Current exposure to my LEDs - Page 1
Forum section: electronics / beginners

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 17, 2022, 10:15:21 pm
content: I have a project wherein I'm using 9 LEDs (5mm white) in parallel and powered by a USB cube. The draw according to my meter is 0.1 amp or 100mA when the leads are placed in series with the red wire from the cube. Does this mean that all LEDs are being exposed to 100mA of current? Or does each LED draw 100/9 (11mA)? I know the safe limit generally is 20mA. Thank you

author: IanB
author_rank: Super Contributor
timestamp: February 17, 2022, 10:59:21 pm
content: You only know the total current is 100 mA (probably more, accounting for the internal resistance of the meter).What you don't know is how that current is distributed between the 9 LEDs. Some may be drawing more current than others.In general it is not a good idea to wire LEDs in parallel unless they have individual current limiting resistors. If they do have current limiting resistors then it is fine, in which case the current should split fairly evenly between them.

author: TimFox
author_rank: Super Contributor
timestamp: February 17, 2022, 10:59:48 pm
content: A batch of LEDs connected in parallel (without individual resistors in series with each) will pull quite different currents from each other, since the current is an exponential function of the voltage and a small difference in the co-efficient will make a relatively large change in the current. In general, one shouldn't connect diodes directly in parallel for that reason. If possible, it is better to wire such devices in series, since they will all have the same current (although the voltage across each will be slightly different.

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 17, 2022, 11:28:29 pm
content: I should have mentioned that I'm hacking an existing mass-produced retail product in which the parallel wiring is already there. The 9 LEDs prematurely (in my way of thinking) burn out one by one. I was hoping that using the resistor might eliminate that. Changing the product from parallel to series is not an option for me. I'm currently leaving them on 24/7 to see how they do.

author: Terry Bites
author_rank: Super Contributor
timestamp: February 17, 2022, 11:45:50 pm
content: No. The current in each parallel branch cannot be exactly the same. LED charcteristics are not that well matched.Nearly all premature LED fatalities are casued by poor thermal management.

author: BrokenYugo
author_rank: Super Contributor
timestamp: February 18, 2022, 12:21:09 am
content: As I explained last time, the easy way out of this problem is to use one big LED (readily available mounted on a PCB) instead, e.g.https://www.ebay.com/itm/122604065816The ONLY way to make those flashlight board work how you want will be to massively under drive them (like 20mA to the board, not 100), because as was explained above, and last time, you can't just put diodes in parallel and expect them to share current evenly.

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 18, 2022, 12:49:22 am
content: https://www.ebay.com/itm/122604065816Can you help me with selecting the best one for the job?

author: IanB
author_rank: Super Contributor
timestamp: February 18, 2022, 12:56:55 am
content: Quote from: billbyrd1945 on February 17, 2022, 11:28:29 pmIs this "existing mass produced retail product" a flashlight that uses 3 AAA cells, by any chance?In which case, you cannot replace the 3 AAA cells with a USB power supply. It is not an equivalent replacement. The original design with the AAA cells was already marginal and destined for premature failure. As soon as you replace the batteries with the USB supply you are guaranteed to destroy the LEDs sooner or later.Quote from: billbyrd1945 on February 17, 2022, 11:28:29 pmWhat resistor? You didn't mention a resistor above. Even so, you can't put LEDs in parallel, with or without a resistor. It doesn't work.Have 9 resistors, separate the LEDs, and put one resistor for each LED, and then you have a chance of a reliable design. The 9 resistors will now balance the current between the 9 LEDs.

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 18, 2022, 01:30:00 am
content: "Even so, you can't put LEDs in parallel, with or without a resistor. It doesn't work."I'm sure you know waaaaay more about electronics than I do. But-- one can definitely put LEDs in parallel. I was born in 1945. Christmas lights for Christmas trees came in long strings and were wired in series. Stupid even for the time. Every time a bulb burned out, the whole string went out. Then, one by one, you changed out the bulbs with a known good one until you found the one that was burned open. Then they wised up and started making them in parallel. In my little project, the LEDs go out one at a time. If they were wired in series, the whole system would go dark with one failure. As to the value of the resistor, I simply kept trying different values until the current was the same with the USB as it was with the battery pack. Why would I use the battery pack as a yardstick? Because the LEDs don't burn out when using the battery pack. (The resistor turned out to be 5.6R)

author: IanB
author_rank: Super Contributor
timestamp: February 18, 2022, 01:52:30 am
content: Quote from: billbyrd1945 on February 18, 2022, 01:30:00 amSure, of course you can wire them up that way. But you missed out the important part.QuoteThere. That's the important part. In a proper design, the LEDs don't go out. They last forever.QuoteOh, they will burn out eventually. It will just take a bit longer than you have waited so far.

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 18, 2022, 02:41:45 am
content: Oh, I understand what you're saying. I know I'm playing with trash. One can learn a lot from trash. That doesn't keep me from wanting to solve the riddle. And the riddle is: one full week, 24/7 with batteries, no failures. Eighteen to 24 hours with USB power, 4 LEDs fail. I repeated the exercise and got the same result. I fully realize that coincidences occur. Maybe that's what it is. Maybe I should go for three tries.

author: BrokenYugo
author_rank: Super Contributor
timestamp: February 18, 2022, 03:28:49 am
content: YouSHOULDN'Tput diodes of any type, including LEDs, in parallel. As has been explained by multiple people before, diodes do not have perfectly matched electrical characteristics, and it is the nature of a solid state diode (LED, rectifier, whatever) to conduct BETTER when hot, forward voltage drops with temperature. So one diode will always end up taking most of the current, and if that's enough current to kill it, you get a cascading failure until they all fail. So again, the only way to make these boards work long term is to act like only one LED is present and limit current accordingly, it won't be very bright and might not even light all the LEDs if they're extra crappy, but it will hold up long term.As to what is best for your application in that ebay link, that depends on what you want, I can't say beyond "with PCB", and "1W" will probably do it if you consider the born to die flashlight boards you have to be of adequate brightness.

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 18, 2022, 03:57:04 am
content: "... I can't say beyond "with PCB", and "1W" will probably do it if you consider the born to die flashlight boards you have to be of adequate brightness."Thank you. Understand completely now that you describe the weakest-link aspect. Is there any reason not to get the highest wattage of the ebay selections?

author: tooki
author_rank: Super Contributor
timestamp: February 18, 2022, 06:10:08 am
content: Quote from: billbyrd1945 on February 18, 2022, 02:41:45 amYou’re seeing the effects of the lower voltage of the batteries, plus the batteries’ internal resistance.

author: mariush
author_rank: Super Contributor
timestamp: February 18, 2022, 07:26:55 am
content: LEDs must be current limited.The easiest way to do that is to add a resistor in series with the LED to limit the current going through both components.You use the Ohm's law : Voltage = Current x ResistanceSo if you have a LED with a forward voltage of 3v and a 5v power supply and you want the led to work at 20mA (0.02A) you put these into the formula :Input voltage - (number of leds in series ) x (forward voltage of led) = Current x Resistanceso with the values above : 5v - 1x3v = 0.02 x R so from here : R = (5-3 ) / 0.02 = 2/0.02 = 100 ohm.You say you have 9 leds consuming 100mA and that's powered from 3 AAA batteries and you say there's a 5.6 ohm resistor to limit the current. You're not including the resistance of the battery which can be half a ohm or more.If the AAA batteries are rechargeable, those are 1.2v each so you have 3.6v input voltage. If they're alkaline, you have 3 x 1.5vLet's assume those leds have 3v forward voltage, let's put that in formula : 4.5v (3xAAA alkaline) - 3v (forward voltage, just one because leds in parallel not series) = 0.1A x RSo R = (4.5 - 3) / 0.1 = 1.5/0.1 = 15 ohmIf we go with rechargeable batteries we have (3.6 -3 ) / 0.1 = 6 ohmThe internal resistance of a AAA alkaline batteries is 0.15-0.3 ohm .... so with rechargeable alkaline your flashlight has around 6-6.5 ohm of current limiting resistance, keeping those leds to around 100mASo when you're using 5v, you don't have the resistance of the battery, and your current limiting resistor remains around 6 ohm, so now your leds do (5v-3v ) / 6 = 2/6 = 0.33A which means each of the 9 leds gets 37mAA simple solution would be to add 1 or 2 diodes in series, because each diode would drop around 0.7v ... basic 1n400x diodes would work.edit : also pay attention on how you're measuring those 100mA - multimeters have a burden voltage , they measure the current by placing a resistor in series with the circuit, so you basically have an extra resistor between the battery / power and the circuitOn small ranges, that resistor inside the multimeter is quite large (could be 1-10 ohm) so that could affect the reading. On big ranges (ex 10A scale), the internal resistor is typically smaller, like 0.1 ohm or something like that, so the multimeter won't affect the reading that much but you get less digits on the multimeter screen.

author: golden_labels
author_rank: Super Contributor
timestamp: February 18, 2022, 07:53:22 am
content: You certainly can put LEDs in parallel. It’s not about if you can, but if you should. You have already mentioned they were in parallel, without any current limitation oringally and theyburned out one by one, in a cascading manner. You were also told that mismatch between LEDs in parallel will make each of them to see different current. Can you connect the dots?Semiconductors experience thermal runaway, where higher temperature leads to lower resistance, which in turn produces even more heat. That’s opposite of what incandescent lightbulbs did. Your christmas lights were stabilizing current on each branch. Even LED X-mas lights may happen to work this way, because the characteristics get averaged over dozens of LEDs, so each branch sees more or less the same situation. That is not the case with single parallel LEDs.In some circuits designers do not use any current limiting to reduce costs and get away with it for two reasons. First is that those devices are not expected to last, so if 1/3 of them ends up in a waste bin within a year no one cares. Second is that around the knee the I/V curve of LED is not that steep. If the supply voltage is low enough, changes in current are slight and countered by power supply’s internal resistance. In some cases there is a sweet spot in which that operates for some time.Forward voltage of white LEDS is likely in the 3.5–4V range. Three cells have a nominal voltage between 3.6V and 4.5V (NiMH and alkaline, respectively). If the knee of the original LEDs was sufficiently smooth, you can see how they could survive for some time. But 5V from USB is way too much to stay in that “lucky region”. Your 5.6Ω resistor is dropping over half a volt at 100mA, which brings the voltage back to that range. But this is not how things should be designed.

author: IanB
author_rank: Super Contributor
timestamp: February 18, 2022, 02:27:40 pm
content: Quote from: billbyrd1945 on February 18, 2022, 02:41:45 amThat's certainly a riddle. At the top of the thread you suggested the current draw was 100 mA. Now the capacity of a AAA battery is about 1000 mAh. So we should expect the batteries to be depleted within a day if running continuously. You changed out the batteries every day, or more often than that?Also bear in mind that an alkaline cell has a voltage around 1.5 V only at the very beginning of its life. It goes down to about 0.9 V at the end, so for most of the run time the light is not seeing 4.5 V, is is seeing an average of only about 3.6 V. Which also means it is not seeing 100 mA, but much less than that.

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 18, 2022, 02:58:58 pm
content: Quote from: tooki on February 18, 2022, 06:10:08 amThe batteries are 4.5v, the USB is "fivish". I placed an external resistor to mimic the battery resistance.

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 18, 2022, 03:06:16 pm
content: Wow! You put a lot of work into that response. Please know I will study it closely and know that I really appreciate it.

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 18, 2022, 03:10:08 pm
content: "You changed out the batteries every day, or more often than that?"Yes. Every time they got noticeably dimmer. I'm retired, have a ton of batteries, nothing better to do. I wanted the LEDs subjected to optimal power available in that setup.

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 18, 2022, 03:35:28 pm
content: "But this is not how things should be designed."Remember... I didn't design these things. Just trying to improve the design to the extent possible by someone with extremely limited electronics knowledge so I can get away with using them in a project. But I get it. I'll order proper components as soon as I can zero in on that. I've been given a suggestion (maybe brokenYogo) to order something from eBay. I just need to decide which of the 3 wattages they offer.And here's an addendum to the whole thread (actually it came up the last time I posted about this project a month or so ago. I just didn't know where to go with it). There's yet another component I'd forgotten about until last night when I found my loupe. I'm talking about 5R1.And like I said, someone mentioned it last time, but I forgot about it. I'm attaching a picture. When I try to measure across it (power turned off), I get zero continuity on any scale. What is it? When I use the diode scale, it beeps in one direction but not the other. So is it a diode? Duh! Connecting dots here. And why would you put one there? Probably obvious to you guys and it probably doesn't change anything about the destiny of the project. I'll still go with the eBay components and forget about these flashlights. Just the same-- wanted to provide full exposure.

author: mariush
author_rank: Super Contributor
timestamp: February 18, 2022, 03:55:47 pm
content: It's a 5.1 ohm resistor.Your multimeter may not beep or say "continuity" if the resistance is higher than some value ... let's say 1 ohm.There is some tolerance because the leads of your multimeter have some resistance (typically under 0.25 ohm) and the tips of your probes could also be dirty enough to introduce some resistance. So in continuity mode, a meter would beep and show continuity when the resistance is below some threshold.5.1 ohm may be just big enough to not register as a short in your multimeter and won't trigger the continuity.In diode mode, your meter outputs a specific voltage from one probe to the other and it measures the voltage drop on the other probe. There is a voltage drop across the 5.1 OHM resistor, so your meter may detect the resistor as a diode... but it will show some voltage drop regardless of how you position the probes.Put your multimeter in the resistance mode and touch probes together and the meter, if it's good enough, it should show around 0.1 ohm.. 0.25 ohm ... my own probes are 0.2 ohm (uni-t ut61e)Here, I made you a video with my phone ... may have to look at it sideways, too lazy to do it again or edit it :https://drive.google.com/file/d/13uWUNq53IWjEbKiaqxT54Pyy9pO4Pw0s/view?usp=sharingresistance of probesresistor is 1 kohm , showing you meter reads it at 996 ohm (good enough)then you see the meter in diode mode,and reading again resistor it shows 1.06v drop across the resistor, so the meter thinks it's a diodereverse probes and measure again, same valuethen see diode and the big pads...measuring voltage drop across diode shows 0.18v (typical for a schottky diode)the probes reversed you see OL or 1.7v (that 1.7v is consequence of how circuit is, there's a capacitor or inductor in parallel with the diode messing the measurement)then going back to meter in continuity mode, to show the meter won't show continuity across resistor because 1 kohm is too much.

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 18, 2022, 10:33:59 pm
content: Thank you VERY much for going to all that trouble. I see in the video that we are using the exact same multi-meter. I got someone without tremors to recheck the resistor and to check my leads. My leads shorted together measure 0.4 ohms. Measuring across the resistor measures 5.46 ohms. So you were exactly right. Thank you!

author: mariush
author_rank: Super Contributor
timestamp: February 19, 2022, 06:44:28 am
content: Cool ... so your resistor is around 5 ohm, after discounting the multimeter leads resistance.But, you also have to account for the resistance of that spring (under 0.1 ohm but not 0) and the resistance of the copper traces on the round circuit board, and the resistance of the wire going to the back of your flashlight to the other end of the batteries.So all these together add up to some resistance value, let's say 5.2 ohm and you add up to around 0.3 ohm internal battery resistance of the AAA batteries, ending up with around 5.5 ohm.So if you then use that ohm's formula and assuming you have a nominal voltage of 4.5v (3x1.5v for non rechargeable batteries) and the leds have 3v forward voltage, you're looking at4.5 - 3 = current x 5.5 ohm => current = 1.5 / 5.5 = 0.27A which is 30 mA per led, if you have 9 leds.Keep in mind that at 0.27A, the battery voltage will sag a bit so in practice the current will be a bit smaller, and the battery voltage will go down a bit.The flashlight is probably assuming you're gonna use rechargeable batteries, that will peak at 1.35v and provide their power down to around 1.1vWith 3 x 1.2 v you'd have a current of 0.6/5.5 = 110 mA and around 12mA per led.

author: billbyrd1945
author_rank: Regular Contributor
timestamp: February 24, 2022, 12:48:33 am
content: I decided to take BrokenYugo's advice and order one of these:https://www.ebay.com/itm/122604065816Obviously, I don't know what the hell I'm doing and didn't know which watt to order and didn't know which 'k' value to order. I finally decided to order the 3w with 'cold white' which had the highest number 'k' value.Would one of you be so kind as to tell me how to power it? I'd like to use USB if possible, but a battery pack would be just fine if that's what I need. And I'd also need some advice on which resistor.Thank you again for all the explanations.PS: For the life of me I cannot get my account or profile configured so that I get emailed when someone responds to my posts. I've tried several times but can't get it to work.

author: tooki
author_rank: Super Contributor
timestamp: February 24, 2022, 06:23:37 am
content: Quote from: billbyrd1945 on February 24, 2022, 12:48:33 amFYI, what you’re calling the “‘k’ value” is the color temperature, given in Kelvin (so capital K, not lowercase k which stands for kilo). Higher color temp = more blue, lower color temp = more yellow/orange. Around 4000K is “natural” white, anything much lower is warm white, anything much higher is cool white.