Hooking up 4 x 3v LED string lights to a 6v PSU - Page 1
Forum section: electronics / beginners

author: bob21
author_rank: Regular Contributor
timestamp: February 01, 2020, 05:01:57 pm
content: Hi,This is a total n00b question, and one which I am sure has probably been answered but I am not exactly sure what it is that I am searching for, so searches turned up a lot of results that didn't seem to fit.So here goes:I have 3 x cheapo LED string lights. The kind you get from bargain stores that run on 2 x AA cells, so totaling ~3v DC per set. Like this:https://imgur.com/a/CrOQbYII want to hook all 3 of these up to a PSU I have here that is 9v DC. So the question is: can I just splice all 3 together and connect to the PSU? Or is this the worst idea ever?My logical head says the load will just balance and each string of LEDs 'should' get its required ~3v. But then I wouldn't be posting this question if I wasn't totally sure it wouldn't catch fire to the house, cat and everything in the vicinityWill this work? If not, what is the best way to go about this? If I need to grab some components and solder something, I can do this. But I wouldn't have a clue what I would need to buy and solder to what.Any input would be very much appreciatedEdit: Changed title to reflect current aims.

author: TerminalJack505
author_rank: Super Contributor
timestamp: February 01, 2020, 06:46:45 pm
content: Are there some components other than the switch in the base? I can't tell from your picture. It is likely that there is some type of boost circuit in there, such as a Joule thief. How you proceed will depend on what type of circuit is being used to drive the LEDs.

author: tunk
author_rank: Super Contributor
timestamp: February 01, 2020, 07:00:43 pm
content: Looks very much like a LED string I have and there's no electronics inside.If it's 9Vdc +/-0.1V, then in principle it should work if you connect thethree strings in series, i.e. 3+3+3V. As a precaution you could add a lowvalue resistor in series, maybe 5-10ohm.

author: Zero999
author_rank: Super Contributor
timestamp: February 01, 2020, 07:43:35 pm
content: Those LED strings rely on the internal resistance of the batteries and LEDs to limit the current.What voltage does the 9V power supply actually measure, when you connect a meter to it?

author: bob21
author_rank: Regular Contributor
timestamp: February 01, 2020, 09:35:54 pm
content: Thanks all for the swift responses!Quote from: TerminalJack505 on February 01, 2020, 06:46:45 pmThe picture I linked to was just an example. I have checked the 3 lights I am using, and they seem to be just a switch between the cells and the string. Like tunk's. They are cheapo units from B&M, Home Bargains, Poundland etc.Quote from: tunk on February 01, 2020, 07:00:43 pmSeries isn't possible, the lights move away from each other in different directions. Sorry I should have mentioned this a bit better. All battery units are within a few cm of each other, but the other end of each string is not. It's a bit like this:https://www.dummies.com/wp-content/uploads/312189.image0.jpgIs there no hope of a parallel connection? I have small ABS boxes that can house some components if needed. Basically, the whole idea of this is that the lot is plugged into a smart plug at the end. It's a bind turning on and off each lot every time I use them. I want to just ask Alexa to do it instead and they chew cells if I leave them on too.Quote from: Zero999 on February 01, 2020, 07:43:35 pmI hoard old PSUs so I have some choice. I've just checked the 9V PSU I originally mentioned and it's reading at: 8.9v and is supposedly 1.7A. Please note my meter is a cheapo and not a nice Fluke as I would like one dayI have others that may be better suited, these are all the others below 12V:6V~300mA x 28V~400mA5V~1A4.5V~1A6V~800mA5V~800mA

author: tunk
author_rank: Super Contributor
timestamp: February 01, 2020, 09:49:42 pm
content: When I said series I meant:- power supply positive (to resistor) to positive first string.- negative first string to positive second string- negative second string to positive third string- third string negative to power supply negativeThis should work with your setup; assuming the PSU is 9.0V or below,and the strings are identical.Edit: that 8V PSU may be even better, again assuming it's really 8V.

author: bob21
author_rank: Regular Contributor
timestamp: February 01, 2020, 10:29:48 pm
content: my error, I understand what you mean now.The strings aren’t identical in appearance (some are small globes some are large, some are stars), but they may have the same amount of LEDs etc. I will need to check. They are certainly all powered by 2 x AA cells. I suppose worst case, they were very cheap, but I know that’s not the point.I think it’s probably worth just giving it a crack with the resistor as you suggest and the 8V PSU.Do you mean something like this for the resistor? eBay: 202893151256

author: Zero999
author_rank: Super Contributor
timestamp: February 01, 2020, 11:31:26 pm
content: Yes those will do but the price is silly. Rapid Electronics sell a pack of 100 for £0.94, including VAT. Delivery is a lot more, so unless you're ordering other things to make the value up to enough to qualify for free delivery, the rip-off eBay price might work out cheaper.https://www.rapidonline.com/truohm-cr-025-6r8-carbon-film-resistor-0-25w-pack-of-100-62-0324

author: tunk
author_rank: Super Contributor
timestamp: February 01, 2020, 11:37:05 pm
content: If it's really 8V then it should in theory work without a resistor.I guess you have to test it, wire it up and connect it for a secondor two and see if it works.

author: Brumby
author_rank: Unknown
timestamp: February 02, 2020, 03:11:32 am
content: If all 3 strings are electrically the same (or very similar), then this simple series connection should work fine - but I'll always endorse the addition of a current limiting resistor for cheap protection.In this case, what I mean by "electrically the same" is that the current drawn by each when running off the 3V of batteries is much the same for all 3 strings. When connected in series, the same current will pass through all 3 strings and if there is a notable difference in the stand-alone current draw of each, you might encounter some undesired effects - such as differences in brightness.

author: bob21
author_rank: Regular Contributor
timestamp: February 02, 2020, 09:38:55 am
content: Quote from: Zero999 on February 01, 2020, 11:31:26 pmYes, I see what you mean. I wonder if CPC have something, they offer free delivery after about £8. I am sure I could find something to bulk that out.Quote from: tunk on February 01, 2020, 11:37:05 pmThis morning I have checked a few things.. small curve ball. I was just checking exactly what the 8V PSU outputted, then I noticed it's actually AC:> 8V~400mA is ACnotDC, so is the 6V~800mAEdit: I also have another set of small globes that I could add and use the 9V?So the highest DC voltage I have (sub 12V) is the 9V 1.7A, or then it's the 6V~300mA which is an old Jabra chargerQuote from: Brumby on February 02, 2020, 03:11:32 amI have measured the strings and counted the LEDs:1. Large Globes:a. Total wire length: ~176cmb. No of LEDs: 10c. Space Between LEDs: ~16cmd. Just a switch between cells and leds?: Yes2. Small Globes:a. Total wire length: ~190cmb. No of LEDs: 10c. Space Between LEDs: ~15-16cmd. Just a switch between cells and leds?: Yes3. Stars:a. Total wire length: ~140cmb. No of LEDs: 6c. Space Between LEDs: ~24cmd. Just a switch between cells and leds?: YesI am not so concerned if some of the LEDs aren't as bright as some of the others, as long as they're light up 'reasonably'. I'm kinda thinking it'll be worth just hooking them up to the 6V and seeing what happens.Update:I connected the LEDs in series as tunk said using the 6V 300mA PSU that I mentioned above. And unfortunately there wasn't even a flicker of lightI realise 6V is less than 8V, but I would have expected something.So, before doing something else, or reattaching all the disconnected AA cell boxes, I tried connecting each of the lines directly to the PSU. Low and behold, they are working, evenly lit, and brighter than they were with fresh AA's.... on a 6V~300mA PSU... my brain is hurting. Each AA pack was outputting almost 3V, yet a 6V PSU with them connected in parallel is brighter. I really don't understand how this works. With that said, it's working. So, Is it safe to leave like this?

author: Brumby
author_rank: Unknown
timestamp: February 02, 2020, 11:04:02 am
content: Quote from: bob21 on February 02, 2020, 09:38:55 amNO!

author: Brumby
author_rank: Unknown
timestamp: February 02, 2020, 11:16:16 am
content: By having each string connected to 6V, you are overdriving the LEDs something extreme!!!Connecting the strings in parallel means theyeachget 6V - which is far too much! Each string does not know about the others - they just see the voltage provided by the power supply. At 6V, they will be sucking an excessive amount of current for them to survive for very long - but they will be bright!Just connect one of the strings to the 6V supply and you should see the same, excessive brightness. Adding more strings in parallel does not change this, as long as the power supply can handle the current drawn.The power supply, on the other hand, sees the current going out as being the sum of the current for each individual string - and as long as this is within its capability, then it will pour this current into the strings.If you really want to use this 6V supply, then use ONLY 2 strings and put them in series. If you get a fourth string, then you can make a second set of two strings in series and connect that second set in parallel to the first.

author: bob21
author_rank: Regular Contributor
timestamp: February 02, 2020, 11:51:50 am
content: Please, don't worry - This is why I asked the questionI have not left them plugged in like that. I checked and updated the post.So, let me just try to summarize what I think you've said. I have drawn a crappy diagram too.I can use the 6V PSU with 2 x 3V LED lines in series. Then if I were to get a fourth set (I actually already have one) and connect to the 3rd in series, I can then connect this in parallel with 1 and 2?I also have a 4.5V 1A PSU - or would that still overdrive them too much? What if I found a 3V PSU? Would parallel be ok then? I guess not...?I am not trying to be obtuse, I am just trying to learnEdit: Just noticed an error in my drawing where the strings connect to the PSU - pos and neg incorrect, have updated drawing

author: mariush
author_rank: Super Contributor
timestamp: February 02, 2020, 12:48:52 pm
content: You have to understand a very simple thing: LEDs are current limited, not voltage limited.This means the amount of current that goes through the LED must be limited, otherwise the LED will be damaged, will blow up.So once the voltage is above the forward voltage of a LED (a value which varies depending on the chemicals used to make the LED, and it's between around 1.7v for red leds, 2..2.2v for amber / green leds and 2.8v..3v for blue/white leds) the led turns on and lets any amount of energy flow through it. If you let too much energy go through it, the led overheats and gets damaged.You can limit the amount of current going through the led with various methods, but the most simple one is to put a resistor in series with the led.So for example, if you have a single white led with 3v forward voltage which can consume up to 20mA safely and you want to power it with 5v from a usb charger (power bank, phone charger, doesn't matter), you want to pick a resistor in such a way that the led will only "see" 20mA go through it. So you use the formula:Input voltage - Forward voltage of led = Current x Resistance value.so if we put the values in our example in the formula: 5v - 3v = 0.02A (20mA) x R ===> R = 2v / 0.02 = 100 ohmNow in your string of leds powered by 2 AA batteries.You have to understand that there's a difference between a battery powered circuit and a circuit powered by a power adapter (doesn't matter if it's AC or DC).A battery has an internal resistance, it can only push so much current before the battery itself overheats and the voltage drops. So you have to imagine the battery as a power adapter PLUS a resistor in series with the power adapter.In the case of AA batteries, these typically can output up to around 1-2A (2000mA) of current ... so you can look at that chain of leds and count how many leds are there.... i count about 20 leds.So if you had ideal batteries that can output continuously 2000mA of current, these 2000mA of current would be evenly split across the 20 leds, so each led would get 100mA.So you already have an inherent current limitation without adding a resistor to each led, to make sure it doesn't blow up.You also have to keep in mind that the wires are not ideal, they have some resistance... so you see that chain of leds has 20 leds and it's maybe let's say 2 meters long... that means that between the + and - poles of the battery there's going to be 4 meters of wire.The resistance of the wire will vary depending on the thickness of the wires... let's say that those wires are AWG24 ... if you look at an AWG table, you can determine that's a resistance of around 85 mOhm per meter of wire :https://en.wikipedia.org/wiki/American_wire_gauge#Tables_of_AWG_wire_sizesso with 4 meters of wire, you have around 0.4 ohm of resistance in the wires alone. Keep in mind that the resistance value is for copper, while these cheap led strips may use a mix of copper and steel wire, or steel and aluminum wires, because steel and aluminum is cheaper... so let's go with 0.5 ohm of resistance in the wires alone.So the battery behaves like there's a resistance inside it, and the wires also have some resistance, and all this resistance prevents the battery from pushing too much current in the chain of leds, so those 10-20 leds will receive maybe a bit more current than what it's recommended when the batteries are fully charged, but the leds will survive because the wires will behave like a heatsink, sucking the heat from the leds and dissipating it over the long wires.This design works for cheap stuff like those 10-20 leds powered by 2 batteries sold for 1-2$, because nobody will request refunds or complain if they die 1-2 months after Christmas.If some led dies, the other will continue to work because they're all in parallel, but keep in mind that the rest of the leds now receive more current, so they're more "stressed" and the risk of another one failing increases. If too many die, the other will also die, because there's no resistor for each individual led, to protect it.However, if you use a power adapter, this power adapter will have much lower internal resistance, instead of let's say 50 ohms of internal resistance, the power adapter will have maybe 5-10 ohm of internal resistance, so this resistance along with the resistance of the wire will not be big enough to limit the current going to the leds and the leds will blow up.In chains of leds designed to be powered with an adapter, you will often see a tiny resistor at the base of each led, or in the socket where the led is placed.Or, you may see a resistor hidden inside the wire going to groups of 3 leds, for chains of leds designed to be powered by 12v (3 x 3...3.2v = ~ 10v, 12v is used to account for losses in the long wire all the way to the last led)So how can you convert that 10-20 led chain of leds (in parallel) powered by 2 batteries to work with DC adapter.Well, you have to add the resistance that's hidden in the battery.Pick the amount of current you want for each led and determine what voltage you're gonna use to power that chain of leds where the leds are all in parallel.Let's go with 5v and 10mA for each led that's white and has a forward voltage of around 3v ... so if you have 20 leds on the chain, you would have 20x 10ma = 200mA or 0.2ASo going back to that formula : 5v - 3v = 0.2 A x R so R = 2 / 0.2 = 10 ohmSo if you add a 10 ohm resistor to this chain of leds, you can power them from a 5v adapter, and they should not consume more than 200mA. In reality they'll consume a bit less, because those meters of wire also have a tiny resistance (less than 1 ohm)If you want to power 3 such chains from the 5v adapter, you can put one 10 ohm resistor at the start of each chain, and connect them in parallel to your adapter. Each chain will only consume up to your calculated current value (the 200mA in my example).If you want to be more efficient you can re-arrange the leds to have groups of leds in series, then these groups you can place in parallel .For example, let's say you have a 12v power adapter, or you want to power some leds from a computer power supply.You know that the forward voltage of a single white led is around 3v, so you can realistically only put 3 leds in series (3 x 3v = 9v) because 4 leds will be exactly 12v, and your power supply may not always be exactly 12v.So you can make a group of 3 leds and put a resistor in series with this group of 3 leds and that resistor will protect the group of 3 leds:12v (input voltage) - 3 leds x 3v (forward voltage of led) = 0.01A (10mA current for the group of leds) x R => R = 3v / 0.01 = 300 ohm, so I'd use one of the standard values like 270 ohm (which would give a tiny bit more current through the group) or 330 ohm (a bit less current)Now, you can take these bits of 3 leds + resistor and connect them in parallel , as many as you want, and the total power consumed will be number of groups x ~ 10mA (current for each group of leds plus resistor)

author: Brumby
author_rank: Unknown
timestamp: February 02, 2020, 01:08:52 pm
content: Quote from: bob21 on February 02, 2020, 11:51:50 amYes, that's the idea.But even then, there is a weakness in the setup - andmariushhas gone to a lot of trouble writing it up.In short, it's thecurrentthat needs to be controlled for LEDs and series resistance (whether internal resistance of the source of power, resistance of the wiring or a separate resistor) will play an important part in that. Too little resistance and the LEDs will draw too much current - and die early.Take a moment and consider this chart of current vs voltage for an LED. Don't worry too much about the actual volts on this chart, it will be for a particular type of LED (not yours) - but the same general trend will be found onanyLED (including yours!).This LED would be sold as operating at 20mA with 2.7V applied:Notice what happens to the current when the voltage creeps up by just 0.1V ... the current is 2.5 times the rated current. With the voltage up by just 0.3V, it is 6.5 times higher!! Letting 130mA run through an LED that is meant to handle 20mA is going to kill it a lot quicker.But, yeah, it will be bright.

author: Brumby
author_rank: Unknown
timestamp: February 02, 2020, 01:17:53 pm
content: You will also notice from that chart that when you drop the voltage by 0.2V from the nominal 2.7V, the current drops down to practically zero. When that happens, you won't get any visible light output. (Well, that isn't strictly true, but it won't be particularly useful and you might have to darken the room to see it.)LEDs be crazy.

author: bob21
author_rank: Regular Contributor
timestamp: February 02, 2020, 01:52:22 pm
content: Wow, thanks so much for such a time consuming write up Mariush, and thank you to you also Brumby for all the info. I totally appreciate the time people are taking to help me with this.I am still re-reading everything and trying to get my head round it, but, I have added a 4th line to the circuit like this:So if I understand correctly, each parallel circuit (let's call them A and B) has almost the same amount of LEDs.If I understand correctly, a 300mA 6V PSU should be dishing 150mA to A and B. 150mA / 20 = 7.5mA, so each LED is receiving 7.5mA - or am I totally off target here?I think I am definitely going to grab some resistors to make this safer, so I have also marked the diagram with coloured dots.Would I just need one resistor where the purple dot is? Or would I need 2, where the green dots are? I am thinking the answer to this is that both A + B need one?Would the aforementioned resistors from eBay / Rapid still be suitable?I'm really sorry if it seems like I'm not getting this. But I'd like to say again how much I appreciate you putting up with meUpdate:Ok, I have re-read all of this and I think I understand I need a resistor where the green dots are, at the beginning of each series.I have (lets's say) 20 white 3V LED's on each series chain (A and B), and they should get no more than about 20mA.The PSU is 6V 300mA running 2 parallel circuits, so 150mA a circuit. So looks like I only have 7.5mA available to each LED with this PSU assuming all are working, so lets limit to that.6V - 3V = 0.0075 (7.5mA) x R ===> R = 3V / 0.0075 = 400 ohmSo I could use the 'common' 330 ohm resistors, place where each green dot is in the diagram and still be ok?

author: Brumby
author_rank: Unknown
timestamp: February 02, 2020, 02:45:34 pm
content: Quote from: bob21 on February 02, 2020, 01:52:22 pmNo.This is a very common misconception. The current rating of a supply is themaximumcurrent it isratedto provide. Theactualcurrent drawn will be a function of the load. If the load only takes 100mA (for example), then that is all the supply will dish out. This is the same for any and all power sources that you would generally come across.** Ohms Law describes this for the simple case of a resistive load.** This is the "constant voltage" type. There are supplies that operate on a "constant" current or current limiting basis - but we're not talking about these just here.QuoteOne where the purple dot is would be enough to provide basic protection - but one for each of the green dots would be considered a better option as it would allow independent "finding the operating voltage" for each arm.QuoteNeed? - Not really - but that would be the "better" way, in my book.

author: gf
author_rank: Super Contributor
timestamp: February 02, 2020, 02:49:36 pm
content: Line 3 (6 LEDs) and line 4 (15 LEDs) in series is not a good idea, since the two lines likely require different drive currents (assuming the LEDs are all the same type).

author: bob21
author_rank: Regular Contributor
timestamp: February 02, 2020, 02:57:37 pm
content: Quote from: Brumby on February 02, 2020, 02:45:34 pmOk, understood. So I need to work on the assumption that each series could receive the full 300mA load. Like if there is a loose connection or all LEDs in a series fail.In which case, that would be a max of 15mA per LED6V - 3V = 0.015 (15mA) x R ===> R = 3V / 0.015 = 200ohmSo I need to place a 200ohm resistor on each series for the best possible safety?Quote from: gf on February 02, 2020, 02:49:36 pmHow not good are we talking? Broken LEDs or Fire?These things are near furniture which Is why I am trying to make this as safe as possible. If the LEDs just fail, they were cheap. If they're going to burn the house down, that's different.

author: mariush
author_rank: Super Contributor
timestamp: February 02, 2020, 03:34:36 pm
content: First be sure that your 6v 300 mA is actually 6v. And is it DC or AC? How are you measuring the voltage?If it's AC voltage, you'll have to convert that to DC using a bridge rectifier and a capacitor.You'll get a DC voltage with a peak of around 1.415 x Vac - ~ 1.5v ,,, so if you have 6v AC power adapter, you'll get around 7v DC.The current will be lower, around 0.62 x Iac ... so the 6v AC 0.3A becomes ~7v DC 0.2AIf it's DC voltage... some cheap power adapters will have a slightly higher output voltage when there's no load (or very low load). For example, that 6v adapter may be 6v when you measure using a multimeter because the multimeter is a very low load, it consumes less than 1mA measuring the voltage. However, if you connect something that consumes a bit more power (like an incandescent 6v bulb for example), the voltage may go down a bit, let's say to 5.5vIf your chain of leds has no resistors or anything to control the current going through them, you WILL damage the leds.Such a chain will such as much current as the power supply can provide so you'll want to limit the current in some way.If the chain is basically multiple leds in parallel like I assumed in my previous post, a single resistor at the start of the chain could be enough.You can use that formula: Voltage input - Forward voltage of leds = current x resistor.With 20 leds in parallel, each consuming 10mA, you have 200mA or 0.2ASo with 6v DC and 3v for the led, you have6v - 3v = 0.2 x R => R = 3 / 0.2 = 15 ohmWhat I forgot to mention is that you also have to be aware how much power is lost in the resistor, so that the resistor won't burn.The formula for that is Power = Current x Current x RIn the above example, the power is 0.2 A x 0.2A x 15 = 0.6 watts, so you'd want a resistor rated for at least 1w otherwise the resistor will be super hot and may even burn up.If you have 2 chains each with 20 leds in parallel... you could connect them together to form a single chain of 40 leds in parallel.BUT, you'll have to keep in mind how will the extra long wire will affect things and you'll also have to adjust the resistor to account for that.So for 40 leds, you'd have R = (6v-3v) / 0.4 A = 7.5 ohm (you'd probably want to use standard 6.8 or 8.2) , and the power on resistor P = 0.4 x 0.4 x 7.5 = 1.2 watts, so a 1w resistor is no longer suitable, you'd have to use a 3w rated resistorThe extra long wire could cause some of the leds towards the farthest end to be less bright than the leds closer to your power supply.

author: Brumby
author_rank: Unknown
timestamp: February 02, 2020, 03:55:48 pm
content: Quote from: bob21 on February 02, 2020, 02:57:37 pmI'm going to stop you right there.You seem fixated on the 300mA rating of the supply.Forgetabout that while you play with all the other numbers.The 300mA ONLY comes into consideration when you work out everything else and come up with the total current that your circuits will want to draw. Then all you need to do is compare that load current with the 300mA.... and here is theONLYtime you use the 300mA rating of the supply:If the load current <= 300mA then all is well.If the load current > 300mA then you will be overloading the supply.The load circuit determines the current drawn. For a particular circuit, the current it will draw from a supply of a given voltage will be the same, whether that supply is rated at 300mA or 300,000A. If that circuit requires 1mA, then that's all that a power supply will put out.

author: bob21
author_rank: Regular Contributor
timestamp: February 02, 2020, 05:39:40 pm
content: Quote from: Brumby on February 02, 2020, 03:55:48 pmYou'll have to forgive me for being as thick as two short planks when it comes to this. I am trying to digest as quickly as I can though, so forget the 300mA for now. Got it. The only reason I was going back to the 300mA is because I logiced that if 300mA / 41 LEDs = 7.31mA, then I could not use the below calculation with a single LED current of 10mA, as the final currrent draw would be 410mA, which is overloaded.Quote from: mariush on February 02, 2020, 03:34:36 pmThe PSU is a 6V DC and I checked it by switching my Multimeter to '20' on the DC side and the reading was 6.01V.Because I think the LEDs at the end of a super long chain of 40 would suffer (as you've suggested), I think it may be better to place 2 resistors at the green dots and have the 2 x chains connected individually as per the last picture I posted (unless you mean this isn't possible?). If it is possible, I need to find 2 resistors - one for each green dot, right?I am just repeating this so you can see my thinking, I want to make sure I understand.So with each LED consuming 10mA:Circuit A - 20 LEDs: 6v - 3v = 0.2 x R => R = 3 / 0.2 = 15 ohmCircuit B - 21 LEDs: 6v - 3v = 0.21 x R => R = 3 / 0.21 = 14.28 (so safe to just use 15?)For the resistors losing power and getting hot, I would like them to be as cool as possible as these are near a sofa. So:Circuit A - 20 LEDs: 0.2 x 0.2 x 15 = 0.6 watts (as you've said)Circuit B - 21 LEDs: 0.21 x 0.21 x 15 = 0.66 watts (still ok to use 1W and remain cool?)So I need to place 2 x 15 Ohm Resitors rated at least 1W at each green dot point in my diagram?Just a side question, would it be possible to use a 15ohm @ 3W on these smaller chains to keep the heat down even more? Or will this not work? I am thinking about minimum heat due to proximity to fabrics.Now going back to the PSU mA rating - can this load cope with it? 10mA x 41 = 410mA, so no...So is it not at all possible to just adjust the calculation so it fits within the confines of the PSU (Yes I know I am meant to be forgetting this... but this was my thought process)If I have a 300mA PSU and 41 LEDs, I do not want each LED to draw more than 7.31mA. So recalculating the above using that value:Circuit A: 7.31mA x 20 = 146.2mA or 0.146ACircuit B: 7.31 x 21 = 153.51mA or 0.15351ACircuit A - 20 LEDs: 6v - 3v = 0.146 x R => R = 3 / 0.146 = 20.54 ohm (so safe to just use 20?)Circuit B - 21 LEDs: 6v - 3v = 0.15351 x R => R = 3 / 0.15351 = 19.54ohm (again, just use 20?)For the resistors losing power and getting hot:Circuit A - 20 LEDs: 0.146 x 0.146 x 20 = 0.426 watts - so 1W well and truly safe?Circuit B - 21 LEDs: 0.15351 x 0.15351 x 20 = 0.494 watts - again, 1W well and truly safe?So this is why I kept going back to the 300mA.. 41 LEDs x 7.31mA = 299.71mAQuestion is, is it wise to run it at it's limit? Would it not be better to take it down a notch and say, for example:Each LED 7mACircuit A: 7mA x 20 = 140mA or 0.14ACircuit B: 7mA x 21 = 147mA or 0.147ACircuit A - 20 LEDs: 6v - 3v = 0.140 x R => R = 3 / 0.140 = 21.4285 ohm (so 22ohm?)Circuit B - 21 LEDs: 6v - 3v = 0.147 x R => R = 3 / 0.147 = 20.4081 (so 21ohm?)For the resistors losing power and getting hot:Circuit A - 20 LEDs: 0.140 x 0.140 x 20 = 0.392 watts - so 1W surely not getting hot at all now?Circuit B - 21 LEDs: 0.147 x 0.147 x 21 = 0.453 watts - again, 1W well and truly safe?Now we come back to the 41 total LEDs each drawing a maximum of 7mA = 287mA so we aren't running the PSU to the max.So I need to put this on the 21 LED string:https://www.reichelt.com/gb/en/metal-oxide-resistor-1-w-5-22-ohm-1w-22-p1799.html?PROVID=2788&wt_guka=92495611608_406948187429&PROVID=2788&gclid=EAIaIQobChMI15_ok7Oz5wIVSLTtCh3rvgI9EAQYASABEgJIjfD_BwE&&r=1Can't seem to find 21 ohm 1W resistors easily, so would a 22ohm just work on both? Would be way easiser to grab one pack.If I haven't understood it now, I don't think I ever will.

author: mariush
author_rank: Super Contributor
timestamp: February 02, 2020, 06:58:52 pm
content: You can slightly adjust the resistor values to get something easy to buy. You'll just get a bit less or a bit more current through the leds.Reuse the formula just put R instead of current, for exampleCircuit B - 21 LEDs: 6v - 3v = 0.21 x R => R = 3 / 0.21 = 14.28 (so safe to just use 15?)Rewrite it like thisCircuit B - 21 LEDs: 6v - 3v = Current x 15 => Current = 3 / 15 = 0.2A (200mA / 21 leds = 9.52mA per led)The PSU is a 6V DC and I checked it by switching my Multimeter to '20' on the DC side and the reading was 6.01V.Yes, well that's no load.Repeat the measurement with 20 leds connected to it and see if you still get 6v. You may find out that the closer you are to its maximum advertised output of 300mA, the voltage may go down a bit. So you may see 5.5v at 250mA or something like that.The power supply may be conservatively specified, you may find out that it can actually output 500mA while the voltage goes down to around 5.5v ... not saying it would a good idea to use such a power supply for long time above its advertised capabilities... it's up to you.If you go above 300mA, it may or may not output more, but there's no guarantees.The power supply may lower the voltage a bit (so the total power will be the same, for example 6v x 300mA = 1800mW or 5.5v x 325 m = ~1800mW, same amount of power)... The power supply may have some protection mechanism built in, where it would constantly reset itself for brief moments of time.Yes, you can use resistors rated for higher wattage than needed. You can also parallel resistors to spread the heat across a larger surface. When you parallel two resistors, you get half the resistance but 2x the area to dissipate heat.So for example, instead of using a 10 ohm resistor, you could have two 22 ohm resistors in parallel ending up with 11 ohmYou can use 3w resistors but they're big and ugly. I'd look into paralleling 2 1w resistors first.

author: gf
author_rank: Super Contributor
timestamp: February 02, 2020, 08:14:25 pm
content: Quote from: bob21 on February 02, 2020, 05:39:40 pmWell, it's a matter of the desired brightness. You can even go down further if you don't need that much brightness. Just try different currents and check visually. But don't exceed the maximum current given in the datasheet of the LED chips! And, since the thermal conduction of the bulbs is likely unknown, I'd even stay quite below the maximum current given in the data sheet, in order to prevent overheating the LED chips. Overheating is one of the major causes for a premature death of LED chips.For the calculation of the current limiting resistor I would alsomeasurethe actual forward voltage of the LEDs at the desired operating point, and I'd also measure the PSU voltage under the actual load (it may drop when loaded, and even drop significantly if it gets overloaded and a current limiter happens to joins in).QuoteIs it a SMPS or a traditional PSU?I.e. does it provide clean DC (filtered, or even stabilized), or just pulsating DC directly from the rectifier?Do you read any voltage with the multimeter switched to AC mode?[ Note, in case of pulsating DC, the calculation of the value and power dissipation of the current limition resistor is no longer that simple, but requires integration of the rectified sine wave over the time interval where the waveform is larger than the LED's forward voltage. ]

author: bob21
author_rank: Regular Contributor
timestamp: February 02, 2020, 09:35:40 pm
content: Quote from: gf on February 02, 2020, 08:14:25 pmIs it daft to say, as bright as possible? I obviously want them to look good, but if they're going to burn out then they will of course have to come down a bit. As mentioned before, I don't so much care if they die at this point (as this is a learning curve for me) but what I would care about is if they caught fire. So if by 'death' you mean they go off and never come on again ergo 'you've lost £3' then I'm annoyed, but not that much vs them overheating catching fabric and setting a blaze to the place because I didn't keep a constant eye on them with a fire extinguisher in hand.Quote from: gf on February 02, 2020, 08:14:25 pmThe PSU is a traditional PSU. It's an old Jabra brick Model: ACGN-22BThe label says: Output 5-6V max 5W (6V 300mA)Quote from: gf on February 02, 2020, 08:14:25 pmYes, I have just checked and I read 13V AC when red probe is to pos and black/common probe is to neg. I don't get a reading if I swapped them round (in DC I would see a negative value ie. -13, I didn't see this).Quote from: mariush on February 02, 2020, 06:58:52 pmUsing a friends meter, which cost a bit more than mine, but is still no Fluke I did this:1. Took a voltage reading with nothing attached to PSU: 6.19V2. Took an Amp reading by connecting the red probe to the unfused 10A port and switching the dial to 10A: it fluctuated .44-.49 so WAY over the quoted measurement if I understand that .3 would equate to 300mA.I then repeated, by hooking up all 41 LEDs for max load as the above diagram (Circuits A and B comprised of 1+2 for A and 3+4 for B):1. Took a voltage reading by bearing the cable at the point the purple dot is shown and equivalent on the neg wire. The reading was: 5.93V2. Took an Amp reading using the same method as above, all the LEDs go out upon probe contact and the reading is the same: .44-.49Quote from: mariush on February 02, 2020, 06:58:52 pmGot it, thanks, that makes sense.Quote from: mariush on February 02, 2020, 06:58:52 pmWell, it seems from the testing I have just done that you have hit the nail on the head. It's way over stated spec. I don't want to use something or create something unsafe.Quote from: mariush on February 02, 2020, 06:58:52 pmWill this generate more heat than a single 3W?Quote from: mariush on February 02, 2020, 06:58:52 pmThis may be less of a problem, because I could end up housing this in a small ABS box that's non-offensive. But I will keep both methods in mind.So what's the verdict on this PSU? It's not fit for this task....? Any recommendations for a replacement?

author: Damianos
author_rank: Frequent Contributor
timestamp: February 02, 2020, 11:10:47 pm
content: Quote from: bob21 on February 02, 2020, 01:52:22 pmFrom this point and next, it cannot be assumed that the LEDs consume the same current.Explaining this as a kid: in the midpoint of the branch B "arrives" a current equal to 6*ILEDbut "departs" a 15*ILED, how is this possible?It is late at night and I am not able to post now some thoughts about possible solutions.

author: gf
author_rank: Super Contributor
timestamp: February 03, 2020, 12:01:28 am
content: Quote from: bob21 on February 02, 2020, 09:35:40 pmHmm, did you hook up the Amp meterin parallelto the LEDs, i.e. shorting the LEDs with the meter?In this case you would of course measure a higher current than flowing through the LEDs w/o the meter attached.For current measurements, the meter must be connected in series (and for voltage measurements in parallel) to the DUT.

author: Brumby
author_rank: Unknown
timestamp: February 03, 2020, 02:15:45 am
content: Quote from: gf on February 03, 2020, 12:01:28 amOh dear....For your typical multimeter:Voltage measurement - the meter is just peeking in on the circuit:Current measurement - the meter becomespart ofthe circuitYes, this means you have to break the circuit to do a current measurement - which is one of the things nobody likes. This also introduces a voltage drop - called "burden voltage" - in the circuit which would not be there in normal operation. Thiswillaffect the numbers and itcancause problems. Just be aware of the burden voltage for your meter and the range it is set on. (It is usual that this varies between ranges on a given meter - and, of course, it can vary between meters.)

author: Brumby
author_rank: Unknown
timestamp: February 03, 2020, 02:30:39 am
content: Quote from: Brumby on February 03, 2020, 02:15:45 amThis is when you use atypicalmultimeter. There are other options using a clamp mechanism.Using the clamp style, AC current measurement is easy, which is why you see these everywhere. DC requires a different approach which is a bit more involved.The most commonly cited meter is the UNI-T UT210E. This is a clamp-style meter which has DC current measurement - and it is not expensive!All you have to do is place the clamp over ONE conductor and it will measure the DC current. Resolution is 1mA, but accuracy is less as you need to deal with the fluctuations in the reading due to other magnetic fields. This meter would still be quite OK for the sort of measurements you are doing.

author: bob21
author_rank: Regular Contributor
timestamp: February 03, 2020, 06:47:43 am
content: Yes, that’s how I tested. So the Uni-T looks like a must have for the toolbox.Ok, so how does this sound for a plan:1. Order the Uni-T2. Use to test the draw of the entire circuit3. Reattach each AA cell pack to its respective string and test the current draw of each so we know what they each draw (if this is possible)Would this provide helpful information?I’ve only ever used a non-clamp cheapo meter just to test cells and continuity etc. so I welcome something better for the toolbox.

author: Brumby
author_rank: Unknown
timestamp: February 03, 2020, 06:56:21 am
content: The clamp goesaroundONEwire. The meter does not have to touch any wires at all. All clamp meters work on the magnetic field generated by flowing current.Don'tput both wires coming from the supply through the middle. If you do, the magnetic field from one wire will be cancelled out by the other.

author: bob21
author_rank: Regular Contributor
timestamp: February 03, 2020, 07:12:26 am
content: Just edited my post because I noticed you said attach to one wire, but I did actually think it would have to make contact. This sounds awesome! Amazon have them and my friend has prime. Watch this space ;-)

author: Brumby
author_rank: Unknown
timestamp: February 03, 2020, 07:31:02 am
content: Here's one example of wrong and right...The same rule applies for DC as it does for AC.

author: bob21
author_rank: Regular Contributor
timestamp: February 03, 2020, 08:20:08 am
content: And that seriously works for small wires? Wow. I thought it would have clamped in the jaws, glad you pointed that outJust waiting to hear back from my friend to see if I can transfer money and use his prime. If it’s a no go (can’t see this being the case though) are there any high street alternatives? Or is this pretty much one of those price to functionality gems that no one else produces?Edit: ok, scratch that. Got green light for amazon so Uni-T ordered,but also a neighbour who works for a telecoms company has lent me this:https://www.screwfix.com/p/kewtech-kt203-ac-dc-clamp-meter-400a/4048P?kpid=4048P&ds_kid=92700048793290430&ds_rl=1244066&gclid=Cj0KCQiApt_xBRDxARIsAAMUMu8DA0ohmKcjeAFfc-vQHwNb4LW314LOOg-5oqWlW5T9buupCpmoQWUaAnAnEALw_wcB&gclsrc=aw.dsthat I can borrow until my Uni-T arrives. Is this suitable? It seems to mention DC current?Nevermind. It's completely lifeless... I guess that's why he didn't mind lending it to meI'll wait for the Uni-T.

author: Brumby
author_rank: Unknown
timestamp: February 03, 2020, 10:31:43 am
content: Quote from: bob21 on February 03, 2020, 08:20:08 amOh.Seems it does do DC current via clamp so, yes, it will be able to measure DC current without having to cut into wires.The lowest range, however, is 40.00A (the maximum value displayable on a particular range) - which means the resolution (last digit) is 10mA. This should be enough for you to get some idea of what's going on, but the Uni-T has a 1mA resolution on the lowest range which will provide more information.Here's a trick: if you create several loops of wire, the current reading shown on the meter will be multiplied by the number of loops. Here is an example of 5 loops. The meter reading will be 5 times higher than the actual current flowing - because there are 5 wires carrying current through the clamp.So to get the correct current in this example, you divide the displayed value by 5. Using the 30.00 A value displayed in this image, the actual current flowing through the wire would be 6.00 A.Change the number of loops and you change the number you divide by.

author: bob21
author_rank: Regular Contributor
timestamp: February 03, 2020, 11:30:45 am
content: Makes sense. ThanksI am guessing using a mean calculation with more wraps like that is better for smaller wires?I actually can't wait for this Uni-T to arrive now.... want to get stuck back in.

author: bob21
author_rank: Regular Contributor
timestamp: February 03, 2020, 12:38:48 pm
content: I am just grabbing some other stuff for another project from CPC, and they have a ton of resistors. Is it worth me grabbing a small array for the toolbox? Maybe some of these:https://cpc.farnell.com/search?st=22ohm%203whttps://cpc.farnell.com/search?st=22ohm%201whttps://cpc.farnell.com/search?st=10ohm%203whttps://cpc.farnell.com/search?st=10ohm%201wI also want to connect the wires with plugs so it won't be a right faff to move the whole lot. I was thinking drill 3 holes in the ABS box, mount something like this:https://cpc.farnell.com/cliff-electronic-components/fm686803/circular-receptacle-pin-3-way/dp/CN16599?st=male%20female%20plugthen have the corresponding plug on the PSU and lines. I can't seem to find these things sold in male and female sets, anyone know what they are called?

author: Zero999
author_rank: Super Contributor
timestamp: February 03, 2020, 01:00:28 pm
content: A kit of resistors is always a nice thing to have. At bare minimum I'd recommend an E12 kit of resistor values from 10R to 1M, preferably 0.5W metal film, but carbon film is fine for most applications.Note, resistors come in standard values to make manufacture and specifying them easier. See the Wikipedia article.https://en.wikipedia.org/wiki/E_series_of_preferred_numbers

author: Brumby
author_rank: Unknown
timestamp: February 03, 2020, 01:28:59 pm
content: Quote from: Zero999 on February 03, 2020, 01:00:28 pm+1 for this.You will end up using just a few values from this on a regular basis (that you will end up getting more of those values), a few more every now and then and most will be sitting there 10 years from now. BUT - it's always worth having a range on hand if you are going to be experimenting and for the price they go for, it's a no-brainer.

author: bob21
author_rank: Regular Contributor
timestamp: February 03, 2020, 02:31:38 pm
content: Thanks both. Do you mean something like this?https://cpc.farnell.com/unbranded/mf0w2ffe012kil/resistor-kit-0-5w-1-e12/dp/RE03416It's quite a lot to drop on a starter project though.Also, @Zero999 is "Through-hole Resistor with metal oxide film element and high safety standard high purity ceramic core." what you mean by metal film?Ok, for like a third of the price of that kit, I can get these:Code:MCKNP03WJ0100AA9MULTICOMP PROMULTICOMP PRO Resistor, Wirewound, 10R, 5%, 3WMOR01SJ0100A10MULTICOMP PROMULTICOMP PRO Resistor, 1W 5% 10RMOR03SJ0220A19MULTICOMP PROMULTICOMP PRO Resistor, 3W 5% 22RMF50 10RMULTICOMP PROMULTICOMP PRO Resistor, 0.5W 1% 10R, 50 PackMF50 12RMULTICOMP PROMULTICOMP PRO Resistor, 0.5W 1% 12R, 50 PackMF50 15RMULTICOMP PROMULTICOMP PRO Resistor, 0.5W 1% 15R, 50 PackMF50 18RMULTICOMP PROMULTICOMP PRO Resistor, 0.5W 1% 18R, 50 PackMF50 22RMULTICOMP PROMULTICOMP PRO Resistor, 0.5W 1% 22R, 50 PackMF50 27RMULTICOMP PROMULTICOMP PRO Resistor, 0.5W 1% 27R, 50 PackMF50 33RMULTICOMP PROMULTICOMP PRO Resistor, 0.5W 1% 33R, 50 PackMF50 39RMULTICOMP PROMULTICOMP PRO Resistor, 0.5W 1% 39R, 50 PackMF50 56RMULTICOMP PROMULTICOMP PRO Resistor, 0.5W 1% 56R, 50 PackMF50 68RMULTICOMP PROMULTICOMP PRO Resistor, 0.5W 1% 68R, 50 PackMF50 82RMULTICOMP PROMULTICOMP PRO Resistor, 0.5W 1% 82R, 50 PackMF50 100RMULTICOMP PROMULTICOMP PRO Resistor, 0.5W 1% 100R, 50 PackMFP1-12R JITT ELECTRONICS / WELWYNTT ELECTRONICS / WELWYN Resistor, 1W 5% 12RMOR01SJ0150A10MULTICOMP PROMULTICOMP PRO Resistor, 1W 5% 15RMFP1-18R JITT ELECTRONICS / WELWYNTT ELECTRONICS / WELWYN Resistor, 1W 5% 18RMOR01SJ0220A10MULTICOMP PROMULTICOMP PRO Resistor, 1W 5% 22RMCKNP01WJ033KA10MULTICOMP PROMULTICOMP PRO Resistor, Wirewound, 0R33, 5%, 1WMFP1-39R JITT ELECTRONICS / WELWYNTT ELECTRONICS / WELWYN Resistor, 1W 5% 39RMFP1-56R JITT ELECTRONICS / WELWYNTT ELECTRONICS / WELWYN Resistor, 1W 5% 56RMOR01SJ0680A10MULTICOMP PROMULTICOMP PRO Resistor, 1W 5% 68RMFP1-82R JITT ELECTRONICS / WELWYNTT ELECTRONICS / WELWYN Resistor, 1W 5% 82RMOR01SJ0101A10MULTICOMP PROMULTICOMP PRO Resistor, 1W 5% 100RI've gone for 1% variance and MOR where I can.Just gotta find those plugs nowLooks like eBay wins for that: 223071120027

author: Zero999
author_rank: Super Contributor
timestamp: February 03, 2020, 04:51:58 pm
content: Yes, that looks suitable. Repeat that pattern up to 1M and you'll have a nice stock.I personally have E24 from 10R to 1M and E12 from 1R to 10R and 1M to 10M, but it's a bit overkill for the hobbyist.

author: bob21
author_rank: Regular Contributor
timestamp: February 04, 2020, 05:47:25 pm
content: It's here!I switched to'2A~'and with the entire circuit hooked up I get:1 wire: ~0.1585 wraps: ~0.820This is measured from1of the wires of the PSU using the clamp, before the series/parallels start (at the purple dot).I did notice when doing the single wire, it was quite an exact immediate number. With 5 wraps it took a while to climb and stabilise.More to follow. Doing this around making dinner etc.Edit:Ok, got it:Line 1 (10 small globes)Single wire: 41mA5 Wraps: 198mALine 2 (10 Large Globes)Single wire: 36mA5 Wraps: 256mALine 3 (6 White Stars)Single wire: 37mA4 Wraps: 146mA (could only manage 4 wraps as wire very short before 1st star)Line 4 (15 Globes)Single: 41mA5 Wraps: 245mALines 2 and 4 seem a bit off when dividing the wrap figure (it doesn't give exactly the single wire figure). Whereas 1 and 3 are almost spot on. Is this normal?I suppose the important thing to note is that the total figures are adding up for the single wires:Sum of singles: 155mA, and is 158mA measured from PSU wire.But I have probably screwed up my wrapping a bit:Sum of wraps: 881mA (obviously taking into account that line 3 needs to be 182mA for this calc), and is 820mA from PSU wireDo I need to retest?Also, I noticed that the draw from 1+2 (41+36) is almost the same as 3+4 (37+41)... so circuit A is pulling 77mA and circuit B is pulling 78mA..... does this mean they are [kind of] correctly balanced?Edit2: Btw, an extra thanks to Brumby for the DMM recommendation. This thing is great. It's about 1/3rd the size of the old piece of junk I had it's auto ranging is something I didn't have before along with a zero and hold button etc which is awesome and it's super intuitive to use.Edit3: Just got to pop out, when I get back I'll see if I can work out what resistors I need for each circuit.

author: gf
author_rank: Super Contributor
timestamp: February 04, 2020, 08:32:19 pm
content: Quote from: bob21 on February 04, 2020, 05:47:25 pmWhich current limiting resistor values did you use for each these measurements?What do you mean with "5 Wraps"? Do you mean 5 lines (of the particular lamp type) connected in parallel?

author: bob21
author_rank: Regular Contributor
timestamp: February 04, 2020, 08:58:24 pm
content: Quote from: gf on February 04, 2020, 08:32:19 pmI'm sorry, but what do you mean?If you mean what was limiting the current, I hooked each led string back up to it's original AA pack and measured using the newly purchased Uni-T clamp meter using the clamp. I checked each string individually as if i had never changed anything (ie. reconnected to each 2 x AA cell pack) and I also took readings with them all hooked up as per my rubbish diagram with the purple dot and 2 green dots where I have labeled them Lines 1-4 and Circuits A-B.There was nothing else added to limit current. Or do you mean something else?Quote from: gf on February 04, 2020, 08:32:19 pmI mean wrapping the wire round and through the clamp as Brumby illustrated a few posts back.

author: gf
author_rank: Super Contributor
timestamp: February 04, 2020, 09:45:40 pm
content: Quote from: bob21 on February 04, 2020, 08:58:24 pmOK, using the original battery pack - w/o resistors.What is the voltage across the LED string (or across the battery) under load? Still about 3V?QuoteAll clear.I just noticed:QuoteWhy AC and not DC? (particulary since it is connected to the battery pack)

author: bob21
author_rank: Regular Contributor
timestamp: February 04, 2020, 10:00:33 pm
content: Quote from: gf on February 04, 2020, 09:45:40 pmI didn't check this, I only checked the current using the clamp.Quote from: gf on February 04, 2020, 09:45:40 pmThat is he dial setting for this model meter... it also has the DC sign on it, but I don't know how to type it you just press the blue select button to change to DC.Ok, I have worked out some [possibly very incorrect, and stupid] numbers!Please don't shoot me, I am still learningI didn't even know what a resistor was/did 4 days ago, so bare with meLet's start with the individual calculations of running each circuit inparallel.Line 1 - 10 LEDs = Total Consumption off a 2 x AA pack putting out 2.92V = 41mA read from Uni-T clampResistor needed: 6V-3V = 0.041 x R ==> R = 3V / 0.041 = 73 OhmsWattage of resistor needed: 0.041 x 0.041 x 73 = 0.122W - so an R82 (or R68?) 0.5W should be correct for this?Line 2 - 10 LEDs = Total Consumption off a 2 x AA pack putting out 2.92V = 36mA read from Uni-T clampResistor needed: 6V-3V = 0.036 x R ==> R = 3V / 0.036 = 83 OhmsWattage of resistor needed: 0.036 x 0.036 x 83 = 0.107W - so an R82? 0.5W should be correct for this?Line 3 - 6 LEDs = Total Consumption off a 2 x AA pack putting out 2.92V = 37mA read from Uni-T clampResistor needed: 6V-3V = 0.037 x R ==> R = 3V / 0.037 = 81 OhmsWattage of resistor needed: 0.037 x 0.037 x 81 = 0.110W - so an R82 0.5W should be correct for this?Line 4 - 15 LEDs = Total Consumption off a 2 x AA pack putting out 2.92V = 41mA read from Uni-T clampResistor needed: 6V-3V = 0.041 x R ==> R = 3V / 0.041 = 73 OhmWattage of resistor needed: 0.041 x 0.041 x 73 = 0.122W - so an R82 (or R68?) 0.5W should be correct for this?BUT: We want 1 + 2 in Series and 3 and 4 in Series:1 + 2 = Circuit A3 + 4 = Circuit BCircuit ATotal consumption = 77mAResistor needed: 6V - 3V = 0.077 X R ==> R =3V / 0.077 = 38.96 Ohm (So 39 Ohm)Wattage of resistor needed: 0.077 x 0.077 x 39 = 0.231W - So an R39 0.5W should be sufficient for Circuit ACircuit BTotal consumption = 78mAResistor needed: 6V - 3V = 0.078 X R ==> R =3V / 0.078 = 38.46 Ohm (So 39 Ohm)Wattage of resistor needed: 0.078 x 0.078 x 39 = 0.237W - So an R39 0.5W should be sufficient for Circuit BSo, going back to my rubbish diagram, I need to add 2 x R39 0.5W resistors where the green dots are, and they shouldn't get hot because they aren't under much load?Am I correct in thinking that we don't need to limit the PSU? Each individual line of LEDs is run in series, so a failure would take out the whole line? Would then not the entire 6V be dumped on the remaining 2 lines.... or am I getting this totally wrong?Edit: Oops, I typoed one of the resistor values as R9 instead of R39, fixed.

author: gf
author_rank: Super Contributor
timestamp: February 04, 2020, 11:18:49 pm
content: Quote from: bob21 on February 04, 2020, 10:00:33 pmDo you really mean a trueseries connection, or do you just mean daisy-chained behind each other, but electrically still connected in parallel?When two strings are connected in series then the current through both stringswill be the same, and the voltage drops will add up. This mean, you would need to combine strings 1+4 and strings 2+3 each, since these are the pairs with (almost) equal current consumption. If one string has as voltage drop of say 3V, then two strings in series will have a voltage drop of 6V. When the PSU voltage is 6V as well, this would leave no headroom for an additional voltage drop across a current limiting resistor :-(That's one reason why I asked"What is the voltage across the LED string (or across the battery) under load? Still about 3V?"If the actual voltage drop across the LED strings happens to be less than 3V, then we gain some headroom for the resistor in case of a series connection. And generally it is of course better to know the actual voltage drop instead of guessing "roughly 3V" for the calculation of the current limiting resistor.Btw, I'm still also wondering whether your LED bulbs (or their sockes) possibly have a tiny current limiting resistor built-in, or whether they include only the LED chip and nothing else? Can you see anything in the bulbs which looks like a SMD resistor?

author: bob21
author_rank: Regular Contributor
timestamp: February 04, 2020, 11:38:06 pm
content: I did have a quick look for individual resistors on each LED a few days back and didn’t notice anything. It’s late here now, but tomorrow I will check:1. Each LED in each line for individual resistors2. The voltage drop, ie hook up each individual string to its AA pack and take a voltage reading before and after turning on and post back. How would be best to test this? Probe at start and end of string? Each string only has 2 wires. If you removed 1 LED, the entire string would fail.I’m not sure what you mean by daisy chained. If you mean line 1 pos to psu pos, then line 1 neg to line 2 pos, then line 2 neg to psu neg, then yes. This is how they are connected. Same for 3 and 4.If I need to change the config of the strings so that the numbers are better, I can do this. My end goal is safety first and longevity of the LEDs second.Watch this space.

author: gf
author_rank: Super Contributor
timestamp: February 05, 2020, 12:25:37 am
content: Quote from: bob21 on February 04, 2020, 11:38:06 pmMeasuring both ends is a good ieda, since there might be a voltage drop along the string(at only 40mA I would not expect too much, though).QuoteOnly if you destroy the cable during removal. The LEDs of a string are not connected in series, but in parallel, therefore the removal of a LED is not supposed to break the circuit. Only the current consumption would decrease.Note, if the LEDS of a string were connected in series, then you would need a voltage of about 3V*10=30V in order to drive 10 LEDs - too much for 2x AA cells (unless a switching regulartor would be used to boost the voltage).QuoteWith daisy chained I mean concatenated behind each other, i.e. the second string connected to the end of the first one (+ to +, and - to -), which is electrically aparallelconnection.QuoteOK, you really mean an electrical series connection.Again, this would only be possible for strings 1+4, which draw the same amount of current, and for strings 2+3 which also draw the same amount of current. If you connect two strings in series then thesameamount of current flows throughbothones.If you would connect strings 1+2 in series, then you'll either need to drive string 1 with a lower current than 41mA or strings 2 with a higher current than 36mA, since there is "only one current" flowing through both strings then. The whole amount current which leaves string 1 enters string 2, since there exist no path where the current could escape. At each connection point in the circuit, the sum of incoming and outgoing currents is the same (Kirchhoff's law).

author: gf
author_rank: Super Contributor
timestamp: February 05, 2020, 12:46:35 am
content: Quote from: gf on February 02, 2020, 08:14:25 pmIf the PSU provides just pulsating DC, then one consequence to be expexted will be flickering of the LEDs.

author: bob21
author_rank: Regular Contributor
timestamp: February 05, 2020, 08:29:06 am
content: Quote from: gf on February 05, 2020, 12:46:35 amWell in that case I would have to guess that the DC does not pulsate. The LEDs definitely do not flash when hooked up to this. They are solid on.Quote from: gf on February 05, 2020, 12:25:37 amThis might be tricky as I’ve have to start stripping other ends of cables which I ideally wanted to avoid, but what I can do this morning is take a voltage reading before and after each string is turned on (when connected to its cell pack) and provide the voltage drop figures.Quote from: gf on February 05, 2020, 12:25:37 amAh, I understand now. Thanks.Quote from: gf on February 05, 2020, 12:25:37 amI have both, 1 and 2 are electrically in series, as are 3 and 4. Then these are in parallel to the PSU to share the 6V.Quote from: gf on February 05, 2020, 12:25:37 amGot it. So I need to do some reorganising: 1+4 in series, 2+3 in series, both connected in parallel to PSU.Just out at the moment, will provide voltage drop readings for each unit in a bit.Edit: Changed title to reflect current aims.

author: bob21
author_rank: Regular Contributor
timestamp: February 05, 2020, 09:44:11 am
content: More info:With a 2 x AA cell pack start reading of 2.93V, I hooked up each string of lights and checked the voltage again from the point where the pack was reconnected to the wires.Line 1: 2.86VAny visible resistors in LED heat shrink? Difficult to tell, as opaque, but my guess is no, it's too thin. Ie. it only looks as wide and long as the LED legsLine 2: 2.85VAny visible resistors in LED heat shrink? Difficult to tell, as opaque, but my guess is no, it's too thin. Ie. it only looks as wide and long as the LED legsLine 3: 2.87VAny visible resistors in LED heat shrink? Definitely not, the heat shrink is transparent, and it's just the LED legs and wire.Line 4: 2.86VAny visible resistors in LED heat shrink? Definitely not, the heat shrink is transparent, and it's just the LED legs and wire.The size of each LED heat shrink part is the same size in 1 and 2 as in 3 and 4. I know resistors can be tiny, but I honestly don't believe there's any in there. The heat shrink would be bigger or show the outline of an additional component, it's tight against the LED legs and wire.

author: bob21
author_rank: Regular Contributor
timestamp: February 05, 2020, 11:45:03 am
content: Having worked out some more numbers (again, please do not shoot me, these might all be junk!), we are now using:Line 1 + 4 = Lets call this Circuit CLine 2 + 3 = Lets call this Circuit DUpdated Diagram:Pink and Purple dots are resistor locations? Is this correct?C's total draw is 82mA6V - 3V = 0.082 x R ==> R = 3V / 0.082 = 0.246 OhmWe only have 22 or 27 Ohm resistors, so can we use a 22 for this?D's total draw is 73mA6V - 3V = 0.073 X R ==> 3V / 0.073 = 41.09 OhmWe only have R39 and R47What is the formula to adjust a resistor value and work out what the resultant current will be?Is it:Circuit C: R = 3V x C ==> 3V / R22 = C ==> 0.136mA so if using a R22 resistor on C, it will get 136mA not 82mA.. problem.Circuit D: R = 3V x C ==> 3V / R39 = C ==> 0.076mA so if using a R39 resistor on D, it will get 76mA not 73mA... no big deal?So D is pretty ok at this point - how to go about C? I thought about using 2 x R47's for 1/2 R so 23.5R:R = 3V x C ==> 3V / R23.5 = C ==> 0.127mA so if using a 2 x R47 resistor on C, it will get 127mA not 82mA.. still a problem.Am I over thinking this, could use some wisdom

author: gf
author_rank: Super Contributor
timestamp: February 05, 2020, 06:09:17 pm
content: Quote from: bob21 on February 05, 2020, 11:45:03 amI think you still didn't get it. If you put two components inseries, then the currentsdon't add up.41mA flow from the PSU into the "+" connector of line 1, 41 mA come out of line 1's "-" connector and flow into line 2's "+" connector, and again 41 mA come out from line 2's "-" connector and flow back to the PSU.So C's "total draw" is still41mA.But the voltage drops of the components add up. If one string has a voltage drop of 2.86V, then two strings in seris will have a voltage drop of 2 * 2.86V = 5.72V.This leads to the following calculation for your sub-circuit C:Vpsu= 6 V (measured)Vled= 2.86 V (measured)Iled= 0.041 A (measured)Vresistor= Vpsu- Vled- Vled= 0.28VR = Vresistor/ Iled= 0.28 / 0.041 = 6.8293 Ohm ==> 6.8 OhmPresistor= Vresistor* Iled= 0.011480 WThe problematic thing here is that after subtracting 2 * Vledfrom Vpsuthere are only 0.28V are left for the resistor (which leads to the low resistor value of only 6.8 Ohm).Under these conditions the resistor is not supposed to ensure a very stable working point for the circuit.Note, however, the above calculation applies only if the PSU output is 6VcleanDC!For "pulsating DC" (i.e. full-wave rectified sine-wave) we will get different numbers!In this case 6.8 Ohm will be ways too low and would lead to a significantly higher (average) current than 41mA!I don't know details about your PSU, but I have the feeling that itdoes not deliver clean DC.

author: bob21
author_rank: Regular Contributor
timestamp: February 05, 2020, 06:38:05 pm
content: Quote from: gf on February 05, 2020, 06:09:17 pmYou are correct, I didn't get it, but I do now --> Current does not add up when in series. Thanks for being patient with me.Quote from: gf on February 05, 2020, 06:09:17 pmJust to clarify, this is a dropto2.86Vfrom2.93V so a loss/drop of 0.07V. Or have I somehow measured this incorrectly? If I have measured this incorrectly, and I need a PSU with more voltage and current, I have a whole box of them here. I am looking at one here now that is 12V 1.5. Would that be more suitable?Quote from: gf on February 05, 2020, 06:09:17 pmHow do I test if it provides clean DC? You said the LEDs would flicker if unclean, but they do not......

author: gf
author_rank: Super Contributor
timestamp: February 05, 2020, 06:50:47 pm
content: Quote from: gf on February 05, 2020, 06:09:17 pmNow the same calculation for unfiltered, pulsating DC with anaveragevoltage of 6V:(The calculation below is Matlab/Octave code. I'm doing the integration in a very simple way numerically.)Code:t = [0:99] * 0.01 * pi; % 100 steps 0 <= x < pi for numerical integrationVpsuavg = 6 % measured average voltageVpsupeak = Vpsuavg / 2 * pi % peak voltage of the rectified sine waveVpsu = sin(t) * Vpsupeak; % PSU voltage as function of timeVled = 2.86Iledavg = 0.041Vresistor = Vpsu - Vled - Vled;Vresistor(Vresistor<0) = 0; % set to 0 in the off-time where Vpsu < 2 * VledVresistoravg = mean(Vresistor) % avarage voltage across resistorVresistorrms = sqrt(mean(Vresistor.*Vresistor)) % RMS voltage across resistorR = Vresistoravg / IledavgPresistor = Vresistorrms * Vresistorrms / RThis gives the following results:Code:Vresistoravg = 1.4235 VVresistorrms = 2.0477 VR = 34.720 OhmPresistor = 0.12077 WattSo we would now require 33 Ohm (while we had only 6.8 with clean DC).

author: bob21
author_rank: Regular Contributor
timestamp: February 05, 2020, 07:07:42 pm
content: Ok, but see my post above. Did you think the voltage drop was by 2.86V or to 2.86V? Because it was a drop to the values listed, so:Line 1: loss =0.07Line 2: loss =0.08Line 3: loss =0.06Line 4: loss =0.07I appreciate your time explaining the differences in ohms when the DC supply is dirty, but all of this is going way over my head now. If the PSU is not suitable, I would rather:1. Test my existing PSU to see if it really is / is not clean - you specifically mentioned a few posts back that unfiltered DC would cause flickering - and I replied that my PSUdoes not cause even slight flickering2. Try another PSU (I have tons here), most are from gadgets like phones, power banks, speakers etc.. surely they can't all be dirty?3. Buy a PSU that is clean - is this an example?https://tinyurl.com/uutxd9vI just want the simplest easiest way to power these 4 strings off a wall outlet that won't catch fire to my house.If I need to buy something else I will. I've just bought a new DMM and a ton of resistors. What else do I need?I am an electronics n00b. I knew nothing about any of this 4 days ago and my brain hurts. I am trying my best to take all this in as quickly as possible. I've even tried calculating values myself and not just expecting other people to do it all for me. Yes, some have been wrong, but I am posting in the beginners section, after all.

author: gf
author_rank: Super Contributor
timestamp: February 05, 2020, 08:21:45 pm
content: QuoteOK, I see, you primary aim are obviously quick results.Here's one of several possible approaches:Don't connect strings in series, but 4 all strings in parallel to the PSU.Use aseparatecurrent limiting resistor in series witheach string.Determine resistor values experimentally, for one string at a time:For string #1 start with 82 Ohm (>= 1/4 W), connect to the PSU and measure the currentIf it is lower than 41 mA reduce to 75 Ohm and measure the current againUse the same resistor value for string #4For string #2 start with 100 Ohm (>= 1/4 W) and measure the currentIf it is lower than 36 mA reduce to 91 Ohm and measure againIf it is still lower than 36 mA reduce to 82 Ohm and measure againUse the same resistor value for string #3EDIT:Note, if yourenounce the desire to connect two stings each in series, there is significantly more voltage headroom for the resistor to establish a stable working point. Additionally, clean vs. "dirty" DC does no longer make such a huge difference, so that resistor values in the range 82...75 Ohm should fit with the 41mA strings, and 100...82 Ohm should fit with the 36mA strings. When you determine the value experimentally, then start with larger resistor values, measure the current, and decrease R until you hit the string's current target (+/- 5% or even +/- 10% is perfectly OK).

author: bob21
author_rank: Regular Contributor
timestamp: February 05, 2020, 09:39:04 pm
content: Thank you.This is exactly the information I was after. I am more than happy to connect each in parallel and forget series if this is overall a better approach.My resistors have been dispatched, I am hoping they will arrive tomorrow.I will post again with an update when I know more and have had to chance to experiment.Ps. I want to say again that I appreciate the time you have taken to explain a lot of this stuff to me, but being from a non-electronics background, I need some additional digestion time before the next project. Maybe then we can go into deeper stuff like dirty DC.

author: bob21
author_rank: Regular Contributor
timestamp: February 07, 2020, 08:07:24 pm
content: Done!I used some male and female jack connectors in an ABS box that are just connected to the PSU, so each port can provide the 6V 300mA that the PSU is designed to provide.I then did as you suggested, and have soldered the resistors in series to the positive side of the cable going to each line. I heat shrinked the whole lot.I considered putting the resistors in the ABS box, but then if a line were plugged into the 'wrong' port, it would receive the incorrect current/voltage. So this way, any line can be plugged into any port. I figured human error should not be ignored. I did a quick google, and it seemed to suggest that heat shrinking 0.5W resistors would be perfectly safe as in this particular application they aren't drawing anywhere near their max load.Here is a pic of the box (Yes, I know the wiring isn't the prettiest, but it's definitely secure and soldered well):The final values were:Line 1:R39 + R33 (in series) both 0.5WResultant Current: 45mAResultant Voltage: 2.85VSufficient wattage resistors? Yes: I x I x R = 0.145Line 2:R82 + R10 (in series) both 0.5WResultant Current: 35mAResultant Voltage: 2.84VSufficient wattage resistors? Yes: I x I x R = 0.112Line 3:R82 + R10 (in series) both 0.5WResultant Current: 35mAResultant Voltage: 2.86VSufficient wattage resistors? Yes: I x I x R = 0.112Line 4:R39 + R33 (in series) both 0.5WResultant Current: 38mAResultant Voltage: 2.67VSufficient wattage resistors? Yes: I x I x R = 0.103Note: Line 4 is the longest wire (and 15 LEDs). I assume this is why the mA and V readings are lower. I suppose I could reduce the resistance on that at some point maybe try a single 68, but they are sufficiently bright for now.So what's the verdict? Did I do ok? Is it safe to heat shrink the resistors? Do the values look ok?Edit: I am considering doing something similar for my sons bedroom (he has a secret den with a lot of battery powered LED lights and it would be great to have them hooked up to one plug that is voice controlled) so any input on the safety and/or appropriateness of this would be greatly appreciated

author: bob21
author_rank: Regular Contributor
timestamp: February 08, 2020, 07:48:55 am
content: Anyone?Safe? Unsafe? Ok? Not ok?

author: Brumby
author_rank: Unknown
timestamp: February 09, 2020, 01:00:22 pm
content: First comment - useIfor current. It might seem strange, but that's the normal notation. Blame the French guy - Ampère - whose name is used for the unit, which is most often abbreviated toamp. He used I for current.Next: I haven't checked you math, but if things are working OK, then you are halfway there. As for heat-shrinking resistors, this is completely ok, subject to thermal considerations. Heatshrinking reduces the cooling effectiveness, but if you can comfortably hold a resistor between your fingers for 10 seconds, then they should be fine.

author: bob21
author_rank: Regular Contributor
timestamp: February 09, 2020, 10:07:16 pm
content: I'm not opposed to such peculiarities. I'm sure there was a reason'I' for current. Right you are. Edited previous post to reflect correct usage (I x I x R)And no, absolutely no hotness at all. If I were being totally picky, I would say perhaps the 'stone coldness' (to use a very nontechnical term) had gone in favour of 'not stone coldness' (that really is the best way to describe it). I'd hardly call it luke warm, body temp, warm, and definitely not too hot after 10 seconds. It's just no longer stone cold.A bit off topic now, but what is the best way to determine the wattage of these resistors? If they weren't written on the bag, I wouldn't know. I used a colour calc to confirm tolerance and R rating, but how to reliably know what wattage? Let's say they all get mixed up. I can use the coloured striped to determine R value and tolerance, but some 0.5W look the same as the 1W... seems odd to me?

author: Brumby
author_rank: Unknown
timestamp: February 09, 2020, 11:50:55 pm
content: Sounds good.Having four separate parallel circuits will have one negative aspect: each resistor will be dropping more voltage than with your series/parallel idea which means more power wasted in heat.BUTfor the levels of current involved here, that is not a real concern unless you wanted to max out the number of LEDs you want to hang off this one supply.There are, however, a number of positives!1. There will be no need to match the current draw of pairs of LED strings2. If one string fails, then only that string will stop working and you will know which string to check out. In series connections, an open circuit failure will result inallLEDs in that arm will stop working and you will have to go hunting. (Brings back memories of chasing a blown globe in a string of mains powered Christmas lights in my teen years.)3. If the LEDs get a short circuit, then the resistor (being a larger value) will not getashot as one would for the series arrangement. It might still get hot and it might fail if this happens, but it has a better chance of surviving. Less heat in a failure situation = better safety.4. You can add, remove or change LED strings (up to the capacity of the supply) without having to re-work any of the other strings.As far as other safety considerations, there are two I would suggest:1. If the supply has a particular capacity, then set yourself a lower limit for your circuit to use. For example, with your 300mA capable supply, keep it under 200mA. Things that aren't pushed to their maximum all the time will last longer - this applies to most things, from power supplies to motor vehicles.2. Add a fuse. The rated current of the fuse should be:(a) No more than the rated current of the supply(b) Subject to (a), sufficiently high enough to avoid nuisance blowing. For your setup, I might suggest a 250mA rating.Also, for other characteristics:(i) Fast blow is fine for LEDs(ii) Any fuse type will be suitable for the voltages and currents involved here. No need for ceramic or anything fancy - a basic glass M205 would be absolutely fine.

author: Brumby
author_rank: Unknown
timestamp: February 10, 2020, 12:04:56 am
content: Quote from: bob21 on February 09, 2020, 10:07:16 pmThat is often a challenge even for those of us who've been playing with them for years. If you have been working with some particular types a lot, you might be able to identify their power rating purely from experience. Most of us aren't that well versed.Physical size is just one factor. The materials from which they are made is another. It is a well known fact that you can have a half watt resistor that is smaller than a quarter watt resistor. If you do the math, you will find out that the smaller half watt resistor will, at the limit of its rating, produce twice theheatof a quarter watt resistor - and when this happens in a smaller package, thetemperaturegoes up significantly higher. Here's where the materials matter - where the smaller resistor is made from materials than can withstand this temperature.

author: Zero999
author_rank: Super Contributor
timestamp: February 10, 2020, 05:07:28 pm
content: Another limiting factor to the power dissipation of a resistor is the surrounding materials, such as the PCB, solder and any nearby components. A high temperature ceramic resistor might happily be able to sit at 300oC, but it's no good if it melts the solder, falls off the board and burns a hole in the plastic case!

author: bob21
author_rank: Regular Contributor
timestamp: February 10, 2020, 08:24:10 pm
content: Ok, understood. Thanks again both for the information. I feel like I am starting to understand more and more of this.So the resistors seem fine as they are, but I will bear heat dissipation in mind for any future projects.I will also do as you suggest Brumby and add an M205 glass fuse to the PSU wire (pre-ABS Box) as an additional safety component. Just to be clear, do you mean this:It seems to suggest the voltage can go up to 250V, but I am guessing the important part here is the mA of 250? One of these in a basic fuse holder solder in series to the positive wire will do the job?That’s an au site, but CPC have this, same thing?https://cpc.farnell.com/multicomp/mc000893/fuse-5x20mm-glass-quick-blow-250ma/dp/FF03065?st=250ma%20fuseThanks to all for the time spent helping this n00b out!

author: Brumby
author_rank: Unknown
timestamp: February 10, 2020, 10:01:12 pm
content: Quote from: Zero999 on February 10, 2020, 05:07:28 pmYeah ... thatwouldbe undesirable.Quote from: bob21 on February 10, 2020, 08:24:10 pmAbsolutely correct. The voltage is a maximum rating. The figure of 250V means it can be used as a mains fuse pretty much anywhere in the world ... or at any voltages lower.A common glass fuse is fine for the power levels involved here - all the way up to domestic mains power situations for a lot of equipment. Ceramic and high rupture capacity (HRC) fuses are complete overkill here, but if that's all you had access to, then one of those would be OK.QuoteYes.

author: Brumby
author_rank: Unknown
timestamp: February 10, 2020, 10:24:44 pm
content: To be fair - so that you don't get too complacent with your knowledge of fuses - there is a whole industry that is built around fuses. It does become quite involved, especially when you get into high current and high energy circuits (yes, they are different things). Do a Google search on "types of fuse" or something like that when you have a couple of days to get totally lost. I would also suggest you check out the challenges breaking DC circuits (particularly high voltage) as against AC ones.Fortunately, for the vast majority of situations you will find yourself in as a hobbyist, this is one example of a table that covers the basics:It is possible you could encounter ceramic and HRC fuses (in a multimeter, for example), so being aware of those is useful.Just one more word of caution ... if you come across any fuse that is thicker than your average cheapie ball-point pen, don't mess around with it unless you areabsolutelycertainof what you are dealing with. I'm not just talking about the equipment being protected, but also, the capability of the power source, for example: how far a spark could jump and how big aspannerbus bar*it could vaporise.* Edit: Change automotive reference to a power systems one.

author: bob21
author_rank: Regular Contributor
timestamp: February 12, 2020, 06:45:27 pm
content: Understood. Thank you for the additional information. I have some fuses on order. I've never seen a fuse larger than a standard UK plug, I wouldn't be comfortable playing with anything that large anyway. Sounds to me like high voltage AC stuff, which isn't my cup of tea.I will keep that image for reference though.Thanks guys