[buscht]

I bought a 28VDC 15amp power supply for my Xylotex board. I have a 24VDC fan to cool the Xylotex board.

Will power supply burn up the fan? Is there a way to limit the voltage going to the fan? Or should I just find a 28VDC fan?

Please answer as if you are talking to a five year old. Anything doing with electronics is over my head.

Thanks
Trent

[ViperTX]

Limit the current to the fan.

I suspect that if you view the limits of the supply it's probably 24 VCD +-10 percent.

Other alternative is to let it run and see if it does die.....

[MrBean]

You could series a resistor with the fan to limit current. Not sure what value though. So I guess that's not so much help.

Regards Terry.....

[OCNC]

You could series a resistor with the fan to limit current. Not sure what value though. So I guess that's not so much help.

Regards Terry.....

Using the current rating on the fan you can calculate the value of the resistor. You know you need to drop 4v (28v-24v). R=E/I or R= 4/Ifan . The power dissipated in the resistor will be P=(I^2)*R in watts. So if the 24v fan was rated at 1.4a the resistor would be ~3 ohms. The power lost to heat would be 6 watts which would be the minimum rating for the resistor. If the fan was scrounged I would be inclined to see if it ran at 28v.


Also you're limiting (reducing) voltage not current. With a series resistor the fan will still draw it's rated current.


Chris

[buscht]

Thanks for the replys.
Chris, I understand that I could add a resistor in series to limit the voltage. I don't understand your formulas at all. (other than the 4v drop).

R=E/I or R= 4/Ifan . The power dissipated in the resistor will be P=(I^2)*R in watts. So if the 24v fan was rated at 1.4a the resistor would be ~3 ohms. The power lost to heat would be 6 watts which would be the minimum rating for the resistor.

The fan in question is rated for 24VDC .08a. I don't know an ohm from Adam.
or what is R, E, or I?

How about a simple tutorial?

I guess in your example, I'd want a 3ohm resistor rated for 6 watts minimum. The watts is so the resistor doesn't get overloaded and the ohms give me the correct current limiting.

But in my linear mind, does this mean that the 24 volts is irrelevant, only the 4 volt drop is important (along with the total amps)?

Actually, I'm more interested in how to calculate this stuff myself. I can afford to burn up a $3.95 fan to experiment.

Thanks
Trent

[MrBean]

Also you're limiting (reducing) voltage not current. With a series resistor the fan will still draw it's rated current.


Chris

That always gets me.

Series a resistor with a stepper motor and it limits the current. But....
Series a resistor with a fan (still a motor) and it's limiting voltage.

How does a resistor know which one to limit. voltage or current. Perhaps they are intelligent.

Could someone explain this further. Thanks.

Regards Terry.....

[Al_The_Man]

Most of these type of fans have a fair tolerance so If it were mine I would run at 28v.
All resistors basically have the same effect, they limit current, which in turn causes a volt drop or potential to appear across the resistance. Saying a 'current limiting resistor' and 'voltage reducing resistor' is like expressing the same thing in a different way.
Actually using a resistor to drop voltage is very wastfull as you can see by the wattage (heat) dissipation of the resistor.
Al.

[Al_The_Man]

The fan in question is rated for 24VDC .08a. I don't know an ohm from Adam.
or what is R, E, or I?
How about a simple tutorial?
thanks
Trent

Trent, R,E & I stand for Resistance, EMF (voltage) and Current and are typically used in 'ohms Law' which in a DC circuit is express by I=E/R E=I*R R=E/I .
and watt are expressed W=IxE.

Series a resistor with a stepper motor and it limits the current. But....
Series a resistor with a fan (still a motor) and it's limiting voltage.

How does a resistor know which one to limit. voltage or current. Perhaps they are intelligent.

They are actually achieving the same thing, but you have two very different loads, so an understanding of the loads themselves helps explain it.
With the (dc) fan motor all you want to achieve is to run it at the proper voltage so the load is known and is constant, so it is easy to calculate a way of dropping the voltage, if neccessary by ohms law.
With the stepper motor the load is very different, firstly the motor speed is never constant. So why do you have a motor that is rated at 3.5v we will say and build a 24vdc power supply for and wire-in dropping resistors?
The stepper motor has high torque when stationary and is supplied with its rated voltage of 3.5v at this it is drawing its rated current, so if we have a supply of 24v, we have to limit the current (drop the voltage across a resistor) as the speed of the stepper increases the load (stepper) is seen as an inductance with a progressively increasing impedance ( in ohms), if the supply were kept at 3.5v this increasing impedance will no longer allow the optimum current through the motor, but if we have a larger supply with a resistor in series with the motor, as the impedance increases the current through the resistor will drop (decreasing the volt drop across it) and allow a much higher voltage across the motor winding, and thus maintaining torque as speed increases.
This is more of a dynamic load because motor characteristics change with rpm
Al

[ViperTX]

That always gets me.

Series a resistor with a stepper motor and it limits the current. But....
Series a resistor with a fan (still a motor) and it's limiting voltage.

How does a resistor know which one to limit. voltage or current. Perhaps they are intelligent.

Could someone explain this further. Thanks.

Regards Terry.....

Terry, it just a matter of semantics. They are both the same.....draw a circuit with a 12 VDC battery and a 1 ohm resistor (represents the resistance of a some unknown motor coil/phase). Use ohm's law to calculate the current...I=V/R what do you get 12 amps...now add another 1 ohm resistor in series (call this your current limiting resistor or voltage dropping resistor)...now calculate the current I=12/2 = 6 amps.....so you have 6amps flowing through the circuit....the voltage drop at each resistor is Vdrop = RI and that is 6 VDC.....so in this simple circuit the Sum of the Voltage Sources = Sum of the Voltage drops. I hope that helps.

Paul

[MrBean]

Thanks guy's. That's cleared that one up, for me at least.

It's kind of like how a flask knows wether to keep something hot, or cold. One day it could all go wrong. You put in some nice cold drink for a day at the beach, only to find the flask thought it should be keeping it hot. Time for a new flask I think.

Regards Terry.....

[OCNC]

Thanks for the replys.
Chris, I understand that I could add a resistor in series to limit the voltage. I don't understand your formulas at all. (other than the 4v drop).

R=E/I or R= 4/Ifan . The power dissipated in the resistor will be P=(I^2)*R in watts. So if the 24v fan was rated at 1.4a the resistor would be ~3 ohms. The power lost to heat would be 6 watts which would be the minimum rating for the resistor.

The fan in question is rated for 24VDC .08a. I don't know an ohm from Adam.
or what is R, E, or I?

How about a simple tutorial?

I guess in your example, I'd want a 3ohm resistor rated for 6 watts minimum. The watts is so the resistor doesn't get overloaded and the ohms give me the correct current limiting.

But in my linear mind, does this mean that the 24 volts is irrelevant, only the 4 volt drop is important (along with the total amps)?

Actually, I'm more interested in how to calculate this stuff myself. I can afford to burn up a $3.95 fan to experiment.

Thanks
Trent

Ohms Law: E= I*R : E is voltage in volts; I is current in amps; R is resistance in ohms.

In a simple DC series circuit the voltage goes from the high side of the source value (in your case 28v) to zero (the low side of the source). Along the way the voltage drops as it passes across each component until it reaches the low side of the source. Initially in your case the only component in the circuit is the 24v rated fan. Since the source is at 28v the fan (if it remains the only component in the circuit) will have to drop the full 28v. The idea then is to insert a component in the circuit ahead of the fan that will drop 4v. This will leave the fan looking at 24v instead of 28v. The fan will then drop it's rated 24v and we'll be back at the low side of the source with 0v. The sum of the various voltage drops in a simple series circuit always needs to equal the initial high value of the source voltage. The current in a simple series circuit is the same in each component regardless of the value of the voltage drop across the component(this is an electrical fundamental- in a simple series circuit you can have different voltage values across different components but you always have the same current value through each component ). We know from the fan specs that the current in the fan is .08a at 24v. We also know that this is a simple series circuit and we can therefore conclude that the current in the 4v dropping resistor which we intend to add to the circuit will also be .08a. Then using Ohms Law E=I*R we can caluate the resistance of a particular resistor that will drop 4v when .08a of current is flowing through it. E=I*R : 4v =.08a* R : R= 4v/.08a : R= 50 ohms.
Proceeding from here you need to calculate the power rating for this resistor. We're passing .08a of current through 50 ohms of resistance. Using P=(I^2)*R. P= (.08^2)*50 : P=.32 watts.
Therefore the power rating of the 50 ohm dropping resistor (dropping in this case 28v to 24v) should be equal to or greater than .32 W. The closest standard resistor rating that qualifies is a 1/2W 50 ohm resistor. This should be readily available and is not a huge power drain.

Did any brain cells light up? :cheers:

Chris

[buscht]

Chris, thank you very much. I'll print this out and put it on my reference wall.
Trent