Driving this dual coloured COB LED?  - Page 1
Forum section: electronics / beginners

author: LooseJunkHater
author_rank: Frequent Contributor
timestamp: February 02, 2023, 07:34:50 pm
content: Hello, I want to use 2x potentiometers to control the brightness of the warm white (WW) & cool white (CW) LED’s (image 1;https://www.aliexpress.com/item/4000381180509.html), but I’m not exactly sure how to drive this LED. I know that I need both CC (constant current) and CV (constant voltage) control for LED's, but since the ground is separated for this LED design, it confuses me. I’m assuming that this style (image 2;https://www.aliexpress.com/item/1005002303002760.html) of boost converter WON’T work because the ground are commoned together??? Maybe I could get this boost converter to work if they are isolated from each other, but maybe I’m thinking about this the wrong way?I was thinking of just using a single boost converter and then using a mosfet to limit the current of both the WW and CW LED’s, but this design is obviously nowhere as efficient as simply using two separate boost controllers.Mosfet to limit current circuit:What is the intended way/best way of driving an LED like this? Thanks!

author: ledtester
author_rank: Super Contributor
timestamp: February 02, 2023, 08:15:46 pm
content: It appears that the COB LED looks like this internally:The "constant current source" presented in the video operates on the low side - technically it is a constant current "sink". To control both LEDs in the COB you can use two copies of the constant current control like this:Each CC control unit would have its own potentiometer.

author: LooseJunkHater
author_rank: Frequent Contributor
timestamp: February 03, 2023, 05:57:18 pm
content: Is there any other way to control the LED brightness aside from using the MOSFET for an inefficient constant current control?

author: Zero999
author_rank: Super Contributor
timestamp: February 03, 2023, 06:36:29 pm
content: What's the supply voltage vs LED forward voltage?The current needs to be controlled, not the voltage.The LEDs being common anode doesn't help matters.

author: LooseJunkHater
author_rank: Frequent Contributor
timestamp: February 03, 2023, 06:47:12 pm
content: Quote from: Zero999 on February 03, 2023, 06:36:29 pmVoltage: 36-39V, current:600mA+600mA. I understand that the LED needs current control.

author: Zero999
author_rank: Super Contributor
timestamp: February 03, 2023, 07:30:12 pm
content: Quote from: LooseJunkHater on February 03, 2023, 06:47:12 pmIs that the supply, or LED forward voltage?

author: LooseJunkHater
author_rank: Frequent Contributor
timestamp: February 04, 2023, 07:03:23 pm
content: Quote from: Zero999 on February 03, 2023, 07:30:12 pmNo idea, I'm listing what's on the AliExpress page that I had listed above.

author: LooseJunkHater
author_rank: Frequent Contributor
timestamp: February 04, 2023, 07:36:04 pm
content: So I thought this up last night before falling asleep, but would this work (as seen in the image)?Having the voltage boosted to around 37V (absolute max of XL4015 is 40v) and then using the constant current function of the XL4015 to limit the current flowing through each individual LED and using the constant current pot on the XL4015 module to control the brightness?In my head it seems like this would work but I'm not sure? What may be the problem of this design?Product link:https://www.aliexpress.com/item/1937996219.htmlEdit: This LM2596HVS module supports 50V and appears to support constant current as well, for those worried about the max voltage of the XL4015?https://www.aliexpress.com/item/1005005083519953.html

author: timenutgoblin
author_rank: Regular Contributor
timestamp: February 04, 2023, 09:27:51 pm
content: Quote from: LooseJunkHater on February 03, 2023, 06:47:12 pmQuote from: LooseJunkHater on February 04, 2023, 07:36:04 pmThe LM2596HVS has a minimum input voltage requirement of 4.5V which means that if you have your COB LED connected in series with a forward voltage of 36V then the boost converter will need to be outputting >40.5V to support the operation of the COB LED.You may need a boost converter with a higher output voltage capability.

author: timenutgoblin
author_rank: Regular Contributor
timestamp: February 05, 2023, 12:30:49 am
content: Since the LM2596HVS has a minimum output voltage of 1.2V, you will need to connect a load resistor to complete the circuit otherwise the COB LED won't work. It might flash briefly, but won't stay on. Since the COB LED has a forward current of 600mA, you will need a 2 ohm 1W-rated resistor connected at the output of the LM2596HVS buck converter. Either a pair of 1 ohm resistors in series or a 2.2 ohm and 22 ohm resistor in parallel should work.

author: LooseJunkHater
author_rank: Frequent Contributor
timestamp: February 05, 2023, 04:32:19 am
content: Woops, didn't even consider a minimum input voltage for the buck converter. However taking that into consideration, should I go the route of using a boost converter and then constant current buck converters (very complex and more to fail), or instead a mosfet (simpler but likely less efficient?) for constant current regulation? Maybe someone could help me with efficiencies math, since I'm not good with math.What sort of power dissipation should I expect if using a mosfet for use of constant current regulation of a 25W LED? No idea how I'd do the same calculation for the buck converter?

author: Zero999
author_rank: Super Contributor
timestamp: February 05, 2023, 09:44:56 am
content: Quote from: LooseJunkHater on February 04, 2023, 07:36:04 pmNo, connecting the LEDs between the output of a boost and input of a buck converter won't work. A buck converter doesn't work like that. A buck converter regulates the output, not input current. A current limited buck converter with a short circuited output will hardly draw any current from the input. If the output of the buck is tied to 0V like that, the input current will be very low.The LED needs to be connected to the output, not the input of the buck converter. You could put a MOSFET in the negative of each LED, each driven with the opposite signal, so either one LED is on, or the other. If you need to turn them both on, then PWM at 50%You still haven't stated your power supply voltage. It's pretty important to know that, before going any further.

author: mariush
author_rank: Super Contributor
timestamp: February 05, 2023, 10:34:47 am
content: Ok.... easiest would be to get a power supply that outputs more than the forward voltage of the leds, and then use a couple BUCK led drivers to control the current going through each led, using a potentiometer for each driver. Or you could use a dual potentiometer / stereo potentiometer (but these are logarithmic so the brightness won't change as linearly)The page says 36-39v forward voltage, up to 600 mA for each led ... but knowing the quality of leds on Aliexpress I'd say don't go over 450-500mA ... anyway.Example of step-down / buck led drivers :AL8862:https://www.digikey.com/en/products/detail/diodes-incorporated/AL8862QSP-13/10670150max 55v in, 1A 0.4v .. 2.5v on CTRL pin to adjust brightness (so a potentiometer and a resistor to form a voltage divider and you're good.ILD6150:https://www.digikey.com/en/products/detail/infineon-technologies/ILD6150XUMA1/4897530max 60v, 1.5A ... same deal, you can set max current with the Rsense resistor and then use a potentiometer to adjust brightnessDatasheets contain example circuits with values of components, example application circuits..Let's say you need at least 42v and 1.2A so a total of 50 watts.Easiest to obtain that would be by either with a 48v power supply (you can adjust the output voltage a bit to bring it down a bit for less losses if needed) or by making your own power supply using a classic transformer.For example find a toroidal transformer or a classic transformer with two 18-24v AC secondary windings (or a single 36-48v secondary winding), add a bridge rectifier and a big capacitor on the output and you have your 42v+ power supply. If you want to get the full 50w on the led, you'll need at least a 75VA or higher transformer, otherwise you could go with a 50 VA transformer (and expect maybe 400mA on each led, still quite bright)Here's some examples :50w toroidal for 18$https://www.aliexpress.com/item/1005004459527878.html100w toroidal for 28$https://www.aliexpress.com/item/1005004459735891.htmlA bridge rectifier converts AC voltage to DC voltage, and result is a DC voltage with a peak equal to Vdc peak = 1.414 x Vac - 2 x voltage drop on rectifier diodes.So for example, with 18v AC secondary windings for a total of 36v AC you would get Vdc peak = 1.414 x 36v - ~ 2v = 49v DC peak.Now you determine how big of a capacitor to use to keep the minimum voltage always above some threshold, let's say 42v in our case ...C = Current / ( 2 x AC frequency x (Vdc peak - Vdc min) ) = 1.2 A / 2 x 60 x (49 - 42) = 1.2 / 120x7 = 0.00142857 Farads or 1428.6 uF ... so you'd need at least a 1500uF 63v rated capacitor to keep the voltage above 42v all the time. I'd use 2200-3300uF.

author: Zero999
author_rank: Super Contributor
timestamp: February 05, 2023, 11:27:46 am
content: Quote from: mariush on February 05, 2023, 10:34:47 amThey almost certainly won't be designed for both LED strings to be operated at full power simultaneously.QuoteThe trouble is, the LEDs are common anode, which makes them incompatible with those drivers, because they both have the sense resistor in the positive line.

author: timenutgoblin
author_rank: Regular Contributor
timestamp: February 05, 2023, 11:50:56 am
content: Quote from: Zero999 on February 05, 2023, 09:44:56 amThis COB LED is a common anode design which requires a current sink to drive them independently, as opposed to a current source like you're describing. If the buck converter (CC module) is connected between the boost converter output and the common anode, it will not be possible to independently regulate the current in each LED string. The buck converter (CC module) needs to be connected between the cathode and ground to independently regulate the current in each LED string.Alternatively, it's possible to power the COB LED using a negative voltage or negative current source. If the common anode is connected to ground and a negative voltage is applied to the cathodes then that should work.Common anode to ground and -36V to each cathode at 500mA (as @mariush suggested).

author: Zero999
author_rank: Super Contributor
timestamp: February 05, 2023, 12:25:11 pm
content: Quote from: timenutgoblin on February 05, 2023, 11:50:56 amI wasn't suggesting controlling them independently. Use one buck converter and connect the LEDs to the output via MOSFETs and switch them alternately. The LEDs won't be able to take the full current simultaneously on both channels anyway.QuoteUsing two of the buck converter modules suggested by the OP won't work, because they're current sources and the LED is common anode.QuoteNow go and find a buck converter module with a negative output current.

author: Psi
author_rank: Super Contributor
timestamp: February 05, 2023, 12:26:44 pm
content: MEANWELL have some mains input CC led drivers that will do 36V leds.And some have 3 types of brightness control input, 0-10V, PWM or external potentiometer.Check out something like this one for $28https://www.digikey.com/en/products/detail/mean-well-usa-inc/XLG-50-AB/9858488

author: Zero999
author_rank: Super Contributor
timestamp: February 05, 2023, 02:21:54 pm
content: Quote from: Psi on February 05, 2023, 12:26:44 pmI've used those type of of Meanwell LED drivers before and they're good.It's definitely an option, if it's running off the mains. It has an isolated output, so the LEDs being common anode isn't a problem. I'd probably choose a lower current model though, as the LEDs will smoke if both are run at 530mA simultaneously.

author: LooseJunkHater
author_rank: Frequent Contributor
timestamp: February 05, 2023, 06:17:15 pm
content: Quote from: Zero999 on February 05, 2023, 09:44:56 amYou're right. I plan to use this module [https://www.aliexpress.com/item/1005002303002760.html] to create the main ~36-40V input. It can output up to 50v.Quote from: mariush on February 05, 2023, 10:34:47 amI have 0 intention of using this COB LED at full power for most of its life as I'm aware that LED's are less efficient when being driven harder plus their lifespan significantly decreases.Quote from: timenutgoblin on February 05, 2023, 11:50:56 amThis is a brilliant idea but unfortunately from my research creating above -15 volts with constant current control for cheap is quite difficult (and expensive).Quote from: Psi on February 05, 2023, 12:26:44 pmAnother great idea but sadly for me (I didn't mention this earlier; my apologizes) I plan to make it a portable light likely running off of high power lithium cells (hence the boost converter).___Based off of everything that I've read so far it seems like my idea of using a mosfet for constant current control is the simplest and cheapest idea as it seems like the other ideas listed so far either won't work (like my buck converter idea) or are more expensive than I'd like?Could I maybe get some help with calculating how much power I expect each mosfet to dissipate?

author: Zero999
author_rank: Super Contributor
timestamp: February 05, 2023, 07:13:16 pm
content: What's wrong withmy suggestion? The MOSFETs will hardly dissipate any power, above the on resistance, multiplied by the current. If both LEDs are required to be on simultaneously, PWM the input at 50% duty cycle, which will make each half as bright, but won't change the overall brightness, just the colour temperature. You'll now be able to set the brightness by changing the buck converter's current and the colour temperature by PWM.If you go for a boost converter+linear current sink, then the power dissipation in the transistors will simply equal the current, multiplied by the voltage drop. Set the voltage of the boost converter to a little over the voltage dropped by the sense resistor + LED + transistor. If it's just a one-off, then you could find out the actual LED forward voltage, at the desired current, rather than relying on the worst case, given on the data sheet. I wouldn't bother with a MOSFET, because it's prone to oscillation. A power transistor with a high enough hFE such as the BD131 will be able to sink enough current, when driven with the LM358.

author: LooseJunkHater
author_rank: Frequent Contributor
timestamp: February 05, 2023, 07:45:07 pm
content: Quote from: Zero999 on February 05, 2023, 07:13:16 pmTo be honest, I don't quite understand how your circuit works. What's controlling the "input" of the mosets? If it's more efficient and still relatively simple, I love it. I'm not an expert on EE (hence why I'm posting in the "beginner" section).

author: Zero999
author_rank: Super Contributor
timestamp: February 05, 2023, 08:22:08 pm
content: Quote from: LooseJunkHater on February 05, 2023, 07:45:07 pmThe input just controls which MOSFET turns on. A1 is a NOT gate, but it doesn't have to be an actual gate. It was just to show the opposite signal must be applied to M2's gate, to M1's gate. This means both LEDs can't be on simultaneously, but that doesn't matter. If you want both on, just flick between the two at a high enough frequency, they eye won't be able to tell, so they'll both appear on.Thinking about this again, it might be simpler to just do with the boost converter+linear current sink. That would save having to have another buck converter module and if the voltage from the boost converter is kept low enough, it'll achieve comparable efficiency.

author: LooseJunkHater
author_rank: Frequent Contributor
timestamp: February 05, 2023, 09:14:49 pm
content: Just for my own knowledge and growing, what is the input controlled by? A microcontroller? How would that provide constant current control (I guess it wouldn't?). I understand the part of frequency + duty cycle for brightness control.

author: Zero999
author_rank: Super Contributor
timestamp: February 05, 2023, 09:53:27 pm
content: Quote from: LooseJunkHater on February 05, 2023, 09:14:49 pmA microcontroller is one way, but the old 555 timer will also work. Another transistor could be used to make the NOT gate.QuoteThe constant current symbol represents the output of your current limited buck converter module, which presumably is controlled with a potentiometer.

author: Psi
author_rank: Super Contributor
timestamp: February 05, 2023, 10:41:43 pm
content: Quote from: LooseJunkHater on February 05, 2023, 06:17:15 pmMEANWELL won't let you off that easy, they have a step-up DC input option for that too.Input 12V (9-18V range)Output up to 43V LEDCC 1.05A (a little under your 1.2A LED spec, but that's good, the LED will have a good lifespan not being driven right on the edge of failure/running hot)Power 45WAnalog input for brightness control (I think on these ones you need couple resistors and regulated source to get the pot to produce a 0.25V to 1.3V signal. (unlike that previous one that can take pot directly on pins)91% efficient.Even cheaper, $21Check out this onehttps://www.digikey.co.nz/en/products/detail/mean-well-usa-inc/LDH-45A-1050W/7704776Datasheethttps://www.meanwellusa.com/upload/pdf/LDH-45(DA)/LDH-45-spec.pdfMEANWELL have a huge range of modules, i only looked for a minute so there might be ones that can do 1.2A if you spend some time looking.The LDH-65-1400 maybe, it does 1.4A, but you could hardware limit it to 1.2A avg by limiting the max scale of the brightness control voltage.$28https://www.digikey.co.nz/en/products/detail/mean-well-usa-inc/LDH-65-1050W/12759739

author: Zero999
author_rank: Super Contributor
timestamp: February 06, 2023, 08:34:28 am
content: Quote from: Psi on February 05, 2023, 10:41:43 pmThe original poster's LED is common anode.There's no mention on that datasheet of isolation, or whether the + outputs of two modules can be commoned. I strongly recommend talking to the manufacture before ordering.

author: Psi
author_rank: Super Contributor
timestamp: February 06, 2023, 09:08:48 am
content: ah true, i missed that.

author: Zero999
author_rank: Super Contributor
timestamp: February 06, 2023, 07:28:14 pm
content: Here's an idea. Pot1 controls the brightness of both the warm and cold white. Pot2 alters the colour temperature, but changing the currents through D1 and D2. It's not perfect, because if a Pot2's wiper goes open, it'll blow up the LEDs. U1 and U2 can be the cheap old LM358.

author: spostma
author_rank: Regular Contributor
timestamp: February 07, 2023, 10:11:13 pm
content: Zero999's solutionhttps://www.eevblog.com/forum/beginners/driving-this-dual-coloured-cob-led/msg4682669/#msg4682669could be simplified a little by omitting the MOSFET on the higher-voltage LED.I sometimes do something similar to make a red/green LED toggle by having a single pull-up resistor on both LEDs,the Cathode of the Green (2V4) LED fixed to ground, and the Cathode of the Red (1V8) LED via a BSS123 controlled MOSFET to ground.

author: LooseJunkHater
author_rank: Frequent Contributor
timestamp: February 07, 2023, 10:13:28 pm
content: Quote from: Zero999 on February 06, 2023, 07:28:14 pmOhhhh I really like this design! However question, why a NPN transistor (BD131) and not instead a MOSFET? Would a MOSFET not be more efficient and easier to drive?

author: Zero999
author_rank: Super Contributor
timestamp: February 07, 2023, 10:52:07 pm
content: Quote from: LooseJunkHater on February 07, 2023, 10:13:28 pmWhat makes you think a MOSFET would be more efficient?QuoteYes and no. A MOSFET has a virtually infinite current gain, so don't take any steady state current from the op-amp's output. The disadvantage is it has a high gate capacitance, which makes oscillation very likely. I chose the BD130 because it should have just enough gain to give a collector current of 600mA, when driven by the LM358.

author: LooseJunkHater
author_rank: Frequent Contributor
timestamp: February 07, 2023, 11:13:10 pm
content: Quote from: Zero999 on February 07, 2023, 10:52:07 pmIsn't this really simple to overcome with a 10k resistor?Quote from: Zero999 on February 07, 2023, 10:52:07 pmActually I think you're right. Since both the transistor and fet would both be doing constant current by creating wasted heat, they'd both be equally inefficient, right? Or would the NPN be even less efficient because of lost current between base/emitter?

author: Zero999
author_rank: Super Contributor
timestamp: February 07, 2023, 11:34:20 pm
content: Quote from: LooseJunkHater on February 07, 2023, 11:13:10 pmThat would make it worse because it would increase the RC time constant.QuoteYes the MOSFET would be a tiny bit more efficient, as it doesn't require any base current, but the difference is small enough not to matter.

author: langwadt
author_rank: Super Contributor
timestamp: February 08, 2023, 12:24:56 am
content: Quote from: Zero999 on February 03, 2023, 06:36:29 pmwould work with an "upside-down" buck, e.g.https://www.mouser.dk/new/texas-instruments/ti-tps92511-driver/

author: LooseJunkHater
author_rank: Frequent Contributor
timestamp: February 08, 2023, 01:28:56 am
content: Quote from: langwadt on February 08, 2023, 12:24:56 amYou know what's funny, almost as soon as you posted this comment, I came across a similar I.C called the MP4700.https://www.monolithicpower.com/en/mp4700.htmlI think these I.C's appear to more commonly be referred to as "low side buck converters". Definitely more complex than driving a transistor/fet in a linear manor, however, potentially a lot more efficient since they'd automagically PWM + duty cycle to limit current and not be visibly noticeable? I think I'm going try to find more alternatives (preferably some cheaper IC's; the MP4700 is about $1 and the tps92511 is about $2) but I'll explore this route as well. Further suggestions, tips, and ideas welcomed!

author: Zero999
author_rank: Super Contributor
timestamp: February 08, 2023, 01:48:37 pm
content: The TPS92511 is limited to 500mA, but that's probably enough, as it's never a good idea to drive LEDs at the maximum current.The MP4700 will need a lower voltage supply, just like my LM358 op-amp suggestion, but that's not a problem, as you can just use whatever is powering the boost converter.A buck converter will be theoretically more efficient, than a linear regulator, but bear in mind, if the output voltage of the boost converter is only a few volts above the forward voltage of the LEDs, it might not gain you much. Suppose VF= 37.5V and VIN= 40V then the efficiency of a linear regulator is 37.5/40 = nearly 94%.

author: LooseJunkHater
author_rank: Frequent Contributor
timestamp: February 08, 2023, 04:17:41 pm
content: Quote from: Zero999 on February 08, 2023, 01:48:37 pmYou have again solidified my decision into using a transistor/fet for constant current. The efficiency gains are not worth the extra hassle for my use case. Thank you! :p

author: Zero999
author_rank: Super Contributor
timestamp: February 08, 2023, 06:31:36 pm
content: Ideally you need a single buck-boost converter, with a negative output voltage and current, that way you can dispense with the boost converter and linear regulator.I had a quick search and found the MAX25610. It's not a beginner friendly IC as it's surface mount and has a thermal pad, which will require hot air and a proper PCB. An evaluation kit is available, but it's not cheap.https://www.analog.com/media/en/technical-documentation/data-sheets/MAX25610A-MAX25610B.pdfhttps://www.mouser.co.uk/ProductDetail/Maxim-Integrated/MAX25610EVKIT?qs=T3oQrply3y%252BPYTt501wHzw%3D%3D

author: LooseJunkHater
author_rank: Frequent Contributor
timestamp: March 02, 2023, 06:54:43 pm
content: Got around to building the circuit on a breadboard, except with a N-channel mosfet (currently waiting for some N-BJT's to come in the mail) but it doesn't work. It simply turns on-or-off the mosfet, depending on how much I turn the potentiometer (circuit attached).When I measure the gate of the mosfet, I can see the voltage varies between 3.680-3.660v, depending on how much I turn the potentiometer.What's wrong with the circuit? I tried to remove the 10k resistor and remove ground from the pot but neither worked. Help?

author: Zero999
author_rank: Super Contributor
timestamp: March 02, 2023, 07:33:59 pm
content: The circuit you've posted is doing what is expected. The op-amp is simply acting as a buffer between the potentiometer and MOSFET. There's no current sense resistor and no feedback from the output current.Refer to the schematic I posted above. The potentiometer sets the reference voltage, which the op-amp compares to the voltage across a sense resistor.https://www.eevblog.com/forum/beginners/driving-this-dual-coloured-cob-led/msg4685614/#msg4685614I wouldn't recommend using a MOSFET. It's too prone to oscillation. Use a power BJT.