LED BLOG I told them in this blog a lot about LED lights but does not hurt a
very brief overview before we tell you otherwise. They are really
excellent, because they have a very long life, between 50 thousand and
70 thousand hours, much more than a regular incandescent bulb, use only
a fraction of the electricity it requires a common light bulb, much
less than even CFLs. And if that were not enough, are small.
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But there is a problem, have not spread by all households, why not? One of the main reasons is that they emit a very white light that you can enjoy a common light bulb or a CFL. But that has come to an end, now there are LEDs with a light as white as any. In physorg is said that scientists in India have made a great discovery, which they named "The Holy Grail of lighting industry" have produced LEDs that emit pure white light suitable for homes, offices, etc. They have achieved by combining a series of materials that may not be those who would have preferred the green, as it contains cadmium which is toxic, although it will take a very small amount, much less mercury that CFLs have almost all current . Now it only remains for scientists to achieve that efficiency of the lamp is practical for everyday use, so we still have to wait a bit for these new LEDs.
| LED FLOURESCENT TUBES LED SAVING had spoken recently of the new solar chargers released by the company Powersafe specializing in energy saving technologies. The company also will make available to the Spanish market, from October, their first tubes LEDs SUPER BRIGHT The LEDs are low-energy lighting and high efficiency as explained in a previous article. In the case of proceeds of powersafe take the form of tubes LED lighting that allow energy savings of up to 80%, which ensures the company as we will see reflected in our electricity bill. According to information from the official website of the company these tubes have a lifespan of 50 thousand hours, five times higher than common fluorescent tube.The tubes come in three sizes, 60 cm, 120 cm and 150 cm and emit super bright white light, hence its name, which has the advantage of not emitting heat or flash. The 60-inch tube consists of 174 LEDs and uses only 8 watts of power. The 120 cm has 342 LEDs and 150 cm in 420, the cost is 15 watts and 18 watts respectively. To compare energy saving fluorescent tube standard 120 cm consumes 36 watts, at least more than double the size of the LED tube. And the LED tube light brings a lot more, spending less electricity. |  |
ENERGY SAVING WITH LEDS
Via LED SAVING blog, I just read an interesting article that discusses the
benefits of replacing traditional lighting of street lamps by LEDs.
An LED is a light emitting diode, that is, a semiconductor device that
emits light when electric current flows through the. Its great
advantage over traditional tungsten filament bulbs, and even off the
energy saving light bulbs, is its energy efficiency:
*
The LEDs do not have a tungsten filament as the light bulbs. This makes
them more shock resistant and lasts longer because it does not depend
on the Filamente burning is finished (When the bulbs are fused)
* The efficiency of LEDs is much higher. While the energy efficiency of
a bulb is 10% (only one tenth of the energy generated by light), the
LED's profit up 90%.
* The equivalent of a light bulb can
be built with about a dozen LEDs. If any break is even possible to
replace it. They are cheap and easy to manufacture.
According
to the article, the city of Raleigh, is conducting a pilot program to
install street lighting based on LEDs, allowing them to save up to 40%
of energy.
Suppose you measured the voltage across the terminals of the LED, and you want to find out the energy needed to light the LED. Suppose you have a red LED and the dap between terminals is 1.71 volts, the energy required to light the LED is E = qx V or
E = -1.6 x 10 exp -19. 1.71 Joule, since Coulomb / Volt is a Joule. Multiplying these numbers gives us E = 2.74 x 10 exp -19 Joule.
FINDING THE RATE FROM THE WAVELENGTH OF LIGHT
The frequency of light is related to the wavelength of light in a very simple. The spectrometer can be used to examine the light from an LED, and to estimate the peak wavelength emitted by the LED. But we prefer to have the frequency of the peak intensity of light emitted by the LED. The wavelength is related to the frequency of light by the formula
F = c /, where c is the speed of light and is the wavelength of light read from the spectrometer (in units of nanometers, ie one millionth of a millimeter). Suppose we observed a red LED with the spectrometer and saw that the LED emits a range in colors with maximum intensity according to wavelength viewed in the spectrometer of = 660 nm. The frequency corresponding to the emission of the red LED is of 4.55 x 10 exp 14 Hertz. The unit of one cycle of a wave in one second (cycles per second) is a Hertz.
BACKGROUND INFORMATION ON LEDs
Most of the characteristics of the LED s are specified for a current of 20 mA, if one is not assured of 20 mA in the role of heat conductivity in the platelet more heat from the LED, and heat flow variations, it is all designed to 15mA. How to achieve 15 mA through the LED:
First you need to know the voltage drop across the LED. It can be assumed with sufficient certainty to 1.7 V not very bright red, 1.9 V for high brightness, high efficiency and low current red and orange and yellow to 2V, 2.1 V for green, 3.4 V white to bright yellow and bright green without most of the blue, 4.6 V for 430 nm bright blue. It is generally designed for 12 mA for the kinds of 3.4 V and 10 mA for blue 430 nm.
You can design a source that delivers more power if you are sure of an excellent heat dissipation in the set. In this case assign the 25 mA LED of about 2V, 18 mA for 3.4 V and 15 mA for blue 430 nm. Under optimal conditions of heat dissipation can be circulated but a higher current LED life is reduced to 50% of normal: 20,000 to 100,000 hours. As the voltage should be somewhat above what was allocated for LED s. Use at least 3 V for low voltage, 4.5 V to 3.4 V and 6 V for 430 nm blue.
The next step is to subtract the voltage of the LED s to the source, this gives the voltage drop is achieved through a resistance. Eg 3, 4 V with a power LED 6 V. doing the subtraction gives 2,6 V drop must be produced by the resistance. The next step is to divide the voltage drop on the LED current, thus obtaining the value of the resistance by dividing V / A gives a resistance value in ohms. If we divide V / mA the resistance is obtained in K ohms.
Another step is to determine the power of resistance. Multiply the voltage drop on the LED current to obtain the power of resistance. S Do not put LEDs in parallel with each other and, while this works is not reliable because the LED s become more drivers with increasing temperature, which becomes unstable current distribution. Each LED should have its own resistance.