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2

I will assume that you have thought through the safety of your product, and won't be electrocuting your users. Bear in mind that if you rectify 110V AC, you will get a peak voltage of around 155V. the LEDs will not conduct when the rectified AC is below about 100V, and so will only be on for part of each cycle. Once they do start to conduct, your current ...


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Your question is more of a red herring; it's most valuable to know how to effectively drive your LED power supplies' PWM dimming input from your microcontroller. You state that the maximum input on the PWM pin is 8 V, so I assume you're using the 300-700 mA model that has a high logic level of 3.5 - 8 V. It also looks like you're just using a 5 V regulator ...


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I search for this kind of implimantation on the internet but was not able to find anything like this, and it is a fairly simple approach but why are manufacturers avoiding this kind of approach for high count LEDs in Series. Is there something I am missing? Well, yes. First, simply rectifying a 110 VAC and using it is wildly dangerous. It's the ...


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Some Initial Thoughts If your AC mains isn't completely floating (and my experiences are admittedly limited regarding AC mains supplies around the world), then you should consider isolating it with an isolation transformer. And if you are willing to consider it, then you are free to also select whatever secondary voltage that's convenient to both you and ...


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Design Specs As your TV and computer monitor only have 8 bit intensity control per channel, I think your 16 bit control spec is a overly optimistic for optics. I suggest 10 bit or 0.1% is adequate ~ 60 dB range and 12 bit is presuming too much. 1% is often adequate for lighting control and 1% of 1kHz PWM max = 10 us although they do not indicate the ...


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Don't (ab)use an led driver for this purpose. Even though it'll probably work as push-pull-multiplexer (with little disadvantages though) there are dedicated ICs that do exactly what you want. Search for "24 channel SPI PWM controller" and you'll find a device I would recommend.


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You need a power supply that's high enough to run 3 LEDs in series, with a bit of headroom, or in other words, 3 times the voltage of one LED. Different color LEDs run at different voltages, so your supply might be 9V or 12V. Or use separate power supplies for the 1 and 2 LED strings, and choose lower value resistors to suit. Yes, you need resistors, and ...


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A schematic would definitely help. In general, LEDs don't regulate the current themselves, so you'll need a way to bias them. Most simply this is done with a series resistor. For instance in your group A with 3 LEDs and 3V forward drop across them, would give 9V across the LEDs as you mention. You'll need the regulated voltage to be greater than that, ...


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This seems solved by replacing the cheap temporary chinese 24vdc power supply. I now have a meanwell sdr-120-24 powering the LDH module and all led's are completely off!


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With your signal you see the high levels of DC which is the sum of all reflections during the red strip including stray light not focused on the edge. As a result in the short term , the SNR appears to be 10:1 with signal = 82~92 and noise =1. The hysteresis must be greater than the noise yet much less than the drift of the signal of long term. LEDs are ...


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It won’t quite work the way you intend because the op-amp output can only reach about 3.5V and the MOSFET will need some Vgs to fully turn on. If you divide the DAC output down to 0-1V (eg 47K+20K) and replace the 47 ohm resistor with 14 ohms you can then use a MOSFET with 2.5V drive. You can add 33 ohms in series with the LED to make the MOSFET run cooler. ...


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You could throw in a zener to take up the additional voltage, but it should be as low leakage as you can get. You'd be surprised how little current it takes to visibly light some LEDs.


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The meaning of that statement is pretty clear, the minimum output voltage is the input voltage. So to get less minimum brightness you can either use more LEDs in series or lower the input voltage (or both). Too few LEDs and they might be damaged. The topology probably looks like this (from Wikipedia commons): For minimum output, the switch is always ...


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