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48

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 ...


29

Sounds about right. To get white (6500K) using NTSC (colour TV) phosphors, the relative intensities are G=0.59, R=0.3, B=0.11 - most of the energy is in the green, least in the blue. (slightly differently rounded numbers in Wikipedia ) At equal intensity, blue would appear brightest. The actual numbers will differ here (LEDs not phosphors) but the relative ...


17

I do not claim the other answers are wrong, but they miss two important points. One of them that I consider the most relevant. RGB-LEDs are not meant to produce white light. They are meant to reach a certain gamut Wikipedia on gamut, i.e. the colour space that can be displayed by the LED. And they do. If the three channels are driven with an 8-bit ...


17

It's even a little worse than winny indicates: Green LEDs are finicky. One result is that green LEDs emit over a broader distribution of wavelengths instead of being a nearly laser-like single wavelength. And when you map that range of greens to the xy colorspace you're not on the spectral locus anymore. So even your yellow-optimized RGB LED may not get ...


15

According to the datasheet, it has six legs because it contains 3 LEDs, with nothing common: Many RGB LEDs have a common cathode or common anode, so need only 4 leads. For some applications that may not be acceptable. For example, it wouldn't be possible to control two RGB LEDs with common cathodes or anodes in series, or to arrange them for charlieplexing. ...


14

The red led will hog all the current because it might only need 2 volts across it to begin conducting. The green and blue LEDs need a higher voltage but because they are all in parallel the red led dominates. Try measuring their respective volt drops when each operates. It's like putting a 5 volt zener in parallel with a 10 volt zener. The 10 volt zener ...


14

The resistor is there to limit the current into the input pin. The input likely has a very high DC resistance (more than 1 megohm) so negligible current flows (on the order uA) and a negligible voltage drop is produced (on the order uV or mV). The resistor is likely used to slow the slew rate of the connection (the input pin will have some capacitance, so ...


13

The goal of each capacitor here is to smooth out the power supply of each WS2812B. When you power an array of LEDs, there is a good chance that a transient voltage drop will occur and create a flickering effect on your LED strip and most likely on other LEDs that are currently on. If you were directly powering a LED from the power supply, you wouldn't need a ...


12

Fairly simple. Just using npn transistors. Duplicate per color channel. Your transistor can sink up to 600mA continuously. At 20mA per led, 4 per channel, that's only 80mA. That's more than enough. But make sure your power supply can support that. 80 * 3, that's 240mA on the leds alone, not including the rest of the arduino, the transistor base, anything ...


11

Those caps are small bypass capacitors, use to filter or buffer the voltage input to an IC. They are cheap and useful and you will find them as close as possible to any IC on any manufactured pcb anywhere. They should not be ommited without expecting funny operation. The speed that WS2812B are already driven at and the current draw of the leds already cause ...


10

I'm taking a guess about what you are misunderstanding, but here goes... colours are already just different frequency waves. This is true, but the frequencies involved are very very high. \$f = \frac{c}{\lambda}\$. So for a wavelength of 640 nm (red), you're looking at a frequency of about 470 THz. That's 470 x 1012 Hz. We don't have any technology ...


10

These are "decoupling" or "bypass" capacitors. Their purpose is to stabilize the local supply voltage against fluctuations caused by the interaction of varying current consumption in the circuit with the impedance of the power supply network. "Local" in this case means the supply voltage at each individual LED, which is really an integrated circuit ...


8

The simplest analog circuit I can come up with is this: V1 represents the temperature sensor output value. The values of R1 and R3 may need to be adjusted specially if you use other transistors (you can use variable resistors to find out the correct values then replace them with fixed value resistors). You may also need a voltage divider on the base ...


8

You should use 3 different resistors, one for each color, although the blue and green have the same specs. At 150mA the forward voltage for the red is 2.2v, green is 3.5v and blue is 3.5v. So you should use a 22ohm 1watt resistor for the red, and 10ohm .5watt resistor for the green and blue. You have a bit of wiggle room on these figures, and if you don't ...


8

These diagrams/images might be better to visually see whats going on: Side note @Passerby: Where did you get that schematic image in your post?


8

These RGB LEDs have six pins because all the connections are brought out individually. You didn't ask, but others may wonder why manufacturers do this. The advantages of bringing out all connections individually over tying some together in the package include: One part works in both common anode and common cathode configurations. If the package needs to ...


7

sRGB is a colorspace. That is, it defines a reference system in which a perceived color can be expressed by three numbers: R, G and B. If you want to reproduce a color expressed in sRGB value using three LEDs, you have several computations to perform: First, you have to know the spectrum of the three LED you are using. ( I hope: one green, one red and one ...


7

LEDs of different colors are made with quite different materials and processes and designs. There's no guarantee that they'll turn out to be the same brightness. It makes more sense to put more efficient LEDs in there when they are available rather than degrade the more efficient ones in order to match the least efficient color. Sure they will have to run at ...


7

Composite video signals are usually AC coupled, so you shouldn't need to pull the output below ground. The receiving end will restore the DC level (using the sync pulse as a reference) if it needs to. At the input side the signal may go below ground - or not, depending on the source. To cover all possibilities you should terminate the input with 75Ω ...


7

Using an LTspice simulation with 100 MHz 0.5 V peak sine wave input and the Figure 36 unity gain buffer (+/-5 V supplies), and a 20 pF load, I get about 130 ps. You can see from Figure 5 (in the datasheet) that it is much less than 1 ns.


7

LEDS actually come in all colours. However, what we as humans call colour is dictated by the fact that the cones in the human eye are only actually sensitive to three primary groups of wavelengths which we call red, green, and blue. Different LED technologies produce light in a similar but narrower bell curves around a particular wavelength. When we look ...


6

For a single LED you are correct - which side you put it on does not matter. However, an RGB LED is a rather different animal since one side of all of the elements are tied together. This presents a bit of a problem. If you wire the common terminal directly to a power rail and put 3 resistors on the other side, it will work as expected. However, if you a ...


6

Start by scaling the R,G and B by the intensity value, for your example at 50% intensity you would set 15*0.5, 170*0.5, 230*0.5 (7, 85, 115). The LEDs might have different non-linear responses, so you might need to tweak the scaling for perceived colour In particular, at very low levels, you will suffer from rounding error. There is no easy way around ...


6

You are attempting to map a triplet of values (the responses of the LDR to your RGB LEDs) to another triplet (the RGB values you used to print the colors on the paper). There are loads of environmental values that will influence this mapping. IMO the best you can do is Make sure each LED gives a decent and compareable effect. Depending on your LEDs and the ...


6

I used Firefox Web Developer Eyedropper to grab the RGB from a photo of an Edison Light Bulb. #FACA08 (250,202,8) A lighter pixel This is very close to the color when a 2700K 97 CRI LED was reflected off bright white paper: #F4D4AB (244,212,171) just a little less blue. #F5D483 (245, 212, 131) I do not believe the LED actually illuminates ...


5

A prism with a halogen filament bulb will probably do an OK job as a light source. Many ideas in the images below - each links to a relevant page. A diffraction grating and slit can be used instead of a prism. An unused DVD makes a fair diffraction grating [Note: The links are intended to be to pages of images which are linked to text, and not to pages ...


5

In a rush to get out of here, but hope this helps. Ask question and me or someone will answer :)


5

A linear regulator dissipates heat proportional to the amount of voltage it must drop, and the amount of current flowing through it. Input supply = 12 Volts Option A: Output Vout = 3.3 Volts Vfred = 2.1 Volts Current = (Vout - Vf) / R = 7.5 mA per LED = 60 mA total Power dissipated by regulator, P = (Vinput - Vout) x I = 522 mW Over half a watt is ...


5

The red, green and blue of RGB LEDs do have different perceived luminous intensities, for several reasons. Some of them are listed below, not in the order of importance: Eye color sensitivity The human eye is sensitive to different colors to different levels - and this relationship also varies by light intensity: Both ambient light intensity (bright room ...


5

The 100mA/10% duty cycle is absolute maximum rating at 25°C ambient temperature. It is not something you should be designing to, and note that it only gives you 10mA average current per LED. It should be derated significantly (let's say 30%, 40% or more) to allow for ambient greater than 25°C and to avoid the absolute maximum limit. At 40% derating, you can ...


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