Powering a diy case fan

[sushee]

Hi forum! I have a problem with a PC case I'm working on. I want a fan going at 30 rpm on the outside of the pc front, with a hole going to the inside of the pc. The whole thing is complete, except for one thing. The motor is 6v. What's the best way to power it from inside the pc?

[iamdarkyoshi]

It'd probably run fine on 5V, just connect to red and black on a molex connector. Red is +5v, and black is ground/0v

[sushee]

3 hours ago, iamdarkyoshi said:

It'd probably run fine on 5V, just connect to red and black on a molex connector. Red is +5v, and black is ground/0v

It won't run on 5V. Tested with an arduino uno and the motor doesn't do anything. it however had a little plastic lid that came off. Is it broken?

[Aelita Sophie]

4 minutes ago, sushee said:

It won't run on 5V. Tested with an arduino uno and the motor doesn't do anything. it however had a little plastic lid that came off. Is it broken?

Still try putting it on a Molex connector, as said before. As it probably WILL work. The arduino just simply can't give enough current on 5v for the fan to start spinning. As the initial first few rotations take a lot more energy then when its already turning.

[sushee]

3 minutes ago, Aelita Sophie said:

Still try putting it on a Molex connector, as said before. As it probably WILL work. The arduino just simply can't give enough current on 5v for the fan to start spinning. As the initial first few rotations take a lot more energy then when its already turning.

Still nothing

[Aelita Sophie]

1 minute ago, sushee said:

Still nothing

Then you'll be needing a boost buck converter of some sorts. https://www.banggood.com/DC-DC-3V-35V-To-4V-40V-Adjustable-Step-Up-Power-Module-3V-5V-12V-To-19V-24V-30V-36V-Boost-Converter-p-1166673.html?rmmds=search

Hook that up to the 5v rail of your molex connector. Then simply adjust the voltage (by using a voltmeter) to 6v. Though I still highly doubt that 1 volt makes a difference.

[sushee]

Could I simply not hook up the 12v molex wire to a resistor?

[sushee]

14 minutes ago, Aelita Sophie said:

Then you'll be needing a boost buck converter of some sorts. https://www.banggood.com/DC-DC-3V-35V-To-4V-40V-Adjustable-Step-Up-Power-Module-3V-5V-12V-To-19V-24V-30V-36V-Boost-Converter-p-1166673.html?rmmds=search

Hook that up to the 5v rail of your molex connector. Then simply adjust the voltage (by using a voltmeter) to 6v. Though I still highly doubt that 1 volt makes a difference.

 

Could I simply not hook up the 12v molex wire to a resistor?

[Aelita Sophie]

1 minute ago, sushee said:

 

Could I simply not hook up the 12v molex wire to a resistor?

That isnt that easy. You'll need to calculate how much your fan will initially use and how your PSU catches voltage drops. It wil also emit a lot of heat compared to a voltage regulator. It's simply just a lot easier to use a voltage regulator like I linked you. 

[mariushm]

You could use an adjustable linear regulator, like LM317 for example ... two resistors will set the output voltage to 6v , and you can power the linear regulator from 12v.

 

See picture below, but use 100 ohm or 120 ohm for R2 and adjust R1 according to the formula.

C1 would be a decoupling capacitor, usually it's ceramic. But most likely the circuit will work just fine without it, if you can't buy this part from same place you buy the other stuff.

C2 can also be a different value, doesn't really matter as long as it's 10uF or higher. 100uF, 220uF are common values. Whatever you find cheap, won't make any difference. Voltage rating of capacitor should be higher than output voltage, but for safety just use 16v rated or higher (as it's higher than power supply's voltage, 12v)

 

[Unimportant]

@sushee

You can hook it up on a molex between 5V and 12V (red and yellow) for 7V operation.

[Aelita Sophie]

7 hours ago, Unimportant said:

@sushee

You can hook it up on a molex between 5V and 12V (red and yellow) for 7V operation.

You are joking right? That would break the PSU. Or atleast turn it off. Try it out yourself. Doesnt work that way. You'll be needing additional circuitry to get that sorted out. By then it's easier just to use the module I linked for $2.

[mariushm]

@Aelita Sophie

@Unimportant

 

It will work in most power supplies as long as the components pull a significant amount of current from 5v rail but it's really not recommended.

 

If the fan dies shorted or somehow gets blocked, the power supply can be damaged by the current sent back into it through the 5v wires.

 

It's bad to do this with power supplies which have independent dc-dc converters.

 

-

 

Back to the topic, another option would be to use DIODES in series instead of resistors to reduce the voltage. A basic diode like 1n400* will work ... 1n4001 to 1n4007, the difference is just the maximum voltage they can tolerate, where 1n4001 has a maxim voltage of 50v  and 1n4007 can do up to 1000v,

 

Diodes have a forward voltage which varies depending on diode type/model/series and the amount of current going through it and how hot the diode gets. But in general, the cheap generic diodes like 1n400* will have a forward voltage of 0.7v to 1v when the amount of current typical for a motor (50-100mA or more)

 

Forward voltage = you get a voltage at the input of the diode, you get a voltage lower by this amount at the output of the diode (ex 12v in , 11.2v out for a 0.8v forward voltage)

 

See http://www.vishay.com/docs/88503/1n4001.pdf

 

So as you can see, at around 0.1 A of current going through the diode, the forward voltage will be a bit higher than 0.75v and at 0.2A it's almost 0.8v

 

With this in mind, you could simply connect 8 of these diodes in series and then you'll have 12v - (8 diodes x 0.7v..0.8v ) = 12v - 5.6v .. 6.4v = 5.6v .. 6.4v

You can start with 9 diodes to make sure the voltage will never be higher than 6v and measure the voltage with a multimeter while the motor spins and after around 20-30 minutes (to give diodes the time to warm up to their normal operating temperature and get the forward voltage stabilized). Then if you feel the voltage is too low, you can remove one diode from the series and get a higher voltage.

Diodes are super cheap, and 8-9 of them will be cheaper than a linear regular but you lose the ability the adjust the output voltage using a simple resistor or a simple potentiometer installed in the place of that resistor.

 

-

 

Another option you could use would be to use a single zener diodes.  Zener diodes are a special kind of diodes which only let voltage go through when the voltage reaches a certain threshold. So for example, you can use a very generic easy to find 5.6v zener diode and a capacitor after it to smooth out the voltage would be enough to have around 5.6v at the other end.

5.6v is very common, 6.2v should be fairly common as well ... in between there's fewer choices. For example for 5.6v or 6.2v a store like Digikey has around 1300 different diodes for each, but for values in between only around 10-50 different diodes and not all in stock

 

[Unimportant]

5 hours ago, mariushm said:

 

If the fan dies shorted or somehow gets blocked, the power supply can be damaged by the current sent back into it through the 5v wires

It's bad to do this with power supplies which have independent dc-dc converters.

 

@Aelita Sophie

@mariushm

 

On cheaper PSU's, without independent DC-DC converters, it works fine for small loads, such as a fan. Some cheap fan controllers with a LOW-MED-HIGH setting even work this way. It's not great but for a single fan it's fine. These cheap PSU's have parallel resistors on their outputs (cheap way to prevent a no-load situation, which is bad for a SMPS) which will bleed off enough current on the 5V rail for such a small load. Should the fan fail as a short circuit the PSU's OVP on the 5V line will simply trip.

 

A quality PSU with independent DC-DC controllers would actually be better because a synchronous buck can sink current, which makes the problem go away completely. In fact that's how a BTL (bridge tied load) class D amplifier works.

Should the fan die shorted on one of these PSU's the OCP will trip as on a normal overload.

[Unimportant]

5 hours ago, mariushm said:

Another option you could use would be to use a single zener diodes.  Zener diodes are a special kind of diodes which only let voltage go through when the voltage reaches a certain threshold. So for example, you can use a very generic easy to find 5.6v zener diode and a capacitor after it to smooth out the voltage would be enough to have around 5.6v at the other end.

5.6v is very common, 6.2v should be fairly common as well ... in between there's fewer choices. For example for 5.6v or 6.2v a store like Digikey has around 1300 different diodes for each, but for values in between only around 10-50 different diodes and not all in stock

 

You'll need to add a current limiting resistor before the zener, otherwise it'll simply pop.

It would have to be a smallish resistance value, because it'll also be in series with the load, and because of that, you'll also need a beefy high current zener.

Typically you'd want to use a larger resistance value with a normal small current zener and then buffer the output, with a emitter follower for example...

 

Perhaps you meant to put the zener simply in series with the load (reverse biased)?

[mariushm]

I actually didn't think about the diode being in series but you make a good point, that would also work.

 

I was assuming the motor is a small load, below 0.1A, especially if he wants to run the fan at 30 rpm ... unless he uses some weird gears to reduce the motor speed down, the motor shouldn't use a lot of current... so i was thinking of a 1w Zener diode - very common - should be adequate.

Yeah, the motor could pull some current when it turns on but the diode should have a fair amount of thermal mass and handle such small duration bursts of current.

 

[Unimportant]

31 minutes ago, mariushm said:

I was assuming the motor is a small load, below 0.1A, especially if he wants to run the fan at 30 rpm ... unless he uses some weird gears to reduce the motor speed down, the motor shouldn't use a lot of current... so i was thinking of a 1w Zener diode - very common - should be adequate.

Yeah, the motor could pull some current when it turns on but the diode should have a fair amount of thermal mass and handle such small duration bursts of current.

 

I meant that without a current limiting resistor before the zener, this is just a (12V- Vz) dead short trough the zener diode and it will burn, even with no load connected.

 

As per this image: Rs is required.

 

[mariushm]

Ah yes, i forgot about that.

I quickly made that picture by taking a screenshot of a PDF page from a Microchip app note and then quickly removing the extras from the circuit .. but forgot about the resistor.

 

See page 2 (capacitive transformless power supply) : http://ww1.microchip.com/downloads/en/AppNotes/00954A.pdf

You're right, a resistor is indeed needed.