I have a LED light hooked up to a battery power source of 4 AA batteries. Touch the positive and negative wires together and the light comes on, separate the wires and the light goes off.

I created an airgun bell target consisting of a wood rectangle box with a desk bell mounted to the back wall of the box. I have a metal plate with a 1/2" hole on the front. Shoot through the hole, hit a paddle that swings which hits the plunger of the bell to make it ring. I wanted a visual effect as well as audio so I hooked one wire up to the bell plunger guide and the other up the paddle that hits the bell plunger to make it ring and light the light for a split second. I would like to delay it so the light stays on for a few seconds longer.

I know nothing about electoral engineering so I am already over my head.

 

There are a number of ways to do this. From a hardware standpoint, the simplest is probably to use a six- or eight-pin microcontroller. But that means you have to invest in the equipment to be able to work with such a device and learn how to do so. Unless you have a number of projects that you are interested in doing, that is probably not work the time, effort, and expense.

Probably the next simplest is to use something like a 555 timer IC as a "one-shot" timer that is triggered by the plunger. There are a number of simple ways you could use the motion of the plunger to trigger the circuit.

You could also do it using relays and there are relays that are specifically designed to provide delays of specific amounts of time.

 

True -I have gone with an 8 pin microcontroller for this in thepast. Maybe the simplest for somebody who has not built electronics projects before is to just use a large capacitor to stretch the pulse. Using Google image search I see somebody has already drawn the assembly. I cannot imagine the purpose of the diode and I think it can be left out safely. The large the value of the capacitor, the more slowly the LED will fade out..

 

[edit] Looks like Dick beat me to the punch. I had to build mine and test it. [end edit]

I had in mind a switch that supplies power to some capacitors. From there the LED is powered through a resistor. Here's the rig I just built:

S1 (target switch) strikes momentarily, charging C1, C2 and C3 (each is 100µF @ 25 VDC Electrolytic). The green LED (super bright) comes on full brightness and quickly decays in intensity. Over the period of one second it dims down to barely visible. Adding more capacitance (since I didn't have a 470µF cap handy I just paralleled three 100's (for 300 µF). For longer durations you can increase the capacitance. My circuit uses a 470Ω resistor (actual measurement was 501Ω) so my circuit was running at about 6mA. This super bright LED should handle up to 30 mA, but why push it?! It's bright enough for my tests. If you want a brighter light then reduce the resistance. You'll likely also need to increase the capacitance as well. Higher current draw will shorten the decay rate. Greater capacitance will lengthen the decay rate

 

Changing my capacitors to 1000 µF resulted in a decay rate of about 3 seconds. You may find that pleasing. 1200 µF drove the LED for a good 4 seconds. At 1300 µF the decay rate was around 5 seconds and honestly I didn't think it was as good as the 3 second decay. Of course, it will depend on ambient light present as well. In a dark room the decay will be dramatic, but in a well lit room (or daylight) you'll hardly see it.

 

@DickCappels Why did you include the protection diode? The TS mentioned he was using a 6 volt supply. The protection diode would further reduce the voltage, which may be fine with a 9 volt battery but since the original post said he was using 4 AA batteries. My LED was running 2.88 Vf, dropping the working voltage down to 3.12 volts (assuming a perfect 6 volt battery). To further drop 3.12 volts by approximately 0.7 volts would have resulted in a voltage too low for the LED. Again, a 9 volt battery would be fine. But since the TS is using 6 volts I wonder if the protection diode would be worth while. I respect your clarity on electronics and would appreciate a little info on why you included the protection diode.

 

The diode would prevent LED breakdown if the battery were reversed.

 

Yes. Please note that the diode was in the circuit as I found it on the internet and I said I did not see any use for it. Alec_t just showed that there is a use for it.

 

R1 at 200Ω and a capacitance of 1100µF produced plenty of brightness and sufficient decay rate (my opinion).

As for the protection diode, with a 9 volt battery and a momentary on switch, the likelihood of plugging the battery in backwards and harming the LED is just not likely to happen. Using 4 AA batteries in a holder, again, putting one in backwards is a difficult thing to do. But at least now I have a reason for a protection diode in the circuit. But honestly I thought the LED would do nothing if it were reversed. Wouldn't the LED be somewhat self protecting? Especially if its break down reverse voltage is sufficient?

@DickCappels Yeah, I went back and re-read your post. Sorry for my oversight.

 

R1 at 200Ω and a capacitance of 1100µF produced plenty of brightness and sufficient decay rate (my opinion).

As for the protection diode, with a 9 volt battery and a momentary on switch, the likelihood of plugging the battery in backwards and harming the LED is just not likely to happen. Using 4 AA batteries in a holder, again, putting one in backwards is a difficult thing to do. But at least now I have a reason for a protection diode in the circuit. But honestly I thought the LED would do nothing if it were reversed. Wouldn't the LED be somewhat self protecting? Especially if its break down reverse voltage is sufficient?

@DickCappels Yeah, I went back and re-read your post. Sorry for my oversight.

As I read the first entry the switch would only be closed for a fraction of a second--would that be enough time to charge a several hundred uFcap?

 

Here's a link to a video of how it works as is. I'd like the light to stay on for the duration of the bell ding. From what has been posted it doesn't seem to difficult.

 

As I read the first entry the switch would only be closed for a fraction of a second--would that be enough time to charge a several hundred uFcap?

Do the math.
6V into 1000 uf costs 6000 micro-amp seconds or, 6 milliamp seconds.
A AA battery can supply half an amp for a few milliseconds so 6 ma/500 ma x 1 second = 12 milliseconds of "on" time required.
A human would be hard pressed to push a button for only 12 milliseconds.

 

I'm counting about 3 to 4 seconds of ringing. With the rig I posted, using 2000 µF you should get just that. One thing to keep in mind; if the button is pushed ever so fast the capacitors might not have enough chance to charge long enough to carry the light during the decay of the bell. I've been testing my rig by swiftly wiping two wires together. I'm going to guess that I'm getting a longer charge time than a push button being hit by a slug. But throwing a couple caps in there is worth the try. Very low cost, and if you don't like the results - only a few cents spent to learn this does not do what you want.

The alternative is to use an op-amp and a reference voltage. The problem with that is battery power is constantly being consumed to maintain a reference voltage. The LED would be on full brightness until the voltage drops below the threshold and then would almost suddenly go completely out. Faster with a comparator, but now we're starting to talk more engineering of a circuit to do what you want. It's easy enough, but lets first determine whether you like the Capacitors or not. They might work well enough to suit your taste.

 

The alternative is to use an op-amp and a reference voltage.

Consider an SCR. Delicate balancing of its hold current can allow it to fire a charge into a capacitor and shut off when the only load is the LED. Then again, an SCR wastes almost 1.5 volts, so that might not be a good idea. Then one might fiddle with some transistors to eliminate the A-K voltage burden, but I think the 12 to 24 millisecond push button scenario is viable for this case.

 

Do the math.
6V into 1000 uf costs 6000 micro-amp seconds or, 6 milliamp seconds.
A AA battery can supply half an amp for a few milliseconds so 6 ma/500 ma x 1 second = 12 milliseconds of "on" time required.
A human would be hard pressed to push a button for only 12 milliseconds.

But a human isn't pushing it. A pellet is hitting a paddle of some kind that is striking the bell. Whatever switching arrangement is being coupled to that needs to last long enough to charge the caps.

I was concerned about that, which is why I didn't offer up this suggestion since a simple way of ensuring adequate charge didn't come immediately to mind. If such an arrangement could be devised (and I suspect it can) this approach has the very nice advantage that it only draws a tiny bit of energy from the battery only which it is struck. No need to turn it on and off and the battery will last for years.

 

But a human isn't pushing it.

I was afraid I got the math wrong, but I found another way to miss the target.

 

Consider an SCR.

Interesting idea. A brief charge on the control should be enough to turn it on. However, I'm at a loss to figure out how to turn it off. I'd be curious to see your solution if you have one.

Whatever switching arrangement is being coupled to that needs to last long enough to charge the caps.

Yeah, that's one of my concerns too. However, the way I'm firing my test rig is to take a solid copper wire with just about 2 mm stripped off the end and am striking it in a sweeping motion with the negative lead from the battery to produce a glancing contact for a very swift and short duration. I'm finding my rig working sufficiently. Maybe I'll grab the cell phone and shoot a video then post it to YouTube just so I could post a link.

 

OK, here's a link to a video I just published. Let me know if it's not working.

 

Good demo!

 

Agreed -- the demo definitely shows that it is an approach worth exploring.