wiring new PSU to a flickering LED lamp - Page 1
Forum section: electronics / beginners

author: smile
author_rank: Regular Contributor
timestamp: October 17, 2018, 06:59:59 pm
content: Hello, I have this LED lamp module, image attached.Module parametersvoltage 175.4Vtotal current 26mAmanufacturer specified wattage 5.5Wled amount 16pcsI have bought this PSU for LED that had similar wattage.Product Name: 4-7W built-in bare board LED constant current drive powerConnection:RED / Led + LED: Positive OutputWhite / Led-LED: Negative OutputPIN: Power InputPIN: Power InputInput voltage: 85-265VOutput voltage: 12-25VOutput current: 290mA ± 5%Product size: 34 * 17 * 16MMNote: This driver uses a professional chip, the output voltage will be based on the lamp beads automatically adjust the voltage, the circuit with over-current protection, short circuit protection, open circuit protection.The question is how to properly wire it?

author: JS
author_rank: Frequent Contributor
timestamp: October 17, 2018, 07:42:48 pm
content: If your parameters are true those two pieces arent compatible, you need a PSU which delivers the current you want up to the voltage you need or higher.JSJS

author: Twoflower
author_rank: Frequent Contributor
timestamp: October 17, 2018, 07:51:41 pm
content: That will be hard to implement. Even re-wiring the LED module will probably not work with that supply.The LED emitter has 16 LEDs and need a current of 26mA all in series. But the voltage looks a bit odd, 175V/16 = 11V per LED. It seems that each transmitter has 3 LED chips in series in one package.Nevertheless your supply is a constant current source. It tries to drive 290mA. Because this higher current the output voltage is much lower (12...25V). That is far too high for one LED (26mA), so you need to put some in parallel. So 290mA/26mA = 11,1. So you need to connect 11 or 12 LEDs in parallel. But with the calculation above the minimum voltage of 12V of the supply one LED (assuming the 11V is correct) that's not going to work. You need some additional resistor. And you can't get all 16 LEDs active that way.A simpler solution is probably either using a matching supply or a different module.

author: jose347
author_rank: Contributor
timestamp: October 17, 2018, 07:52:51 pm
content: From where do you plan wiring this lamp? mains? where did you get it from? what I understand from what you have is that you got a DC load(led module), and a AC to DC (PSU module), and if this was the case, you may have problems because from the specs you provide the output voltage will not be enough to power the leds. It will go like this say you are in america your mains will output a ~120Vac, this will go to the input of your PSU polarity at this point can be ignored if you happen to have a third wire then this will be your chassis ground(place on the metal base of your lamp) then from your PSU output (12-25V DC) you will connect the positive with the positive of your leds and negative with negative. problem here is that those 12-25Vdc will need to be 175Vdc to power the leds. but then again that really depends from where you plan powering the lamp and from where do you got your led module.

author: smile
author_rank: Regular Contributor
timestamp: October 17, 2018, 08:19:10 pm
content: QuoteThis is Mains AC 230V LED lamp. It is new high CRI lamp, and it was quite cheap not like some online for 10 to 15Eur.The problem is that PSU used in this lamp flicker very much, that is why I want to use new PSU.QuoteWhat is strange is that :26mA is used by 16LEDS when voltage is 175V ACThen 175 / 16 = 11V (looks like 3Leds in single led chip)But then it gets even more stange is that 26mA / 16 = 1.625mASo a single LED chip needs 11V and 1.6mA of current when my PSU isOutput voltage: 12-25VOutput current: 290mA ± 5%But then I never seen LED lamp PSU with such low current or such low current LEDs for that matter.290mA by PSU / 1.625mA = 178LEDS (Free energy anyone)

author: smile
author_rank: Regular Contributor
timestamp: October 17, 2018, 08:34:57 pm
content: Connected to my LAB PSU (set for 12V and 3mA ) I get this for a single LED chip10.38V and 1mA current

author: mariush
author_rank: Super Contributor
timestamp: October 17, 2018, 08:36:12 pm
content: You have 16 leds in series, so at the minimum your power supply must be able to reach 16 x forward voltage of a led - assuming there's a single led element inside each LED chip, then your power supply would need to produce ~ 16 x 3v = 48vHowever, if what you say is true and the module says it needs 175v DC to function, then most likely each individual LED chip has 3 small LED dies under the phosphorus (the orange stuff) wired in series, because that would make sense : 175v / 16 leds / 3 leds per led chip = ~ 3.64vThe power supply you have is only capable to adjust the output voltage between 12 and 25v and has a fixed current output of 290mA. It won't do.You need a power supply that will allow you to set the output current to 25mA for example (close enough to 26mA and safer, your lamp will still be bright but leds may last longer) and then gradually raise the output voltage all the way up to 175v or whatever voltage gets the leds reach that 25mA current value.

author: smile
author_rank: Regular Contributor
timestamp: October 17, 2018, 09:02:53 pm
content: QuoteEasier said then done, all a I can find are typical CC PSU like 300mA etc.

author: mariush
author_rank: Super Contributor
timestamp: October 17, 2018, 10:26:24 pm
content: Well, you could make your own "power supply", though it would be a bit inefficient.For example, something extremely simple.1. bridge rectifier (4 diodes) ... converts your 230v AC into DC voltage with a peak voltage of sqrt(2) x 230v = ~ 325v2. bulk capacitor ... smooths out the dc voltage and makes sure the minimum voltage will always be above some threshold (value will get back to it)3. voltage regulator Microchip LR8K4 :https://uk.farnell.com/microchip/lr8k4-g/linear-volt-reg-0-02a-440v-to/dp/2448524This regulator is designed for maximum 20mA and datasheet says maximum 30mA output and it ca work with up to 450v input voltage and output voltage can be up to input voltage - 12vIt can also be configured as constant current (see figure 3-4 in datasheet)So bulk capacitor... assuming we go for 25mA (even though we'll limit current through leds to 20mA because that's what this linear regulator claims it can do) we can aproximate capacitace with this formulaC = Current / [2 x mains frequency x (Vdc peak - Vdc min desired)]In US mains frequency is 60hz, our current is 0.025A (25mA) and the minimum voltage desired is approximately 175v + 12v or around 190v DC so.... C = 0.025 / 120 x (325v - 190) = 0.025 / 120x135 = 0.025/16200 or 1.543e-6 Farads or 1.54 uF ... just round it up to a standard value like 1.5uF , 1.8uF, even 2.2uF and a voltage rating of 400v or higher.This will basically guarantee your DC voltage will be at least around 200v and potentially up to 325v at very low currents.and as i wrote in the image... WARNING .... as with any device wired directly to mains, risk of electric shock, you should have an isolation transformer in front of this if you're ever going to handle this light with your hands or you risk getting zapped etc etc

author: smile
author_rank: Regular Contributor
timestamp: October 18, 2018, 09:55:15 am
content: Thank you for your circuitI simulated my 300mA 12V minimum PSU and LED lamp online and got this result:

author: mariush
author_rank: Super Contributor
timestamp: October 18, 2018, 12:23:19 pm
content: What's the point of your circuit?It's a badly designed one ...You basically have 2 strings of leds and in each string, the leds are connected in parallel. this means each string will only require the power supply to provide at least the forward voltage of an individual led. And, because the leds are connected in parallel, the current adds up, so each string because like a single led that needs 8 leds x 20mA = 160mASo, both strings in total would require 320mA.Now, if the leds are arranged this way, you would need to limit the current going into each string with a resistor.You have the formula V = I x R soVin - (forward voltage led x number of leds ) = Current x Resistance where number of leds = 1 because you connected all 8 leds in parallel and current is 8 x 20mA again because parallelso 12v - 10v = 0.160 x R => R = 2v / 0.160 = 12.5 ohmAnd the power dissipated in the resistor would be P = IxIxR = 0.16 x 0.16x12.5 = 0.32w so you'd need a 0.5w rated resistor at minimum, ideally a 1w resistor for EACH string of leds.In your picture both strings receive a limited amount of current ... 12v - 10v = I x 100 ... so I = 2 / 100 = 0.02A or 20mA , so each led would only consume 20 mA / 8 = 2.5mAYou can see here the proper way you should do it :http://tinyurl.com/ybjmzqut- it's an online circuit simulator, you can right click on components to edit the values, and you can hover mouse over components to see voltage, current, power dissipated etcIf you don't want to click on shortened links or in case link dies, go tohttp://www.falstad.com/circuit/and use Import from text option in menu and paste this:The exact current values and power in components in simulated circuit is just a tiny bit different than the math above, it's just the simulated leds consuming slightly different current (~22mA instead of 20mA each)Code:$ 1 0.000005 10.20027730826997 50 5 43162 256 96 256 160 1 10 1 1 1 0.16162 288 96 288 160 1 10 1 1 1 0.16162 320 96 320 160 1 10 1 1 1 0.16162 352 96 352 160 1 10 1 1 1 0.16162 384 96 384 160 1 10 1 1 1 0.16162 448 192 448 256 1 10 1 1 1 0.16162 416 96 416 160 1 10 1 1 1 0.16162 448 96 448 160 1 10 1 1 1 0.16162 480 96 480 160 1 10 1 1 1 0.16162 480 192 480 256 1 10 1 1 1 0.16162 256 192 256 256 1 10 1 1 1 0.16162 384 192 384 256 1 10 1 1 1 0.16162 288 192 288 256 1 10 1 1 1 0.16162 320 192 320 256 1 10 1 1 1 0.16162 352 192 352 256 1 10 1 1 1 0.16162 416 192 416 256 1 10 1 1 1 0.16w 256 96 288 96 0w 288 96 320 96 0w 320 96 352 96 0w 352 96 384 96 0w 384 96 416 96 0w 416 96 448 96 0w 448 96 480 96 0w 480 160 448 160 0w 448 160 416 160 0w 416 160 384 160 0w 384 160 352 160 0w 352 160 320 160 0w 320 160 288 160 0w 288 160 256 160 0w 480 192 448 192 0w 448 192 416 192 0w 416 192 384 192 0w 384 192 352 192 0w 352 192 320 192 0w 320 192 288 192 0w 288 192 256 192 0w 480 256 448 256 0w 448 256 416 256 0w 416 256 384 256 0w 384 256 352 256 0w 352 256 320 256 0w 320 256 288 256 0w 288 256 256 256 0r 208 96 144 96 0 12.5r 208 192 144 192 0 12.5w 256 96 208 96 0w 256 192 240 192 0w 240 192 208 192 0w 256 256 80 256 0w 256 160 80 160 0w 80 160 80 256 0w 144 96 144 192 0w 144 192 144 288 0w 80 256 80 288 0v 80 352 144 352 0 0 40 12 0 0 0.5w 80 288 80 352 0w 144 288 144 352 0But anyway the problem with this kind of layout is that if one led dies shorted, each string is STILL limited to 160 mA so now instead of having 160 mA divided between 8 leds, so 20mA per led, you now have 160mA divided among 7 leds, which means each led in that string with a dead led will receive around 23mA .... If one of those 7 remaining leds is particularly sensitive to over current, it could die and then you have 6 leds which receive 160mA so each led now gets 160mA / 6 = 26.6mA ... you get the idea? You risk a cascade failure, where a tiny number of dead leds in string of parallel leds can kill the whole string of leds.

author: Zero999
author_rank: Super Contributor
timestamp: October 18, 2018, 02:33:04 pm
content: Quote from: smile on October 18, 2018, 09:55:15 amWhy did you simulate that circuit? It's about the most pointless thing ever. It won't tell you how well or badly it will work.The LEDs will all have slightly different forward voltages, so the current sharing will be imperfect. Some LEDs will draw more current than others. The simulator will just use exactly the same model for each LED, which will show perfect current sharing, which is not a true representation of reality.If you plan to power this from the 230V mains, then use a capacitive dropper (Google it) and a rectifier for the LEDs. Of course the LEDs will all be live at mains voltage, so put them in a well insulated enclosure.If the LEDs need to be powered from 120V mains, then use a voltage doubler circuit to power the LEDs. Again, the LEDs will need an insulated enclosure.If you must run the LEDs at a 'safe' voltage, then splitting them into separate strings is the correct thing to do, but put a low value resistor in series with each string, to ensure good current sharing. It will only need to drop half a volt or so to give decent results.You could use your constant current driver board to drive eight strings of two LEDs in series, each with a 22R series resistor, to aid current sharing.

author: smile
author_rank: Regular Contributor
timestamp: October 19, 2018, 01:43:01 pm
content: Thank everyone for replies.QuoteWell my PSU can't output 175V, so I thought that I can connect parallel and drop the 2V, from minimum 12V to 10V and be happy. My simulation was for perfect world I know it. I simulated 20mA current when the actual is 26mA so even if 5 LEDS will fail the circuit should use 26mA only then and should not immediately fail.QuoteBut that will make leds flicker as far as i know. The stock led driver is just that capacitor based.QuoteI rewired the 16LEDS parallel in one string.I tested this connected to my lab psu. I set voltage to 10.4V and was increasing the current. The CC switches to CV at 35mA I thought parallel would make it want to use 300mA? I tried to give more current like 100mA but it was limited to CV and obviously current did not increase and the LED brightness was low.

author: mariush
author_rank: Super Contributor
timestamp: October 19, 2018, 02:58:49 pm
content: Your voltage of 10.4v may actually be too small and your leds may not be fully turned on.Why don't you actually determine the forward voltage of your leds first?Your power supply may not have enough precision at such low currents, often they're something like +/- 5mA precision or something like that.Set the maximum current output of your psu to something small like 50mA and then you can fine tune the current limit by simply connecting the probes of the multimeter (set on mA current range) to your power supply - the current will go through the current shunt inside your multimeter so the circuit is complete and the multimeter will show the actual current. Now if you have the meter set on a small range like 100mA or something like that, you can easily fine adjust the current limit of your power supply to 15mA or whatever desired value you want.Note that there's going to be a tiny voltage drop on the current shunt resistor inside the meter but it's relatively small (on sub 1A range on a meter, typically you have a 1 ohm current shunt, so at something like 20mA you have V = I x R = 0.020A x 1 ohm = ~ 0.02v drop inside the meter... practically nothing.The probes of your multimeter will also have some resistance but that's usually small, maybe 0.2 ohm or thereabouts ... so in general it's small enough to ignore it.OK, once you set your current limit to known threshold on your power supply, like 15mA or 20mA for example, you can lower the output voltage to a value you know it's way below the forward voltage of a single led, and then you can connect the led in series with the multimeter.Once that's done, you can slowly raise the voltage and watch your multimeter... the led will start to be somewhat bright and use a bit of current (between fully off and fully on) and at some point when the voltage is high enough, you'll see the led hitting that current limit you set.

author: smile
author_rank: Regular Contributor
timestamp: October 19, 2018, 05:38:11 pm
content: QuoteYes that seems to be the case.QuoteThe DP832 is more precise then my meter I think.I increased voltage to 10.98V and current is now 300mA, quite an increase from 10.4v at 35mAHowever the PSU in my first post seems to be bad quality in that it's not silent it produces typical whine noise of low quality inductors make. Hmm, now if only it was silent then that would be OK. I measured 8% flicker compared to stock PSU at 123%. The capacitor LED PSU is about 20-40% flicker.

author: smile
author_rank: Regular Contributor
timestamp: October 19, 2018, 05:42:29 pm
content: The LR8K4-G chip is 20mA is there anything for 300mA at 11V or adjustable? LR8K4-G is linear so it should not flicker right?

author: mariush
author_rank: Super Contributor
timestamp: October 19, 2018, 06:07:33 pm
content: If you read the specifications, it's 20mA typical , maximum 30mA.So "your mileage may vary" ... you can set the current to 25mA or 30mA and hopefully it will work, if not it will limit at lower current.The limit is also because of thermals .. for example if you have 400v in and you output 10v then you have 390v differential... at 20mA that's 390v x 0.02A = ~ 8w of power dissipated in the chip.I recommended that particular regulator chip because it's capable of working with high voltages, up to 450v so you can use the leds as you have them on the circuit board, in series - majority of linear regulators are designed to accept only up to 60v ("high voltage" versions of some linear regulators) or fr the majority of linear regulators, up to around 37v.Of course there are plenty of linear regulators which can output 12v or whatever voltage you want at higher currents, but the input voltage will have to be smaller.BUT the thing with leds is that you want to limit the CURRENT and adjust voltage based on current , because the forward voltage of the leds changes slightly as they heat up or cool down. It's also a bad practice to have multiple leds in parallel as i explained above.If you have a smaller voltage,then it would make more sense to actually use LINEAR led driver ICs which can monitor the current and automatically adjust the voltage as forward voltage changes while in use (forward voltage varies with heat).. go on digikey.com or other similar sites, go to led driver ICs and filter them by "linear" regulation, select your maximum current, the voltage and then check out datasheets and see if you like a specific chip

author: smile
author_rank: Regular Contributor
timestamp: October 19, 2018, 07:04:07 pm
content: QuoteThank you for your input, I will give the chinese LED driver another go, I have unsoldered the transformer and took it apart to put some liquid epoxy in effort to make it silent. Lets see how that goes.

author: smile
author_rank: Regular Contributor
timestamp: October 25, 2018, 07:59:59 pm
content: So epoxy dried and now the inductor is silentHowever it seems the LEDs are not full brightness even without any resistors. They are capped by PSU output power of 290mA.Is there easy tweak like put bigger output cap or something to make output like 350mA etc?I used this driverhttps://www.ebay.com/itm/300mA-4-7x1W-Led-Driver-4-7W-Power-Supply-AC-85-V-265-V-110-V-220-V-For-Light/132727839097?hash=item1ee7321979:g:-p0AAOSwEfBbYsS0:rk:2:pf:1&frcectupt=true