Question about lighting models.

[Vince]

This is a pretty helpful site, explains a bit about use LEDs:

http://www.eatel.net/~amptech/elecdisc/led.htm

[cschell]

The LEDs in question are superbrights, I guess. 5mm, 10mA

there are 6 "white", and 2 blue superbrights inside the ship, and two regular red LEDs (non-superbright) in the front of the nacelles to light the bussard collectors.

The 5 in the base are white superbrights also, all are 5mm size.

Hi,

I must say I'm surprised that all of your LEDs are even lighting. White/blue LEDs have a much higher forward (turn on) voltage than most red LEDs. With everything connected in parallel your red LEDs should light at a voltage too low for your white/blue LEDs to even turn on. This would mean that all the current from your batteries (probably about 150 mA for alkaline) would flow through your two red LEDs, most likely destroying them. Of course once they were destroyed all of your white/blue LEDs would then light. Are you sure your red LEDs are still working? (Are they shining through red plastic, perhaps the red is lighting from backlight from the nearby blue LED?)

Anyway, connecting a bunch of LEDs in parallel without individual resistors is not generally a good idea. (Too late now I know.) If you want to do this you need to make sure all of the LEDs have matching forward voltages at the current you wish to run them. (Even two LEDs of the same brand / model will likely have varying characteristics.)

Let me try to explain this with an analogy. Think of an LED as a spring loaded trap door. Now imagine a room with several spring loaded trap doors in the floor of that room, and lets say that the doors are all water tight when they are closed. Now you begin filling that room with water. As the water level increases the pressure on the trap doors will rise, until eventually there is enough pressure to open the trap door with the weakest spring. Water will begin flowing out of the open trap door. The trap door will continue to open wider (and allow more water to flow out it) as more water is added to the room. This will continue until one of three things happens.

1) The water level rises to the a point where it opens the trap door wide enough that the rate of water flowing into the room is equal to the rate of flow out of the trap door. At this point equilibrium has been reached and things will continue in this state forever.

2) The water level rises to a point where the pressure on the trap door is too much, causing the trap door or fail catastrophically. To complete the LED analogy I'll have to add at this point that the trap doors have a failure sensing mechanism such that once a catastrophic failure is detected an emergency bulkhead completely and permanently seals the opening created by the damaged trap door. (Sorry, I know it's weak. ) Thus no more water flows out of that trap door again, ever.

3) The water level rises to the point where the second weakest trap door opens, and now water flows out of both open trap doors. However, there is still (and always will be unless the first door fails) more water flowing out of the first trap door than the second.

This process will repeat itself until eventually the system reaches a state where the rate of water flowing into the room is equal to the rate of water flowing out of all the open trap doors. As the water level rises more and more trap doors will open, but the chances of the weakest trap doors failing also increases. If the flow of water is too strong it would eventually destroy all of the trap doors.

Ok, I hope that was fairly clear. Now the analogy is that a bunch of LEDs wired in parallel is equivalent to a room with several spring loaded trap doors in the floor. The water level in the room (or the pressure it places on the doors if you prefer) is equal ot the voltage applied to all of your LEDs. The water flowing out of open trap doors is equal to the current flowing through the system. Catastrophic failures are when you too much current flows through and LED, destroying it and effectively removing it from your circuit. The strength of the spring in the trap door is equivalent to the voltage at which the LED will turn on and start to carry current. (Just like door opens and lets water start to flow.) The more current that passes through the LED, the brighter it glows.

So in your system, you have 13 doors with strong springs (the white/blue) and two doors with weak springs (the red.) The red will either be glowing very, very brightly before the white/blue turn on, or they will have already failed. (Unless you have red LEDs with unusually high turn on voltages.) But even when all of your LEDs turn on, they will not glow with equal brightness, and they will not have the same current flowing through them.

Pairing each LED with an individual resistor is like raising the trap doors to different levels in the room. You place the strongest trap doors on the bottom, and weakest near the top. That way the weakest doors will have less pressure on them and all the doors will open at approximately the same time and the let the same amount of water flow through them.

Ok, end of analogy. Now let's go back to your situation.

Red LEDs typically have a turn on voltage around 2 volts. White/blue LEDs typically are around 3.6-3.7 volts, although some can be higher. Your red LEDs are likely dead, or being overdriven to the point where they won't last that long, so I'm going to ignore them.

To clarify a lot of what has been said in this thread, I'm going to bring in the equation that relates voltage, current, and resisitance (Ohm's law.)

It is simply: voltage = current * resistance (* = multiplication)

So if you have a 5 volts across 1 ohm, you get a current of 5 amps.

Most LEDs can handle 20 mA (.020 Amps) without a problem, although you stated yours were 10 mA, so if that is what they are rated for that's what I'll use.

Once an LED turns on, it starts conducting current through it with a very low resistance. This means that a small change in the voltage across the LED will create a large change in current through the LED. So while 3.6 volts might turn on a white LED with a few mA of current, 4 volts is likely high enough to destroy the LED. In your case, you have 13 LEDs in parallel. They are almost certainly not matched to one another, but let's assume that they are close. Alkaline batteries are not high current devices. AA Alkaline batteries are 4.5 volts, but only at about 150 mA. So your system is current limited and the average current going to each LED is about 150mA/13 = 12 mA. (Probably 8 to some, 14 to others, etc.)

Now, what would happen if you connected a 4.5Volt, 1A wall adapter to this same chain of LEDs? They would probably all be destroyed. So AA batteries are ok because Alkaline batteries can't provide enough current to damage your LEDs, and NiCD/NiMH batteries are low enough voltage (3.6 Volts) that they will most likely will not turn your LEDs on to a point where they will draw enough current to damage themselves. (NiCD/NiMH can source more current than Alkaline batteries).

Your typical AC/DC wall adapter is an unregulated voltage source. This means that it attempts to provide power at the rated voltage, but does not use feedback to ensure that is happening. So a 4.5 Volt, 1A wall adapter will attempt to supply 4.5V as long as the current draw is 1A or less. If you draw more current, the voltage out of the adapter will sag to less than 4.5 volts.

There seems to be an idea in this thread that wall adapters will supply exactly the amperage they are rated for. That is not true. They will supply up to that amount as needed, just like a battery. If you need less, they will supply less. Using a high voltage, low amperage wall adapter to run LEDs directly (without a resistor) is a bad idea. It can work because the adapter will sag to the lower voltage of the LEDs because it can't meet the current requirements to run the LEDs at its voltage rating, but this can cause the wall adapters to over heat because you are using them in a way that they were not designed to be used and basically asking them to supply too much power. (Note that you can get current sources, which supply a fixed current rather than a fixed voltage, but that is not what a typical wall wart does.)

So what I would recommend is that you get a power supply that can supply enough voltage and current to safely drive your 13 LEDs and use an appropriate sized resistor. So lets say you have 13 LEDs all drawing 20 mA. (I know you're aiming for 10, but let's be conservative). That's a total of 260 mA. So a 5V, 500 mA adaptor is more than enough. Now we need a resistor to limit the current flowing into your LEDs. The equation for choosing the resistor value is as follows:

R = [(Power Supply Voltage) - (LED Forward Voltage)]/(Desired Current)

So if your LED forward voltage is 3.6 Volts. (And I DO NOT KNOW THIS, as I DO NOT HAVE THE SPECS FOR YOUR LED) it would go like this:

Power Supply Voltage = 5V

LED Forward Voltage = 3.6V

Desired Current = 10mA

R = (5-3.6)/10mA = 140 Ohms

Resistors also have a power rating. So you have to make sure your resistor can handle the power it will be dissipating. (It converts the energy into heat.)

Power Dissipation = (Voltage Across Resistor * Current through Resistor)

= 1.4V * 10mA = 0.014 watts

Resistors are commonly available as 1/4 (0.250) or 1/8 (0.125) watt resistors. So using one of either wattage resistor should be fine.

So, does that help at all?

And as someone else recommended, I highly suggest trying any power supply setup on an external set of resistors before connecting them to your model. It's a lot better to make a mistake on that set than have to rip th model apart.

Chad

P.S. I kept switching between forward voltage and turn on voltage. They are the same thing. Sorry for the inconsistency but I'm too lazy to fix it.