Two fan variable speed circuit

[RedHerring]

I am new to electronics and circuits, but I have always been currious.

Here is what I want to do:
Connect two identical fans up with a battery and a a switch or two to tun them off and on and from low to high speed. Both fans would be on at

the same time.

Now for the complication; the fan and the battery are different voltages.
-- The battery is lithium-ion, 7.4V, 1500mAh.
-- The fans are 5V, .11A, .6W

My thought is to put the fans in parallel, because in series would just give me the one slow setting (3.7V).

So of course I am going to have to create some way of dropping the voltage. So far I have read that I can do this one of two ways. With resistors

or with diodes. With the diodes I would use the .7V voltage drop, using 4 in series should drop the voltage just below 5V. With resistors it

would be the standard voltage divider setup.

1. What are the advantages and disadvantages of using diodes to knock down the voltage?
2. What are the advantages and disadvantages of using resistors to divide the voltage?

I have read read that resistors will affect the amps on a circuit.

3. What does this affect?
4. Do I need to get those amps to equal the amps needed for the two fans to run(.22A)?

For variable speeds on the fans I only need a high(normal) and low setting. I have read that to do this I just need to lower the voltage. So my

thought was to just use a switch to go between the two.

For the diode setup my thought is to use a double throw switch to flip between no additional diodes for high(5V) and a series of additional

diodes for low(3V).

For the resistor route I think I could get away with a single throw switch and incorporate it into the voltage divider. When the switch is open

only one resistor is used, when its closed two resistors in parallel are used. Providing the low(3V) setting when when open and the high(5V)

setting when closed.

5. What are the advantages and disadvantages of using the diodes?
6. What are the advantages and disadvantages of using the resistors?

Additional info:
-- I put a multimeter across a fan to measure its resistance and it read 890 Ohms

7. Do I need to worry about heat?
8. How long will this last at the high setting?
9. How long will this last at the low setting?

Any other thoughts or pitfalls?

 

[Reloadron]

You have a lot there to digest. Before looking at the options you present there is another option. I may be off on this so welcome other forum members to critique me.

You have a 7.4 volt Li-Ion battery. I believe you don't want to run that battery below about 6.9 volts as it discharges. So here is something to think about. Use a 5 volt regulator like the LM7805 available from any radio shack or parts retailer. Voltage regulators like this are literally a few bucks. They have what is known as a drop out voltage of about 2 volts. This means the regulator to get 5 volts out needs at least 7 volts in. When the input voltage drops below about 7 volts the regulator will drop out. That may be a good thing. The LM7805 will handle up to an amp with a good heat sink and you are well below an amp. Two fans in parallel drawing .110 amp works out to be about .220 amp roughly. This is not an exacting science. Personally I would try a circuit similar to the attached. For the value of the resistor you could try a 10 Ohm which would get the fan(s) voltage down to about 2.8 volts. If that is too slow try to parallel a pair of 10 Ohm resistors (5 ohms total) which will get the fan voltage down to about 3.9 volts. You want a minimum of 1 watt resistors with 2 watt preferred.

As to using a voltage divider. The problem here is once you have the divider worked out the fans will be in parallel with one leg of the resistive divider changing the resistance of that leg. Overall this is just not a good way to do this as forgiving as fans are.

Using diodes would work. You could series 3 for example 1N4002 diodes and drop about 2.1 volts leaving just over 5 volts. The fans wouldn't mind but you would need to watch your battery voltage as you don't want to over discharge the battery pack.

Hopefully another member will chime in on my thoughts with using the 7805 regulator.

As to runtime? Maybe about 4 hours give or take?

Ron

 

[RedHerring]

Thanks for the feedback.

Thanks for the diagram, I like your idea on the circuit.

I read a little about those LM805's, but was concerned about the need of the heat sink and some say it requires capacitors. I was hoping to seal the battery and the circuit inside a small case.

How hot do these LM7805's get?
Will it need air flow or a fan of its own?
How are you calculating runtime? 1500mAh/220mAh?
For volt considerations should I use 7.4V, 5V, or 3V?
Is the amount of amps used indepenent of whether it is in 5V or 3V mode?

You said a 1 watt or 2 watt resistor. A single fan says it pulls .6 watts.

So, will 2 in parallel use 1.2 watts?
Will the watts be the same in both 5V and 3V mode?


Sorry for all the questions, but I just want make sure I am doing this correctly and learn as much as I can along the way.

Thanks again.

 

[Reloadron]

Yes, generally using a LM7805 one uses a few capacitors to eliminate noise for electronic circuits. Since you will be running a few fans I don't see any need for the caps. However, even if we did add a few small caps that are readily available and cheap it would not increase the overall size much at all. You could place a small 10 uF and .1 uF cap across the output easily.

How hot will it get? Well we have a 7.4 volt input and we are regulating at 5 volts so 7.4 - 5 = 2.4 volts. Each fan as a maximum draws .110 amp so both draw a max current of .220 amp. Therefore 2.4 * .220 = .528 watts since power = voltage * current. The LM7805 will not see any heat to speak of dissipating about .5 watt. I doubt for this load it would need a heat sink but even if it did there are tiny heat sinks designed to just clip on it. Again, I see no need for it.

As to the resistor. We know both fans combined in parallel draw .220 amp. With a current of .22 amp a 10 Ohm resistor will have a voltage drop of .22 * 10 = 2.2 volts. Therefore 5 - 2.2 leaves 2.8 volts for the fan. The power dissipated by the resistor would be 2.2 * .220 = .440 watt. I would not use 1/2 watt but at least a 1 watt resistor. If you parallel two 10 ohms for a 5 ohm total the fan would run on about 3.9 volts.

Using the diodes is still a very viable way to go about it and simple. Just make sure when your battery voltage under load gets to about 6.9 volts you stop and recharge the battery pack.

How long? I really winged it on that. You have a 1500 mAH battery. In theory it would deliver 1500 mA for an hour or 750 mA for 2 hours and so on but in reality it doesn't work that way. I figured a 250 mA load (overkill) which should yield 6 hours and dropped it to 4 hours. You may get more. I was very conservative.

Ron

 

[RedHerring]

Well I must of hooked something up wrong, cause the LM7805 started smoking and now will not work.

I thought everything was hooked in the right order (no heat sink and no capacitors). I put one fan in the circuit and turned the power on checked the voltage by the fan and it read at 5.4 volts. So I turned the power off made sure all my connections were right then put both fans in the circuit and the voltage was reading right about 5 volts. Looked away to jot down some notes. Then I started to smell something. I look back and the LM7805 is smoking! I turned the power off.

Good lord was that thing hot to the touch. I waited for it to cool down and took a look at it, no obvious damage to it. So I pluged it back in and now nothing. I am going to get another one and try it again this time with big heatsink and a watch full eye.

Is there another volatge regulator that can handle greater heat/power/watts?

 

[Reloadron]

Make real sure the LM7805 is correctly in the circuit based on the exact one you have. There is no way it should have smoked. It is a pretty forgiving chip. Just make sure in/out and ground are correct.

With a heat sink the LM7805 in a TO220 case can handle an amp. You havent got 1/4 of that and you are only a few volts over on your input. There should be no problem. If I am not mistaken the thing even has overload protection to shut it down.

Ron