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28

BOM optimization is a likely candidate. The value 2k is probably used elsewhere and adding another value would mean one more line on the BOM and one more slot in the pick-and-place machine. Another BOM optimization could be availability. For example two 0603 resistors are easier to source than one 0805 (for power handling). You mention an eight resistor in ...


20

It is a bit hard to see from the picture, but since each LED has six pins, and knowing that it is meant for 12V supply, it is extremely likely that the LEDs might be arranged so that there are three parallel chains of three LEDs in series. The blue LED requires approximately slightly more than 3V voltage drop per LED, and since there is 3 in series, the ...


14

XY problem. Your best efficiency solution from a power and cost point of view to drive a 1A LED load is to use a LED driver IC. There are many to choose from; the Diodes Inc PAM2804 seems likely to meet your needs. It supports PWM dimming. These chips are very inexpensive (about 15 cents Digi-Key price, half that in volume.) You could drive up to 3 of your ...


11

If either side of the LED is shorted to either supply it will probably survive with two resistors. With only one, it probably won't. If the circuit has to be resistant to "knuckleheads" perhaps the 0.1 cent for the second resistor was deemed justified. Edit: The comment from mkeith is also very much valid. If you keep the number of different parts ...


11

Ordinary LEDs can also be used to detect incident light, because they work as photodiodes too. You will need to amplify the incoming signal. Take a look at this projects (Jeff Han is the first one that used LEDs as sensors): https://cs.nyu.edu/~jhan/ https://hackaday.com/2016/06/17/ask-hackaday-whatever-happened-to-led-light-sensors/ Or google "LEDs as ...


11

Blue LEDs require a higher voltage than red and green. If the panels are designed to work on 12 V, there may not be enough voltage to light the blue LEDs when run on 9 V. The blue LEDs must work for the panels to produce white and violet light.


9

I read about a method of limiting current using 2 transistors (schematic attached below). All linear solutions, resistor or whatever, will have the same efficiency. If you run the LED at 1A, it will draw 1A from 5V, so that's 5W. If there's 1.6V across the LED, then there will be 3.4V across the current regulator (whatever it is) so the LED will get 1.6W ...


8

Yes, certainly is possible. For further info, I suggest finding a copy of a MERL reseach report by Dietz et al from 2003 how they did it. It does not in fact produce enough voltage to detect it with the MCU directly. It usually is done in a way that enables the generated photocurrent to discharge the parasitic junction capacitance of the LED which is first ...


6

7000W of LEDs? Are you lighting a stadium? Well, since this project will make you blind, it wouldn't be complete without the risk of electrocution, so let's wire the LEDs in series. As you noticed, the voltage across your COBs varies with current (and also temperature, and between each COB). So you can't drive them with a constant voltage supply, it has to ...


5

Without knowing the schematic im going to just guess these are standard 5050 leds put in a 3 led series setup. For purple you need Red and Blue 100% on. For white you need Red Green and Blue 100% on. Since your source voltage is much lower than the typical 3.2 to 3.6V forward voltage for a blue led diode time 3 ( 9.6 to 10.8) , thats an issue. You need to ...


4

Justme's answer is very possible. Expanded here. I have used photodiodes directly connected to a GPIO pin, along with a small-value capacitor to detect light. A LED should work as detector too. The GPIO digital pins must be able to switch from input to output - most are versatile-enough to do this. It may help if the GPIO pins can generate an interrupt-on-...


4

R2 provides base current for pass transistor Q1. If it is too high then LED current will be less than VBE/R1 and vary depending on the HFE of Q1. HFE is quite temperature sensitive and depends on the transistor. If R2 is too low then the circuit wastes more power than it needs to. R2 is 2K on your simulated circuit which gives more than 10mA drive at 24 V. ...


4

You have too many questions in one post, but I'll try. You calculated the resistor dissipation as if it was the only load connected to 5V supply. The LED will drop about 1.5V so the resistor would not see full 5V. It would still dissipate a lot, about 2.45W. Regarding the transistor constant current circuit, it still has to dissipate the exact same 2.45W. So ...


3

You can halve your losses by series connecting two LEDs. This will give you a combined forward voltage drop of 2.8 to 3.2 V reducing your resistive losses from about 70% to about 40%. How to choose hfe? Your circuit won't be picky about hfe. You just need to ensure that R1 provides enough bias to turn on Q1 to the maximum current required. What values ...


3

Is there any reason why this circuit would need two 2k resistors instead of one 4k resistor? No. I thought maybe it would be for power dissipation but unless I am making a stupid mistake, its only dissipating 1 milliwatt which even small SMD resistors could handle. Subtract the voltage drop of two LEDs from 24V. To be conservative let's say 20V. 20V/4k\$\...


3

The MEAN WELL LSR-350-36 (there is no LSR-250-35) is not a good match for your project. LEDs need a regulated current to operate properly. The LSR-350-36 is a constant voltage power supply. You can approximate a constant current power supply from a constant voltage power supply, but that requires resistors in series with the LEDs and a voltage source ...


3

The resistors are split for better common-mode transient immunity (CMTI), i.e. for better separation of the two supply domains. The two sides are galvanically isolated, otherwise there would be no reason to use optodiodes (=isolators). If there is any noise (more like disturbances caused by motors, or high current fast circuitry) on one of the supplies, then ...


3

It is caused by your eyes, because when the duty cycle is small (for example 20%) and you have a higher frequency (few kHz or more), the LED is on only for small amount of time. Your eyes can't take such high speed changes, so you sees it as a smaller brightness. The LED is actually as bright as it would be in long term on state. You have got 5V and a ...


3

The apparent brightness of a PWM driven LED to human visual system is a very logarithmic curve. It means you can easily see a big difference between PWM signals of 1%, 2% and 3%, but will not see hardly any difference at all between PWM signals of 97%, 98% and 99%. The average brightness of the LED to a light meter will still be linear in respect to the PWM ...


2

You are mistaken, the pins do not have a rated maximum of 40mA. That is the rating of permanent damage happening, and very much outside of normal operating conditions. The rating for normal use conditions for a single pin is 20mA. Also do note that the pin voltage is not any more 0V or 5V when 20mA flows, it is guaranteed to be max 0.8V sinking 20mA and min ...


1

The optoisolators serve several purposes: protecting the MCU from ESD: You'd need to design appropriate ESD protection. freeing the potential of the switches, so that if they get shorted to some other potential, no damage is done: Recall that such devices are installed by technicians who are often under a lot of pressure to get it done quickly, and ...


1

Update 2021apr14hkt2008 The OP has not yet given a user requirement of the venue for the 100 sets of COB LEDs, each set of 70W. This is easy to do, as we can easily find 12~24VDC 70W LED lamps. The following is an example. AliExpress 70W DC 12V Dimmable COB LED Light with Controller Strip Board Outdoor Camping Lights DIY Fishing Rod - US$10 ~ 12 Notes If ...


1

I did this exact thing in a design for heat dissipation. But, my design didn't have a fixed voltage as input like yours, rather it had a wide input range.


1

These COBs (and most of them) are current-driven, which means you need a current driver, not a voltage source. It also means that you cannot connect them in parallel. You could if by adding some small resistor in series to each COB to compensate for the voltage differential, but it is not very safe and they may end up dying after some time. In your situation,...


1

The light output of an LED is dependent on the junction temperature, light output decreases as temperature increases. Even at low currents, the junction temperature will rise slightly. The time constant of your curves is consistent with thermal effects. https://www.lrc.rpi.edu/programs/NLPIP/lightingAnswers/LED/heat.asp Unfortunately, I can't find the ...


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