Lighted Hockey Puck

[Sterling Martin]

I'm trying to put together a lighted hockey puck. I would like to use one AA battery, giving me 1.5 V. to work with. I was wanting to know what would be the best led/resister combination to give me the most amount of light. I'm thinking about sticking up to 30 led's in the puck, and would like the battery to last at least half an hour. I want the led's to stay on steady, not blink, as it is more visible that way. The puck is REAL tough rubber and very hard to work with, so I was wanting some good input, rather than trying it and seeing what happens!

[philba]

1.5V isn't enough to get much illumination from an LED. Basically you will need a boost regulator to get the voltage up to a reasonable level - 2V for red LEDs, much higher for white (check the datasheet) Do you have a specific LED or color in mind? <p>Now, a different way to do this is to use a coin cell which is 3V. They dont have the capacity of AA cells but will be a lot easier to design with.<p>There are circuits for driving a flashing LED from 1.5V using a charge pump. I've got one blinking away on my bench as I write this. It uses a PIC.

[Chris Smith]

A novel way of regulation comes from key chain lights. I have a blue led that runs directly off of two button batteries, and these batteries were chose because their max deliverance of current is 40 mills. So basically two batteries and one led, are in harmony with out the need for any regulation. The led tries to draw the maximum voltage and current, and the batteries deliver their best, which happens to be the correct amount of both. And, the batteries are paper thin.

[jollyrgr]

Did I miss something? The original post wants to use 30 LEDs and one 1.5V battery. Even if you put a 3.6 volt AA sized battery you would still need to put the legs in parallel. Then figure 20mA each LED. That is a 600mA draw. Would coin cells or even a 3.6 volt AA sized battery last long? I don't think so. What did I miss?

[Sterling Martin]

Ok, Ok, Ok, I thought maybe someone would "burst my bubble" and tell me that I can't run 30 leds off of one battery forever! Now then, if that is not possible, about how many leds will I be able to power up for about half an hour? I will be using red leds. Brighter is better! <p>-------------------------------------------------
There are circuits for driving a flashing LED from 1.5V using a charge pump. I've got one blinking away on my bench as I write this. It uses a PIC. !!✰✰♨✰✰♨✰✰♨♨ Just kidding! A couple of inverters, Rs, Cs and a transistor.
------------------------------------------------- <p> I'm interested in the charge pump with one AA battery! I'm a little limited on space, so I don't feel that I can afford two AA's. Blinking would be O.K. too, as long as I can get at least 50% duty cycle.

[Mike6158]

Check this guy out too-<p>Odd Ones LED Site

[Sterling Martin]

I have decided to try and squeeze two AA batteries into the puck. That will give me 3 volts to work with. How many red leds can I stick in it and expect it to last half an hour?

[Mike6158]

A good alkaline AA battery is rated at around 2000mA/hr of capacity. That doesn't mean that you can pull 2.0 amps out of it for an hour. Typical discharge rates for a AA battery are in the order of 50mA which just happens to be what the max If of a 5,000 MCD red LED is. Vf (typical) is around 2.0 V and If (typical) is around 20mA. The problem is the lack of voltage. There isn't enough to run them in series and you can only run 3 in parallel (just barely over the 50mA rate)<p>A 9V alkaline battery is rated at 9V/500mAH/15mA drain. Since you only want 30 minutes I would go with this. 4 LED's in series with a 51Ω dropping resistor would do the trick. If you ignore the drain value you can go with something like 3 series / 5 parallel (total of 15 LED's) with a 30Ω dropping resistor (I know... heresy but it can be done). The problem is that the battery needs to supply 105mA of peak power to supply the LED's and the resistor. That is almost an order of magnitude more than what is recomended for current draw from a 9V battery. Somewhere around here I have a little board with 6 LED's in parallel, with individual dropping resistors and a 9V battery connector. It ran for about 4 hours. The LED's were mixed in pairs. IE (2) Red, (2) Blue, (2) UV. Each color required a different resistor. The UV's were not on when I came home 4 hours later. Red and blue were going strong. That was around 160mA of current draw.<p>Now... if you go with a boost regulator as Philba mentioned then you can drive quite a few more LED's. The problem is the boost regulator, even using SMD's, is going to take up space.<p>Really... A breadboard, a 9V battery, some resistors, the Excel spreadsheet on the N&V site, and a fistful of LED's will tell you all that you need to know. Plan out the series parallel circuits with the Excel sheet. It will tell you what resistor to use. Build the circuit. Plug it in and watch your work shine. It's a good way to learn. If you're real curious insert a mA meter in one leg of the power circuit...<p>Here's some pretty good battery info:<p>Battery Info<p>[ January 24, 2005: Message edited by: NE5U ]</p>

[philba]

you might also look for high efficiency LEDs - lower current draw, higher mCDs. DigiKey's parameteric search can help you there.<p>Also, if you go with a flashing LED style, the battery life will be extended by approximately the amount of off time. Or, you could flash twice as many LEDs for ½ sec on and ½ sec off at double the brightness (er, current drain) for the same amount of time. The effect would be perceived as twice as bright. If you PWM the LEDs at say a 1Khz, the moving puck will make a dashed line and might actually look kind of cool. I think you could do this with a 555. But, as Mike said, you might have trouble fitting the components. If you want to get really fancy, different colored LEDs flashed at different times to create some interesting effects. Hmmm, where's my kids hockey stuff...<p>[ January 24, 2005: Message edited by: philba ]</p>

[philba]

Here's the circuit I've got.
<p>Looking at it, the electrolytic is the biggest thing there. It measures 600 mil high, 300 mil diameter. It *might* squeeze in. C1 is 0.1 uF and R1 is 10 mΩ. This gives around a one second period with a 40 mS pulse. dropping R1 to 1 mΩ should give you close to a 50% duty cycle. I'm not sure this will work for you, though since you want BRIGHT.<p>By the way, the circuit is wrong - the transistor in my breadboard is a 2N4401. It's some what beefier than the 3904.<p>editing trying to get the image url right. sigh...
[ January 24, 2005: Message edited by: philba ]<p>[ January 24, 2005: Message edited by: philba ]<p>[ January 24, 2005: Message edited by: philba ]</p>

[Chris Smith]

Actually Thats two batteries for each led, and the author will already find out the impossible, so no need to proceed past the basics. <p>I figue 4 to 6 leds as being relistic, will fit in a puck and the acompanying batteries will keep the puck under 20 pounds, that 30 leds and the power source would otherwise become????

[Sterling Martin]

Thanks for the great input, philba! I'm going to breadboard your circuit as soon as I can! And yes, I did kinda lead everybody astray with th "up to 30 leds" earlier. I actually just want a hockey puck that would be really visible in low lighting conditions. You are supposed to understand what I mean, not understand what I say !

[Chris Smith]

Sterling, have you considered the G forces on the battery and what type might survive these forces? The internal design of the letter type batteries, A through D,... is a core surrounded by a gel. <p>If the core should be over accelerated into the gel, it will fail.

[haklesup]

for me it all boils down to this.<p>1. Select the most efficient LEDs you can find. That is one with the highest brightness for the lowest current. or at least the lowest current and a brightness you can live with.<p>2. Choose a battery with the highest energy density. Today that would probably be a Li-Ion camera battery. Use a battery website to check the specs for charge/discharge and mechanical shock resistance. (you may need to add high density foam to the battery compartment to protect some batteries)<p>Once you have found these two idealized components than you can calculate how many LEDs you can power for an hour and then select the current limiting R to suit the requirements of the LED<p>IMHO 4-6 LEDs on each side or around the perimiter (since it is likly to flip over often) would probably be enough especially if you diffused the light over the surface of the puck, like with an epoxy layer doped with glitter. (on second thought most epoxies would probably not stick to the hard rubber of a puck for long, perhaps clear silicone bathtub sealer would work or ski wax(melted)) <p>SMT LEDs are quite small and would be easier to embed into the puck surface with a little careful soldering to 30# wire wrap wire fitted into a slit cut with a razor. Choose a combination of series and parallel connections for the LEDs and distribute them so that if one branch of your circuit goes open, most of the lights still remain lit.<p>[ January 25, 2005: Message edited by: haklesup ]</p>

[philba]

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by haklesup:

...
SMT LEDs are quite small and would be easier to embed into the puck surface with a little careful soldering to 30# wire wrap wire fitted into a slit cut with a razor. Choose a combination of series and parallel connections for the LEDs and distribute them so that if one branch of your circuit goes open, most of the lights still remain lit.<hr></blockquote><p>I've had trouble finding really bright SMT LEDs. I suspect that part of it is due to the fact that most don't have a lens to focus the lignt into a tight(er) cone.