Making ambient LED brighter - Page 1
Forum section: electronics / beginners

author: symbianas
author_rank: Regular Contributor
timestamp: August 13, 2021, 02:04:32 pm
content: Hi, my car has ambient lights but they are very dim and I would like to make them brighter. It is possible to adjust brightness, but its for all LED's in the car at once, so instrument cluster is way to bright at max brightness, and ambient light at max brightness is a bit better, but still quite dim.I have disassembled the LED light module. It has 442ohm resistor, LED and another 442ohm resistor. It is powered from 1V (at min brightness) to 12V (at max brightness).circuit:I don't have LED specifications, but it looks similar to this one:https://www.arrow.com/en/products/laa67f-aabb-24-1/osram-opto-semiconductorsNot sure how it can handle 12V? Does 884ohm reduce voltage to <5V?How could I make this led brighter? Would removing resistors make it brighter? If not, where could I find Brighter LED with amber color?

author: Saimoun
author_rank: Frequent Contributor
timestamp: August 13, 2021, 02:28:09 pm
content: Like all LED it has a forward voltage, so you do(source voltage - forward voltage) / resistorto get the current going through the LED. More current means more light (it is not linear tho, imagine a logarithmic curve).(the forward voltage will change a little with the amount of current going through - but for simplicity you can assume about 2V).So at 1V it is probably completely off. At 12V you get 10V over 442 ohm = 23mA. If the LED can handle 50mA like the one from your datasheet, then you can safely solder another 442 ohm resistor on top of the other one.Actually looking at the picture I am not sure how it is wired... Are there 2 LEDs, or is it one with: power > 442ohm > LED > 442ohm > GND ?

author: symbianas
author_rank: Regular Contributor
timestamp: August 13, 2021, 02:56:38 pm
content: Quote from: simonlasnier on August 13, 2021, 02:28:09 pmYes, its one LED:

author: TimFox
author_rank: Super Contributor
timestamp: August 13, 2021, 04:20:59 pm
content: Your photo shows two resistors and two LEDs. Normally, there would be only one resistor per LED for a simple circuit like this.

author: symbianas
author_rank: Regular Contributor
timestamp: August 13, 2021, 04:35:07 pm
content: Quote from: TimFox on August 13, 2021, 04:20:59 pmOK, since the board looks confusing I have made additional photos with different angles:So the circuit is like this:Not sure why resistors are on both sides, instead of one 442*2=884ohm, is there a reason for that?As for LED I have connected it to 5.061V power supply and tested entire circuit current consumption its 3.61mA. Is it possible to get LED forward current and voltage? As I don't have exact specification of LED.

author: TimFox
author_rank: Super Contributor
timestamp: August 13, 2021, 04:39:20 pm
content: Yes, that physical circuit is strange—perhaps the board is designed to handle different versions.If you know the input voltage and current (or voltage across one resistor to determine the current), the voltage across the diode isVD=VIn-IxRtotal

author: symbianas
author_rank: Regular Contributor
timestamp: August 13, 2021, 04:51:56 pm
content: I have tested circuit at 5.061V power:total consumption 3.61mAresistor voltage 1.612VLED voltage 1.836Vresistor voltage 1.609Vbut how do I get max forward current and voltage?

author: Siwastaja
author_rank: Super Contributor
timestamp: August 13, 2021, 05:00:49 pm
content: Figuring out the LED part number would be best.Otherwise, you need to assume.With so much voltage dropped over the resistors, the Vf of the LED doesn't matter much. Whether it's 2V or 3V or 3.7V is a small difference.Maximum current rating for a LED like in the photo is at least 20mA but it can be like 50mA. If you don't know, you need to play safe and assume 20mA.Assuming Vin=14V, Vf=3.0V, current now is 11V/884ohm = 12mA.You can double this quite safely. Try 2*220ohms = 440 ohms for 25mA. If the LED is poorly cooled or in a hot environment (in summer), this may reduce the lifetime, though.

author: golden_labels
author_rank: Super Contributor
timestamp: August 13, 2021, 07:14:13 pm
content: What size are those resistors? They seem to me like 0603 package, likely with 1/8W power rating. At 12mA power dissipated on a single one would exceed the limit. Hence two resistors.That this is also going to be used in a car, where transients in range of hundreds of volts are normal. Using two resitors doubles the voltage rating.

author: symbianas
author_rank: Regular Contributor
timestamp: August 13, 2021, 09:24:56 pm
content: Quote from: golden_labels on August 13, 2021, 07:14:13 pmResistor size is something like: ~3mm x ~1.2mm x ~0.3mm.LED size is something like: ~3.9mm x ~3.9mm x ~3.9mm. It looks like this:

author: TimFox
author_rank: Super Contributor
timestamp: August 13, 2021, 09:32:43 pm
content: The resistor sounds like 1206 (inch) size. The typical power rating is 1/8 W. 0603 is maybe 1/16 W.

author: golden_labels
author_rank: Super Contributor
timestamp: August 13, 2021, 11:25:57 pm
content: In either case requires two resistors to handle that amount of power at 12mA.At 25mA the power dissipated will be in the range of 250–300mW. So you need two 1/4W resistors.

author: symbianas
author_rank: Regular Contributor
timestamp: August 14, 2021, 12:15:18 am
content: I think my LED parameters should be worse as the car is made in 2010. "LA A676" model LED seems to have worse specs.:https://www.osram.com/ecat/SIDELED%C2%AE%20LA%20A676/com/en/class_pim_web_catalog_103489/prd_pim_device_2190949/https://dammedia.osram.info/media/resource/hires/osram-dam-6063130/LA%20A676_EN.pdfSo lets say I have this "LA A676" LED.1. Forward current (TS = 25 °C) max. is 30 mA. Does this mean I can use 30mA?2. Forward Voltage (IF = 20 mA) min. 1.83 V typ. 2.00 V max. 2.33 V. Should I use max voltage also 2.33V?3. Lets say supply voltage is 12V; Forward current 30 mA; Forward Voltage 2.33 V. In such case I need one resistor which would be 322.333ohm?

author: golden_labels
author_rank: Super Contributor
timestamp: August 14, 2021, 01:24:15 am
content: Quote from: symbianas on August 14, 2021, 12:15:18 amThat means you should not exceed 30mA at 25°C. Not exceed = you should keep away from it. In particular since it will not be 25°C.Quote from: symbianas on August 14, 2021, 12:15:18 amNo. That is the maximum voltage you will observe on LED if you pass 20mA through it.Quote from: symbianas on August 14, 2021, 12:15:18 amIfthe supply voltage is 12V and the desired current is 30mA, then the forward voltage will betypically2.05V(1). Therefore:(12V - 2.05V) / 0.03A ≈ 331.67ΩThat is a typical value. As you can see from characteristics the forward voltage varies wildly. So you can’t assume you will get exactly 2.05V: it may as well be 2.4V or 1.9V. There is also no 331.67Ω resistors. In the E24 series 300Ω, 330Ω and 360Ω are reasonably close.But you will likely not have exactly 12V, unless this comes from a voltage regulator. More likely somewhere between a bit under 12V to over 14V. And you do not want to exceed or even get too close to 30mA. Which is why 25mA seems more reasonable, with 440Ω resistance — formed by two 220Ω resistors.____(1)Based on the If-vs-Vf chart on page 7.

author: jmelson
author_rank: Super Contributor
timestamp: August 14, 2021, 02:05:54 am
content: OK, the 1-12 V is not an actual variable DC voltage, it is PWM and your meter is just showing the average. If all these LED modules are in series, it will be hard to change. But, I suspect the modules are in parallel, so reducing the resistor value on the ones you want to make brighter will do it. Don't reduce the value too much or it might overload the PWM controller. So, if the maximum PWM voltage is about 14 V, then one LED will drop about 3 Volts, so (14 -3) / (442+2) = 12.4 mA. So, reducing the resistors to 300 Ohms each will increase the peak current to 18.3 mA, for a roughly 50% increase in brightness.Jon

author: antenna
author_rank: Frequent Contributor
timestamp: August 14, 2021, 05:07:42 am
content: Quote from: symbianas on August 14, 2021, 12:15:18 amYou are correct with that resistance value, but now you need to consider the amount of power the resistor can handle. As you already found out, there will be 12-2.33=9.67v across the resistor with a current of 30mA. Volts x amps = watts. So 9.67*.03=.2901w. You would need a half watt resistor as a quarter watt resistor would be too small and over heat. You will need to get the appropriately rated resistor or split the wattage up between multiple resistors as the manufacturer did with the two 4420's. I would also point out that most automotive alternators put out a tad over 14v, so using the absolute current rating of the LED will leave absolutely no room for errors like that, and even if calculated properly, the LED will have a much shorter life than if you gave it the recommended 20mA.EDIT: somehow my eyes didn't see the last couple posts and golden_label already covered some of what I said. Apologies for repeating.

author: symbianas
author_rank: Regular Contributor
timestamp: August 14, 2021, 11:36:10 am
content: Quote from: antenna on August 14, 2021, 05:07:42 amIndeed, I have checked the voltage (at max brightness) when engine is running and its not 12V, it becomes 14V, I thought maybe the voltage should be regulated for LEDs by car ECUs...Quote from: jmelson on August 14, 2021, 02:05:54 amDoes this mean LED's always gets 14V just for shorter period of time and it becomes ~1V (at min brightness)? Does this mean I should always use 14V as supply voltage even if I want max brightness at 7V (at medium brightness setting)?I have also did the measurement with new max voltage:Power supply 14.009 VVoltage across LED 1.911 VTotal circuit consumption 13.700 mA (does this mean each item on the circuit in parallel gets 13.7mA? So LED forward current is 13.7mA in this case?)Is it possible to assume what LED characteristics are if manufacturer set 13.7mA as max current?Quote from: golden_labels on August 14, 2021, 01:24:15 amSo my best bet is: power supply 14V, forward voltage 2.02V (tried to look at the table), forward current 25mA.(14 - 2.02) / 0.025 = 479,2ohm. So I could use 0.25W resistor 470ohm "safely" with life expectancy like 5years?I am also considering buying newer "LA A67F" which would have better performance, but shipping is quite expensive. Unless I would buy items for 50eur. Does888A 2 IN 1 DIGITAL HOT AIR REWORK STATIONsounds as a good perches and worth 84eur? Or its better to buy china ones from "ebay/ali"?

author: Siwastaja
author_rank: Super Contributor
timestamp: August 14, 2021, 12:26:09 pm
content: You don't need hot air for this, soldering iron will do.Even easier, don't desolder the resistors at all, just use soldering iron to piggyback new resistors (of same physical size) on the top of the existing resistors. For example, if you use the exact same value, you will halve the resistance. The obvious plus side, having two resistors share the power dissipation, they run cooler, albeit you won't gettwicethe power dissipation due to mutual heating but nearly so, anyway.There is also enough room to scrape off soldermask and solder new resistors in parallel next to the existing ones if you wish to do that.

author: golden_labels
author_rank: Super Contributor
timestamp: August 14, 2021, 12:49:17 pm
content: Quote from: symbianas on August 14, 2021, 11:36:10 amPWM means that the circuit is powered on and off at fast rate. At any given time it is either cut off and receives no current, or is exposed to the full supply voltage. Due to LEDs having brightness linearly dependent on current, there is no difference if the LED receives continuous 10mA and emits I light, or is off 3/4 of the time and receives 40mA 1/4 otherwise, hence being turned off 3/4 of the time and shine at 4I intensity 1/4 of it. The average brightness is perceived as the same. Capacitances in the circuit additionally smooth out the curve.Code:40mA ,--. ,--. ,--. ,--. ,--. ,--.| | | | | | | | | | | |30mA | | | | | | | | | | | || | | | | | | | | | | |20mA | | | | | | | | | | | || | | | | | | | | | | |10mA - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -| | | | | | | | | | | |0mA ----' `--------' `--------' `--------' `--------' `--------' `---Quote from: symbianas on August 14, 2021, 11:36:10 amNo. The characteristics vary even inside a single batch. It only means that the manufacturer determined that somerange of currentsis good enough and used it. 13.7mA is in that range.Quote from: symbianas on August 14, 2021, 11:36:10 amThe resistance: yes. But keep in mind this is engineering. What you get from calculations tells you where more or less you are standing. Not that everything will work exactly like that. With a 470Ω resistor that will be (14V - 2.02V) / 470Ω ≈ 25mA. But the resistor may be 447Ω and voltage go to 14.2V, at which case you will see 27mA, or may be 493Ω at 12V, giving you 20mA. And that ignoring other variations, slight temperature dependency, I-V curve of the LED etc.However, the power rating of the resistor is wrong. At 25mA a 470Ω resistor will see nearly 300mW. So you need either a 0.5W or tw 0.25W resistors. The second option is preferred. Not only it allows sharing load, but a SMD resistors with higher power rating may be a bit misleading: the heat must go to somewhere and by using a single resistor you are using only half of the PCB to dissipate that heat.Quote from: symbianas on August 14, 2021, 11:36:10 amIf you want to buy new LEDs, obtain the highest bin instead of simply increasing the supply current. According to the datasheet(1)in LA A676 the difference between the worst and the best bin is threefold, in LA A67F about 2.5×.Since the intensity of emitted light scales approximately linearly with the supply current, by increasing the current twice you only get light perceived as about ~40% brighter.(2)Threefold increase in brightness by choosing a better bin is perceived as as ~75% increase.Quote from: symbianas on August 14, 2021, 11:36:10 amAs Siwastaja said, you need no hot air for that. But if you are willing to buy a hot air station at price that low, search for it locally. Those are all cheap imports from China and there is no reason to spend much more money simply to have a “no-brand brand” label pasted on them. If you search locally, be aware they may be sold as plastic welding devices too.____(1)Page 5.(2)Human light perception is not linear.

author: symbianas
author_rank: Regular Contributor
timestamp: August 14, 2021, 03:51:29 pm
content: Quote from: golden_labels on August 14, 2021, 12:49:17 pmWhat do you mean by "bin"/"highest bin"?I should defiantly go for "LA A67F" as it should be ~75% brighter?

author: Siwastaja
author_rank: Super Contributor
timestamp: August 14, 2021, 04:00:28 pm
content: Binning means, manufacturer makes a large batch of LEDs, they end up with different efficiencies - i.e., different amount of light given same current - so they sort them to differentbinsand sell the better ones at higher cost.Though, in reality, the "highest bin" is often unobtanium. Typically distributors like Digikey only stock one or two bins.Sorting for efficiency is what you want and nothing prevents using a different part altogether if you can get it physically fit and somehow soldered in place. Better efficiency would be the best as you wouldn't need to increase current, wouldn't need to change the resistors, and amount of heat would stay the same yet you'd still get more light output.LED efficiency massively vary all over the place, assuming white LEDs for lighting purposes, anything from 50 lm/W to 200 lm/W is available.

author: symbianas
author_rank: Regular Contributor
timestamp: August 14, 2021, 04:57:36 pm
content: Would thisOSO5PA3131A OPTOSUPPLYLED be much, much brighter? It has Typ 8000mcd?I need to search for highest "mcd" to have bright LED with 20-100mA current?

author: golden_labels
author_rank: Super Contributor
timestamp: August 14, 2021, 05:02:46 pm
content: To clarify, bins are what Siwastaja said. For example in the LA A676 and LA A67F datasheets they are listed page 5 as “Brightness Groups”. Those are groupings within as single model of a device. LA A676 and LA A67F are different devices.Quote from: symbianas on August 14, 2021, 04:57:36 pmYes, but likely not in a way you want. It produces a very narrow beam of light compared to the Osram’s parts you mentioned earlier.

author: symbianas
author_rank: Regular Contributor
timestamp: August 14, 2021, 06:52:38 pm
content: I have found such bright LED in my local shopLA G6SP-DAFA-24-1However, it seems to be too powerful (140mA; 4500...14000.0mcd), will my power supply PWM module and 0.35-0.5mm^2 wire will cope with that?I would need 2W resistor?

author: james_s
author_rank: Super Contributor
timestamp: August 14, 2021, 07:48:12 pm
content: The mcd ratings are not very useful for your application. What you want is lumens, that means the actual amount of light. What mcd tells you is intensity, that is the brightness level. If you focus a light source into a small spot the mcd will greatly increase but the lumens will not. At the extreme end, a laser pointer produces a very high mcd number but is worthless for lighting up a room because it only illuminates a tiny dot.Once you get up to such high currents a resistor is a silly way to try to drive the LED from such a high voltage, you're burning up far more power in the resistor than you are getting to the LED. You would be much better off using 2-4 LEDs in series at a lower current. That will result in far more power being turned into light than you get with a single LED. Keep the current low and make use of all that excess voltage you have available.

author: TimFox
author_rank: Super Contributor
timestamp: August 14, 2021, 07:53:40 pm
content: Yes, candela (hence mcd) represent the luminous fluxper solid angle in a given direction, while lumens represent the total light emitted in all directions.The old definition is no longer politically correct: One candlepower was the light produced by a pure spermaceti candle weighing one sixth of a pound and burning at a rate of 120 grains per hour.

author: symbianas
author_rank: Regular Contributor
timestamp: August 14, 2021, 08:26:12 pm
content: Quote from: james_s on August 14, 2021, 07:48:12 pmWhat about LED which would use forward voltage 2.15V and current 50mA, and the resistor 277ohm 0.693W? Still too much current for resistor?Should I take 14V as car power supply for calculation or 15V? 16V?

author: Siwastaja
author_rank: Super Contributor
timestamp: August 15, 2021, 07:57:06 am
content: 0.7W is quite a lot in a small PCB, so in addition to using much larger resistors, the whole PCB assembly could still heat up too much, depending on how thermally insulating the actual location in the car is. On a lab table, 0.7W is likely no issue at all.Spending the effort in finding highest efficiency LED would pay off, then you wouldn't need to go to 50mA.The elephant in the room is, driving a 2V LED from 12V is highly inefficient; 10V is totally wasted in the resistors and even if the LED was magically 100% efficient the efficiency is just 2/12 = 17%.Put two LEDs in series and you get double the light output with the same total power use, and actuallydecreasethe dissipation in the resistors. But this requires mechanical modifications to be able to fit multiple LEDs and still make them illuminate whatever they are illuminating.You can go further and put 3-4 in series but at some pointcurrent regulationof such simple resistor scheme suffers too much.Calculate current at Vin=15V and Vin=11V (if you need the LEDs to be visible with engine turned off), and you'll see what I mean if you compare the solution with Vf=2.0V versus Vf=4*2.0V=8.0V.

author: symbianas
author_rank: Regular Contributor
timestamp: August 15, 2021, 10:33:09 am
content: Quote from: Siwastaja on August 15, 2021, 07:57:06 amYes, ideally I would like to use more than one LED. I could make custom "amateurs PCB", but the issue is that I need to focus the light into a small point (~2.5mm x ~2.5mm circle) where the light goes into a plexiglass tube. Does some kind lens or 2to1 (3to1, 4to1) plexiglass part exist? Any ideas how to focus multiple LED's to one point?

author: Siwastaja
author_rank: Super Contributor
timestamp: August 15, 2021, 11:23:01 am
content: Lenses are possible but not easy or practical.If you have multiple such modules in different places, rewiring the modules in series could be easier.Switch-mode LED supply is always an option but way too complicated for such simple case, and problematic because the supply itself is PWM'd.I'd guess you get the best result the easiest way by finding an LED which gives you more light output for the same current, and then possibly increasing the current a tad, say to 20-25mA but no further.

author: golden_labels
author_rank: Super Contributor
timestamp: August 15, 2021, 04:21:01 pm
content: symbianas:As a supplement to the earlier mention, of how one should treat calculations as the general indication of how a circuit operates, see the attachments. Those are experimental data for two jellybean parts: a LED and a resistor in series. If you would predict the voltage at 25℃, merely putting the device in a cold car in winter shifts it down by 5%.(1)I think your other questions were answered by people above, but I have a suggestion. Since you are moving further and further away from a trivial modification of an existing circuit and move towards a completely new build, perhaps simply buy a suitable 12V LED strip?____(1)The value being the same as the resistor precision is a coincidence. The primary source of variance here is the temperature, while the components themselves happened to be very repeatable.Data collection details: 270Ω no-brand, carbon THT jellybean resistor; 5mm no-brand jellybean red indicator LED. Temperature measured at a PCB onto which the components were soldered at the beginning and the end of each temperature group. The datapoints are a cartesian product of 10 resistors and 10 LEDs: the datapoints are independent only if interpreted as a random choice of component pairs from one’s stash, not unpaired components from the population. Measurement accuracy: voltage ±0.5%+digit, current 1.5%+digit. Data taken after letting the circuit settle (typ. 5–7s). The bubble chart indicates counts of each Vf/If observation.

author: symbianas
author_rank: Regular Contributor
timestamp: August 15, 2021, 05:50:28 pm
content: Quote from: golden_labels on August 15, 2021, 04:21:01 pmIndeed, modification becomes complex and complex... But still I would like to keep same power source. Also the main thing is that plexiglass pipe is embedded into the door and I don't want to mess with it. I want to keep factory look.But I found some interesting lens which seems promising:1294.1001 MENTORI am planning to use it with these 3x LED's:LA M67F-V1AB-24-ZUf = 2.2V*3 = 6.6V If = 20mA. So resistor would be 420ohm 0.168W, I hope that's ok.Though the shipping of LED's from "digikey" will be expensive... or I need to think of what other components I need additionally as you need to buy >50eur to get free shipping.

author: Siwastaja
author_rank: Super Contributor
timestamp: August 16, 2021, 06:47:05 pm
content: Very unlikely you get any benefit from such lens. It's not designed to increase light output by combining multiple LEDs. As the datasheet explains, the purpose of the product is to combine different colors of indicator LEDs. Optical efficiency is unspecified and likely poor enough so that the three "combined" LEDs produce less light than just one LED at the optimal position coupled to the original light pipe directly without extra lenses.

author: symbianas
author_rank: Regular Contributor
timestamp: August 16, 2021, 09:44:16 pm
content: Quote from: Siwastaja on August 16, 2021, 06:47:05 pmWell in the description it writes: "Thus, monochromatic, bichromatic or trichromatic displays are possible as well as corresponding blended colors thanks to the multiple-drive technique. "I thought it blends "monochromatic" one color LED's too? Well, money wasted if I misunderstood that...

author: james_s
author_rank: Super Contributor
timestamp: August 17, 2021, 06:24:27 am
content: Quote from: symbianas on August 16, 2021, 09:44:16 pmIt will blend two LEDs of the same color, it's just unlikely to do so very efficiently. You will get more light but it probably won't look all that much brighter to the eye.

author: Siwastaja
author_rank: Super Contributor
timestamp: August 18, 2021, 01:52:33 pm
content: If it was supposed to be used for increasing light output for producing more light, they would definitely specify optical efficiency numer as that's the key parameter.Problem in such optics is getting the coupling from LED to the lens just right. Even if the lens is perfect, the coupling loses light.High-efficiency LED lenses are designed for a specific LED part number, running the optical simulation of the complete LED + lens system.Efficiency less than 33% is well possible meaning you get less light out of three LEDs. Even if you got 50% optical efficiency, the gain is not enough to warrant this complication IMHO.

author: symbianas
author_rank: Regular Contributor
timestamp: August 18, 2021, 09:17:58 pm
content: Quote from: Siwastaja on August 18, 2021, 01:52:33 pmWell on page 2 ofMENTOR-LL.pdfthey claim:MENTOR light guides are efficient, precision-designed optics that transfer lightfrom PCB-mounted LEDs to a remote light-radiating zone with little or no loss ofluminance and no light bleed between adjacent elements.Technical data is on page 67, but its not much of it. So its "fake news"?

author: james_s
author_rank: Super Contributor
timestamp: August 19, 2021, 01:17:03 am
content: Having not seen the actual application I can only guess, but if you are trying to squirt light into a narrow light guide then there might actually be benefit to an LED with a high MCD rating as it will have a narrow beam. There also are or used to be power LEDs available that had multiple dies in them so the forward voltage (and light output) was considerably higher than of a standard LED.If you really want to go all out, I've built LEDs around the PT4115 driver IC and found it to be relatively simple to use and perform quite well. It has a dimming input that IIRC is capable of accepting either an analog voltage or PWM signal. It would be within the realm of a beginner who wants to design a PCB to build a whole replacement module and using a chip like that you can get quite good efficiency driving a power LED up to hundreds of mA. If that sounds like too much work, you can buy small modules designed to be integrated into Chinese MR16 bulbs that are based on that or similar chips. You can adjust the current by replacing a single sense resistor.

author: Zero999
author_rank: Super Contributor
timestamp: August 19, 2021, 11:56:41 am
content: I wonder if you can get a multi-die LED, consisting of several chips, which should be much brighter and much more efficient. Failing that, you could bodge some smaller LEDs on to a tiny board, soldered to the edge of the main board.

author: Siwastaja
author_rank: Super Contributor
timestamp: August 19, 2021, 03:05:03 pm
content: "little or no loss of luminance" is just complete marketing bullshit.It's like marketing LM358 as "low noise opamp". It says nothing.But of course, depends on what they measure. Their plastic (likely PMMA) mayabsorbnext to nothing, so if you integrate all the light hittingsomewherethe efficiency number can be great. It's just that likely 25% of the light is reflected back to the LED where it disperses everywhere around the LED, and another 25% is lost when you again couple to the existing light pipe in the car. Light is not absorbed in the lens plastic, but ends up in wrong places.Optics is always difficult if efficiency matters. As I said, it requires careful coupling of the light source (you could say optical "impedance" matching) and optical simulation with the exact LED part number.

author: symbianas
author_rank: Regular Contributor
timestamp: August 19, 2021, 05:37:20 pm
content: Quote from: Siwastaja on August 19, 2021, 03:05:03 pmYes, its seems like money wasted... So combining LED's wont work...I have found this one which seems more promisingOS5OGA5111A75000mcd.I have also foundthis postwhich solves similar problem and states that narrow angle of 15 is even better.I have also started thinking about programmable RGB LED'sBut have not found them with high MCD values... So might need to stay with one color.

author: james_s
author_rank: Super Contributor
timestamp: August 19, 2021, 09:22:37 pm
content: RGB LEDs (programmable is not needed, they make plain ones too) are another way of getting a bright white LED by running all three chips at once. Unfortunately most of them have only one common lead for all three so you can't wire them in series. There may be exceptions though.

author: antenna
author_rank: Frequent Contributor
timestamp: August 20, 2021, 06:17:25 am
content: This discussion is missing something (or I missed it), and that is "how much current can that little dimmer knob supplying the wiring harness handle? You guys are talking about PWM's and Jiggawatt LED lights and half watt resistors, so I'm curious how long it will take the dimmer pot to cook and turn it all off...12v LED strips, you know, those ones with the remotes for controlling the color that people like to stick above their kitchen cupboards and along stairs.. Run one of those, separate power, and maybe change the discussion to integrating the two so they activate automatically before the car burns down. If you could change the color to red, the stress on the eyes at night will be nearly non-existent - while still having the option to read a book and attract all the moths you want.

author: symbianas
author_rank: Regular Contributor
timestamp: August 20, 2021, 06:36:10 am
content: Quote from: antenna on August 20, 2021, 06:17:25 amDo you mean its something wrong with this LED?QuoteWhich parameter is wrong? Its Vf = 2.1V If = 20mA. Manufacturer of the car also used similar SMD LED Vf = ~2V If = ~20mA.

author: james_s
author_rank: Super Contributor
timestamp: August 20, 2021, 06:40:41 am
content: Quote from: antenna on August 20, 2021, 06:17:25 amIt's not just a pot, we already know it's a PWM driver. It will still have a current limit, but it's unlikely to be spec'd so close to the limit that any reasonable LED upgrade will damage it. I mean we're talking a modest increase here of maybe a watt or so vs a few tens of milliwatts. I would also expect it to have some sort of protection, automotive electronics typically do. Easy enough to use an external driver and just tap off the PWM signal anyway if you want a LOT of light.