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LEDs don't work like ordinary (incandescence) light bulbs.
Main differences (a bit simplified for very beginners):
They have a polarity, hence they must be powered using DC respecting that polarity. Reverse the polarity and they won't work. You may also damage them if you apply more than ~4V-5V in the reverse direction (these are safe values; the exact ...
answered Nov 18 '17 at 12:25
Lorenzo Donati -- Codidact.com
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From here this is what the leadframe looks like.
This is what they look like when they are molded in small scale production:
After this, the parts are separated but the die is made so that it does not weaken the lead, which means it has to be a bit wider in that area. So the remaining nubs are artifacts of the production process, like ejector pin marks on ...
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It is not possible to produce white light without an efficient blue LED, either using RGB LEDs or a blue LED + yellow phosphor.
The breakthrough was the invention of the high-brightness Gallium-Nitride blue LED by Shuji Nakamura at Nichia
in the early 1990s.
It still took a while to get the overall efficiency up to the level of fluorescent bulbs, and it's ...
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Those resistors aren't doing anything at all. All the pins in a row are connected together.
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I agree with some of the others here... you're trying too hard.
As others have mentioned, the forward drop of an LED varies with its bias current, but for almost every application a hobbyist will get in to this isn't something you have to spend a great deal of time worrying about.
Almost every handheld multimeter has a diode setting. It will tell you the ...
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Don't look to the arduino designs as examples of stellar electrical engineering.
However, there can be a legitimate case for doing this. This part contains 4 resistors. If it was already there for another reason, especially if several more of them are used on the same board, then using two of the resistors that would otherwise go unused in parallel to ...
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Using a regulated current source to light them, wire the LEDs in series and short out the segment which you want to be dark.
simulate this circuit – Schematic created using CircuitLab
You can possibly use a buck-boost converter to make the 30V if you don't already have a suitable voltage.
Here's a simple way to build one using a LM2596S module:
...
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To address the question, first a distinction needs to be made between phosphor LEDs (#1) (e.g. white LEDs, possibly some green LEDs) and direct emission LEDs (e.g. most visible color LEDs, IR and UV LEDs).
Direct emission LEDs typically have a turn-on time in single-digit nanoseconds, longer for bigger LEDs. Turn-off times for these are in the tens of ...
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Any ideas?
Yes. You are feeding the LED with an AC supply. That means that every second half-cycle it will be reverse biased. Most LEDs can withstand about 5 V in reverse. You're applying 24 V AC which will have a peak voltage of \$ 24\sqrt 2 \ \text V \$ so I'm surprised you got any length of time out of it.
You have at least four options:
simulate this ...
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What value resistor would you include? For what voltage? For what brightness? How would you make the resistor when it uses a different process than the LED does? How would you dissipate the heat?
Having said all that, they do exist.
Figure 1. 6.5 V Red LED 3mm Through Hole, Kingbright L-934SRD-5V contains an internal resistor.
The resistor LED lamps, from ...
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The simple answer is that they are using near IR. LED manufacturers have a good handle on how to make them so they are affordable.
Their center frequencies may be invisible to the M-1 eyeball (i.e. human eye), but unless they put a filter in front of the LEDs (which cause them to produce less illumination) there will be some of it that you can see.
The ...
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The existing answers miss the core of the question.
An LED needs to be a diode, specifically because the way the charge carriers recombine in the forward-biased diode junction releases the correct amount of energy to create photons in the visible range. Passing a current through a chunk of semiconductor with no diode junction in it would simply produce heat....
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I see Lorenzo has already answered your question directly (+1). Here is what you can do to light your LED and see what you've got.
LEDs are diodes, so conduct only in one direction. Unlike a ordinary light bulb, orientation matters. If the LED doesn't light one way, flip it around and try again.
To safely experiment with pretty much any LED, use a 5 V ...
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This is where the measurement scientist has to go into full sceptical and investigative mode.
First thing. Fibre, as a passive material, is lossy. It absorbs power. Therefore the power arriving at the end of a length of fibre will be less than was launched. Period. No arguments. We don't do over-unity here.
So what causes your observations?
Single mode, ...
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Most probably because of short wavelength of your green LED and not as monochromatic green as you might expect (x and y coordinates closer to the center). If you take a look at the CIE 1931 curve and plot your red and green x and y coordinates (listed in the datasheet from serious manufacturers, otherwise assume the pure wavelength on the outer rim or move ...
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Properly designed, built and used, today's LEDs have incredibly long lives and the wearout mechanisms are not catastrophic in nature, so instead of using incandescent lamp MTTF statistics, a luminosity percentage (70%) is often used to define the lifetime- this doesn't mean that the LED burns out, it means that the light out put is only 70% of what it was ...
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Through-hole components are generally hand or wave soldered. These apply on local heating to the pads and not to the component itself.
SMD parts on the other hand are generally reflow soldered. This involves putting the entire part in a hot oven for an extended time. The components are generally made from porous materials which by their nature absorb ...
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6.8 volts seems awfully high for a single LED. Are you sure that 6.8 is not the number for all four LEDs? That would make it 1.7 volts per LED, which is more reasonable for a red LED. And that would mean that you are currently pushing 172 milliamps, or almost 3 watts through your resistor.
If that is the case, you should lower your power supply to less than ...
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It's still the case that MCU I/O pins often have weaker drive sourcing current than sinking current.
In a typical CMOS MCU output, when they drive LOW, they turn on an N-channel MOSFET; and when they drive HIGH they turn on a P-channel MOSFET. (They never turn both of them on at the same time!) Because of the differences in mobility that apply for N-channel ...
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What's a step by step way of thinking about the flow of electrons in such a circuit? How can we tell it would provide constant current for instance?
The first step is to forget about flow of electrons. Think about conventional current flow from positive to negative or ground. This is the way we (nearly) all do it and why we draw the positive rail at the top ...
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This is almost certainly within tolerance of expected behavior for these LEDs.
A quick way to see this is to do a back-of-the-envelope sensitivity analysis.
What's the current if the LED forward voltage drops are exactly 3.2 V each (the datasheet "typical" value)?
$$\frac {24 \ \text{V} - 7 \cdot 3.2 \ \text{V}} {82 \ \Omega} = 19.5 \ \text{mA}$$
...
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DC Level 0 Diode
The level-0 approximation of a diode is simply an assumed voltage drop across it when forward-based and infinite impedance when reverse-biased. When reverse-biased, no amount of reverse-biased applied voltage can break through the device --- it can stand off an infinite voltage. (For this level and the following levels below, I will not ...
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I have two possible explanations:
The 'knuckles' are there intentionally to avoid the pins going all the way through the PCB.
Most of the times it is not desired that the pins are going all the way through the PCB.
They are a remnant of the production process (see picture below).
First the pins are a part of one single metal sheet and are cut after the dies ...
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Voltage and current are intimately related. If you attempt to increase the voltage across an LED, the current will increase. Likewise, to increase the current through an LED, you must increase the voltage across it.
It is not possilbe to have the correct current through an LED, but too high a voltage across it.
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I see your problem. Your circuit shows how you're driving a single LED segment. (I presume you then have 7 of these circuits, one for each segment.) The datasheet shows 4 LEDs in series, covering the segment.
Where you've gone wrong is assuming there's 6.8V forward voltage drop per LED. There is no such red LED. Typically a red LED will be around 1.6V-...
40
You seem to be getting confused between the efficiency of the LED and the efficiency of the circuit to drive the LED.
In terms of light output per unit of energy used by the LED they are an efficient way to generate light. In absolute terms they aren't great, they are around 10%[1] efficient in that respect however that is still far better than the ~1-2% of ...
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Yes, this is widely known. Anyone who has tested it knows that. The die manufacturers certainly know it.
They don't specify LEDs for more than 5V reverse voltage because it would not measurably increase sales (ie. very few need that capability) and would require them to actually consider each LED type and what voltage it might withstand (maybe 12V for some,...
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Your PSU meter is only has resolution to 0.01 A (10 mA). The actual current could be anywhere between 5 mA and 15 mA for the single LED.
Switch your yellow multimeter to mA range, connect the leads into the correct sockets and wire the multimeter in series with the one LED and get a more accurate measurement.
Paralleling LEDs in this fashion is not ...
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The actual LEDs are probably blue and the encapsulation contains phosphors to generate the rest of the spectrum. The phosphors used will appear yellow.
Figure 1. The blue peak is direct emission from the LED. The hump in the green to red part of the spectrum is generated by the phosphors. Image source: Physics.SE.
Note that if the LED is not powered there ...
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Physics explanation
Light bulbs
Incandescent light isn't really a light source so much as a heating element. Any current through a wire heats it up a bit (Joule's Law of Heating); once the wire is above room temperature it emits net energy via black-body radiation. The rate at which this energy is emitted depends on the fourth power of temperature, i.e. the ...
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