Double voltage from lithium battery

[Azhrarn]

I'm trying to create a very simple circuit where you have the following main components:

1x 3V lithium battery,
1x 3-positions switch (ON1-OFF-ON2),
1x fan (which can run from 1.5 up to 3V).

I want to create a circuit where my fan can have a high speed (ON1) at around 2.5V and a low speed at around 1.8V (ON2).

The problem is the low-speed position since if I'm using resistors divider to obtain these voltage values, as soon as my battery will starts to drain (even if it is a lithium battery, the voltage will go down at one point in time), my "low-speed" position at 1.8V will drop and if it reach 1.6V my fan will stop.

So I wanted to use a Zener diode of 1.8V (yes, they do exist!) for my low-speed. But my fan would need to be in parallel to this but it also needs to be in series with my resistor for the high-speed position...

Yes...I thought about using a resistor to lower my overall voltage to 2.5V right in the beginning but than my zener circuit would become unstable since I would not have a lot of voltage difference between my zener break-off voltage and my initial voltage...

Am I making simple things too complex?
What can I use instead of a zener then?

Thanks!

 

[MacGyverS2000]

Why are you trying to run a fan off of a 3V Lithium? You also seem to be wasting a lot of power by using resistor dividers and zener diodes.

On top of that, I don't think you understand the operation of a zener... while the zener will regulate a point to 1.8V, you'll be burning quite a bit in the resistor leading up to that point, particularly since the fan will need a good amount of current to run. Since the battery voltage is so low, your overhead is extremely small, making any form of current/voltage regulation with such simple components a challenge, at best.


Dan - Owner

http://www.Hi-TecDesigns.com

 

[IRstuff]

Is there some reason you're not revealing the power requirements?

Maxim makes a number of potential solutions in DC-DC switching regulators:

http://para.maxim-ic.com/en/search.mvp?fam=dcdc_all&tree=powersupplies

TTFN

FAQ731-376

 

[itsmoked]

The best solution would be two regular 1.5V cells. Then your switch puts them in series,(high speed 3V), or parallel,(low speed 1.5V). Anything else will cost more money, be less efficient, and be more fragile.

If you must use a lithium battery then the best solution is a pulse width modulation scheme to essentially turn your 3V on and off rapidly. Like 300 times a second so the average voltage is whatever value you want it to be.

Think 'variable speed drill'.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[Azhrarn]

Thanks for all your input, but the issue here is not of power consumption:

- The 3V lithium battery could possibly be replaced by 2x 1.5V BUT that would be even worse in terms of voltage stability since the alkaline batteries are very bad at that.

- The parallel/series suggestion is not applicable here...as I mentioned earlier, the fan needs at least 1.6V to run...below that, it simply stop.

- I'll go have a look at DC-DC regulator BUT for our needs, the use of a trigger logic signal (ON=1, OFF=0) is too complex for the simple needs of 2-speed fan. Unless this 1/0 signal can be connected directly connected to the power source (3V lithium) where a fraction of that voltage would be considered as 1 and a very-low voltage as 0. We cannot involve a more complex circuit involving a clock generator (ex. 555)or the like. I know this is not complex electronics, but everything is relative.

- I have seen some voltage regulator (ex.:MIC5320 from Micrel) but was wondering how to get a simple (1/0) signal since a switch would need to be used to toggle between Vout1 and Vout2 on the fan.

Thanks for the input though, it's really appreciated...makes the gears working...

 

[MacGyverS2000]

When a 555 can be had for 10 cents in bulk and add 20 cents to a product's total cost, it seems silly to fight with other methods...

Dan - Owner

http://www.Hi-TecDesigns.com

 

[Azhrarn]

It's not a matter of cost, it's a matter of space...

 

[MacGyverS2000]

So we come back to my original question... what is this for? The more information you give us the more useful our answers can be.

Dan - Owner

http://www.Hi-TecDesigns.com

 

[OperaHouse]

You can put a diode in series for a 0.6V drop. That is simple.

 

[ScottyUK]

If it is a medium-to-high volume product then look at chip-on-board technology if space is an issue. You won't get much smaller than that without going to something more exotic like thick film hybrids.


----------------------------------

If we learn from our mistakes I'm getting a great education!

 

[Azhrarn]

It's a micro-fan for air displacement that needs a OFF / Low Speed / High Speed control.

Power of the fan is about 0.25W and it's nominal current consumption is 0.08A.

Thanks for all your inputs!

 

[analogkid2digitalman]

PWM for speed control?

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"It's the questions that drive us"
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[ot1]

The forward voltage drop for a silicon diode is .7v, so using two in a series would be below your minimum. One diode would give you 2.3v...but are you sure you even need to have a two speed fan? Did you measure the RPM at your specfied voltages, to see if its worth if? Maybe the fan is regulated internally already, and is insensitive to voltage input change(seems like it it be). Or what about using 4 germanium (.4V drop)signal diodes parallel to give you a little more current handling and still maintain .4v drop. Unless you use PWM for your custom voltages, anything else will waste power. Use the diode banks in lieu of a PWM circuit for semi custom voltages.

 

[LionelHutz]

This doesn't sound like a really difficult application. Still, I would not use a zener regulator. They aren't very good regulators when the current is very low. Besides, they waste more energy than other methods.

You should be able to just pick a resistor that gives you an acceptable voltage drop. Say, a low battery voltage of 2.8V so subtract the minimum voltage of 1.6V = 1.2V drop required then divide by 0.08A = 15 ohms. Wattage is 1.2V x 0.08A = 0.096W so a 1/4 watt resistor should be good.

If you don't like this, then just use 2 small diodes in series to drop the voltage.

If you still don't like this then look to a small fixed PWM regulator and switch it on for the low speed.

 

[LionelHutz]

Actually, I forgot one. I'm sure you can find a linear voltage regulator that would work too.