Hi, I recently started making LED hula hoops. I have no electronic skills what so ever!! (i have been learning on the job)

I have been wiring up to 30 leds in parellel with a 3.6v battery my led supplier tells me the leds are 2.5-6v input and 20-40 ma. (not much help eh!)

Mostly I have been doing okay and had no problems. Untill someone asked me for a red, orange, yellow and WHITE! hoop.
(I have a face book page 'BIG ROUND HOOP' if you want to see pics.)

The design requested has 2w, 10 mix of r,o,y ,2w 10 mix r,o,y........the whites do not light.
I put the whites on a separate parallel circuit attached to the first and they light very dimly
On my bread board i put 2 r,2w,2o,2w,2y,2w etc and they all light with the white dimmer than should be......after half an hout=r the whites went out.

suggest more power needed.
BUT I can light 24 white leds brightly with 3.6v.....for how long I dont know


So my question is how the hell do you wire different voltage leds from 1 3.6v battery?

the red,orange and yellow leds are prob 1.9-2.1v and the white 3.2-3.4v.

Is it possible?

 

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calculate the current limiting resistor you will need for each LED. Then make a parallel circuit as follows. Assuming both LEDs are 20 mA, then
=> white LED needs a resistor that is (3.6V-3.4V)/0.02A = 10 ohms.
=> second LED needs (3.6V-2.1V)/0.02A = 75 ohms

+ +
| |
R1 R2
| |
D1 D2
| |
- -

 

I have a flashlight made of white LEDs and no current limiting resistor. Upon examination, I find that the batteries in this flashlight (torch to British people) have enough internal resistance to limit the current. If I put a red LED in among the white LEDs, its voltage requirement would be so much lower that the white LEDs would not light until the red LED smoked, and it will smoke.

 

Did you really mean 2W LED's? Those are NOT 20-40mA..
With a 2.5Vf thats 2/2.5 = .8 Amps
With a 6V vf thats 2/6 = .33 Amps

multiple 2W LED's in a battery powered hula-hoop.. not gonna happen. The battery required to have any lifespan out of it will be much too large.

-"wait whats that flash of light"..
-"well that was the LED's turning on and then the battery going dead almost instantly" -"well shoot"

 

Thank you for your replies
Sorry for posting in the wrong place this is my first time on a forum.
GopherT, I will give that a go. I have 22ohm and 82 ohm resistors will try them and order in the ones you recommend.
MCGYRV-- 2 White, not watts sorry was being lazy with my typing!
the batteries usually last 2-4 hours depending what LEDs they have. They are not ordinary LEDs I use strobe and ribbon and some rgb not sure if that makes any difference? They're just very pretty!!

when i have figured out how to make the hoops rechargeable (plug in) It will be easier to have 2 circuits but at the moment the size constraints of the hoop and my lack of knowledge are holding development back!

Thanks again

 

Tried the resistor thing, at first worked really well with white bright! but they went dull in 15mins :'(
Think it's going to have to be 2 circuits but to get 2 batteries in my hoop they will have to be recharge able as they will need to stay in place with 1 either side of the hoop for balance......more head aches!

If I do rechargeable what will i need? will i need something to stop batteries over charging? pcb?

 

The extremely low resistance value is WRONG because your LEDs are not 3.4V, they have a range of voltage and they might be only 3.2V. Then if your supply voltage is exactly 3.6V, the current limited by a 10 ohm resistor is (3.6V - 3.2V)/10= 40mA. With 22 ohms the current is 18mA for each LED.
But your LEDs might be 3.6V then they will not light or might barely light.
Different colour LEDs have different voltages and use different value resistors.
Each LED needs its own current-limiting resistor.

If each LED operates at 18mA then 12 LEDs draw 218mA.
You will have two AA cells and 12 LEDs/resistors on one side of the hoop and another two AA cells and 12 more LEDs/resistors on the other side.
AA Ni-MH rechargeable cells are 2300mAh from Energizer. They will last for 2300/218= 10.5 hours but the LEDs will be pretty dim.

You need a Ni-MH battery charger. Energizer sells them complete with 2 or 4 AA Ni-MH cells.

 

yes its MUCH easier for you to simply allow replacement of the batteries vs building an on-board charger w/fixed (non-removable batteries).

and here.. instead of little snippets of how to do it read/understand this.. It should help you out quite a bit..then google "ohms wheel" and try to understand it.

How to determine the resistor for a LED

Resistance = (Supply voltage - LED forward voltage) / current
where current is in ohms..

Then wattage of the resistor is
W = (current^2 * resistance) * 2
The *2 is for safety factor.. Some specs require *3

So lets show an example
Supply voltage = 3.6V
LED forward voltage = 2.1V
LED current is .020 Amps

Then
R = (3.6 - 2.1)/.02
So R = 75 ohm

Now solve for the wattage that resistor needs to be
W = (.02^2 * 75) * 2
So W = .06W (so a 1/16W or greater) wattage resistor can be used

We can "back-check" to see if its all correct by doing
(3.6-2.1) / 75 = .02 Amps.. voila..perfect

Now... what could be a problem with using a constant voltage setup like this for LED's.
Its really not a problem with small LED's but if you plug in values for larger/more current demanding LED's the numbers can be a serious problem..enter "constant current" supplies (another chapter).
Lets say we "thought" our LED had a forward votlage of 2.1V but its actually 2.5
So now using the "back-check" formula we get
(3.6-2.5) / 75ohms = .014 Amps.. ok so the LED will just be a little dimmer than we thought..no problem

But what if the Vf is lower..lets use 1.6V
(3.6-1.6) / 75ohms = .026 Amps .. ok so the LED will just be a bit brighter and not last as long
again not a real problem with small LED's but could mean the difference between a good LED circuit and a dead LED with higher power LED's

Now what If out LED forward voltage is 3.8V..(use the first formula)
R= (3.6-3.8)/.02
R= -10 well we can't have a negative resistance so the LED will not light at all as there isn't enough supply voltage to overcome the forward voltage so no "work" can be done.