USB Fan Project - Need Help Bringing Voltage Down from 5volts to 3volts...

[audioguru]

Ever seen an propeller plane starting off in a war documentary film?

They give the propeller a push to get it going.

 

[kchriste]

Another thing to consider is that the USB spec states that any device is only guaranteed 100ma on the USB port root hub. A request for 500ma can be made from the host controller but it may be denied.
So, if you have many devices already on that root hub, you may be close to the 500ma limit already.
In Windows, you can check how much power is available by opening the "Device Manager" and then "Universal serial bus controllers" and then "USB Root Hub" and then check the "power properties"
You will see your devices listed and the power they have reserved for them. It will also tell you the maximum current that the root hub will supply. You need to subtract the current from the listed devices from 500ma to get what is left available to your fan motor.

 

[techhappy]

So what you're saying Chung is that I need to measure the motor in action with the fan blade spinning, to measure the motor under load pressure, then I can use that current rating with Ohm's law to find the proper Ohm rated resistor?

I'm glad the 2.5volts thing has been figured out as optimal as far as the motor current is concerned. Still a little grey about the techno talk, but some good news is I found a local electronics store that sells boatloads of resistors, etc. So I'm going to bring in my modded fan and have them take a look at it with their gear to help me figure out the best resistor or diode to get.

I sure hope this fan actually lasts a long time, with all of this tinkering and research!

On a side note, I rigged a extra 90mm computer case fan with a spare usb cable. It's a fan rated 12v max and it's been working fine for the last 3 days. I'm assuming the reason it's working well if because it has a much higher voltage roof and despite it not spinning as fast as it does with 12 volts, it works pretty good.

Can I expect long term usb damage with this? It seem pretty harmless, like it's only taking whatever voltage it can get, in this case, 5v. Regardless, this is the nerdiest/coolest thing ever, I need to get a cheap metal fan grill for it, so it's a big safer. One wondering how much air flow a 120mm fan can get on 5v usb power?

My eyes will open for any gadgets that need 5v power, heehehe, fair game for a usb powered mod.

 

[eblc1388]

So what you're saying Chung is that I need to measure the motor in action with the fan blade spinning, to measure the motor under load pressure, then I can use that current rating with Ohm's law to find the proper Ohm rated resistor?

Exactly.

As pointed out by others, the USB voltage is made available via a controller IC and high load current or short circuit might damage the IC internal circuit with the risk of taking out the whole IC and not just the offending circuit, leaving you with no USB support afterall. This is a risk which you should know.

 

[Dr.EM]

Actually, a case fan like that may be more effective? They will run off 5v, but might need a push to start. You can get 7v case fans designed for silent PC systems, they should be happy with 5v, though are never intended to spin very fast. The current at the rated voltage is around 100-200ma, which is more reasonable, and at lower voltages this can be expected to drop correspondingly.

 

[techhappy]

Okay, I got a multitester and I'm ready to measure current:

**broken link removed**

Here is the underside of the fan, which shows how the usb wires will be soldered in, etc.

**broken link removed**

Here's an overview of the fan and multimeter:

**broken link removed**

Basically, I have no idea what settings to set the multimeter at or where to put the prongs to measure current. Any tips?

 

[Hero999]

Connect the meter inseries with the batteries and select the current setting. The best way to do this is to connect the meter across the switch when it's in the 'off' position.

 

[techhappy]

I can't access the internal switch, it's a cheapy $1 fan, it's sealed inside. I did measure the negative and positive of the battery from the battery compartment. I think I did it right, here are two different settings I wrote down from the multimeter, one was the 6:00 position(200, look at pic I uploaded) which gave me 36 consistently. I switched over to the 9:00 position(200m) which gave me 58 consistently. I have no idea what these numbers mean.

Did I even measure this correctly? I had the motor running with the fan blade, to keep the underload variable accurate.

Will I be able to get the resistor rating I need using one of these numbers with ohms law?

 

[OutToLunch]

did you get a user manual with your meter? if so, a brief read thorugh may be useful.

Take a look at your meter - specifically at the three input jacks at the bottom. The far left and the far right one have a silkscreened line connecting them and the one on the Right is labeled "A" for amperes. Plug the Red lead into the far right jack and leave the black lead in the left jack. Now put the leads in series with the motor making sure that you have the meter lever turned to measure current and turn the motor on.

 

[techhappy]

Okay, now I'm getting somewhere. I did what you said, by plugging the positive lead into the Amp plug. Now the only question I have, please define "in series", does this mean that I need to route the current so the multimeter becomes part of the circuit?

which method would be a simple way of doing this, with the space and wiring I'm working with. So far, I have been putting the positive and negative prongs on the underside where the battery cage is, to the left side, while batteries are being used. I made sure to just touch the metal contacts, but not the batteries themselves while I did this.

With the multimeter I turned the dial to 200M(200M with the green paint on the multimeter) and measured a steady 0.5-0.6. I'm assuming these are amps?

Is this an accurate reading that I can plug into Ohms law to get the resistor rating?

 

[Hero999]

**broken link removed**
http://www.google.com/search?client...ter+tutorial&sourceid=opera&ie=utf-8&oe=utf-8

 

[Hero999]

, does this mean that I need to route the current so the multimeter becomes part of the circuit?

Yes.

which method would be a simple way of doing this, with the space and wiring I'm working with.

You some how need to break the and connect the multimeter up with it (like you say) so it becomes part of the circuit. This could be between the two batteries or anywhere, try wedging the multimeter probes between the battery terminal s and those where the switch is, the motor should start even with the switch off.

With the multimeter I turned the dial to 200M(200M with the green paint on the multimeter) and measured a steady 0.5-0.6. I'm assuming these are amps?

200M or 200m?
The former is likey to be M ohms and the latter probably is either m omh or mA, but the latter is more likely. Start off with the highest setting, which is probably 10A in your case, then work down and make sure it's set to DC and not AC.

Is this an accurate reading that I can plug into Ohms law to get the resistor rating?

No, even if you were to measure the resistance of the motor and work out the current it would give the current when the motor is jammed not when it's running.

Can you post a more clear picture of your multimeter? One where you can actually see all the markings on the dial.

 

[techhappy]

Okay, a quick stop at radio shack and I got some insulated test jumpers and a AA battery holder to create my in series circuit.

Here is a pic of what it looks like, based on the link that Hero999 gave me about how to create a circuit for multimeter testing:

**broken link removed**

Now here are the results. I wanted to be thorough about this current reading, so I tested 3 different battery brands to see what the different current readings would be. Here is a pic of the multimeter, what setting I chose and the results of the 3 different readings:

**broken link removed**

Now are these finally the right numbers I need or do I need to cycle through the settings?

When calculating Ohms law, based on this current readings, the variables are a 5volt usb cable to a 3 volt motor, I want the curent to drop 2.5 volts from the usb 5 volts. So I need a resistor that can do that job. I figure this way, according to everyone's advice, the motor will last longer at 2.5 volts than running at it's max 3 volt max.

So the voltage to the motor should be 2.5 volts after it goes through the resistor.

Am I getting closer to the holy grail?

 

[AllVol]

Isn't the red lead supposed to be in the center jack?

 

[audioguru]

The right jack in the meter is used to measure up to 10A when the meter dial is set to the red 10A setting. If the meter wire is in the center jack then the max current it can measure is only 200mA and the motor's current will probably be higher.

I think all those wires have too high a resistance and will limit the current during the current measurement.

 

[techhappy]

Okay, I tried the middle jack and audioguru is right, the fan doesn't power and there is no measurement of signal.

So now what? Do I use the measurements I have with ohms law?

 

[eblc1388]

Would you please redo the motor current test with the meter range selector set at the red 10A position and post the results?

I assume you are using two batteries to give a total of about 3V.

 

[techhappy]

Okay, I redid the test with the selector on red 10A position, the batteries are a total of 3V(2 AA) and these are the numbers for the different batteries I used:

panasonic - 0.70
energizer - 0.80
sony rechargeable - 0.89

 

[techhappy]

Are we getting close? Ohms law with these current numbers?

 

[eblc1388]

That set of readings are now seem perfect.

Usually, a good meter would indicate the correct value right away on the display without user to do any conversion or decimal changes.

Now your fan load takes about 0.8A at 3V and so let's assume it will take about 0.7A at 2.5V.

To use it on a 5V supply you will then need a resistor to drop 2.5V at 0.7A and that works out to be 3.3 Ohm or 3.9 Ohm(one can't easily get values in between). The power rating is 3W.

However, that current value has exceeded the current rating of USB!!!