Sparky Advice- Re- Fridge Fan Dc Adapter

Australia & New Zealand Homebrewing Forum

Help Support Australia & New Zealand Homebrewing Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

Barley Belly

Head Brewer - Barley Belly Brewery
Joined
25/5/08
Messages
710
Reaction score
4
I've just bought a fermenting fridge and I was wanting to install a small PC fan for air circulation.

The fan I'm looking at is 12V 0.1amp

I have a collection of old DC 240v adapters from old phones, modems etc and was wanting to use one of those.
The prob is I don't have a 12V one, but I do have a 9V or 15V.

My question is:- Should I use the 9V or 15V to run the fan adequately?
 
I've just bought a fermenting fridge and I was wanting to install a small PC fan for air circulation.

The fan I'm looking at is 12V 0.1amp

I have a collection of old DC 240v adapters from old phones, modems etc and was wanting to use one of those.
The prob is I don't have a 12V one, but I do have a 9V or 15V.

My question is:- Should I use the 9V or 15V to run the fan adequately?

Then there is AC and DC.
 
I would use the 9V. It should stil be enough to drive the fan. It will just run a bit slower. The 15V one would drive the fan faster and it wont last as long.

Kabooby :)
 
If you drop your voltage then your current will increase, which could burn the fan out over time. Plug packs are under $20 these days, just buy a new one.
 
If you drop your voltage then your current will increase, which could burn the fan out over time.

That's completely wrong. I know fans aren't ohmic, but humour me for a moment: V=IR; R=V/I; I=V/R.

A 100mA fan at 12V has an "effective resistance" (impedance) of 12/0.1 = 120Ω.

If you assume the impedance of the fan doesn't change (it does because the fan has an inductor in it, but this is a reasonable approximation), then what happens when we drop the voltage to 9V? I = 9 / 120, which is 75mA.

When thinking of these voltage, current and resistance, try using the water analogy: Voltage is how much water pressure there is in the pipe. Current is the speed of the water. Resistance is the friction from the "thinness" of the pipe. If you reduce the water pressure in a given water pipe, then the speed of the liquid is obviously going to fall.

The disadvantage of dropping the voltage, as kabooby points out, is that you will reduce the power of the fan. P = IV, or V/R. Again, pretending that the impedance is constant, you will drop the power from 12/120 = 1.2W to 9/120 = 675mW. That means you'll only get about half the power out of the thing. That should be fine.
 
Yep as already said, 9v will be ok but it will run slower. It will still be 100% better than no fan. As MartinS calc'd ensure it is at least 75mA.
 
I was trawling a Red Cross shop yesterday and they had a box of what must have been 100 of these old DC 240v adapters,$1.00 each.Perhaps try you local op shop.

Mine runs 24/7 as I want an even temperature throughout the freezer all the time,after all this is the whole idea of the exercise.

Batz
 
That's completely wrong. I know fans aren't ohmic, but humour me for a moment: V=IR; R=V/I; I=V/R.

A 100mA fan at 12V has an "effective resistance" (impedance) of 12/0.1 = 120Ω.

If you assume the impedance of the fan doesn't change (it does because the fan has an inductor in it, but this is a reasonable approximation), then what happens when we drop the voltage to 9V? I = 9 / 120, which is 75mA.

When thinking of these voltage, current and resistance, try using the water analogy: Voltage is how much water pressure there is in the pipe. Current is the speed of the water. Resistance is the friction from the "thinness" of the pipe. If you reduce the water pressure in a given water pipe, then the speed of the liquid is obviously going to fall.

The disadvantage of dropping the voltage, as kabooby points out, is that you will reduce the power of the fan. P = IV, or V/R. Again, pretending that the impedance is constant, you will drop the power from 12/120 = 1.2W to 9/120 = 675mW. That means you'll only get about half the power out of the thing. That should be fine.

I'm not an expert on small DC fans, but, the speed of the fan is inversely proportional to its torque, the torque is proportional to its power, the power is proportional to its current. ie if the speed drops, the torque required to drive the motor increases, along with the power and current.
All that being said, the fans will probably work fine, probably. :blink:
 
It's best not to run them at a higher than rated voltage. I bought a 12v fan the other day to put in the chesty, but connected it up to a 15v power pack and after running it for some time, the fan was very warm and smelled. Won't be doing that again.

I'll be getting one of these from Jaycar
 
It's best not to run them at a higher than rated voltage. I bought a 12v fan the other day to put in the chesty, but connected it up to a 15v power pack and after running it for some time, the fan was very warm and smelled. Won't be doing that again.

I'll be getting one of these from Jaycar

+++1 ;)
Or look around the house for used toy electric toy powers adaptors.
(AC/ DC Output 12V 100mA)
9V will work and will not burn the fan out but as previous stated will not give it full power.

don't go over 500 mA !! for longevity I think :huh:
 
Back in the day, the nerds with noisy nerd boxes used to modify the fan connectors, so that instead of connecting to 0 and 12V they connected to 5 and 12V (i.e. some people ran 12V fans off 7V) It definitely makes the fan go slower and reduces the noise (not that that matters for a fan in a fridge). This was of course before it was common for computers to adjust their fan speed automatically based on their temperature.
 
I'm not an expert on small DC fans

Cool - probably best to start with that when giving advice. It can be hard to tell on forums who knows stuff, and who's just making stuff up or badly remembering things they heard somewhere once.

the speed of the fan is inversely proportional to its torque, the torque is proportional to its power, the power is proportional to its current. ie if the speed drops, the torque required to drive the motor increases, along with the power and current.

I think you're getting confused between torque supplied, and torque required.

All that being said, the fans will probably work fine, probably. :blink:

Not just maybe - it will work. Voltage regalation is the most common way to control the speed of these fans. It's not rocket science, and dropping the supply voltage is not going to cause any harm.
 
All you need to do is get an old computer power supply and an old computer fan or two or three.
Hook them up like that.
Old computers are like armpits, everybody has em.
Grab one and strip it down for parts....Too easy
 
I Got bored this morning so I decided to run a little experiment (I will be making a beer later, so no one panic). I hooked up a 12v fan with a variable plug pack and got the following readings (note that this will vary from fan to fan).

Volts: Current : Locked rotor current
12v : 40.7mA : 76.5mA
9.07v : 31.3mA : 58.1mA
6.05v : 23.29mA : 39.3mA
5.08v : 20.78mA : 33.37mA
4.5v : 30mA : 30ma (fan barely moving)
3.2v : 5.35mA (fan not moving)

I was actually surprised at just how linear it is down to 5v.

So it would appear that as long as the fan is spinning free and smoothly then it will be fine. (as has been said by others).
 
I was actually surprised at just how linear it is down to 5v.

This is a little off-topic, but it's interesting to calculate the resistance for each of your test-points:

V: Rv/Ro
12.0v: 295Ω/157Ω
9.07v: 290Ω/156Ω
6.05v: 259Ω/154Ω
5.08v: 244Ω/152Ω
4.50v: 150Ω/150Ω

I suspect your 3.2v reading is actually a bad reading caused by the way the fan switches the electromagnets on and off, and that if you took the reading again, but moved the fan blades to a slightly different location then hold it steady, you'd still get a 150Ω reading.

Internally, you'd expect the fan to look like an RL circuit with a simple mechanical alternator, and your measurements back that up. The actual resistance of the fan is probably about 150Ω (and you should be able to measure that with a multimeter without the fan plugged in, but may need to move the fan blades to get a good connection across the inductor). The rest of the working "resistance" comes from the effort of turning the fan blades, and inefficiencies in building up and tearing down the magnetic field.

150Ω is plenty to protect the fan from any damage. At 5v, you're only going to generate ~15-20mW of heat. At lower voltages, you're going to generate even less (at 1V, it'll be ~7mW)
 

Latest posts

Back
Top