Led driver and component requirements (Remote phosphor light) - Page 1
Forum section: electronics / projects

author: steveggz
author_rank: Contributor
timestamp: October 08, 2017, 01:45:04 pm
content: I am building a remote phosphor light for videography. Although I understand the fundamentals I am a novice when it comes to working with LEDs. I would like to know what else I will need and make sure my setup is correct and that there isn't anything I am missing.The light requires 120 Cree XT-E royal blue LEDs driven by 3 Meanwell HLG-240H-C1750B drivers. I also need to use one potentiometer to control the three drivers. Each driver will power 40 Cree LEDs is series.The Cree LEDs operate at a max current of 1500mA and a forward voltage of 3.1V max. The constant current Meanwell drivers deliver 1750mA and 71-143v.I understand the driver's current will be regulated by a resistor and potentiometer but how do I regulate the voltage? 40 LEDs X 3.1v = 124v. The driver delivers 143v, am I incorrect in assuming I need to regulate the voltage output to 124v?Also the max current of the LEDs are 1500mA. The driver outputs 1750mA. I will need a potentiometer to dim the lights but I have to add a additional resistor to regulate the current to 1500mA max correct? What combination size and spec resistor and potentiometer do I need?

author: Siwastaja
author_rank: Super Contributor
timestamp: October 08, 2017, 03:17:24 pm
content: You only regulate the current, and make sure your maximum voltage is more than the led voltage (calculate with the "maximum" Vf in the datasheet) + some leeway, so that you never run out of voltage.Note that if your LED string pops open circuit for any reason, the driver gives its maximum output voltage, so you might not want to go to extremes. 143V DC (just like 124V) is beyond safety voltage specs in all legislations AFAIK so you need to design properly.This is analogous to the more well-known constant voltage supply, where you need the right voltage for the load, and must haveat leastthe minimum current, but you can have bigger current rating, it is just not used. Just swap "current" and "voltage" and you have your led supply.Of course, using a LED driver with excessive voltage spec (for example, using that 143V supply to power one single 3.1V LED) would be stupid since you are paying for a big supply and using only a tiny part of the power, but it wouldworkjust fine. Similar to using a 12V 10A supply to power a 12V 50mA tiny load.

author: steveggz
author_rank: Contributor
timestamp: October 08, 2017, 07:34:40 pm
content: Thank you for clarifying that, I wasn't sure if diodes worked opposite of resistance loads in the way that they use whatever voltage they need but require the current to be regulated.In trying to figure out the correct current limiting resistor. Do I go off the LED's max forward voltage? The LED's have a forward voltage of 2.85v - 3.1v.Thanks

author: Siwastaja
author_rank: Super Contributor
timestamp: October 08, 2017, 08:14:06 pm
content: The Meanwell does the current limiting. You don't add any resistors to the LED chain, they would just waste power doing nothing. LED Vf is irrelevant, too, and doesn't affect the current (as long as you have enough max voltage available).The Meanwell datasheet shows how to configure the current: For the 1750mA driver, if you want to drive the LEDs at 1500 mA, it's 86% of 1750mA; this is configured using a 86kOhm resistor between the DIM pins.But please note that typically, getting both max current and good lifetime/reliability from LEDs requires extraordinary care to cooling, so you may not want to run at 1500mA.

author: steveggz
author_rank: Contributor
timestamp: October 08, 2017, 09:46:44 pm
content: Thanks!Just for my understanding, I'd love to learn as I am building this, how did you know to use a 86k Ohm resistor, because it's 86% of the current?I was using guides on the internet (using Ohm's law) and came to a 12.66 Ohm (or 13 Ohm) resistor rated at 30 watts. To find that I used 19 volts of unused voltage divided by 1.5 amps to get 12.66 (or 13 Ohms). And then multiplied 19volts by 1.5A to get a resistor rated for 28.5 (or 30) watts. Can you please help me understand why that isn't correct?Thank you!!

author: jbb
author_rank: Super Contributor
timestamp: October 08, 2017, 10:55:01 pm
content: Safety warning: the raw blue light output might be an eye safety hazard[\b]. Intense blue light can - in extreme cases - damage your retina (even if it isn't laser light). You'll need to do some calculations to check (or maybe find an experienced person to help).The MeanWell unit is set up to control the current internally (internal control chips adjust PWM duty cycle for you, I guess). The 86k resistor is for configuration so that the Mean Well knows what to do. (Note: I haven't read the data sheet.)

author: Siwastaja
author_rank: Super Contributor
timestamp: October 09, 2017, 05:47:11 am
content: Quote from: steveggz on October 08, 2017, 09:46:44 pmDirectly from the Meanwell datasheet. There were other options, too, such as using a 0-10V control signal. You should definitely read it... (RTFM)QuoteBecause this is not a voltage source at all -- it's a proper LED driver, already internally doing what you are wanting to DIY here, but in a modern, efficient, and more stable manner. Basically, it will guarantee that it only gives out the current you tell it, never more. (It can givelesscurrent if it runs out of voltage - that's why it's good to have some excess in the rating, as you do!)If you had a voltage source and needed to use a power resistor to dissipate excess voltage, then those calculations would be correct. Now you don't need one - 30 watts ridiculous power wasting saved!OTOH, you won't be learning much about electronics, since you are just using a commercial product that does everything you want internally, as is. You only need toconfigurethe current as instructed in the manual. (That 86k resistor.)But building the LED chain will teach you about thermal design. Be sure to google about that. It's not trivial. These LEDs will require significant effort for cooling design at 1500 mA. Consider driving them at 1000 mA. Even then the cooling must be right.

author: steveggz
author_rank: Contributor
timestamp: October 10, 2017, 03:47:26 am
content: Ah yes, that schematic which looked like hieroglyphics now makes sense after you explained it. Thank you!!!What power rating or watt 86Ohm resistor would I need?And since I'll be using one potentiometer to control all three drivers what resistance, power rating, and type potentiometer will I need?Thank you.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: October 10, 2017, 08:47:10 am
content: Quote from: Siwastaja on October 09, 2017, 05:47:11 am"Never more" requires slight qualification - you need to make sure you don't connect the LEDs to the PSU while it is switched on, as when un-loaded, it will be outputting its maximum voltage, so connecting the LEDs to a running PSU will produce a momentary high-current pulse, whose length and available power will depend on the amount of internal capacitance in the PSU.And at 120-odd volts, you will almost certainly get some sparking which would produce inductive transients from the cable inductance.

author: Zero999
author_rank: Super Contributor
timestamp: October 10, 2017, 08:57:47 am
content: Quote from: steveggz on October 10, 2017, 03:47:26 amI thought it was a 86k resistor? If so, the power rating won't be an issue. The voltage across the resistor will have to be very high for that to be a factor. Try running some figures through Ohm's law.QuoteDoes it not give the value of the potentiometer on the data sheet?If it's voltage controlled input (0 to 10V for example) the value of the potentiometer will not be important. It will need to be low enough to drive the voltage input, but not too high to load down the logic voltage output.

author: mikerj
author_rank: Super Contributor
timestamp: October 10, 2017, 11:41:07 am
content: Quote from: steveggz on October 10, 2017, 03:47:26 amYou must carefully read and understand the datasheet for the driver before jumping in, otherwise you are likely to damage many parts.The required resistance value is described by a chart at the bottom of page 4, and note that the resistance value includes an 'N' term which is the number of drivers you wish to synchronise the dimming control with. 86kOhms will give you 1500mA with one driver operating, for three drivers the resistance value will be a third of this.

author: Zero999
author_rank: Super Contributor
timestamp: October 10, 2017, 12:17:52 pm
content: I've just taken a quick glance at the data sheet.The dim function consists of a constant 100µA current source connected to the dim+, followed by a low pass filter. Varying the average voltage across dim+ and dim- from 1V to 10V, controls the output current from 10% to 100% of the driver's rating. If a voltage source is connected between dim+ and dim-, 100µA will flow into the voltage source. Due to the low pass filter, it's the average voltage which counts, hence a 0 to 10V 50% duty cycle square wave, will give an average voltage of 5V and half the maximum current. If a resistor is connected between dim+ and dim-, the voltage can be calculated using Ohm's law, for example with a 50k resistor: V = IR = 50×103×100×10-6= 5V.You need to control three modules. If the dim+ and dim- terminals are all connected up in parallel, then the currents add together, thus the resistor values need to be divided by three. If you want to go from 10% to 86% brightness, then you need a potentiometer with a resistance which varies from 31/3k to 282/3k, which you won't be able to get. The best you'll get is a 25k potentiometer, in series with a 3k3 resistor. The problem with potentiometers is the tolerance is poor, which might mean you get more than 8.6V ant 86% of the rate current.

author: steveggz
author_rank: Contributor
timestamp: October 10, 2017, 09:31:40 pm
content: Thank you all for your input and help.Quote from: mikerj on October 10, 2017, 11:41:07 amThank you for pointing that out. With three drivers I would need a 28.6 Ohm resistor. Please excuse my ignorance, would that require three 28.6 Ohm resistors? One for each driver connected between the DIM terminals?Quote from: Hero999 on October 10, 2017, 12:17:52 pmIf my above question is correct then would having them connected in parallel look like the attached picture?Out of the 3 dimming options I want to dim using a resistance potentiometer instead of 1-10 volt.

author: Corporate666
author_rank: Unknown
timestamp: October 10, 2017, 10:49:06 pm
content: How are the LED's being mounted and connected?At 1.5A, these LED's have a forward voltage of 3.125, and with 40 of them, you're putting almost 190 watts into your LED engine. Royal blue LED's are pretty efficient, but at full power, you're still going to have to deal with about 90 to 100 watts of waste heat. That's a massive amount of heat to dissipate. The datasheet suggests you can run them up to 40C ambient provided the thermal resistance from the LED junction to ambient is 25C/W. The LED has 5C from junction to pad, so you're talking some pretty massive heatsinks with fans to get rid of all the excess heat from this.If you just bolt the LED's to a piece of metal, they are going to die very quickly, unless you're pulsing them for a very short duration, like for a flash photography light.

author: steveggz
author_rank: Contributor
timestamp: October 11, 2017, 08:12:42 am
content: They're on MCPCB's and going to be attached to a aluminum heatsink with thermal paste.I've seen people use this method for their reef lighting. I'm assuming they keep those running constantly although maybe not at full power.I was going to use finned aluminum heat sinks with fans. Is there something I'm doing wrong? Please let me know a more effective method if so?Thanks

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: October 11, 2017, 09:26:20 am
content: Quote from: steveggz on October 11, 2017, 08:12:42 amDepends on how densely packed they are. You may want to look at heat pipes if you need to get a lot of heat out of a small area. but a fan-cooled heatsink is a good start - you'll need to do measurements to see if it's adequate.

author: Zero999
author_rank: Super Contributor
timestamp: October 11, 2017, 06:01:25 pm
content: Quote from: steveggz on October 10, 2017, 09:31:40 pmYou mean 28.7kOhms? Please be more careful with the decimal place.QuoteThat's fine.Note that these modules are voltage controlled. The brightness control has a built-in current source, which converts the resistance to a variable voltage.

author: steveggz
author_rank: Contributor
timestamp: October 13, 2017, 01:48:34 am
content: Please bear with me, I'm sorry It's taking so long for me to get this but once it does everything else will click.So I'm using 3 Meanwell drivers, each driver will power it's own set of LEDs (40 LEDs in series each) but be controlled by one potentiometer.Going off Meanwell's datasheet that would mean I would need a to divide the resistance by 3 which comes to 33.3. So a 33.3K potentiometer if I connected the DIM leads in parallel to the potentiometer. (So far could someone please correct the above if wrong?)But the drivers output 1750mA and I only need 1490mA (LEDs are rated 1500mA max). That's a reduction of about 14.9% of the current so I would need a 14.9K resistor in addition to the potentiometer and since 3 drivers 14.9/3= 4.97kohm. One 33.3Kohm potentiometer and three 4.97Kohm resistors between each DIM lead like in the attached picture? Or 3 4.97Kohm resistors like in the picture and a 18.4Kohm potentiometer?

author: Zero999
author_rank: Super Contributor
timestamp: October 13, 2017, 08:16:16 am
content: Your schematic shows three 4k97 resistors, in parallel with a 33k3 potentiometer. Look up parallel resistors, using a search engine. It will give a much lower value than you need.Also note that resistors only come in standard values, typically the E24 and E96 values and that it can be difficult/expensive to get non-standard values. Potentiometers also have a poor tolerance, often as bad as 50%, so they tend to only be sold in E3 values (multiples of 1, 2.2 and 4.7) or rounded E3 values (multiples of 1, 2 or 2.5 and 5).https://en.wikipedia.org/wiki/E-series_of_preferred_numbers

author: steveggz
author_rank: Contributor
timestamp: October 14, 2017, 03:35:03 am
content: Referencing the HLG-240H-C1750B, would adding a resistor between DIM + and DIM - increase the voltage of the driver?It's rated to output 143v at 1750mA. But if I added a resistor between the DIM terminals decreasing the current to let's say 1500mA would I get more voltage? If so how much?

author: steveggz
author_rank: Contributor
timestamp: October 14, 2017, 02:23:32 pm
content: I decided to switch from linear dimming to PWM using a power supply and Mean Well LDD-1500H buck drivers. The LDD-1500H out put 1500mA but need to reduce the current to 1480mA. Would adding a current limiting resistor in series with the LED's mess with the PWM slew rate or pulse frequency? I need the LED's to pulse in the kilohertz frequencies without any disturbance and don't want the fast slew rates of the buck to be negatively effected.

author: Zero999
author_rank: Super Contributor
timestamp: October 17, 2017, 01:04:50 pm
content: Quote from: steveggz on October 14, 2017, 02:23:32 pmHow could a series resistor possibly help to reduce the current through the LEDs, driven by a constant current supply? It won't. All that'll happen, is the power supply will increase its voltage, to compensate.I doubt there will be any problems with operating at 1500mA vs 1480mA. The output current of the driver probably has a looser tolerance than that.Connecting a resistor in parallel with the LEDs could be used to reduce the current through the LEDs, by diverting it around them, at the expense of dissipating extra power.

author: steveggz
author_rank: Contributor
timestamp: October 21, 2017, 09:31:03 pm
content: Thanks Hero999! I appreciate your help.