LED Knowledge Base

  • A light emitting diode (LED) is a semiconductor device that emits visible light when electrical current passes through it, it is a special kind of diode.
  • Most LEDs are monochromatic. Light color is associated with the light wavelength. LEDs made with different semiconductor materials emit lights in different wavelengths. LED light wavelengths range from 400 nanometers (blue) to 800 nanometers (red). The colors of popularly available LEDs in the market are red, orange, amber, yellow, green, blue and white.
  • Yes and no. LEDs have increasingly been used in many different applications to replace old incandescent light bulbs. Examples include LED indicator lights for all kinds of electronic devices such as cell phone, calculators, automotive dash panel, LCD back-lighting, etc., LED seasonal decoration lights for Christmas and holidays, LED traffic signs and other LED direction signs, LED flashing lights, and so on. As currently available LEDs have limited brightness, in most cases, they can not be used to replace incandescent or fluorescent light bulbs for general lighting purposes.
  • There are several obvious advantages LEDs have over traditional incandescent light bulbs, they are as follows: Low power consumption - energy saving,
    Long lasting,
    Cold lighting,
    Ruggedness,
    Small size and weight,
    Fast switch times,
    Simple to use.
  • Yes. Currently available LEDs are more energy efficient than incandescent bulbs, but less than fluorescent bulbs with the same light output. The power consumption of popular LEDs ranges from 30mW to 200 mW.
    People can save money & energy by using LEDs instead of equivalent incandescent decorative bulbs.
    The rated average working life of LED is 100,000 hours compared with 1,000 hours of incandescent bulbs.
    LEDs emit cold lights as working LEDs generate very small amount of heat, so LEDs are much safer than equivalent incandescent light bulbs in terms of danger of fire. LEDs don't have filaments to heat up in order to emit lights just like the case of incandescent bulbs. Lights are emitted from LEDs as a result of energy exchange occurring in the different semiconductor materials an LED is made of.
    LEDs are usually shielded with solid transparent plastic materials so they are more rugged than incandescent and fluorescent bulbs that are usually sealed with glass.
    LEDs can be powered by either AC voltage or DC voltage. The circuit that is required to appropriately drive LEDs is much simpler than that for fluorescent bulbs.
  • Yes, the most significant disadvantage is the light output limitation of LED. Currently available LEDs emit limited amount of lights at a relatively small angle range, while incandescent and fluorescent light bulbs illuminate in all directions and give out much more brightness of light. And this is the reason why LEDs currently can not be used for general lighting applications. The second significant disadvantage is the high prices of LEDs. The currently available LEDs in the market are 3 ~ 10 times more expensive than equivalent incandescent light bulbs. This is why some customers still hesitate to buy LED products even though LEDs have so many obvious advantages over incandescent bulbs and people actually can save money by using LED products in the long run.
  • The most popular LED products available in volume at the market place are various kinds of LED decoration light products. And in fact, they are ideal for replacing conventional incandescent decoration light products, because when compared with incandescent light products, these LED products save up to 95% energy, have much longer working hours, are more rugged, and safer in terms of danger of fire and electric shock. At present, people can buy beautiful LED light strings, LED rope lights, LED icicle strings, and individual round LED bulbs with changeable colors, LED rope lights, LED pipe lights, and more other LED products will be put into market in the near future.
  • LEDs are more reliable. Most available LED light strings have series configuration like old incandescent light strings, where lights in the string are connected in series with each other, so a disconnection of LED light from the string will always make the entire string fail just like incandescent light strings. However, in the case of light burning, an incandescent light bulb will always burn open, namely, the filament of the incandescent bulb breaks, so the whole string will always fail, while an LED light may either burn open like an incandescent light bulb or burn short where a short circuit is formed inside the LED, so the burned LED will stop lighting but all the rest LEDs in the string still light normally. And LEDs usually burn short, therefore, LEDs are more reliable than incandescent light bulbs in that respect.
  • Yes, there is. Various LED string products are more reliable than those LED strings in series connection in market. OUR strings use a unique patented connection configuration that ensures an LED string will work normally even one or several LEDs are disconnected from the string while keeping the low prices and all the features of LED strings.
  • The reason is that different semiconductor materials are used for different color LEDs, and some semiconductor materials are more expensive than others; another reason is that manufacturing costs are different. White LEDs are the most expensive because red, green and blue LEDs are combined together to make a white LED.
  • It depends on the application. Diffused lens contains diffusers which distributes the light output more evenly and makes wider viewing angle of a LED. Water clear lens, without diffusers, narrows the viewing angle, creates more focused beam, and increases intensity of a LED. Color-transparent lens has same effect as water clear lens. The purpose of tinting is to prevent mixture of different LEDs during mass production.
  • There is no direct conversion between the two units. Luminous Intensity measured in Candela or Millicandela is the light output in a particular direction. While Luminous Flux measured in Lumens is the total light output in all direction. Our factory can measure Luminous Flux (Lumens) with special requests, which takes about 1-2 weeks to finish.
  • No, LEDs can only be operated at DC and require specific anode and cathode connection.
  • I'm interested in driving my IR LED to the maximum power without significantly reducing it's life span. I've heard that i can drive an LED above its rated power if i do it in pulses? If you actually need 50% duty factor, you will find yourself limited to just double the rated continuous current for the LED, usually about 50mA. For best results, get an AlGaAs IR LED with as much directionality (narrow beam) as you can tolerate. At 100mA, it will drop about 1.5Volts, so choose limiting resistance accordingly or use a current source (Voltage level drives NPN base, emitter resistor sets current). Contrary to what others have said, you can easily get 50% df from a CMOS 555. Just connect timing capacitor from 2,6 to ground, and connect a timing resistor from 3 to 2,6. Use pin 7 (open drain FET) to drive the load. The C555 makes a Schmitt oscillator this way. Don't forget to use bypass capacitor across pins 1 and 8. If you can operate with less than 50% DF, you can jack the peak current up proportional to reduction in DF, but don't try to go past about 2 Amps, since the wirebond in most IR LEDs can't handle more. In fact, some will fail at significantly lower currents. The most fragile LEDs tend to be the ultra efficient visible units. A few IR units are rugged enough to handle about 10Amps peak. If you are operating the LED in bursts with quiet periods in between, you can increase peak current somewhat. Exactly how much is beyond the scope of this reply. EMail me with further questions.
  • There is a common belief that it is best to drive LEDs with square waves. This idea seems to be related to the use of tuned receiver circuits and the observation that, of all rectangular waveshapes, the squarewave (50% duty factor) has the most energy at the fundamental. However, for LED transmission/reception, much higher SNR is usually possible with smaller DF. This has to do with the nature of the LEDs themselves (you can achieve higher peak power by reducing DF) and with the nature of the dominant noise sources (shot noise in photodetector junction(s), op-amp input noise, and ambient light fluctuations (especially 2x mains freq), which add up to a noise spectrum that tilts up at low frequency. Consequently, a great many photoelectric data channels are operated at low DF with matched filters (or, for cheap systems, peak detectors) which, unlike traditional tuned circuits, do not have maximum output when all the energy is at the fundamental.
  • The primary approaches to driving large arrays are series strings, parallel connections, and matrix addressing (multiplexing). Series Strings of LEDs: When driving a large array of LEDs, you need to be concerned about current distribution and power efficiency. Series strings of LEDs are often used to improve both of these factors. Depending on the type of LED and the operating current, forward voltage can range from 1.1V (IR emitter at low current) to about 10V (some blue emitters at high current). However, when you operate LEDs in series, you can be sure that all of them in a string have the same current. Also, if you are working with a relatively high supply voltage, you can improve efficiency by connecting strings of appropriate length. For example, operating 6 IR emitters in series at 50mA DC will require about 8V, depending on temperature and the type of LED. For a 12V supply, this leaves 4V for switching and current regulating bias. Series strings have a couple of disadvantages: any LED failing open circuit will disable the entire string, 2) Any LED failing shorted will reduce the forward drop for the string, possibly affecting current regulation, 3) Compared to parallel connections, circuit board layout can be more complicated.
    Connecting LEDs in Parallel: You can also connect LEDs in parallel. However, variations in the forward voltage requirements of individual LEDs will result in non-uniform current distribution. To minimize these effects, you can use any combination of several approaches: 1) Use individual current limiting resistors or regulating circuits, 2) Use LEDs chosen from the same production lot and/or matched for forward voltage, 3) Connect series strings of LEDs in parallel. This last approach has the effect of averaging out the forward voltage over several LEDs. Matrix Addressing (Multiplexing): If you need to control the pattern of driven LEDs, as required for graphical and character displays, matrix addressing should be considered. Also, many ready-made LED arrays are designed for matrix addressing. (Lest you think that this is your only choice, be advised that may LED displays are actually driven by long shift register chains with individually current regulated parallel outputs for each LED.) Depending on the nature of your logic or microcontroller port outputs, you can choose common anode or common cathode connections. For example, common anode arrays have groups of LED anodes connected together, with each cathode available for a separate switching and/or regulating circuit. You could use NPN common emitter (open collector) switches or NFET common source (open drain) switches on each cathode. Further circuit details are beyond the scope of this topic. However, the important thing to note about multiplexing is this: Since each LED is only on part of the time, you will need to supply higher peak currents to achieve the same brightness compared to non-multiplexed DC applications. Among other things, this will mean that the forward voltage for each LED is higher, possibly much higher. For example, suppose that each LED in a multiplexed array is on 10% of the time. To achieve the same brightness as with 50mA DC, you will need to supply approxmately 500mA peak, and you can expect forward voltage to rise, perhaps as much as a volt. This may mean that the LED gets significantly hotter. For information about driving LEDs at higher currents, see How Hard Can I Drive LEDs?

    Some additional details, cautions:
    1. To maximize array output, obtain 'superbright' or 'extra super bright' LEDs. 2. LEDs have a negative tempco of light output and a negative tempco for forward voltage (usually ~-2mV/K). To some extent, these may be self compensating if you use fixed resistors for current limiting, but be aware that the above setup may draw more current as things warm up.

  • If you wish to produce a collimated beam, put a point near the center of curvature of the LED 'dome' at the nominal focal point of the lens. With visible LEDs, judge the results by eye. If you're working with IR, you can get feedback on the focusing process by monitoring signal level at the detector.....just work at optimizing it. I've listed a couple of other IR optimizing methods in the Beam Optimization Section below. Regarding lens quality: don't be too concerned for a system of this type. Inexpensive molded plastic lenses work fairly well provided that you don't try to go lower than f2 or so.
  • 1. Switch to red LEDs and use your eyes to optimize. This can be a temporary or a permanent switch, depending on whether red light will work in your application. Very high output red LEDs are available. If you're going to switch back to IR after the experiment, be sure to choose the most similar red LED in terms of angular specs.

    2. Obtain an IR viewing card. Radio Shack used to carry them....don't know if they still do...Edmund Scientific, Kodak. This will make your IR visible if you use enough LED drive.

    3. Obtain a cheap monochrome CCTV camera, e.g. a security camera. They generally have good IR sensitivity. If you put a visible blocking filter (Wratten 88 or 89 or similar) over the lens or CCD chip, the camera will only pick up near IR. Use this with a monitor to see the beam!

  • 1. LED dice are generally 0.3mm to 0.5mm square, with a thickness of about 0.3mm. The smaller dice are, therefore, approx. cubic. 2. Most types emit from all facets. However, the base facet is usually metallized and therefore opaque. A few special purpose types have unusual metallization patterns (see below).

    3. Since substantial amounts of energy emanate from the side facets, a 'die cup' serves to reflect such light in the same general direction as the light from the top facet, thus improving the forward power output or (for visible LEDs) luminosity. In the most popular bullet-shaped 'T1' and 'T1-3/4' plastic packages, the die cup is coined from the leadframe material, and it is necessarily relatively shallow. Silver plating improves optical and electrical characteristics. The fact that you can see the glowing die from the side of such packages is evidence that the cup is not entirely effective. The ultimate die cups are deep parabolic bowls with very smooth gold plating.

    4. The bullet housing is an 'immersion lens' which works with the die cup to form the 'beam'. Owing to its directionality, the beam has much higher radiance or (for visible LEDs) brightness than a naked die. The immersion lens also provides a less abrupt transition from the index of refraction of the compound semiconductor material to the air, thereby reducing attendant losses.

    5. When bullet LEDs are used in conjunction with objective optics, you have a choice about what sort of image you want to produce. You can: a)(approximately) focus on the immersion lens, or (b) (approx.) focus on an image of the die apparently located some distance behind the bullet package. Setup (a) gives you a fairly uniform 'condensed', round 'spot' at the focus distance, with a diameter equal to the bullet diameter times the magnification. Setup (b) gives you a square image of the glowing die with dark areas in the region of the wirebond and bond wire. This square will be surrounded by a 'halo' that is the image of the 45 degree angled die cup rim. The size of the square image will be approx. 7 times larger in each dimension than you would predict by multiplying the magnification of the objective by the size of the die. In other words, the immersion lens increases magnification by about 7x. This effect is easy to observe with your eyes.

    6. You can also obtain many LEDs which substitute a flat top or window for the immersion lens. If you are trying to deliberately produce a small spot, this works best unless the dark zones resulting the wirebond and wire are a problem. Image size will be objective magnification times die size.

    7. Leadframes are often poorly centered in the immersion lens package, making beam aiming with respect to the package a hit and miss affair.

    8. LED mfgrs. commonly offer several parts which use the same die type and lens shape but which vary in the axial location of the die cup relative to the top of the immersion lens. If you want to produce a round condensed spot using an auxiliary objective lens (see #5), roughly match this angle to the angle subtended by the objective lens at the chosen focal distance.

    9. The most common top metallization pattern is a disc approx. 0.12mm diam. 0.025mm diam. Au wire is used to make the top contact. Bottom metallization covers the entire bottom facet, and attachment is either eutectic (solder) or by means of silver-filled epoxy. Given the internal junction structure of most high performance types and the interference of the metallization, higher radiance may be available from the side facets. Consequently, some special purpose parts are mounted on an insulating substrate and have ball bonds to each side of the junction, thus positioning an edge toward the 'top'. Hammamatsu have a part like this with a 'microball' lens positioned on the edge facet that produces spectacular radiance values. Much less exotic are fancier top metallization patterns such as Xs and grids which attempt to spread current over the junction at the expense of increasing the opaque area.

    10. Compound semiconductor dice are tiny and brittle and the wirebonds are delicate. You can increase the peak output by driving them with short pulses, but you cannot pursue the tradeoff between amplitude and time past a certain point. Even for very low duty factors, most LEDs cannot handle more than 2Amps peak without greatly reduced reliability. Some are damaged by 500mA or less, regardless of pulse duration and freq. A few of the larger units double wirebonds can handle 10A pk. Roughly speaking, the visible types that have the amazing luminosity ratings are the most delicate. To put it another way, you can run most LEDs at their rated average current with much higher peak currents provided that peak current is an amp or less. Since most LEDs are rated about 40mA continuous, don't push the duration/amplitude tradeoff past about 4% duty factor without careful investigation.

  • LEDs are used more and more as incandescent light replacements. There are at least 4 possible advantages to using LEDs compared to incandescent sources: a) longer lifetime, i.e., years rather than months of service, b) power savings in some applications (more on this later), c) reduced heat output in some applications, and d) deep saturated hue (color) of output without filters. Does this mean that LEDs are generally more efficient, that is, can they be used to save power? With the current state of the art, incandescent bulbs actually put our more visible light per watt than most LEDs. So, if you want white light or if you don't care about color of illumination, incandescents are somewhat more efficient than any combination of LEDs, and they remain more efficient than the newest 'white LEDs'. However, if you want single color illumination, LEDs can be much more efficient, since, in that case, much of the light output of the incandescent source is absorbed by a filtering system. If we include fluorescent fixtures, here is the general picture of light source efficiency:

    1) As of today, fluorescent fixtures remain the clear winner in terms of output in lumens/watt. Halogen incandescents are second. LEDs are third.
    2) LEDs win the race if what you want is light of a single color available from efficient LEDs. The losses involved in filtering fluorescents and incandescents are such that LEDs win out here. Examples include stoplights and auto taillights.
    3) LEDs provide the longest lifetime by far.
    4) If your supply is some low DC voltage, driving circuits for LEDs are extremely simple and cheap compared to same for fluorescents.
    5) White LEDs, although they do not presently provide as many lumens/watt as incandescent bulbs, can be useful because of the long life and low heat output.

    Don Klipstein has done some lumen efficiency tests of LEDs.

    For reference, fluorescent fixtures put out about 30 lumens per watt. Incandescent bulbs put out about 8 lumens per watt.

    In summary, if I were equipping a remote cabin with lighting, I'd be using low voltage fluorescent fixtures at this point in time.

  • The easiest way to measure total power output is to position the LED very near to a large photodetector. You should be able to get large area photodiodes from: Vactec, EG&G, Hammamatsu, IPI Centronic, Optodiode, Silonix, among others. The last one is a Canadian co. Solar cells OK, but you should probably do a rough calibration. Here's one way to do it: Noonday sun at your latitude irradiates the surface of the earth with about 800 Watts/sq-m. This is equivalent to 800/10,000 = 80mW/sq-cm. Since Si efficacy for sunlight is about 0.5Amps/Watt, you should get around 40milliAmps for each square cm of solar cell. Supposing that you get a smaller value, then you can assume that your particular solar cell is less sensitive than 0.5A/W and calculate a calibration value appropriately.

    Example: you obtain a 1"x1" solar cell, i.e., about 6.5sq-cm. At midday on a clear day, you measure about 200mA short circuit current max as you orient it toward the sun. You calculate 80 * 6.5 = 520mW incident on the cell, so you were expecting more like 260mA. This means that your cell has more like 0.38A/W efficacy. I've glossed over some of the fine points, but all the above info is reasonably accurate.

  • Visible LEDs seem to be specified in millicandelas, while IR LEDs are specified in milliwatts. What are millicandelas (mcd) and how can I compare them to milliwatts? Candelas, candles, foot-candles, lux and related visible light output units are all photometric units, meaning that they are measured using devices that mimic the human eye response. Watts are radiometric units, meaning that they are used to measure power without regard to the response characteristics of the sensor.. The human eye has its maximum sensitivity in the yellow green part of the spectrum, where it has easily 50 times more sensitivity than it does for deep red light. In other words, the sensitivity of the human visual system varies strongly as a function of wavelength. LEDs emit light over a range of wavelengths, and human eye sensitivity is different at each wavelength. If you want to get an estimate of how many milliwatts are output by an LED with a given millicandela rating, there is a procedure for Using a triple convolution integral, it is possible to convert from mcd to mw, there's an easier way. Most silicon photodiodes convert light to current with an 'efficacy' of about 0.45Amps/Watt. So, procure a large area photodiode and connect it to a low impedance current metering device. Good quality digital multimeters work well, but cheapies won't do. You need to present nearly a dead short load to the photodiode. If you have any trouble getting a big photodiode, see about getting a single solar cell (same thing). You need a photodiode that is at least as big and preferably bigger than your red LED diameter. Fire up the LED, position it quite near to the photodiode and aimed squarely at it, and measure the current. Adjust the LED current (vary voltage, series resistance, etc.) until you get about 2.5 milliamps photocurrent. Since 2.5/0.45 ~ = 6 you know that about 6mW got converted to about 2.5mA! The large photodiode is monitoring total power. The photodiode should not have any magnifying lens over it...just a flat window or no window. If you can't obtain a big one, you can estimate the amount of current you would get if it were larger by multiplying measured current by the ratio of beam 'spot' area to photodiode area. For example, suppose that the red LED makes a 2 cm diameter spot at 2 cm distance away from itself. Such a spot has about 320 sq mm area. If your photodiode is only 2mm x 2mm, you're only get about 1/80th of the power, so multiply measured current by 80 to estimate what you would measure with a big enough photodiode. It's very important to verify that the photodiode is operating linearly, which only happens if it has very little forward voltage. If possible, connect a voltmeter across the photodiode/ammeter combination and verify that the voltage is < 200mV. Otherwise, your measurement will underestimate power. Good quality selected AlGaAs IR emitters will put out 10mW total power at about 40mA drive. The very latest super efficient 660nm red emitters have approx. the same efficiency. So, obtain some T1-3/4 super red parts and you'll very likely get your 6mW at 25mA or so. Highest eff. parts come form HP, Stanley, Sharp, Toshiba, and Mitsubishi Cable.
  • Candela: Systeme Internationale (SI) unit of luminous intensity. The modern calibration reference is a platinum radiator heated to just below its melting point: each square cm of such an object has 60.0 candela luminous intensity! (In old-fashioned parlance, each sq cm of the hot platinum emits as much light as 60 standard candles.) A point source of 1 candela radiates 1 lumen into a solid angle of 1 steradian. A 1 sq-cm source of 1 candela produces an illumination level of 1 foot candle at a distance of 1 foot.

    It gets a little confusing at times, because there are 3 ways of measuring the light output of sources: 1) Total luminous flux, 2) Luminous intensity or surface brightness per unit area (Summing this up over a surface gives total flux, but a point source has no area so you can't really calculate its intensity.) See below. 3) Illumination level at some distance away, sometimes measured in foot-candles.

    Luminous intensity is surface brightness, sometimes measured in candelas. Note that a big surface can have a high luminance at the same time that it has a low luminous intensity and vice-versa. An example will make this clear (I hope): A 40W fluorescent tube is a lot easier to stare at than a 40W incandescent. Why? The light output of the incandescent is spread over a much smaller area, so it has a much higher luminous intensity. The 40W fluor. puts out about 2000lumens spread out over about 2500sq-cm. Meanwhile, the incand. puts out only about 750lumens, but it's spread over only about 150sq-cm (assuming a frosted bulb). So, the surface of the incandescent will have about 6 times greater luminous intensity. Imagine how bright the filament itself is, with its much smaller surface area. So, as you look at a CRT or other display, its surface brightness can be measured in candelas. It it has 60candela brightness, it's as bright as the Pt standard. Another way to look at it is that each sq-cm would emit the same luminous flux as 60 candles! However, the intensity of a candle flame is another matter: it depends on the size of the flame!

    Here's an example of a practical problem:

    What's the luminous intensity of a white screen illuminated by 9 ft-cd lighting? (This is like slightly dimmed office lighting..I chose the value 9 to make calculation easier, see below.)

    [Here's the significance of this question: In order to see images projected onto the white screen under such conditions, the display projector must produce at least a comparable level of luminosity.]

    A. There are about 900 sq-cm per sq ft. For 9 ft-cd illumination, it's as if each sq ft of the screen were illuminated by a 9 candle source at a range of 1 ft. Those 9 candles of flux are spread over 900 sq-cm of area, so each sq-cm has an intensity of 10 millicandelas! In order to be seen clearly, the display device should provide a luminous intensity significantly higher than 10 mcd.

    0) Standard candle - a point source of light having a particular radiance spectrum (power vs wavelength) and total power output which are approx. that of an 'ordinary' candle. Point sources radiate uniformly into a sphere.

    1) 1 Lumen - the amount of visible light emitted by a standard candle through a solid angle of 1 steradian. Since a sphere has 4pi = 12.57 steradians, the standard candle emits a total of 12.57 lumens. (Of course, 1 Steradian is just that solid angle over which the subtended area is exactly equal to the radius of a sphere, so, 1 sq-ft of surface area on a 1 ft sphere covers exactly 1 steradian.)
    Foot candle = 1 lumen/sq-ft, which is the illumination from 1 standard candle at 1 foot range.
    2) 1 Lux = 1 meter-candle, i.e., illumination from 1 std cdl at 1 meter range. Since there are 10.76 sq ft per sq meter, it follows that there are 10.76 lux per foot candle.
    3) Illuminance of noonday sun - ~ 10,000 ft-cd, which is 107,600 lux. Since the sun is about 491,040,000,000 ft away, it's visible energy output must be about 241,120,281,600,000,000,000,000 candles!
    4) Illumination levels in various other settings:
    Office illumination: 50 to 200 ft-cd Art galleries 10 to 50 ft-cd Hospital operating table 1000 ft-cd Moonlight, full moon 0.01 ft-cd Major league baseball 100 ft-cd
    5) Lumen outputs of various sources:
    Incandescent lamps, inside frosted 20 lumens/Watt Fluorescent lamps, cool white 50 lumens/Watt
    6) Calculating illumination level: In general, if the light from an x lumen source is spread over y square feet, the average illumination level will be x/y ft-cd.
  • Most modern automatic focus cameras use the parallax resulting from the differing locations of a beam generating emitter and an imaging detector as the basis for estimating the range to the target (subject). The LED is usually a powerful AlGaAs type emitting at 880nm, and pulse currents as high as 10 Amps are not uncommon. Because the detector is positioned at an offset (also called baseline) from the emitted beam, the position of the image of the beam (as reflected by the photographic subject) varies as a function of range. Many of these modules incorporate Position Sensing Diodes (PSDs, also called Lateral Effect Diodes), but dual photodiodes work just as well. The 2 signals provided by the detector are used to derive and estimate of range to the target or photographic subject. The most commonly used signal processing method is called ratiometric range measurement. Hamamatsu, Sharp, and Ricoh are the big suppliers. Sharp offers a module for use outside of cameras. Their US organization is based in Camas, WA.
  • Consider a spot image center with position varying over a pair of photodiodes producing signals A and B. The magnitude of A and B vary with target range AND reflectance, so the difference A-B is ambiguous with respect to range. However, the quantity (A-B)/(A+B) is independent of target reflectance. That is, we normalize the difference A-B by the total reflected signal. So, we can write: f(R) = (A-B)/(A+B) The sum and the difference terms are easily done with op amp circuits, but the division operation requires fancier circuits or digital computations. One popular signal processing approach is to logarithmically transform the terms, take the difference, then exponentiate to obtain f(R). The Ham. camera chips work this way. In addition, the chips use pulse timing to subtract off ambient light currents, i.e., the terms A and B represent only the additional signal produced by a briefly flashing LED. Since people don't expect their cameras to take millions of pictures, the camera makers get some of their performance by really blasting the LED...as much as a 10Amp pulse! The simplest chip models produce only a step output to drive a 4 to 7 position focus solenoid. The more complicated ones produce more steps and/or a voltage proportional to f(R). If you want a linear distance instead of f(R), it's best to calibrate each unit, although the geometric derivation of an expression of the form R = g[f(R)] is straightforward. Most of us simply correct f(R) by means of a lookup table or best-fit polynomial. Sharp Corp. also offer a little rangefinding module, suitable for shorter distances and higher update rates. It costs about $13 in 100 qty. Sharp USA is in Camas WA. Their Optoelectronics books are filled with great stuff.
  • I need to know what color filter (1-2mm thick, plastic) will allow the minimum amount of sunlight to penetrate, yet the maximum amount of infra red light to go through.

    Lexan is the trademarked name for polycarbonate thermoplastics made by General Electric Co. USA. GE offers a series of color formulations (many more than 2) which are visibly black but transmit near IR. The transmission spectra of all these absorptive colorants depends strongly on material thickness (Beer's Law), so, for optimum results, you should start with a thickness requirement, then determine which colorant you need. GE offer sample chips which have a thickness step molded in.

    It is difficult to find stock in these materials, so you may be up against a large minimum order quantity. Contact me via email if you can use 3mm thickness sheet optimized for IR LEDs and Si photodetectors.

    The other alternatives include: Kodak Wratten 88 series, developed but unexposed transparency film, a sandwich of theater lighting 'gels', and, in fact, a variety of plastic sheet materials that fit the following description:

    Lustrous black, no dull characteristic, when held up to a strong incandescent light, slight reddish, purplish, or greenish transparency.

    If a plastic material matches the above description, its black color is caused by absorptive dyes rather than carbon or other black fillers, and it is a good candidate for IR filtering.

    None of the absorptive filter materials will filter out the longer wavelengths (beyond LED emissions) which also contribute significantly to ambient photocurrent. If you really want maximum rejection of ambient, you need to look at dielectric filters, or a combination of dielectric and absorptive filters. You can possibly save yourself all the trouble by selecting a photodetector with an integral filter. Most photodetector manufacturers have such. Finally, there are many circuit techniques for rejecting the predominantly low frequency components of photocurrents which are generated by ambient sources (e.g., sunlight is mostly 'DC').

    Q. I am attempting to build a circuit where I require an Infrared link and also a visible led link. What I would like to know is, is it possble to somehow blank out the infrared signal from the visible led phototransistor so that the phototransistor only receives the visible signal and not the infrared.

    If you are asking for a short wavelength pass filter (pass visible, block IR), there are 2 kinds: dichroic 'hot mirror', and 'heat absorbing glass'. An example of the 1st is that building glass that looks like a pink mirror. Experimenters can get pieces from Edmund Scientific. For a few more bucks, you can buy hot mirrors from OCLI, OCA, and other mfgrs. The latter is used in slide and movie projectors, and you can get a piece from a pro photographers shop.

    If you want to do just the opposite: block the visible, then you need IR transparent black material. Many lustrous black plastics are quite transparent in the IR.

    Rosco makes a clear film, ThermaShield, that runs about $40 to $50 for a 12"x12" sheet. It reduces most of the infrared but it's probably not "optically" clear.

  • You are liable to be disappointed about green LED output power. Since the human visual system is so sensitive to green, high radiance green LEDs haven't been much of a priority. Of course, some are brighter than others. If you're trying to achieve high visual brightness, have a look at units from Stanley, Sharp, Toshiba, all of whom have 'extra super bright' greens in the 'true green' and 'yellow green' parts of the spectrum. If you're generating a beam to be detected by Si photodetector, you'll get better coupling with yellow-green types from Stanley and HP. In any case, choosing a part with a narrow coverage angle will give higher intensity by concentrating the power.
  • Q. In my mistreatment of the LEDs, I ended up with one which puts out no light at 10mA, but will light up OK at 100mA!

    A. You have probably 'cratered' the ball-bond.

    Another poster mentioned 'dark line defects'. This is, indeed, one mechanism for declining output. However, many of the more modern LED processes are not prone to DLDs. Here are some other defect types:

    1) Cratering: A crack develops under the ball bond metallization zone. If you pull on the bond wire (not possible in encapsulated LEDs), a chip pulls out, leaving a 'crater', hence the name. Symptoms: If you apply pressure to the ball (sometimes just by pressing on the top of the LED, light output momentarily increases (or perhaps, is restored, in the case of total failure). Decapsulate and test bond/chip integrity. If you check the VI characteristic with a curve tracer, you may see 'breakover' characteristic if you apply sufficiently high bias Voltage.

    Causes (singly or in combination): a) incorrect ball bonding parameters such as too much pressure, bad capillary, contaminated pad, etc. b) tension on bondwire, related either to incorrect looping, vibration, or shock c) power density of input pulses exceeds device capabilities

    2) Die attach migration shunts junction and/or reduces optical transmission: This is more likely to happen with LEDs that use silver-filled epoxide die attach materials, as opposed to eutectic (solder) attachment, but it can occur in either case. The silver can creep up the side of the die, eventually shorting it out. Symptoms: Can be seen using appropriate visual inspection techniques. Usually shows up looking like a parallel resistance on curve tracer VI characteristic, i.e., current starts to flow at low bias voltage, whereas the healthy junction shows little current flow until the bias is near the threshold for the material, usually between 1.1 and 1.8 Volts for LEDs.

    Causes: The mfgr. is using too much die attach material if this happens. However, the problem is aggravated by high temperatures and pulse energy levels. Cures: Get vendor to control process properly. Reduce drive levels and/or temperature.

  • The most prominent manufacturers of high performance visible LEDs are: Stanley Electric, Hewlett Packard, Sharp, Toshiba, and Siemens. Notable manufacturers of high performance infrared emitters include all of the above, plus these additional manufacturers: Hamamatsu, Honeywell, Mitsubishi, Rohm, Optodiode, Dowa Mining, and Mitsubishi Cable (a different company). Some of these manufacturers are represented in the LINKS section below. Others will be added as time permits.
  • Most visible black fabrics are surprisingly reflective in the near IR. We went to carpet stores with an IR viewer and kept looking until we found true IR black carpet. It's easy to make an IR viewer from a monochrome (B&W) closed circuit TV system: you put an IR transmissive filter over the lens. However, you can also simply measure the detection range using your sensor.

    Flat black lacquer spray paint works well if you take care to apply light coats. Its pigment is carbon black, which is absorptive at most all wavelengths. You can also purchase excellent 'flock paper' from Edmund Scientific.

    Oh yes, circuit boards pipe light around. Testors flat black model paint, available at hobby shops, is quite opaque. However, black electrical tape is semi-transparent to IR.

  • A superior technique is called 'background suppression'. You set up 2 detectors connected in inverse parallel and arrange them such that optical parallax keeps one of them from 'seeing' the emitter image until a certain range. Beyond that 'cutoff' range, the 'background' detector suppresses the signal from the 'foreground' detector by cancellation.

    For backgnd suppression, separate the LED several inches from the pair of photodiodes. Aim one of the photodiodes somewhat inclined toward the LED. Aim the other directly perpendicular to the circuit board plane. The inclined detector will pick up the spot through some range of overlap btwn its field of view and the beam of the LED. The perpendicular detector will have a region of overlap beginning at a greater distance. Connect this latter detector with a negative polarity with, respect to your demodulation scheme. Its output will suppress response to the background quite effectively. You can adjust the angles to achieve various cutoff ranges. This works best when you use auxiliary lenses, which are actually quite easy to apply.

    If you have access to a technical library, you might have a look at Juds, S. 'Photoelectric Sensors and Controls'. I am involved in revising this 1988 book now, but it is still quite useful.

  • Of course phototransistors give you more signal than photodiodes: they are the equivalent of a photodiode and a transistor. However, their characteristics are more variable from unit to unit and over temperature. Consequently, we usually use separate components for production products. Also, you can't use phototransistors connected in inverse parallel as described above.

    You can readily get photodiodes from mfgrs and mfgrs reps. An excellent source is Infineon Opto in Cupertino CA. Also, with any of the 3 leaded phototransistors, just use the base and collector terminals.

  • Q. The last time I used the facilities at our local airport, I noticed that the toilets (or urinals) flushed automatically when I moved away. OK, I figured they must be ultrasonic or infrared. The question that really puzzled me was "How do power the devices?" There are no wires to the sensor unit, and I can't believe that someone has to come around and change batteries regularly. Is it possible that these devices are powered off of miniature generators taking power from the moving water stream? Or do they hide the wires inside the plumbing?

    If you're talking about the type that has a black plastic dome over the flush valve regulator (the chrome piece that usually has a bat handle sticking out), they are powered by 4 AA alkaline cells. Service life is over 50,000 flushes. The design is a marvel of energy efficiency. Active infrared sensing (modulated LEDs), very low duty factor, and magnetic latching pilot solenoid valve. If you do a web patent database search on 'flush valve sensor' you can learn more.

  • Considering that LEDs emit their energy in a narrow spectral band, it may seem surprising that White LEDs are available. There are presently 2 kinds of LEDs that have a visual white appearance. The earlier type actually consists of red, green, and blue LED chips packaged closely together. This kind of array can produce many colors, but with the appropriate currents in each chip, the combination appears white.

    A newer kind of white LED consists of a blue LED combined with special phosphors that give off a broadband glow when excited by blue light.

  • Compared to DC operation, it is sometimes possible to significantly increase visible LED brightness by pulsing. There are 3 reasons why pulsing can increase brightness:

    1) LEDs have more output at lower temperatures. Low duty factor pulsing can, in some cases, lower the operating temperature of the LED.

    2) The human visual system is non-linear. With the right choice of duty factor and pulse rate, perception will correspond more to the peak brightness than to the average brightness. This is especially true at low pulse repetition rates. However, there are a couple of potential problems with using low rep rates:

    a) Visible LEDs may appear to flicker, especially at frequencies below 30 Hz.

    b) Flickering sources may be a hazard to epileptics. For more information, see the following sources:

    c) Many LEDs have non-constant luminous efficiency. Luminous efficiency is the ratio of light output per milliamp of input. At very low and very high current levels, you get less light per milliamp than you do at intermediate currents. For an example of this phenomenon, see Figure 4 in the following Agilent (formerly HP) App Note:

    From the above considerations, we can make the following generalizations:

    1) If you are trying to operate visible LEDs at maximum possible brightness levels, you will always get better results using DC operation. This is because luminous efficiency declines at high currents.

    2) If you are trying to achieve a compromise between battery life and LED o, utput, i.e., you are not operating the ,, , LED at , maximum power, pulsing is probably advantageous. If the LED datasheet does not provide an efficiency curve, you will have to experiment to find the best compromise.

  • Q. It seems that the optical output IR LED's is generally given in milliwatts and phototransistor/diode sensitivity is given in flux density, lux or millicandelas. The only connection that I can find between the two is that 1 W = 683 lm at 555nm.

    A. First of all, you only need the efficacy of Si photodetectors to predict photocurrent from incident power. Of course, it varies as a function of wavelength, but at 880nm, it is about 0.5 Amps/Watt. Simply multiply the incident power produced by the IR LED by this value to get photocurrent. At other wavelengths, use the appropriate values from the efficacy curves published in datasheets. However, you won't see much variation from photodiode to photodiode. Silicon detectors all have pretty much the same curve, reaching a peak around 900nm and falling off roughly linearly toward the shorter wavelengths, crashing precipitously beyond 950nm or so.

    It can be difficult to calculate incident power from datasheet values, particularly for visible emitters that are specified in photometric units. It depends strongly on geometry. If an emitter is spec'd in milliwatts per steradian, you multiply the spec by the number of steradians covered by you detector. Steradians are a measure of solid angle. By definition, a sphere has 4*pi steradians. This is handy, since the surface area of a sphere with radius r is 4*pi*r*r. It follows that a small area A on a sphere of radius r subtends a solid angle T:

    T = (4 * pi) * A /(area of sphere with radius r)

    = (4 * pi) * A / (4 * pi * r * r)

    = A / (r * r)

    For a detector with active area A at a distance R from an emitter with steradiance S, the power P incident on the detector is approximately:

    P = A * S / (R * R)

    Obviously, the units of A and R have to be the same, e.g., if A is expressed in square mm, measure R in mm. The above approximation applies only to the case where a directional emitter is aimed at the detector with the detector facing the emitter. This is because those are the conditions under which the datasheet value S were measured.

    An example will make this more clear. For Siemens IR emitter SFH484-2, S = 900mw/sr at 1 amp. For Siemens photodiode BPW34, A = 7 sq mm. At R = 1000mm, we calculate the incident power:

    P = 7 * 900 / (1000 * 1000) = 6.3 microwatts

    From this figure, we predict about 3 microamps photocurrent. Voila!

    If you need to do the same calculation for visible LEDs that lack the steradiance spec, you can divide the luminosity (in millicandelas) by the value (in lumens per watt) of the standard observer curve at the wavelength of the LED. Note that a 1000mcd source has a total luminosity of 4*pi lumens, i.e., 1 lumen per steradian. Since there are 4*pi steradians over a sphere, this definition of lumens makes it easier to work with steradians. The standard observer curve is published in many places. Just look in the back of most any LED databook. Since LEDs output a band of wavelengths, you are making an approximation by using a single value from the curve, but it will generally give you a good estimate. Again, an example will make this more clear. Siemens red LED LS5420 is rated 100mcd (i.e., 0.1cd) at 10ma drive with dominant wavelength 628nm. Standard observer value at that wavelength is about K = 300 lumens per watt. Its steradiance at 10ma is therefore estimated by:

    S = [0.1cd * (4*pi lumens/cd)]/ [300 lumens/watt * 4*pi angle per sr]

    = 0.1 / 300 watts/sr = 333 microwatts/sr at 10ma.

    For comparison with the SFH484-2 IR emitter above, note that its steradiance at 10ma would be about 100x lower than the 900mw/sr used above, making the IR emitter about 25x more intense than the red emitter at the same current! However, note that the visible LED emits into a much broader angle (+/- 24 degrees) than the IR emitter (+/- 8 degrees). The red LED covers 9 times the area (at 3x the angle). We can infer that the red LED puts out about one third of the power of the IR emitter at the same current (25/9 = ~3).

 

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