Need some basic electronics help please. RE: "pots"

[alien resident]

I have a project in mind.
I play a lot of Paintball/woodsball. My big problem is high humidity overwhelms the puny helmet/goggle fans that I have. I bought two 25mm x 25 mm fans to redo the air ventilation in my helmet. They are rated at 5V dc, .06A..... the battery pack I intend use is a rechargeable 6V 1400 mAh type, it actually holds 7.2V at the start when fully charged.
I want to get a potentionmeter to control the rpm of the fans and limit the voltages going to it. I only have Radio Shack near by and I'd rather not order one online. Is there a "pot" that someone could recommned that is carried by RS that would be very ideal in this situation.

I also want to rig the fans in parallel NOt series.

[ctal]

I'm looking at a somewhat old RS catalog, but the info is probably still valid. I'd suggest part number 271-265, which is a 25 ohm 1.2W pot, in series with either a 10 or 15 ohm 1/2W resistor, part numbers 271-1101 or 271-1102 (buy both and see which one best works to your liking). With the 10 ohm resistor in series, you run the risk of overvolting the fans somewhat, if you turn the pot to its end while the battery is at 7.2V. With the 15 ohm resistor you won't be able to get the voltage to the fans as high as you might want when the battery is at 6V.

The relevant equations are I (current) = E (volts)/R (resistance), which is Ohm's Law, and P (power) = I x E = Isquared x R. The two fans in parallel (which is the correct way to connect them) will draw 120 ma at 5 volts, which corresponds to a resistance of 42 ohms. With the pot at minimum resistance (assumed to be 0), the 10 ohm resistor would result in 42/(10+42) times the battery voltage being applied to the fans, which would be 5.8V for 7.2V battery voltage, or 4.8V for 6V battery voltage. With the 15 ohm resistor, you would have 42/(15+42) times the battery voltage being applied to the fans, which is 5.3 and 4.4 for the two cases.

With the pot set for maximum resistance, with the 10 ohm resistor and 7.2V from the battery, the voltage applied to the fans would be 42/(42+25+10) x 7.2 = 3.9V. The 15 ohm resistor at 7.2V would result in 42/(42+25+15) x 7.2 = 3.7V. At 6V, these numbers would be 3.3V and 3.1V.

Regards,
-- Al

[alien resident]

wow ! thats a very detailed response. ! thank you.

I do have two questions. Why the 25 Ohm pot in series with the 10 or 15. What is that first pots purpose ?

and how did you get "120 ma at 5 volts" EDIt ; nvm two fans of 60ma at 5V each..got it

[ctal]

The 10 and 15 ohm resistors are not pots. They are fixed resistors, which would prevent the voltage to the fans from going too high if the 25 ohm pot were turned to the minimum resistance position.

Each fan draws 60 ma (.06A) when 5V is applied to it. So two of them in parallel would draw 120 ma with 5V across them. 5V/120ma = 42 ohms, so the two fans in parallel can be thought of as a 42 ohm resistor. It's not quite as simple as that, because the fans have inductance and capacitance as well, but for these purposes (where the only current flowing is DC) the assumption of a purely resistive 42 ohms is appropriate.

Regards,
-- Al

[ctal]

This is probably clear already, but just to be sure: The two fans would be connected, as you said, in parallel with each other. The battery, the fixed resistor, the pot, and the parallel combination of the two fans would all be connected in series. The order in which the series elements are connected does not matter electrically and could be whatever is physically most convenient.

Regards,
-- Al

[alien resident]

please see the pic. do I have it right. ?

edit : crap I can see the mistake but cant fix the image. hold one I'll try to redo it.

[alien resident]

ok how about now ?

[ctal]

Nope. First, you didn't connect the negative terminal of the battery. Second, the fans are connected to the pot incorrectly. Use only two terminals of the pot, the wiper and one end terminal. (In fact, the pot I suggested is listed as a rheostat, which I think might mean it only has two terminals; if the pot has three terminals don't use one of the end ones). Connect the plus end of the two fans to one of these terminals. Connect the minus end of the two fans to the minus terminal of the battery. Connect the other terminal of the pot to the switch, as you showed it.

As I said, the order in which the series elements are connected doesn't matter, but that is one way of doing it.

Regards,
-- Al

[ctal]

Originally Posted by alien resident

ok how about now ?

Much better! Obviously, my post above refers to your first diagram. Your second diagram looks good!

-- Al

[alien resident]

very mcuh for the help. I'll post back here later with pics if I can

[ctal]

You're welcome! I'd suggest trying the 15 ohms first, because that is safest for the fans. Only go to the 10 ohms if you find that the fan speed doesn't stay high enough for you as the battery voltage goes down. And if you do go to 10 ohms, try to keep the pot a little away from the minimum resistance position when the battery has a full or nearly full charge.

Regards,
-- Al

[Deer Slayer]

Well, if you're not happy with it, you could try a variable voltage regulator. Those little fans are OK to get an overvoltage, they won't last as long, but they will work, usually up to double their rated voltage. I wouldn't worry about feeding them 7.2v, it probably won't hurt them, and they'll move more air. As for the voltage adjustment, you could use an LM317 adjustable voltage regulator, and a pot. You'd have complete adjustment over the fans down to 1.5v. Usually they won't crank out the full input voltage, it's usually 1.5v difference minimum between the input and output. So, you'd be down to 5.7v max even with the full 7.2 in, so no need to add anything to limit the output voltage. I think you use a 5k (it says what resistance on the back of the package) pot between negative, and the center terminal of the regulator. Positive from the battery goes to one side, and the positive wires from the fans to the other. The negative from the fans goes to the negative on the battery. There's a simple schematic on the back of the regulator's package from radio shack that shows the parts you need. I think everything is under $5.

[Drufuss]

Okay, ctal and Deer Slayer already answered your question sufficiently.

I'd prefer Deer Slayers method, first because i like to fiddle with electronics, and second because this way you may save some mAh when using a regulator instead of "burning" precious mWs.

Greetings

Dru

[ctal]

Yes, I think that Deerslayer offered a good alternative approach. I think a fixed resistor would be needed in addition to the pot, using an LM317 or similar regulator. Here is a link to the datasheet:

http://www.national.com/ds/LM/LM117.pdf

(The LM117 and LM317 are similar except for their specified operating temperature ranges).

I have no feel myself for how well the fans would be able to handle overvoltage conditions over the long term.

With all due respect to my good buddy Dru, I don't think that use of an LM317 or similar voltage regulator would save any power, or prolong the battery life between charges. In fact it probably would do slightly the opposite. A much more elaborate switching power supply, such as a computer power supply, provides the kind of high efficiency he is thinking of. But a simple 3-terminal series regulator such as an LM317 works by dropping whatever voltage is not dropped across the load across itself, while passing the entire load current through itself. Which is exactly what the pot/resistor combination would do, resulting in a total power dissipation around the circuit path equal to the load current times the battery voltage. And in fact a little more than that, because the regulator would consume a little bit of current internally, as well.

Regards,
-- Al

[Deer Slayer]

A PWM circuit would save power and vary the speed. Being on your head, the whine it produces might be annoying. A PWM circuit gets a little complicated though. You could build this circuit, allthough I'd suggest using the TIP29A transistor, because it doesn't need to handle a lot of current. The TIP31 is probably more common and easier to find. The mosfet isn't necessary, but would also work. Honestly I don't think any of the transistors will make much difference in the way it works, they just change the size of the load it can handle. Anyways, if you really want to fiddle with electronics, you could build this circuit. I think it will run on 7.2v, and it will allow you to overspeed the fans a little if you want, and adjust down until they stall.

Pulse Width Modulation 2