You need to indicate the polarity of the LED in your schematics because it very much matters.Yesterday I got a 470μF capacitor at RadioShack (the biggest one they had that fits on the bottom of my hovercraft). I tested it with an LED and it is definitely more noticeable.
I tried wiring my circuit (with the new capacitor) on a breadboard, and #12's answer to my original question doesn't seem to be working. I have attached a diagram of my new circuit layout. The capacitor doesn't seem to be getting charged at all, and has no effect even if I remove the fan from the circuit. What am I doing wrong here?
EDIT: Attached photo of actual breadboard setup (breadboard_circuit.jpg)
@WBahn: I like your suggestion, but the size and weight of the components needed make it impractical for this very small hovercraft. It can't even hold the weight of the 9V battery that powers it, so I have the battery and switch connected by a 5 or 6 foot long wire. I really should get a lightweight lithium ion rechargeable battery like the ones in RC helicopters, but they're expensive.
Thank you for your detailed answer. The LED polarity is such that it is on when the 9V battery is connected, so now I understand why the circuit wasn't working.You need to indicate the polarity of the LED in your schematics because it very much matters.
The problem that you have in your circuit is that, LED polarity aside, your capacitor can discharge through the resistor without any current going through the LED. If the LED polarity is such that it is on when the 9V battery is connected, then it will be reverse biased when the battery is removed. If it is such that it is not on when the battery is connected, then as soon as you remove the battery it will allow the capacitor to discharge quickly through the low fan resistance (depending on how quickly the fan dies and stops generating a back EMF).
As an aside, if it takes the fan a few seconds to wind down, you might be able to use it to generate the power to light the LED, which will bring the fan to a stop quicker which might be something you would like.
Let's build up something that might work for you. Keep in mind that I haven't tested, even in simulation, any of this. It's train-of-thought kind of stuff.
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The next thing we would like to do is add a timer so that the LED shuts off after a second or so in order to conserve battery life. There are a number of ways to do this depending, in part, if we would like the LED to fade out or do we want it to turn off fairly abruptly. Also, we need to decide if we want it to turn on or fade on when the external battery is connected. Let me know what you prefer and we can go from there.
You can do a pretty simple test to see if the fan can power the LED. Just connect the LED and resistor in parallel with the fan but connected backwards so that the LED is off when the 9V battery is connected. Then disconnect the battery and see if the LED lights and, if so, for how long.Thank you for your detailed answer. The LED polarity is such that it is on when the 9V battery is connected, so now I understand why the circuit wasn't working.
The option you mentioned that sounds most interesting to me is to use the fan to generate the power to light the LED as it winds down. I think that would be really neat. How would it be done?
I would really like to do that, but if it doesn't work well, then I could try a coin cell with a ~4.8 second timer.
I just did that and the LED didn't light up at all. I don't know if it matters that the fan uses 5V and 0.12A. I would have thought that it would at least have a very faint light that could be seen in pitch dark. I tried taking the resistor out, because I didn't think the LED would blow out when it was wired backwards, but that made no difference.You can do a pretty simple test to see if the fan can power the LED. Just connect the LED and resistor in parallel with the fan but connected backwards so that the LED is off when the 9V battery is connected. Then disconnect the battery and see if the LED lights and, if so, for how long.
Sorry for the confusion. When I wrote the original post, I thought the fan used 12A (quite a big difference!). I edited it to correct that now.Was worth a try.
In your first post you said that the fan took 600 times the current as the LED, but 0.12A is only 6 times 20mA the LED current you mentioned in a later post. I was thinking that you were using a 9V supply that was delivering about 10A to the fan (and had forgotten about this claim by the time you mentioned that you are using a 9V battery (which I'm assuming is the ubiquitous rectangular 9V battery that used to be called a "transistor radio battery"). Also, powering a 5V fan with nearly twice its rated voltage may not be too good, but if it's a pretty cheap fan it can probably handle it for what you are doing and may simply not last as long as it would otherwise.
How long does it take the fan to stop spinning when you disconnect it (no LED connected)? Does this seem to be shorter with the LED in the circuit? What if you disconnect the battery and immediately short the fan wires together? This should make the fan stop much quicker. If not, then the fan is not a simple permanent-magnet motor and so it can't operate as a generator when overhauled by the inertia of the fan blades.
But apparently not the kind that an LED needs.A brushless DC fan usually has an internal oscillator to drive the motor. Just give it enough voltage and current, and it converts the power to the kind the motor needs.
It's not that, it's that there is driving circuitry between the power input and the motor and that circuitry does not allow the power produced by the motor when if overhauls to make it back to the input terminals. It would actually take a bit of effort to make that happen.But apparently not the kind that an LED needs.
Your fade with a 470uF cap and 470Ω resistor is only going to be about 0.2 seconds.Sorry for the confusion. When I wrote the original post, I thought the fan used 12A (quite a big difference!). I edited it to correct that now.
I tested the fan all three ways twice, and didn't get more than a ±0.06 second difference in the results, so apparently the fan isn't the right type. It says on it "brushless", whatever that means, and I think it has ball bearings inside.
Apart from buying a different fan, I guess I have only three choices:
- To use one or more coin cells with a ~4.8sec timer.
- To use your first diagram with the 470μF capacitor I just got and a diode. This would produce a noticeable, but not very long lasting, fade.
- To just use no capacitor at all.
True, I wasn't thinking about the oddness of specifying approximately a precise number. I'm not exactly sure what range would be acceptable. I suppose there could be a lot of variability in the time it takes for the fan to wind down. It might be >5 seconds for a fresh 9V battery, and only ~3 for one that's dying. So maybe a good range would be 3-6 seconds or so, but there seems to be so much potential variability that I really don't know for sure. There could be a problem with the timing of things if the coin cells were brand new but the 9V was almost dead.Your fade with a 470uF cap and 470Ω resistor is only going to be about 0.2 seconds.
So let's see if we can build a simple timer.
When you give a number like 4.8 seconds, it implies that the .8 is significant and that 4.7 seconds or 4.9 seconds is not acceptable. When you say ~4.8 seconds it implies that perhaps these are acceptable but that 4.6 seconds or 5.0 seconds probably isn't. After all, why specify it at ~4.8 seconds if, say, 4 or 4.5 seconds was acceptable and if 5 seconds was acceptable, then why wasn't it specified as ~5 sec? See what I'm getting at? What would you consider the minimum and maximum times that would be acceptable for your needs? That will give us an idea of whether we can use a circuit that is sensitive to the transistor gain, which can vary over a factor of three or four.
Also, is it okay if the LED fades on when the 9V battery is connected and, if so, what is the maximum time that is acceptable for the LED to turn completely on?
Is it okay to design around a 10mA LED current, or do you want 15mA or even the original 20mA? The smaller the better?
Does it need to be a blue LED (with it's 3.2V drop) or will a red LED with about a 2.2V drop be acceptable?
Okay, so let's shoot for 5 s keeping mind that you can tweak values a bit if you need to. We'll stick with a blue LED and go for 20 mA of current. Again, you can tweak things later.True, I wasn't thinking about the oddness of specifying approximately a precise number. I'm not exactly sure what range would be acceptable. I suppose there could be a lot of variability in the time it takes for the fan to wind down. It might be >5 seconds for a fresh 9V battery, and only ~3 for one that's dying. So maybe a good range would be 3-6 seconds or so, but there seems to be so much potential variability that I really don't know for sure. There could be a problem with the timing of things if the coin cells were brand new but the 9V was almost dead.
I guess it would be OK if the LED fades on, since that would correlate with the fan gradually reaching full speed.
Doesn't current affect brightness? I wouldn't really want to reduce the LED's brightness, since its purpose is illumination, but if it had to be reduced a little bit that might be OK.
Sorry, I really want the LED to be blue. Blue is my favorite color, especially of light.
Please remember that the circuitry must be compact. It needs to fit underneath the foam plate, which is about 6in in diameter and ½in deep and has a 2in-diameter hole in the middle for the fan to blow through.
I am in awe! Being a beginner, I never saw such complex circuit diagrams before.I broke down and downloaded LTspice and learned how to use it. I wanted to use a voltage controlled switch but it complained about not finding the model. Due to the diodes, it shouldn't matter.
View attachment 77772
I picked an LED from the available parts that should give about 3.2V forward voltage drop and it looks like I came pretty close at 3.29V.
View attachment 77771
The current rises to 20mA in about 10s and falls in about 6s or so. It appears that my paper design came out pretty close, but some of the fine details haven't been explored such as when one battery is strong and the other is weak.
Note that I cut the pulse frequency in half and lengthened the simulation time compared to what is shown in the schematic.
Ah, I think I got it now! Not the diode (the LED I already had in the circuit), but a diode (which I need to add in order to have the current flow in the right direction).Place a diode in the top line between the fan and the capacitor.
by Jake Hertz
by Duane Benson
by Jake Hertz
by Duane Benson