I want to power a ring of 10 LEDs white super bright (6mm) in parallel.
I have this "travel adapter" 220V to 5V at 2A. (see image)
When I power with it, i put in series 100ohm resistor to bring down the 5V. The voltmeter is showing a value on it only when i close the circuit with a Led. I played with 10Leds AND the resistor, until i get 2.8V general (on all 10Leds).
If I take out Leds, the voltage is rising UP, and the brightness is also visible changing.
I want a circuit that will use my travel adapter source power, and the LEDs will maintain luminosity indifferent is 1led or 10 or 20.
My best guess, is that 2A is not true; and the amperage is very low - my amperemetre is not functioning for some reason.
I have an idea using a transistor and power all the leds through it... but if its an Amp problem... i have no idea how to fix it.
Maybe you have a solution?
I tested 1 to 40 leds with a wire transformer 3V /500mA and they were functioning as I wanted. All were at same brightness indifferent on the number of leds i added or subtract. It tells me they were fed with sufficient Amp.
Please confirm to me if I am right with what i guess.
Thank you ! - and see the pictures:

 

A schematic would SURE be helpful!

If you have 10 LEDs in parallel and a single 100 Ω resistor, then the resistor will be setting a total current through the LEDs of about (5 V - 2.8 V) / 100 Ω = 22 mA. That doesn't sounds like the current needed for 10 superbright LEDs. So instead of making use guess what your circuit it, please post a schematic (even just a picture of a hand drawn sketch on a napkin) so that we know what the circuit is you are working with.

 

I would use a 56 ohms resistor in series on each led , and wire them all in parallel across the 5V, assuming they have a vdrop of 3V, that should give you approx 35mA per led.

 

Ohms law is not so much obvious but actually you can learn it by playing with LEDs and resistors, firstly try a single LED.

You need a resistor for each LED as pointed out, or a voltage booster, the MT3608 could be used just on the limit for 10 LEDs in series.
Each LED typically has 20 Ohms on its own so you get approx 200.
This is why you get several volts margin with LED chain, they dont burn out, and voltage coverter can be used.
One reason might be you dont want to wire 10 resistors + dont want them around either.
Directly theres only one way, 10 resistors.
Could be you can only get 8 or 9 to be very bright with the MT3608.

 

I dont have the specification file for my leds , but i used my best judgement and i find this helpful page before I start anything.
https://info.pcboard.ca/led-specifications/3mm-led-technical-specifications/
it say there for Bright White: Typical: 2.8v Current: 18mA - so you are close when you say 3V vdrop. and 35mA.
I like your idea actually.

I put that 100ohm resistor in series to bring down the voltage but he got very hot very quickly and it started to smell. I put 3x100ohm resistors all in parallel - like a pack resistor, to simulate a FAT resistor that will dissipate the hotness more than a single one could. And it worked - now is just warm and is good enough for me. The voltage rise UP a bit doing it like this but it actually helped to get closer to 2.8V per total. I wish to avoid the temperature if possible... but is not that much of a priority. Important is that is working.

I will try put resistors in series with each LED as you suggest and test what is happening. Hopefully the voltage will not fluctuate !
Thank you!

 

Another important question is this:
- How can I lower the overall voltage, from 5V to 2.8V?
In this case, if I have the optimal voltage from start, I will not need each resistor serialized with the leds.
No?

 

if the resistor gets hot for small LED current is too much
Also the LED should remain cool or only heat up very slight
Otherwise, it will burn out quickly.

Because I used many kinds of LEDs for growlight, including panels from small LED,i know
Parallel single LED, is problematic

 

to Dodgydave:
I tested with 2leds so far. My total voltage is 5.4V so i had to put 220ohm resistor for each led. The vdrop is 2.8V per led from measuring. I was curious and I calculated the current: i=v/r >> i=24mA. Oau, it's very weak. Similar with mr. WBahn finding.
Yes - this is working as I want! Thank you very much!
- It is not possible to lower main voltage a little bit? (with -2.6V) 5.4-2.8=2.6. I want to get rid of the resistors for each led from the scheme. With a zenner diode or something?
I will use this circuit in the end since i get the best result so far... but i really want to make it much better.
Thanks again to all !

 

You need the current limiting resistors. LEDs are current operated devices, not voltage. While they do have a typical voltage drop, what is important is to control the current. If you were to have an LED with a 2.2V drop, and put it on a 2.2V power supply, chances are you would destroy the LED due to over current. Those demos running an LED directly on a coin cell with no resistor rely on the internal resistance of the cell to limit the current.

 

Hi

You could use a small 5 to 25v boost converter and wire 10 LEDs (each 2.2v@20mA ) and one 150 ohm resistor in series.
Boost converter is about $10 USD.

eT

 

I would use a 56 ohms resistor in series on each led , and wire them all in parallel across the 5V, assuming they have a vdrop of 3V, that should give you approx 35mA per led.

Your comment was straight to the point and all worked perfectly. Thank you very much !
Here is my entire project (thanks to you):

 

The light makes a big difference. Thank you for posting the final result
Often we do not get to see that.

 

Couple issues I've noticed: First, you spoke of a resistor getting too hot. That happens when you don't use the proper wattage rated resistor. However, you can't just put a bunch of resistors in parallel because you affect the resistance. Three 100Ω resistors in parallel (or a bunch like you said) has a total resistance of about 33Ω. But before we get into that lets determine some facts. Facts are always a great place to start.

Assume you have a 5 volt supply and just one LED who's forward voltage is unknown. Assume it's 3.0 volts forward (or fV). Next, lean upon Ohm's Law and do some calculations. Start with the known voltage of 5 volts and subtract the suspected fV of the LED. 5 - 3 = 2. Assuming you're working with 2 volts we want to shoot for some known values. Assume your super bright LED will work efficiently on 20 mA (0.02 amps).

2 (volts) ÷ 0.02 (amps) = 100 (ohms). So using a single super bright white LED running at 20 mA with a 100Ω resistor should work just fine. It's a good place to start taking some measurements. Measure the voltage across the LED. It may be 3 fV or it may be slightly different. Expect some difference. But concerning the resistor, we need to know the wattage flowing through that resistor. That's easy too. Wattage is voltage times current. So, 2 x 0.02 = 0.04 (or 40 mW). In this case, 40 mW is quite low. You can use a 1/8 watt resistor and it should handle the load easily. A 1/8 watt resistor handles 0.125 watts (125 mW).

Now, with the known average forward voltage of all the LED's intended to be used you can figure out how much resistance to use and what size resistor. I know you don't want to put 10 LED's with 10 resistors, I get that. But here's the reason why you need to follow that advice: If (or when) one LED burns out the rest of the LED's will be getting too much current. AND LED's don't like to share equally. They're greedy. One LED will take most (or all) of the current. If you provided enough current for ALL the LED's and only one takes all the current then it's going to burn out and the next most greedy LED will then start sucking up the all ready too high current and it will burn out. Like domino's each LED will die, one after the next. But if you put a single resistor on a single LED then you don't have that problem because each LED is limited by its own resistor.

You COULD go with 9 LED's and put 3 in series, but you'll need a higher voltage supply. Likely a 12 volt supply. Three LED's in series with one resistor and you get your parts count down. If you are pinned to the 10 LED idea then go with two LED's in series. But again, you're going to need a higher voltage source. On the earlier assumption that your super white LED's are 3 fV, two in series is going to eat up 6 V. In every case you're going to have to know what the fV is in order to select the right resistor. AND remember to calculate the wattage as well. Otherwise you will get very hot and smelly resistors. For a moment. But the smell will linger a while longer.

Here's where you need to engineer your circuit. YOU need to know what path you want to take. Five sets of 2 LED's (total 10) means you're going to have to get a larger supply. Especially if you go with 3 LED's in series. (three sets of 3 for a total of nine LED's)

Once YOU know what you want then we can help you achieve your desired results. And yes, it's always a good idea to include a schematic of what you want to build. That way YOU have a road map and WE can tell you if you're on the right path.

 

A schematic would SURE be helpful!

If you have 10 LEDs in parallel and a single 100 Ω resistor, then the resistor will be setting a total current through the LEDs of about (5 V - 2.8 V) / 100 Ω = 22 mA. That doesn't sounds like the current needed for 10 superbright LEDs. So instead of making use guess what your circuit it, please post a schematic (even just a picture of a hand drawn sketch on a napkin) so that we know what the circuit is you are working with.

The solution is to use one resistor to set the current for each group of LEDs, and also to get a better supply.