Have you considered electro-luminescent wire? It is a very low voltage, low current application. 2 AA batteries gives an hour or more of light. Very east to form, and very high tech-looking.
Here is a link to check out. It will be more cost effective too, considering the amount of time you will put into the build. $12.99us for 10ft with controller.
http://www.bewild.com/newpoeleledl.html

 

To measure the voltage of a battery it must have a load.

To measure the forward voltage of an LED then simply connect the voltmeter across the LED. The positive lead of the voltmeter connects to the positive wire of the LED and the negative lead of the voltmeter connects to the negative wire of the LED.

Your LEDs have an extremely narrow viewing angle (6 degrees). That is how the manufacturer exaggerates the brightness spec. Are they useable? Can you see them if they are not shining directly at you?

We recommend not connecting LEDs in parallel. We don't know the minimum and maximum forward voltages of your LEDs so we cannot calculate a safe current for parallel LEDs.

Look at the datasheet of a little 9V alkaline battery to see that it has a very low capacity.
I posted a graph showing the voltage dropping with a load of 250mA:
1) When brand new its voltage is 9V.
2) In 15 or 20 minutes its voltage is 7V.
3) In almost 1.5 hours its voltage is only 6V.
Maybe you can use it if you don't mind the LEDs dimming.

 

To measure the voltage of a battery it must have a load.

To measure the forward voltage of an LED then simply connect the voltmeter across the LED. The positive lead of the voltmeter connects to the positive wire of the LED and the negative lead of the voltmeter connects to the negative wire of the LED.

Your LEDs have an extremely narrow viewing angle (6 degrees). That is how the manufacturer exaggerates the brightness spec. Are they useable? Can you see them if they are not shining directly at you?

We recommend not connecting LEDs in parallel. We don't know the minimum and maximum forward voltages of your LEDs so we cannot calculate a safe current for parallel LEDs.

Look at the datasheet of a little 9V alkaline battery to see that it has a very low capacity.
I posted a graph showing the voltage dropping with a load of 250mA:
1) When brand new its voltage is 9V.
2) In 15 or 20 minutes its voltage is 7V.
3) In almost 1.5 hours its voltage is only 6V.
Maybe you can use it if you don't mind the LEDs dimming.

You say one have to connect the voltmeter(multimeter?) across the LED to measure forward voltage, is this in a closed or open circuit?[MIXING RESISTANCE WITH VOLT, OPS]

In a open circuit, i.e. connecting the leads from the multimeter to the legs of the LED, with no power source, the LEDS lights up dimmly, and the multimeter shows a random number, from 0.8 to 1.9 for a second, before the display goes blank, except "1" to the left. I read somewhere that this is because the diode function on some multimeter isn't sensitive enough to measure LEDs, but made for other kinds of diodes.[I MIXED RESISTANCE WITH OHM, SORRY. I AM TALKING ABOUT RESISTANCE HERE]

In a closed circuit, measuring across the LED displays 2.04V, for almost every LED.

The LEDs are much more stronger if looked directly at, but can be seen as dim light from the sides. This is not necessary a bad thing in medallions, because they can remind of gold or diamonds that only reflects from some angles.

The specs says 20V and 20mA. If it's possible to measure volts in a closed circuit, those measurements are pretty spot on with 2,04V.

If I drop from 20mA to 10mA, it should be little risk running them in parallel for 5 minutes? I am thinking about doing some other LED projects, and will then use a plugged source, with the luxury of running LEDs in more stable circuits that can last for long times.

A 75/3 design requires (9V-6V)/25 legs X 0,01A = 12 Ohm. That's perhaps below the thresold of acceptable resistor value for LEDs? I can else live with the dimming, as it's only for 5 minutes.

It's acceptable if the batteries get a little warm, but not hot and on the verge of exploding.

 

I suspect the specs for LEDs here, at 20mA is a maximum value, and that those LEDs are effective at 10 to 15mA, too. 18 LEDs in parallel sharing only 20mA(3V supply and a single 50 Ohm resistor) are ok lit, more than dim. Tried one LED, 3V and 100 Ohm, giving about 10mA, and the difference isn't that huge from 20mA. I am getting light spots in my eyes after staring into it(6 degrees, clear water glass LEDs).

The resistor sets the current, not the LEDs. For 3V at 0.01A (10ma) you need 100Ω.

So a setup with 76/2, 10mA in each leg, draining 0,38A could work too, or even a 75/3, draining 0,25A? It would be safer, too, help avoiding the popcorn effect you mention in your article?

D type batteries is fine if better. As long the players can move freely, it does not matter what kind of source they use. But this leaves out car battery and wall plug. But what about a 9V E block alkaline? The size of 2 AA batteries, but 9 VOLTS! Can that work here?

9V batteries are a really bad idea. They have less power than AAA, AA, or D cells, and would discharge quickly.

Is it a limit how thin the wire should be here? I checked limits of AWG wires, and looks like at least 24 is required, with limits of 0,577A for power transmissions. I got a 32 wire here, that is pretty easy to wrap around the leds and quickly solder, and worked with 18 LEDs, but looks like a bad choice according to the charts. Will a 32 wire still do ok for 10 minuttes?

Yes, would like to see a schematic of 72/2 if it don't take much timeto make one.

Wire is the least critical part. 24 gauge and up are fine.

I am not sure how to measure the forward voltage of the LEDs, but after some reading, my guess is that one have to measure voltage before and after the LED. Before it's 0,04, and after, it's 2,07. At the V+, the battery, it's 2.99V. The circuit got a 52,6 Ohm resistor. 2,07-0,04 is 2,03 Vf?

I have uploaded a drawing of the circuit and where voltage was measured, in case this don't make sense. If it's a better or another proper way to measure Vf I would be happy to know. The multimeter don't work very well for this.

What is the minimum of Ohms in resistors for LEDs for good protection, if one can ask such a question?

You measure the voltage across the LED while lit, this is Vf, or forward dropping voltage.

As mentioned, it is the resistor, not the LED, that sets the current.

Knowing the Vf is important, critical even.

This illustration....

What is the resistance? 32.6Ω? If so then the LED is drawing 30ma, way too much. (1V/32.6Ω) Did you actually measure these voltages?

I suspect 100Ω is where you need to be, but would like confirmation about the numbers. 2Vf seems low to me, but it could be correct. This kind of math doesn't need to be exact, the first 2 digits are all that is important.

 

The resistor sets the current, not the LEDs. For 3V at 0.01A (10ma) you need 100Ω.

9V batteries are a really bad idea. They have less power than AAA, AA, or D cells, and would discharge quickly.

Wire is the least critical part. 24 gauge and up are fine.

You measure the voltage across the LED while lit, this is Vf, or forward dropping voltage.

As mentioned, it is the resistor, not the LED, that sets the current.

Knowing the Vf is important, critical even.

This illustration....

What is the resistance? 32.6Ω? If so then the LED is drawing 30ma, way too much. (1V/32.6Ω) Did you actually measure these voltages?

I suspect 100Ω is where you need to be, but would like confirmation about the numbers. 2Vf seems low to me, but it could be correct. This kind of math doesn't need to be exact, the first 2 digits are all that is important.

I was unclear about resistance and volt. I know very well that resistors decide resistance, not the LED. My confusion was that I somehow had it that one could measure Vf with the multimeter, using a diode function. What I did not understand was why it was in the resistance section of the multimeter, but once I got the explaination how to measure Vf, it all got clear to me. Of course one can't measure volt without some power, dang. I need some time to get used to this Volt, Resistance and Current triangle!

I have trouble using this diode function on the multimeter with LEDs, as explained in two posts up, guess I will figure out that later.

Ok with 9V, makes sense. Thanks for confirmating wire gauge.

The Ohm in the resistor from the drawing is 52.6, making it 19mA(1V/52.6Ohm), unfocused it looks like 32.6 yeah!

Curious why you think Vf is low? I found a PDF file with all the specs for those LEDs, as I understand it, those LEDs got 2Vf.

http://www.everlight.com/upload/product_pdf/383_2UYC_S530_A3.pdf

 

Have you considered electro-luminescent wire? It is a very low voltage, low current application. 2 AA batteries gives an hour or more of light. Very east to form, and very high tech-looking.
Here is a link to check out. It will be more cost effective too, considering the amount of time you will put into the build. $12.99us for 10ft with controller.
http://www.bewild.com/newpoeleledl.html

Thanks for link. Must be a lot of fun going to raves nowdays

 

You sound like you have it. Any other questions feel free.

I suspect it will be a let down, since it is so simple, but here is the schematic you asked for...

You want to know why I wonder about the 2.0 Vf? It means these LEDs are very old units. If you read LEDs, 555s, Flashers, and Light Chasers this paragraph sums it up...

The forward dropping voltage, or Vf, of an individual LED is very stable. Go below this voltage and the LED stops conducting. This LED is assumed to be 2.5V, pretty standard for a modern red unit. The target current is 20ma. Going though the math (using Ohm's Law) the resistor is 325Ω. Since 330Ω is the nearest standard resistor value 330Ω it is.

Here is the approximate Vf of most LEDs:

......... Older Generation ... Newer Generation
Current ....... 10ma ................... 20ma
Red ............ 1.5V .................... 2.5V
Yellow ........ 2.0V .................... 3.0V
Green ......... 2.0V .................... 3.0V
Blue ....................................... 3.5V
White ..................................... 3.5V

 

You sound like you have it. Any other questions feel free.

I suspect it will be a let down, since it is so simple, but here is the schematic you asked for...

You want to know why I wonder about the 2.0 Vf? It means these LEDs are very old units. If you read LEDs, 555s, Flashers, and Light Chasers this paragraph sums it up...

Thanks for the design. The simpler the better, and I will go for that one! The only let down is the amount of resistors, but I have learned much about circuit design here. The batteries will have to deliever 750 mA with this design. Will the two AA batteries manage this without getting dangerously hot? You wrote the batteries would become hot, or was that for a different solution? That's my only concern.

 

If you have the hand for it, you can use surface mount resistors to take up less room, or even replace the leg of the led with the resistor.

 

Increase the voltage so more LEDs can be in chain, and you can reduce the number of resistors and the amount of current. The design above will draw around ¾A. However, if you were to use this schematic...

uses 5 AA batteries but only pulls ¼A. This is because the LEDs can share current.

With 6 batteries in series (9V) you could have 4 LEDs in each chain, requiring 0.19A. More voltage, fewer amps. A 9V battery drops its voltage really fast, so it would still be a poor choice for this design. C cells would give hours of entertainment, AA probably under an hour or so.

 

Increase the voltage so more LEDs can be in chain, and you can reduce the number of resistors and the amount of current. The design above will draw around ¾A. However, if you were to use this schematic...

<drawing>

Did you post a new schematic here, I am in a electronic store, and a bit curious about your new suggestion? Only see <drawing>?

 

Look again. I was actively drawing when you saw that.

 

BTW, with 6 batteries, you go back to 100Ω

 

BTW, with 6 batteries, you go back to 100Ω

Thanks, I am amazed at the help I have gotten here! That circuit will be handy in case. With a better understanding of LED circuits, I am ready to assemble all of this together. Bought 300 resistors, at 100Ω today. Will post images when done.

After buying a new fuse(F 200ma/250V) to the meter, it's behaving different. Measuring currents of a simple circuit with the meter at 200mA, with one LED with a 3V source, Vf at 2.03 and 52,6Ω resistance, the meter shows 17.6mA, while it should be 18.4mA(3V-2.03V/52.6Ω=18.4mA). At 20mA, the meter says it's 15.3mA current in the circuit. Looks like I pushed the limits too far when frantically testing everything possible in a simple LED circuit with this cheap 10$ MAS830L meter. At least I will be much more careful with the next meter.

 

Good meters can be had for $5. Everyone has blown one one time or another. You can measure current by measuring voltage across the resistors. This is basically how current meters work. For 100Ω, every volt equals 0.01A.

 

Good meters can be had for $5. Everyone has blown one one time or another. You can measure current by measuring voltage across the resistors. This is basically how current meters work. For 100Ω, every volt equals 0.01A.

Thanks for another very handy tip!

 

It wasn't that hard to assemble those LEDs with resistors. Again, thanks for the help!

Now I started to think about making those LEDs go nut and flash. It's easy with one LED, but again, I want to push some limits here, and flash 76 LEDs.

I found some very helpful schematics at http://www.techlib.com/electronics/flasher.html

I have adjusted the schematic in figure one, to fit a 3V source, instead of 9V, and made it flash about 600 times a second, by trying different resistors.

My two questions, how do I know how much mA the LED is eating when it's flashing? It's hard to tell while flashing, and I only get 1-3 mA with the multimeter. Not sure if the LED receives 10-20 mA, or the multimeter not is quick enough.

How do I calculate the voltage divider at the base of the PNP transistor when it's a LED in the circuit? I know the formula of adding ohm of all the resistors, divide, multiply with voltage, and subtract from source voltage, but a bit confused which resistors to add up and what to do with the voltage forward of the LED? It's ok if this question is too general, and I need to do more reading. I want to know, because I want to raise the voltage thereshold at the base of the PNP transistor so the capacitor charges up more and the LED get more current.

See attachment for schematic and how this circuit looks with breadboard(the large cylindre component is the 100uF capaciator, but guess you experts can see it easily).

To flash 76 LEDs, I plan to use the schematic to the left(see link in beginning of this post), in figure two, the one with a power transistor. I found a power transisor with 1.5A, that should handle the .76A of current for 76 LEDs in parallel. I plan to use a 100uF capaciator.

Is this a good plan, or is it laughable for some reasons? I am also curious if it's possible to use a smaller capaciator with other resistors, as the 100uF capaciator is a bit big to fit at the back of a medallion. I have tried with a 100nF capaciator, but didn't have any success with any visible flashing. Else, 100uF is doable, just curious if I can go smaller.

 

I don't think there is much benefit going this route, you can extend the battery life, but the added complexity isn't worth it.

The math I've tried to teach still applies for the instantaneous current, which is also the max current an LED uses. By turning an LED on and off very quickly you reduce the overall brightness and reduce the average current, which could have been done with the resistor.

There is a another variation called a joule thief, that uses less voltage than the LED drops. It boosts the voltage feed to LEDs, also switches, and by it's nature sucks every last bit of energy out of the battery before going dark. Most batteries actually have a lot of energy left in them when they stop working, a joule thief uses that energy too.

Here is a thread that has a good example of what I'm talking about.

http://forum.allaboutcircuits.com/showthread.php?t=31208&highlight=joule

Complexity is not your friend in a project like this, the original design you were looking at doesn't have a good benefit. Were you going to have one for each LED?

Or are you after special effects? I think I pointed this out to you early on, it covers these basics too...

LEDs, 555s, Flashers, and Light Chasers

The other side of this, have fun. If it is something you want to do then go for it. One minor detail, the breadboard didn't seem to have an insulation on those wire, a very bad idea if so.

I think you may be getting the electronics bug. It's how I started, building what I used to call do nothing circuits.

If I contradict myself during this post know I just got up. I'm a terrible grouch in the mornings, what passes for morning for me at least.

 

Sorry, I forgot to say the flashing is for special effects, not saving energy.

It's my guess the LED flash 600 times a second, and the goal is to make it look manic. The flashing very visible. Looks like fast strobing.

It's cool to try to follow the current in a circuit, so part of reason why I did it with capacitors and transistors I found here. Electronics are fun yeah I did reread your LED/flash/555 article this morning, and discovered now that the local store also have 555 timers, and, wow, they are cheaper than I thought. As I only have 1 week left to figure out how to add flashing to this project, it looks much easier to try a 555 timer solution this time. I will use the 100Ω resistors+LEDs in parallel, and only add the flashing effect to this circuit if possible without altering the orginal design.

In chapter 10: Transistor Drivers, you have an interesting schematic in figure 10.1, the common-emitter mode, that I am considering to use. The NPN transitor I got here can handle 300 mA, that's below 0.76A. So would it work if I replaced a ordinary BJ transistor with a power transistor?

The emitter of the NPN power transistor will then be connected to the ground, and the base to the 555 timer, and the collector to the LEDs in parallel, just like in the schematic in figure 10.1 in your article. The 555 timer will work like a gate opener, so the power transistor acts like a switch, conducting 760 mA of current from the two AA batteries to the 76 LED circuit I have built with 100Ω resistors attaced to every single LED.

I am also considering adding a seperated power source for the 555 timer oscilliator, perhaps a single AA battery or something smaller in physical size, so I don't have to worry about sharing the two 2 AA batteries that are going to lit all those 76 LEDs.

The transistor I have in mind is a BD137, with specs NPN, 60V, 1.5A, TO126.

Will this work? I somehow have a feeling that it can't be this simple, or is it?

Btw, the uninsulated wires on the breadboard are due to it beeing hard to get hold of insulated solid wires around here, and know it's not preferable, yes.

 

The minimum supply voltage for an ordinary 555 is 4.5V so use 6V.

The base of the transistor needs a current-limiting resistor in series with it to prevent the 555 from blowing up. With a 6V supply use 68 ohms.

600 times per second will look like the LEDs are continuously turned on. 20 times per second is very fast flashing. Maybe you want 10 times per second.