Need serious help with LED + voltage problems

Oki here's my situation. I'm currently making a Beatmania IIDX controller, and I ordered LED microswitches from a certain company. The LEDs on the microswitches run on 3.2-3.8 volts per LED. I have 9 buttons (7 for gameplay, 2 for the start and select), and so they are all soldered to the PCB, and to each other.

This is the diagram i'm using for this project.

http://home.ripway.com/2004-1/54890/12.jpg

Now, there's a lot of things I need to know concerning the voltage and how everything should connect. I made this diagram after looking at another diagram that was made for putting in a light kit for microswitches on a pop'n music controller. It works the same way as a IIDX controller, but I'm still unsure. Now for the questions.

1) Since all LEDs are chained together does that mean my power source will have to be in the range "OF THE TOTAL VOLTAGE OF ALL THE LEDs" combined?

2) Previously, I connected all the lights together, which altogether their total voltage should be in the range of 28.8-34.2 I gave it a power source voltage rating of 34.5, tried 3 of the LEDs and they burnt out. I'm going to have the LEDs replaced, but they are going to be replaced with bulbs of a different voltage rating. 2 bulbs will be 1.7-2.4 volts, and the other bulb is 3.6-4.0 volts. Since these bulbs have a different voltage rating than the other lights, should they have resistors connected to them of some sort? I heard of using them but I don't know how they work.

3) If there's any connection problems that you see in the diagram, please tell me because I have to know.

You're schematic makes my head hurt. I can't tell if the diodes are connected to ground. You're using a special chip to drive the LEDs? This is probably unneccesarily expensive. If you look at the voltage and max current rating of the LED, just calculate R=V/I_max where V is your powersupply subtracted from the voltage of the diode. Then just put in a resistor about equal to R in series with the diode. Problem solved. And no, you don't need high voltage for this. 9 volts should work nicely.

Speak up if I need to clarify any of this.

That's not a special chip to drive the LEDs. That's the PCB that the microswitches are wired to. Can you explain to me why it would only need 9 volts? I don't know much about electronics.

I'll try to explain this but it'll probably be a bit scattered.

LEDs are current driven. Given too much current they will burn out. This is what happen to your first diodes. The notable property of a diode is that it acts like a short when connected one way, and acts like an open when connected the other. They arn't perfect however and you will always get some constant voltage across them when they are suposed to be shorting. for LEDs this is typicaly around 2 volts. If you connect that straight to a battery, the battery will try to shove as much current through that it can, and this burns out the LED. That is why you need a resistor in series with it, to limit the current.

Now to the power supple. Assuming a perfect power supply, no matter how many times you tap it, it will give you the same voltage and as much current as the circuit is made to draw (remember Ohms Law V = I*R). Batterys are not perfet power supplies but are probably close enough in your set up. LEDs are current driven devices if you recall. Therefore it isn't your voltage you need to control, but rather the current. A battery has an internal resistance that is always in series with a circuit. So if you try to pull too much current, the internal resistance will start to noticably drop your voltage. From the looks of it though, you only have about 9 LEDs. I've used a nine volt battery to light up a lot more than that. Now the reason you need so much less that with the light bulbs is that LEDs are low power devices (at least in comparison) and light bulbs need a whole lot of current to light them up. LEDs on the other hand can be made to be extremely bright on just a little bit of current. If you find that your batteries are dieing too fast, you might want to put another 9 volt battery in parallel with the other one (this is where you want to use 2 new batteries and always changes them together) or consider using an AC adaptor.

LEDs are diodes, which have a forward voltage that is proportional to the current through them. At no point is the voltage constant. While it is true that batteries contain a thevenin resistance, it is usually quite small, an ohm or less.

To the OP, what you need to focus on is the necessary current for your drive. The answer of how much will depend on your specific LEDs as manufacturers make several various types. Unfortunately, your schematic is horrible and i can't make sense of it.

Can you provide a cleaner drawn circuit?

>>5
Unfortunately, no because I have no real knowledge with electronics :( If I find out how I probably will.

>>4
So. . . considering if I had everything set up like the diagram (with the 3 LEDs replaced), I could use just 9 volts to power it with resistors?

Can you link us to the specs for your LEDs?

>>5
The internal resistance is about 5 ohms for a 9 volt battery. Just incase you wanted to add to your list of random knowledge.

Also, yes it's true that the forward bias voltage does vary with the current, it's usually pretty constant for an LED in it's operating range. Most specs will give you a typical operating voltage.

http://www.controllica.com/spare_parts/spare_parts.html

LED specifications are there. As I've stated before, 2 of the LEDs will have 1.7-2.4 volt bulbs and one will have 3.6-4.0 bulb. the other 6 will have the normal ones.

>>9
Ok, I think it's all starting to come together for me as to what's going on.

So let me know if this is right. You have 6 LEDs with the operating voltage of 3.2-3.8V, 2 at 1.7-2.4V, and 1 at 3.6-4V. All of them are LEDs. Man I had that all screwed up.

Something you may want to look at is the luminous intensity spec of your LEDs. This is a measure of the brightness of the LED. Not all LEDs are created equal in this reguard. It's more of an asthetics thing though if you want to match them.

Ok, now for the nitty gritty. Do you want to use batteries for this or a power supply that you plug into the wall? Is there an existing power supply in the controller? In either case, I think your most reliable bet is to connect all of the LEDs in parallel. Each different type of LED will need a different current limiting resistor of a value depending on your supply voltage. A general schematic will look like this http://www.pitt.edu/~wdw5/LED.jpg

So for LED1 (the one you have 6 of) the operating forward voltage is between 3.2-3.8V. For simplicity, estimate 3.5V. So the volatage across the resistor R1 is (V_in - 3.5). You want a current across that resistor of about 30 mA (the maximum continuous forward current for that LED) to give you the maximum brightness of the LED (since that will mean that 30 mA is also going through the LED. So it all goes back to our good old friend Ohm's Law V=I*R. In this case [(V_in - 3.5)=(30 mA)*R1]. Solving for R1 and changing mA to A, you get

R1 = (V_in - 3.5)/(0.030)

If you were to use a 12V power source, this would solve to R1 ~ 283.3 Ohms. Using a realistic value (or one you can easily get at Radio Shack), 330 Ohms should do nicely for R1 (rounding up is safer than rounding down).

Oki so. . . I've decided to scrap that power jack connected to the whole thing, and I'm going to add resistors to the whole thing. I have some resistors good for 4.7k-Ohms. Will that work?

How many volts do you suggest for powering this thing?

4.7k is probably way too big. They might light, but not much. I highly suggest getting some resistors close to what you need to get the most bang out of youtr LEDs. Radio Shack carrys enough of a variety that you should be able to get what you need there. As for your supply voltage, I'de suggest either 9 or 12. If you're going to use an AC adapter, Make sure it can source at least 500 mA.

Alright thanks a lot. I'll post some info when it finally comes togeter.