]

If so, I would expect the meter to read 24V. What you have is again a voltage divider with the Meter being a very high resistance, probably 500k if it's an analog meter or several megohms if it's digital.

O.k., I guess. Once again, electronics voodoo. I thought a voltmeter, you know, measured voltage. But not always, it seems. I have no idea how you guys ever get this stuff down! LOL.

Exactly!

Great! So I need a 2.7Ohm resistor that can handle 100W assuming a safety factor of about 2 for burnout. That should take care of the problem.

But I will still look for a PC power supply with 12V first, rather than spend $10 on a resistor. What about the current draw for the fan with a PC supply?

• Do I need a 100W and 12V power supply for this too?

Thank you for all your help!

 

[Tom.G]

I thought a voltmeter, you know, measured voltage.

It does. It reads the voltage across its terminals. As an experiment with some high value resistors, say 1Meg or higher, put some in series with the voltmeter like you did with the 1k resistor. If you can get enough resistance so the meter reads one half of the supply voltage then the total of the resistor values equals the input resistance of the meter. (Your old friend the voltage divider pops up again!)

(24+) ------ (1MΩ) ------ (1MΩ) ------ (1MΩ) ------ (Meter+) ------ (Meter-) ------- (ground)

Do I need a 100W and 12V power supply for this too?

That would be adequate, but I doubt you'll find one that small. I haven't seen one below 250 Watts.
EDIT: But they don't dissipate much power so you won't get a lot of heat out of them. :END EDIT

 

[davenn]

The fan is marked as 12 V and 4.3 Amps. It is a little duct fan.

are you REALLY sure the fan is 12V 4.3A that's a seriously powerful fan !
check your decimal point and see if it isn't really 0.43A ... the typical sort of rating of small ducted fans that are used in computers and other power supplies etc

A resistive voltage divider is a really, again I sue that word, really bad idea for controlling voltage, as it has a serious detrimental effect on the power supply's ability to supply enough current, as you have discovered ... yes you dropped the voltage but you have reduced the available current to a miserable trickle.

Please confirm the ratings of the fan ... a buck converter voltage regulator is much better suited for this job
2 good choices

1) if the fan is indeed 12V and 4.3A use a suitably rated buck regulator ... a few $$ ( less than $10) on eBay
2) if the fan is really less than 1A, say the 0.43A, then a standard linear regulator chip, eg a 7812, is ideally suited for the job

This is all info that should have been in the first couple of posts after your OP

cheers
Dave

 

[davenn]

Great! So I need a 2.7Ohm resistor that can handle 100W assuming a safety factor of about 2 for burnout. That should take care of the problem.

But I will still look for a PC power supply with 12V first, rather than spend $10 on a resistor. What about the current draw for the fan with a PC supply?

• Do I need a 100W and 12V power supply for this too?

It does. It reads the voltage across its terminals. As an experiment with some high value resistors, say 1Meg or higher, put some in series with the voltmeter like you did with the 1k resistor. If you can get enough resistance so the meter reads one half of the supply voltage then the total of the resistor values equals the input resistance of the meter. (Your old friend the voltage divider pops up again!)

(24+) ------ (1MΩ) ------ (1MΩ) ------ (1MΩ) ------ (Meter+) ------ (Meter-) ------- (ground)

That would be adequate, but I doubt you'll find one that small. I haven't seen one below 250 Watts.
EDIT: But they don't dissipate much power so you won't get a lot of heat out of them. :END EDIT

please guys ... let's get away from resistive dividers ... bad bad bad

read my last post Dave

 

[Tom.G]

Come on Dave, I took the OP at his word and answered his query. I didn't see a need to bend his requirements to something else.
However, if he did indeed slip a decimal then your approach would certainly be a better approach.

 

[davenn]

However, if he did indeed slip a decimal then your approach would certainly be a better approach.

no come on's about it ...
it's irrelevant for what ever the 2 possibilities of current needed are

and look where it got the 2 of you ... he's ended up asking if he needed a 100W PSU just so he can
burn of 3/4 of that in a high wattage resistor ... what a waste of power

voltage dividers are just really bad for anything with a significant load, you don't get any sort of stable voltage regulation at all and as I stated in my other post, you totally screw up the current supply capabilities ... as the OP discovered by his first post comments

there are MUCH better ways and switching or linear regulators are the best ways

do the job properly the first timeDave

 

[davidwinth]

A resistive voltage divider is a really, again I sue that word, really bad idea for controlling voltage, as it has a serious detrimental effect on the power supply's ability to supply enough current, as you have discovered

I only went to this because it was a circuit I remember from physics, and because I couldn't measure a voltage drop due to one resistor alone (as posted above) so I thought that tack wouldn't work.

are you REALLY sure the fan is 12V 4.3A that's a seriously powerful fan

Yep. http://www.attwoodmarine.com/store/product/store/turbo-4000-ii. As I mentioned, I need to run this to blow air through a 1/2" line reduced down from 4". The stamp on the side says 4.3 Amps, although the tech sheets on the website put it at 2.5 Amps for some reason. Either way, it should do the job if I can get the thing spinning! I (very) briefly touched it to the full 24V just to make sure it wasn't a dud after the failure of my voltage divider. It just about left the counter top.

So you suggest a linear regulator rather than using a 2.7Ω resistor? Is there some advantage over the resistor?

• How do I spec one of those? I have used basic circuit parts like resistors, capacitors, LEDs, but nothing like that!

Thanks.

 

[davenn]

Yep. http://www.attwoodmarine.com/store/product/store/turbo-4000-ii.

ohhh WOW

The stamp on the side says 4.3 Amps, although the tech sheets on the website put it at 2.5 Amps for some reason.

4.3 A is probably the max ... motors in particular require a higher start up current, once they reach optimum revs the current drawn settles down

So you suggest a linear regulator rather than using a 2.7 Amp resistor? Is there some advantage over the resistor?

yes ... well in this case with the higher current go for my option #1 -- a switching regulator ... they are VERY efficient compared to a linear reg.

you probably meant 2.7 Ohm resistor ... yes are better as losses will be worse in the resistor and it won't provide any sort of voltage regulation

How do I spec one of those? I have used basic circuit parts like resistors, capacitors, LEDs, but nothing like that!

let me have a look on eBay and see what I can find for you be back soon Dave

 

[davenn]

to give you an idea for what you are after ...

here's a 5A rated one
http://www.ebay.com/itm/Auto-DC-DC-Constant-Current-Boost-Buck-Converter-Voltage-Regulator-6-35V-to-1-35-/141723797418?hash=item20ff6587aa:g:qPAAAOSw3ydVraCS

I would prefer one rated around 6 - 7 A just to be certain that it can handle that start up current

another 5A one
http://www.ebay.com/itm/DC-24V-to-1...000729?hash=item2ecf55c699:g:LFAAAOxyiOxRyoYa

now a slight overkill but at least current wouldn't be a problem ... drive 2 fans and anything else you like ...
http://www.ebay.com/itm/Voltage-STE...085255?hash=item210b466bc7:g:YhYAAOSwP~tW3ZLgnow these are all eBay, but they are often available from various electronics shops ... don't know where you are in the world so don't know what to suggest
but they are not going to be as cheap as on eBay

Dave

 

[davenn]

ohh ohh ...
love this one ...

http://www.ebay.com/itm/DC-12V-48V-to-DC-5V-24v-15A-Adjustable-Buck-Module-Step-Down-Voltage-Regulator-/361439141933?hash=item542773a02d:g:vLMAAOSw8-tWWs6F

as you can see, they come as a complete module ... no messing around ... you couldn't build one that cheap by buying individual parts

you put your 24V in the input, stick a multimeter on the output and set the adjustment for 12V
then wire in the fan

D

 

[rbelli1]

set the adjustment for 12V
then wire in the fan

An adjustable one is good. You can turn the voltage down to reduce the power requirements if the maximum air handling is not needed in your application.

I second (or third or fourth etc as the case may be) using a separate power supply for the fan. 25-50 extra watts of extra heat and power supply is not something you want to have around or pay for.

BoB