Electronics > Beginners
Hooking up 4 x 3v LED string lights to a 6v PSU
bob21:
--- Quote from: Zero999 on February 01, 2020, 11:31:26 pm ---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
--- End quote ---
Yes, 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 pm ---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 second
or two and see if it works.
--- End quote ---
This 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 |O:
> 8V~400mA is AC not DC, so is the 6V~800mA
Edit: 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 charger
--- Quote from: Brumby on February 02, 2020, 03:11:32 am ---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.
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I have measured the strings and counted the LEDs:
1. Large Globes:
a. Total wire length: ~176cm
b. No of LEDs: 10
c. Space Between LEDs: ~16cm
d. Just a switch between cells and leds?: Yes
2. Small Globes:
a. Total wire length: ~190cm
b. No of LEDs: 10
c. Space Between LEDs: ~15-16cm
d. Just a switch between cells and leds?: Yes
3. Stars:
a. Total wire length: ~140cm
b. No of LEDs: 6
c. Space Between LEDs: ~24cm
d. Just a switch between cells and leds?: Yes
I 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 light :( I 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?
Brumby:
--- Quote from: bob21 on February 02, 2020, 09:38:55 am ---So, Is it safe to leave like this?
--- End quote ---
NO!
Brumby:
By having each string connected to 6V, you are overdriving the LEDs something extreme!!!
Connecting the strings in parallel means they each get 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.
bob21:
Please, don't worry - This is why I asked the question :) I 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 learn :D
Edit: Just noticed an error in my drawing where the strings connect to the PSU - pos and neg incorrect, have updated drawing
mariush:
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 ohm
Now 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_sizes so 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 0.2A
So going back to that formula : 5v - 3v = 0.2 A x R so R = 2 / 0.2 = 10 ohm
So 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)
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