I need to power a small 20mm mini 5v fan. I have a 5v mean well power supply but I want to use a resistor to lower the voltage to 2.80v. I tested the fan at 2.80 volts (with a vaiable power supply) and it was the quietest while giving enough air. Is it possible to just put a resistor before the fan to lower the voltage to 2.80v? Everything online talks about using a resistor for LED's. I don't get it, can you only use a resistor with constant current?

My power source is 5v 6A
My mini fan is rated at 5v .150 but I want it to run at 2.80v (Its draws around 50mA at that voltage according to my variable ps)

 

3 diodes in series with the fan or a 3 terminal regulator with a divider on ground pin.

 

3 diodes in series with the fan

That's what I would try first as each diode will give you about a 0.7 Voltage drop so you get about 5V - 2.1V = 2.9 Volts, if too slow then 2 diodes. Any diode like a 1N4002 should work. Just make sure the fan starts at the lower voltage.

Ron

 

3 diodes in series with the fan or a 3 terminal regulator with a divider on ground pin.

The series diode approach is what I would go with. It's simple, efficient and works fine. Each diode has a voltage drop (typically .6 V, but depends so look at the data sheet) so include as many as needed for your desired voltage.

 

I need to power a small 20mm mini 5v fan. I have a 5v mean well power supply but I want to use a resistor to lower the voltage to 2.80v. I tested the fan at 2.80 volts (with a vaiable power supply) and it was the quietest while giving enough air. Is it possible to just put a resistor before the fan to lower the voltage to 2.80v? Everything online talks about using a resistor for LED's. I don't get it, can you only use a resistor with constant current?

My power source is 5v 6A
My mini fan is rated at 5v .150 but I want it to run at 2.80v (Its draws around 50mA at that voltage according to my variable ps)

Okay so here is the scenario you have a supply voltage of 5V 6A using which you have to drive a load(fan) of 2.8V 50ma.

One crude way of doing it is by using two resistors to form a potential divider that could output 2.8V the other is the proper way of using voltage regulators to provide the required voltage to your fan. I will explain both but the voltage regulator method is recommended.

Using Potential Divider:
A Potential/voltage divider circuit uses two resistors R1 and R2 to convert one voltage range to another. The value of resistors can be calculated using a simple formula. You can use the below link to calculate your resistor values and to know more about potential dividers. I have calculated the values for you below

How to uses and calculate Potential Dividers


So you can use a 4.7K and a 6K resistor to get 2.8V from a 5V supply. Use the link above to know how to build the circuit and stuff. Even-though your fan is consuming only 50mA try using a 1W resistor just to be on the safe side.


Using Voltage Regulator LM317:
The best way to go with this is to use a voltage regulator like LM317. This particular regulator can be used to st output voltage anywhere from 1.25V to 37V and they can power load upto 1.5A so running you fan should be a piece of cake. Again the output voltage of the IC can be set by using two resistors of right values. You can Learn to use LM317 from the link given. Also use an Online LM317 Calculator like the one linked to know the values of your resistors and built your circuit.

Hope this helps you!!!

 

I would vote for the series diode trick. It is easy and cheap.

 

Using Potential Divider:
A Potential/voltage divider circuit uses two resistors R1 and R2 to convert one voltage range to another. The value of resistors can be calculated using a simple formula. You can use the below link to calculate your resistor values and to know more about potential dividers. I have calculated the values for you below

How to uses and calculate Potential Dividers


So you can use a 4.7K and a 6K resistor to get 2.8V from a 5V supply. Use the link above to know how to build the circuit and stuff. Even-though your fan is consuming only 50mA try using a 1W resistor just to be on the safe side.

I don't see a voltage divider working for this application. Figure with 2.8 Volts applied to the motor the current draw was about 50 mA lending me to believe the DC resistance of the motor is about 56 Ohms. Using a divider with the numbers presented the motor will be in parallel with R2 of the divider making the Rtotal of R2 and the parallel motor something less than 56 Ohms. With R1 having a resistance of 4.7 K Ohm the max current, as limited by R1 would be a little over 1 mA. The combination won't work. Placing about a 50 Ohm pot in series with the motor would likely work and adjust the pot for desired fan speed would work but the diodes are the easy and quick solution.

Ron

 

Resistors really only work for constant current flow. You could get away with a resistor as the fan current should be pretty constant when up to speed but the diodes will work well and give you a more defined output voltage.

 

I need to power a small 20mm mini 5v fan. I have a 5v mean well power supply but I want to use a resistor to lower the voltage to 2.80v. I tested the fan at 2.80 volts (with a vaiable power supply) and it was the quietest while giving enough air. Is it possible to just put a resistor before the fan to lower the voltage to 2.80v? Everything online talks about using a resistor for LED's. I don't get it, can you only use a resistor with constant current?

My power source is 5v 6A
My mini fan is rated at 5v .150 but I want it to run at 2.80v (Its draws around 50mA at that voltage according to my variable ps)

You can use the Series Diodes method, or use a Lm317t as a Constant Current source varying the current from 50 to 150mA ,
Or use it as a Constant Voltage varying from 2.5 to 5V .

 

I don't see a voltage divider working for this application. Figure with 2.8 Volts applied to the motor the current draw was about 50 mA lending me to believe the DC resistance of the motor is about 56 Ohms. Using a divider with the numbers presented the motor will be in parallel with R2 of the divider making the Rtotal of R2 and the parallel motor something less than 56 Ohms. With R1 having a resistance of 4.7 K Ohm the max current, as limited by R1 would be a little over 1 mA. The combination won't work. Placing about a 50 Ohm pot in series with the motor would likely work and adjust the pot for desired fan speed would work but the diodes are the easy and quick solution.

Ron

There is no need to use a voltage divider, none at all. One series resistor dropping about 2 volts at 50mA is needed, the nearest standard value to 40 ohms is 39 ohms, (orange-white-black color code)

 

There is no need to use a voltage divider, none at all. One series resistor dropping about 2 volts at 50mA is needed, the nearest standard value to 40 ohms is 39 ohms, (orange-white-black color code)

That's pretty much what I mentioned and why a voltage divider, as suggested would not work. Why did you quote me?

Ron

 

That's what I would try first as each diode will give you about a 0.7 Voltage drop so you get about 5V - 2.1V = 2.9 Volts, if too slow then 2 diodes. Any diode like a 1N4002 should work. Just make sure the fan starts at the lower voltage.

That, the 0.6V drop is what I've always seen, but never used, until now. And what a shock, when reading data sheets. Can't find any listing that number, 0.6V as the "volts forward", which I've also understood as meaning "voltage drop". Where am I going wrong in this? Not trying to hijack the thread but would like a response since it may help the TS.

 

Where am I going wrong in this?

You're not. Normally the forward voltage drop, when a diode is forward biased is about between 0.6 V and 0.7 V for a silicon diode, a Germanium diode comes in lower at around 0.3 V. This is where words like approximately and about come in handy. The actual drop will also be a function of the current flowing through the diode and then we have diodes like Schottky diodes where forward voltage drops of 0.7 to 1.0 Volt are not uncommon. Not a very exacting number. I also have no clue why I wrote Slow verse Low

Ron

 

I say use a 5k potentiometer. While the diode thing has its advantages, you will want the ability to adjust it easily.

 

http://www.onsemi.com/pub/Collateral/1N4001-D.PDF
details the characteristics of a common family of rectifier diodes

Schottky diode forward drop is generally lower than for silicon pn junction diodes.
Here is a datasheet for a family of 1 amp schottky diode
https://www.fairchildsemi.com/datasheets/1N/1N5819.pdf
With schottky diodes there is usually a forward voltage penalty at higher voltages. Silicon shottkys above 100 PIV exist but are rare.

 

A single resistor in series with the fan will work just fine. Some notes:

The current a fan draws is not directly proportional to the voltage across its terminals. For example, a 5 V, 100 mA fan running at 2.5 V (half the normal voltage) will draw more than half its normal current. You can start with Ohm's Law to determine the resistor value for 2.8 V operation, but don't be surprised if you have to decrease the resistance value somewhat to get the operating voltage you want.

Power dissipation is important. Calculate the power dissipated in the resistor using either Watt's Law or Joule's Law. Once you have the calculated value, double it. If the resistor you want is not available in that wattage, get the next higher rated one.

ak

 

@Reloadron , @ebp

Ebp, those are the same data sheets I was looking at.

Ron, so to get the 0.6 voltage drop I just need to put a resistor in series to limit the current through the diode thus giving the correct voltage drop I'm looking for? The diode will only be used as a reference voltage for a window comparator. I was going to use a 1N4002 since I have them but when I looked at the data sheet I was shocked they showed a 1.0 voltage drop at 1amp.

 

Is it possible to just put a resistor before the fan to lower the voltage to 2.80v?

Yes!

 

@Reloadron , @ebp

Ebp, those are the same data sheets I was looking at.

Ron, so to get the 0.6 voltage drop I just need to put a resistor in series to limit the current through the diode thus giving the correct voltage drop I'm looking for? The diode will only be used as a reference voltage for a window comparator. I was going to use a 1N4002 since I have them but when I looked at the data sheet I was shocked they showed a 1.0 voltage drop at 1amp.

While I really don't want to get the thread off topic, here is what I suggest. AAC Forums has a very good write up as Introduction to Diodes And Rectifiers. Give it a read and focus on the typical forward and reverse diode graph. Diode curve: showing knee at 0.7 V forward bias for Si, and reverse breakdown.is a good portrayal of what's going on. Notice how as the forward current increases the forward voltage drop increases. Additionally temperature plays a role making a diode not the best voltage reference.

Ron

 

I say use a 5k potentiometer. While the diode thing has its advantages, you will want the ability to adjust it easily.

Why 5K Ohm? There is a 5 volt supply and we want about 2.8 volts applied to the fan. With 2.8 volts applied the fan current is about 50 mA (0.050 Amp). That leads me to believe the fan motor DC resistance is about 56 Ohms. While a 5K potentiometer will give us between 0 and 5,000 Ohms with the pot at 57 Ohms out of 5,000 Ohms the motor voltage would be about 2.5 volts. The current will be about 44 mA. and with the pot at 100 Ohms out of 5,000 Ohms the current, with 5 volts applied, is now down to 32 mA. and motor voltage down to about 1.8 volts. So all things considered, if using a pot wouldn't something like maybe a 100 Ohm pot be in order?

Ron