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Post by Progenitor A on Feb 19, 2011 9:46:39 GMT 1
I have a light bulb that screws into a 240V fitting that contains about 8 Light Emitting Diodes. Now the pd across one LED is about 3V when it is lighted. So my LED bulb requires only 24V to power it
How does the bulb 'get rid of' the surplus 220V?
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Post by abacus9900 on Feb 19, 2011 11:11:20 GMT 1
I have a light bulb that screws into a 240V fitting that contains about 8 Light Emitting Diodes. Now the pd across one LED is about 3V when it is lighted. So my LED bulb requires only 24V to power it How does the bulb 'get rid of' the surplus 220V? Well, you have to remember that LED's are actually semi-conductor devices and as such work on a different principle from filament bulbs. I forget the exact details but the semi-conducting material in an LED is designed to allow a flow of electrons in one direction and a flow of 'holes' in the opposite direction (a hole is where there is a 'vacancy' for an electron to occupy and jump to) and depending on which orbital level an electron has to jump from, produces this or that colour. So, whereas an ordinary bulb requires 240 to work, a diode requires much less voltage due to the physics of the material it is made off. Presumably, the LED's have to be protected from too much voltage as otherwise damage would occur so there would normally be an arrangement of devices to do this by stepping down the voltage to a suitable working level. Semiconducting materials such as silicon are midway between materials that allow easy conduction of electron flow, such a copper, and insulators which permit negligible electron flow, such as rubber, and possess useful properties which can be used in LED's, transistors, etc.
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Post by Progenitor A on Feb 19, 2011 13:41:15 GMT 1
Thanks That explains why the LEDs need less voltage, but how is the voltage stepped down? There does not seem room enough in the bulb I have for a step-down transformer.
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Post by abacus9900 on Feb 19, 2011 14:21:49 GMT 1
Thanks That explains why the LEDs need less voltage, but how is the voltage stepped down? There does not seem room enough in the bulb I have for a step-down transformer. I presume the plug you plug into the mains socket has a built in step down transformer, naymissus.
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Post by eamonnshute on Feb 19, 2011 14:31:30 GMT 1
LEDs use low voltage DC, so LED lamps contain circuitry to reduce the voltage and convert AC to DC. Simply using a resistor to reduce the voltage would not be very efficient as most of the power would be dissipated as heat, which defeats the object of the exercise!
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Post by abacus9900 on Feb 19, 2011 14:46:33 GMT 1
LEDs use low voltage DC, so LED lamps contain circuitry to reduce the voltage and convert AC to DC. Simply using a resistor to reduce the voltage would not be very efficient as most of the power would be dissipated as heat, which defeats the object of the exercise! Indeed eamonn, which is why I modified my rather ill considered post. Stepping down mains voltage using a transformer is, of course, more efficient. But that is another point, naymissus. LEDs have to have DC in order to operate, therefore there will be components somewhere in the circuit to 'rectify' AC mains voltage to DC voltage to produce DC current.
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Post by mak2 on Feb 20, 2011 14:53:48 GMT 1
LED lamps have a driver circuit that produces the right voltage. It is usually on a small integrated circuit chip.
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Post by Progenitor A on Feb 20, 2011 15:16:49 GMT 1
LED lamps have a driver circuit that produces the right voltage. It is usually on a small integrated circuit chip. Good Whats on it? What do the components do?
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Post by mak2 on Feb 20, 2011 19:53:32 GMT 1
Don't ask me but I found this :-
"The growing adoption of LEDs as a lighting source in different applications has simultaneously driven the demand for LED driver ICs to power them.
To understand the obstacles for the design and manufacture of these LED driver ICs, it is necessary to understand what a white LED requires in order to produce light. A white LED must be driven by a constant current source so that the white point of the light does not shift (that is, it must be uniform).
Furthermore, since the white LED is a diode, its internal forward voltage (Vf) drop has to be overcome. This Vf varies with the current rating of the white LED and will also change with temperature. A typical 20mA white LED has a Vf that varies between 2.5V and 3.9V over the entire operating temperature range.
Most applications use more than one white LED and can also have these LEDs configured in parallel, in series, or a combination of both – for example parallel strings of LEDs in series.
This means that white LED driver ICs must be capable of delivering sufficient current and voltage for the specific configuration of LEDs, and in a conversion topology which satisfies both the input voltage range and required output voltage and current requirements."
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Post by Progenitor A on Feb 20, 2011 20:08:40 GMT 1
Well, thanks for the effort but it doesn't tell us much really does it?
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Post by mak2 on Feb 21, 2011 16:34:51 GMT 1
naymissus,
From the description, it seems that the driver circuit contains some fairly complicated electronics. It does not simply act as a power supply. It compensates for changes in temperature, etc., probably by using negative feedback.
They do not necessarily have to use a transformer to reduce the mains voltage. They might use a voltage divider circuit, for example. The IC will probably incorporate a rectifier, as well.
You do not say whether you are pleased with your LED bulb. Does it give enough illumination and what is the spectral quality of the light like? Do colours look right?
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Post by Progenitor A on Feb 22, 2011 8:55:43 GMT 1
Thanks once more mak2 I am quite happy with my LED light. Not as bright as I would like - only about 8 LEDs but power consumption only 3W. Better than the other economy types in that it gives instant light and should be more reliable (I have had two failures of the other type). The spectrum seems white and cold. I did buy thenm originally for those bathroom spotlight ceiling lights - the ones that are recessed and covered. Using incandescent spotlights these were particularly prone to blow as they would become covered with water vapour and be unable to radiate the heat and it was a real nuisance replacing them. LED lights are more expensive (about £5) but the increased reliability should pay for itself and eliminate hassle
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Post by Mr Red on Apr 27, 2011 12:44:35 GMT 1
Constant Current?
Direct current!
I don't know the exact frequencies but when I was investigating state of the art circuits they were using frequencies of MHz which allowed for small ferrite transformers. The ICs create the frequency & drive. The transformers step down to maybe 5-10 volts. I am pretty sure the diodes would be back to back so that the AC voltage/current could light one LED on each half of the cycle. I would not expect more than two of these back to back pairs to be in series if at all. The rest would be in parallel so that any that go open circuit the resultant voltage would be pegged at one (or two) diode drops. Any reverse voltage on the conducting diode (with the failed in parallel) would thus not exceed a few volts. LEDs have reverse breakdown of not much more than the forward drop. The magnetics would be used to limit the current because any resistive losses are inefficiency.
I am pretty sure that the magnetics plays havoc with light dimmers. Much like EFL lamps. If in doubt - use the halogen bulb within a bulb thingies you can see becoming more prevalent. They waste less than the old tungsten (only) bulbs, therefore can be sold legally.
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Post by jonjel on Apr 27, 2011 13:51:14 GMT 1
Co-incidentally I am planning on fitting some LED bulbs in down-lighters in a new kitchen.
I had a look on line and bulbs seemed as cheap as chips compared with other GU10 fitting bulbs from the DIY stores.
However if you read the small print a lot of those only consume 1.5 or 2 watts. Some research has told me that the rule of thumb calculation is 10:1, i.e., if the LED array consumes 4 watts it will give a light output of around 40 watts.
You can also buy different colour tone bulbs, some a lot better on the eye than others.
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Post by StuartG on Apr 28, 2011 18:32:46 GMT 1
I know nothing about LED light bulbs, but an old dodge used to dim hallways and nurseries, was to use a capacitor [condenser!!] to limit the current through a bulb to dim it and co-incidently to extend its life and 'mains thump' [switch on current surge]. I know it as a 'wattless resistor' and cannot find any reference on searching briefly. reference to it is made in 'Phyisics in Focus' by Michael Brimicombe [He can be found] p454 Thomas Nelson & Sons, 1990. I copied some of His explanation to help illustrate better than I can...... 'Figure 25.29 illustrates how a 220 nF capacitor can be used to limit the current through a pair oif LEDs connected to the 240V rms 50 Hz mains supply. The LED's are rated at 2V, 16mA and indicate that the device is plugged into the mains. The task of the capacitor is to limit the current in the LEDs to a safe level without converting more more electrical energy to heat than is absolutely necessary. The first step in understanding how it does this is a calculation of its reactance. Xc =1/2pi fC therefore Xc = 1/2pi * 50 * 220 * 10^9 = 15kOhm The voltage drop across each LED (2V) will be negligible compared with the voltage drop across the capacitor. So let's ignore the LED's altogether and calculate the approximate rms current in the capacitor. Xc=Vrms/Irms therefore 15 * 10^3 = 240/Irms = 16mA Of course , this current wil be pi/2 out of phase with the voltage, so the average power delivered to the capacitor will be zero.... On the other hand, each LED will behave like a resistor when it is forward biased, with the voltage in phase with the current. The average power delivered to both LEDs is easily estimated. P=VI therefore P led appx.= 2 * 16 * 10^-3 = 32mW An alternative way of limiting the current in the LED is to place a resistor in series with it. You can check for yourself that a 15kOhm resistor will also reduce the rms current to 16mA, but that the power delivered to it will be just under 4W. A capacitor is clearly more efficient than a resistor for limiting the current!' Thanks to Michael Brimicombe. An aside when using on mains filament bulbs use a good non-inductively wound 'ordinary' capacitor [read expensive] rather than a 'dubious rare earth modern funny thing' to avoid disappointment and suicide bombing of the cap. Cheers, StuartG
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