Notes on LEDs
LEDs come in all shapes and sizes, but the 3mm T-1 or 5mm T-1¾ are probably the most common.
The die is an itty bitty cube of semiconductor, the composition of which determines the color of the light given off. It sits in the bottom of the die cup, which has reflective sides to reflect the light emitted by the die toward the dome end of the LED. The epoxy body is shaped to act as an inclusion lens and focus the light into a beam. The distance from the die cup to the domed end of the lens determines how tightly focused is the resulting beam of light. Some LEDs have flat or even concave ends to dispurse the light into a wide beam.
White light is a mixture of all the colors. Color Temperature is a measure of the relative amounts of red or blue - higher color temperatures have more blue.
Remember that this is a measure of color, not brightness, so don't freak out because moonlight is "hotter" than a carbon arc! It just means that the color is bluer, that's all.
The total power radiated as light is radiant power or radiant flux, and is measured in watts. How bright the object appears, however, will depend on two additional factors:
To quantify the first, we must introduce the concept of the steradian, a solid (3-D) angle. Think of a cone with the apex at the emitter.
sr = 2 π (1 - cos(θ/2))
where sr is the solid angle in steradians, and θ is the beam angle.
Luminous flux is measured in lumen, while luminous intensity is measured in lumen per steradian, also called a candela.
The relationship between luminous flux, luminous intensity, and beam angle means is that focussing a given LED into a tighter beam (decreasing the beam angle) will increase its luminous intensity (brightness) without actually increasing the luminous flux (amount of light) it puts out. Keep this in mind when buying LEDs for illuminating purposes - a 2000 mcd 30° LED puts out just as much light as am 8000 mcd LED with a 15° viewing angle. (The angle is half in both width and height, so the beam is four times as bright.) This is one of the reasons that ultra-bright LEDs are often "water clear", to keep the light going in one direction and not diffuse it all over the place. The brightness of LEDs is measured in millicandela (mcd), or thousandths of a candela. Indicator LEDs are typically in the 50 mcd range; "ultra-bright" LEDs can reach 15,000 mcd, or higher.
By way of comparison, a typical 100 watt incandescent bulb puts out around 1700 lumen - if that light is radiated equally in all directions, it will have a brightness of around 135,000 mcd. Focused into a 20° beam, it will have a brightness of around 18,000,000 mcd
1 lux = 1 lm/m²
An older unit, the lumen per square foot, or foot-candle, has become obsolete. If luminous flux F falls at a normal incidence on an area A, the illuminance E is given by
E = F ÷ A
Most light sources do not radiate equally in all directions; it is useful to have a quantity that describes the intensity of a source in a specific direction, without using any specific distance from the source. We place the source at the center of an imaginary sphere of radius R. A small area A of the sphere subtends a solid angle [omega] given by [omega]=A÷R². If the luminous flux passing through this area is F, we define the luminous intensity I in the direction of the area as
I = F ÷ [omega]
The unit of luminous intensity is one lumen per steradian, also called one candela, abbreviated cd:
1 cd = 1 lm/sr
The term "luminous intensity" is somewhat misleading. The usual usage of intensity connotes power per unit area, and the intensity of radiation from a point source decreases as the square of distance. Luminous intensity, however, is flux per unit solid angle, not per unit area, and the luminous intensity of a source in a particular direction does not decrease with increasing distance.
I = 1200 lm ÷ 2π. sr = 191 lm/sr = 191 cd.
The luminous intensity does not depend on distance.
LED Brightness - IR & UV LEDS
Question: How bright is an IR LED?
Rseries = (V - Vf) / If
where Rseries is the resistor value in ohms, V is the supply voltage, Vf is the voltage drop across the LED, and If is the current the LED should see.
Is a series resistor really necessary?
In a word, no. However, neither is a seat belt. Both are "cheap insurance" against disaster.
A series resistor is not necessary if the voltage can be regulated to match the LEDs Vf. One way to do this is to match a battery to the LEDs. If your LED's Vf is 1.2 volts, you can string ten of them (10 x 1.2v = 12v) in series and power them from a 12 volt battery with no series resistor.
However, you must be sure that the battery is capable of supplying the expected voltage - not only do batteries often supply a bit more than the rated voltage (a "12v" car battery for example, reaches 13.8v at full charge), but different types of batteries have different internal resistance, which results in different voltage "sag" under different load conditions.
Here is a small table giving typical internal resistances of different types of battery. Notice how the alkaline AA battery has five times the internal resistance of the NiMH AA battery, and how the alkaline D battery has eleven times the internal resistance of the NiCad D battery.
|Battery Type||Internal Resistance
|Note: internal resistances shown above are at full charge and room temperature.|
| Also, as the battery discharges, the voltage will drop significantly. Because of the steep voltage/current curve (see the graph under "Using LEDs" above), small changes in voltage will result in large changes in current.
Adding resistance to the circuit will help stabilize the voltage across the LED. In a sense, an LED and resistor in series act as a voltage regulator.
In series with a resistor, an LED will see the entire voltage drop across the circuit if it is not conducting. As soon as it starts to conduct, however, its resistance drops to almost nothing - just a few ohms. The voltage drop across the resistor rises, and the voltage drop across the LED remains almost fixed. The voltage drop across the LED cannot drop, as the LED would turn back off, which would raise the voltage drop across it and turn it back on again. Instead, the voltage drop across the LED remains just above the threshhold voltage even as the supply voltage rises. Any further increase in supply voltage increases the voltage drop across the resistor, but not the LED.
Look what happens when the voltage, supplied to a 150 ohm resistor in series with an LED ( , varies from 4.5v to 5.5v.
| You can see how flat the Vled curve is - it varies only 0.03 volts even as the supply voltage varies by 1.0 volt. Even with this small rise in Vled, Iled increases 6mA.
The LED in question has a threshold voltage (Vthreshold) of 1.9v, above which it has a dynamic resistance (Rdynamic) of 4.55 ohms, and draws 20 mA at 2.0v. (This is an example of a real LED, see the graph under "Using LEDs" above) The supply voltage is 5v, and Rseries is 150 ohms. Here are the formulae:
Ve is the voltage above the threshhold, I is the current through the circuit, Vseries is the voltage drop across the resistor, and Vled is the voltage drop across the LED.
LEDs driven by a simple current regulator
Driving LEDs with a 3 terminal regulator
A constant current source, can be achieved by using an adjustable voltage regulator circuit, the LM317.
The current through the LED will be I=1.25/R. The above example will provide a constant current source of 20mA.
If you run diodes in series, it'll take another 4 volts (for white LEDs) or so of input voltage per extra diode to make the circuit work. Don't try to parallel diodes from a current regulator, the diodes will not share current properly unless you add a balancing resistor in series with each diode. In that case, you might as well use the 7805 constant voltage regulator.
You can use this method to drive the high intesity LEDs. For higher currents, you need to make sure that the resistor has higher power rating. ie 1 Watt.
Running these higher power leds using a constant current source, makes sure that Thermal Runnaway is not reached.
Driving LEDs with AC