How to have capacitor power only one part of parallel circuit?

Thread Starter

tjohnson

Joined Dec 23, 2014
611
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.
 

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WBahn

Joined Mar 31, 2012
26,398
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.
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.

First, let's use a diode to prevent the capacitor from driving the fan but allowing the 9V battery to drive the LED.

hoverlight_1.png

The problem with this is that the RC time constant requires too large a capacitor to be feasible.

So let's use a light coin cell in such a way that it only powers the LED once the battery is removed.

hoverlight_2.png

As long as the on-board battery is less than the external battery, D2 will be reverse biased and nothing will be drawn from it. The problems with this circuit are that you want an on board battery that is close in voltage to the external battery otherwise there will be a noticeable drop in LED intensity when the external battery is disconnected (but maybe this is desirable). The other thing is that a coin cell isn't going to power an LED for too long before dying and this circuit will continue to power the LED until the circuit is switched off somehow.

Now, before you balk at the weight of a battery, I'm not surprised that your hovercraft can't handle the weight of a 9V battery, which is relatively large. But do you know what we are talking about when we recommend coin cells? These are watch and hearing aid batteries that are quite small -- considerably smaller and probably lighter than that capacitor. A small 3V DL1216 cell would power a 20mA LED for more than an hour hours and is only half an inch in diameter and only 1/8" thick for two of them stacked on top of each other to get 6V.

So let's modify the circuit a bit so that we can use a 6V battery made up of two 3V coin cells. This will involve using two resistors, one for each battery (though there are other ways to do it, too).

hoverlight_3.png

A blue LED has about 3.2V across it when on and the other two diodes have about 0.6V, so about 3.8V total. To get 20mA from the 9V battery we need R1 to be 260Ω so let's use 270Ω. If we only wanted 10mA we could use either 510Ω or 470Ω). To get 20mA from the 6V battery we need R2 to be 110Ω so let's use 100Ω (for 10mA we could use a 220Ω).

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.
 

Thread Starter

tjohnson

Joined Dec 23, 2014
611
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.

...

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.
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.
 
Last edited:

WBahn

Joined Mar 31, 2012
26,398
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.
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.
 

Thread Starter

tjohnson

Joined Dec 23, 2014
611
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.
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.
 

WBahn

Joined Mar 31, 2012
26,398
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.
 

Thread Starter

tjohnson

Joined Dec 23, 2014
611
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.
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:
  1. To use one or more coin cells with a ~4.8sec timer.
  2. 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.
  3. To just use no capacitor at all.:eek:
 
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#12

Joined Nov 30, 2010
18,223
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.
 

Thread Starter

tjohnson

Joined Dec 23, 2014
611
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.
But apparently not the kind that an LED needs.
 

WBahn

Joined Mar 31, 2012
26,398
But apparently not the kind that an LED 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.
 

WBahn

Joined Mar 31, 2012
26,398
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:
  1. To use one or more coin cells with a ~4.8sec timer.
  2. 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.
  3. To just use no capacitor at all.:eek:
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?
 

Thread Starter

tjohnson

Joined Dec 23, 2014
611
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?
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.

The circuit 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.

EDIT: @WBahn: Thank you for all of your help. I don't want to take too much of your time, because I don't know if I want to implement this for my hovercraft right now. However, I'm interested in how it would be done, because I'm fairly new to electrical circuits and trying to learn how to wire them.
 
Last edited:

WBahn

Joined Mar 31, 2012
26,398
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.
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.

The key to getting long times out of a small capacitor is to use it only for timing and not for power. So you use the coin cells for that. We will use the capacitor to drive the base of the transistor and configure it in an emitter-follower fashion so that as the capacitor voltage decays so will the LED current. This also lets us use a single programming element so that the LED current is continuous when switching between the external and internal supply.

The basic topology is this (incomplete):

hoverlight_4.png
The voltage Vp is the programming voltage and whatever it is, the voltage at the emitter, Ve, will be about 0.6V less. This is the voltage that will appear across Re and whatever current flows in Re will essentially be the same current that flows in the LED (within a fraction of a percent) provided the LED anode voltage, Vo, is high enough to keep the transistor from saturating. To prevent this, Vc must be at least Vcesat greater than Ve. For most small NPN transistors, Vcesat is roughly 0.2V. The coin cells have a cutoff voltage of about 2V (so 4V for us), but this will be pushing things a bit much. So let's use 2.5V (so 5V for us). What we are saying is that we will design the circuit on the assumption that once the coin cell terminal voltage drops to 2.5V that it is time to replace them. Thus, at cutoff, Vo when running from the coin cells will be about 4.4V. We will lose another 3.2V across the LED putting 1.2V at Vc and allowing for no less than 0.2V drop for Vce, we want Ve to be no higher than 1.0V which means Vp needs to be at 1.6V.

With 1V across Re we want 20mA to flow, so Re needs to be 50Ω (so let's go with 47Ω, but 51Ω is another option).

This gives us a collector current of about 20mA and the base current will be much smaller than this depending on the current gain, β, of the transistor we use. We generally assume that β for small signal transistors is about 100 because, in practice, it is safely above that most of the time. But as we will see this circuit, in it's present form, depends on the actual value of β (and is therefore not a particularly good design) so we want to estimate it a bit better. Let's choose a 2n3904 transistor (most small signal transistors will be in this ballpark). The data sheet

https://www.fairchildsemi.com/datasheets/2N/2N3904.pdf

shows a collector current gain at 10mA (and Vce=1V) that is between 100 and 300. Since we are operating at twice this current and about half this Vce, we can expect to be on the lower end of this range (possibly even below it), so using 100 is probably reasonable for this transistor, though it might end up being as high as perhaps 200.

Although we have 20 mA flowing through Re, only 1/β times this is flowing in the base. That means that the capacitor thinks it is connected to a resistor that is βRe is size. Thus our RC time constant is βReC. Our LED will extinguish when Ve gets to 0V which means when Vp gets to 0.6V from 1.6V, or in other words, to 37.5% of it's starting value. Since the voltage will drop almost exactly that much in one time constant, we can use that as our timer value. So to get 5s we need about a 1000μF capacitor. If the β turned out to be 200, then that 470μF would probably do, but then you have both that large capacitor and the coin cells on board, which it sounds like would be too much weight.

So what can we do? There are two possibilities that come to mind and you can try one, the other, or both. The first is to increase Vo using three coin cells and the other is to further decouple the timing from the LED current by using a Darlington-like configuration.

Because the editor is getting very non-responsive with this long a post, I'll follow up on these ideas in separate posts.
 

WBahn

Joined Mar 31, 2012
26,398
Let's explore the option of increasing Vo. To do that, we can simply use three coin cells to give us 9V. But this means that it will be real easy for the coin cell stack to be at a higher voltage than a partially depleted 9V battery, so we can add a second diode in series with D2 to make that unlikely.

hoverlight_5.png
Thus, Vo will start out at about 8.4V and drop down to about 5.4V at the 2V cell cutout voltage. After we lose another 3.2V in the diode and a further drop of Vcesat of 0.2V we are at a Ve of 2V (instead of 1V) which lets us use Re=100Ω. It also means that our Vp can start out at 2.6V and go down to the same 0.6V, or 23% of the initial value, which will take about 1.5 time constants. This lets us reduce the capacitor size to 330μF and, if β is in our favor, perhaps even a 180μF cap. This is an improvement and perhaps even enough. You may well find that using a 100uF cap will get you a fade out time of a couple seconds which is perhaps good enough. You can increase the time by decreasing the LED current; using Re=200Ω will reduce it to 10mA and using something in between will give you, not surprisingly, something in between.

The second approach is to further decouple the capacitor from the LED current by using a second transistor in a Darlington-like configuration.

hoverlight_6.png
This will reduce the current drawn from the capacitor to something like 0.01% of the LED current and would let you use a capacitor that is about 10μF (without Rt). The problem here is that being dependent on the value of β being what we want is even more of a gamble because now we are dealing with β². So instead we can use a larger capacitor, say 47μF, and use a dedicated timing resistor, Rt, to discharge it. Since the discharge is dominated by this new resistor and not the transistor gain, we get a much more stable and predictable time constant. Our Vp voltage needs to be 0.6V higher than before, so now it would need to go from 2.2V to start and down to 1.2V to finish, or 55% of the initial value which takes about 0.6 time constants. If we assume that the β is sufficiently large so that we can ignore the base current into Q2, the our time constant is just RtC, so to get 5 seconds we need Rt to be 180kΩ if we use a 47μF capacitor. With this low of a discharge current, ~10μA, the leakage current of the capacitor will probably come into play. But you can use the much physically smaller tantalum capacitors and use a 100μF capacitor and a 100kΩ resistor which should give you an on time of ~6s, which might be closer to 5s given that once the current falls below some threshold you won't be able to see it as being "on".

A further variant would be to use a third transistor to make the base current draw from the cap negligible and then use a 10μF poly film capacitor (which has negligible leakage current) and a 820kΩ timing resistor.

For grins and giggles, lets combine all of these into one schematic.

hoverlight_7.png

In this circuit we can set Re=100Ω and have it start at 2V which means that Vp will start at 3.8V and decay to 1.8V before the LED turns off. This will take about 0.75 time constants. With a base current from C of only 20nA, we can use Rt=1MΩ and still ignore the base current. To get a 5s period, we can then use a 6.8μF capacitor.

The only thing left is to add the circuitry to get Vp established. I'll do that in the next post since the editor is becoming very non-responsive again.
 

WBahn

Joined Mar 31, 2012
26,398
To establish Vp, we want to use the external battery to set and hold it at our starting voltage and then let that decay through Rt once the external battery is disconnected. The easiest way to do this is to use another diode, D4, to prevent the capacitor from discharging through the fan. This will drop the voltage after the diode to nominally 8.4V. At that point we have a voltage divider formed by Rc and Rt that needs to put us at our 3.8V starting voltage.
hoverlight_8.png

With Rt=1MΩ, that means we need Rc=1.2MΩ. If we use 1MΩ for convenience (so that we don't have to buy two different sizes but rather a small pack of the more commonly available 1MΩ resistors) then our Vp will start out as high as 4.2V which means that our initial Ve might be as high as 2.4V. Given the very low currents in Q3 and Q2, it might even be as high as 2.6V. This means that Vo has to be at least 6.0V once the battery is disconnected which means the cutoff voltage for the coin cell stack of 7.4V which means a cutoff voltage per cell of 2.5V, which is probably acceptable. We could add a second diode in series with D4 and improve this. In fact, it might work out very well to add a third diode as well. The charging time constant is based on the parallel combination of Rt and Rc which, with both equal to 1MΩ, is 500kΩ making the time constant 3.4s. It will take about three time constants for the LED to appear to be fully on (it will be about 90% of its final current), or about 10s.

With both batteries being nominally 9V, you can easily try out the circuit with a single battery by using it as the right hand battery (the coin stack) and then using a wire to temporarily connect to the top of the fan.

Again, keep in mind that this is a purely paper design and has been neither simulated nor breadboarded.
 

WBahn

Joined Mar 31, 2012
26,398
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.
hoverlight_9.png

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.
hoverlight_10.png

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.
 

Thread Starter

tjohnson

Joined Dec 23, 2014
611
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.
I am in awe! Being a beginner, I never saw such complex circuit diagrams before.

Thank you for the time and effort you spent putting them together. I appreciate that. Like I said earlier, I don't want to take too much of your time, because I don't know if I want to implement this for my hovercraft right now.

I think there are several potential issues with this circuit (please correct me if I'm wrong on any of them):
  1. It would be difficult to fit all of the components 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.
  2. There is a lot of variability of the duration in which the LED should/would fade out. This would be affected by: (a) strength of 9V battery, (b) strength of coin cells, (c) transistor gain, which you said can vary by a factor of three or four.
  3. Sorry, but it's far more complicated than what I would want for a small hobby project.
 

Thread Starter

tjohnson

Joined Dec 23, 2014
611
Place a diode in the top line between the fan and the capacitor.
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).

I really want the way the LED fades out to be consistent, and after thinking more about it, I think having it fade out in 4.8 seconds along with the fan is too long a period of time. I think a diode and 1000μF capacitor (which after measuring again, I believe will just barely fit) is probably the easiest way to achieve something close to what I want. I know the fading out of the LED will last for less than a second, but it's definitely noticeable. (I can even notice the 100μF capacitor very slightly when I have it wired correctly, and of course the 470μF one even more).
 
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ronv

Joined Nov 12, 2008
3,770
You can make the resistor in series with the LED larger in value and it will last longer, but not be as bright. But try it. You may not be able to notice the difference.
 
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