You are supposed to use 100 ohms, not 91 ohms.
The spec's for the LEDs do not show the max allowed current like all LED datasheets show. you don't even have a manufacturer's name nor a datasheet. The spec sheet says that the current (Which current? The typical current probably.) is 20mA. If the minimum voltage for the LEDs is 3.2V, the 12V supply is 5% high at 12.6V and the 91 ohm resistor is 5% low at 86.5 ohms then the current in 3 series LEDs is 35 mA which will probably overheat them and might burn them out.

 

Ok
First of all !
Thanks Guys !!!!

2nd
I make a new diagram
with 100ohms 1/4 W
In it you can see that I prefer
50, Now, Another question...

I this Diagram which resistor I need
for the single led string ?

I'm buying the 9V 1A power supply as some of you guys suggested.

Thanks again !!

 

9V won't be enough voltage for putting them in strings of three. You could put them in strings of 2 for that power supply. Then you'd have 25 strings of 2, all the same.

If you go for 20 mA for the LEDs, then that is 25*20 mA == 500 mA total.
500 mA at 9V is 4.5W.

That should work.

If you assume 3.3V for the Vf, and assume that the 9V power supply really does put out 9V, then you need 120 Ohms of resistance in each string for the current limiting.

The total power consumed by LEDs is 50 * 3.3V * 20mA == 3.3W.
The total power consumed by resistors is 25 * (20 mA * 120 Ohms) * 20 mA == 1.2 W.

Therefore, 27% of your power is burning in resistors, causing heat. The other 73% is used in the LEDs, most of it becoming light (and some more heat).

 

9V won't be enough voltage for putting them in strings of three. You could put them in strings of 2 for that power supply. Then you'd have 25 strings of 2, all the same.

If you go for 20 mA for the LEDs, then that is 25*20 mA == 500 mA total.
500 mA at 9V is 4.5W.

That should work.

If you assume 3.3V for the Vf, and assume that the 9V power supply really does put out 9V, then you need 120 Ohms of resistance in each string for the current limiting.

The total power consumed by LEDs is 50 * 3.3V * 20mA == 3.3W.
The total power consumed by resistors is 25 * (20 mA * 120 Ohms) * 20 mA == 1.2 W.

Therefore, 27% of your power is burning in resistors, causing heat. The other 73% is used in the LEDs, most of it becoming light (and some more heat).

But The Diagram indicates 12V 1000mA
As my power source !

 

I'm buying the 9V 1A power supply as some of you guys suggested.

I guess that was a typo ....

 

The voltage of each LED keeps changing and the power supply voltage keeps changing.
I quit!

 

Thanks to all for your patience !!!
It was typo !!
This last diagram represents
the array I like more.
50 LED's.png

Can someone revise it ?
And tell me if it's good ?
If so, which resistor for the
One led string ?

THANKS A LOT Audioguru and everyone who contribute in any way !!

 

Thanks to all for your patience !!!
It was typo !!
This last diagram represents
the array I like more.
50 LED's.png

Can someone revise it ?
And tell me if it's good ?
If so, which resistor for the
One led string ?

THANKS A LOT Audioguru and everyone who contribute in any way !!

Okay, then... why not put the last 2 LEDs on the *same* string and the you have Vf of what? 3.3V?

So for the two LEDs on one string, you aim for total Vf of 6.6V.

The difference from 12V is 5.4V.

5.4V at 20 mA gives a resistance of 270 Ohms for the final string with 2 LEDs.

 

Okay, then... why not put the last 2 LEDs on the *same* string and the you have Vf of what? 3.3V?

So for the two LEDs on one string, you aim for total Vf of 6.6V.

The difference from 12V is 5.4V.

5.4V at 20 mA gives a resistance of 270 Ohms for the final string with 2 LEDs.

Thanks toy you too, sage.radachowsky
I make it this way because I'm a noob at this matter
and it's easy for me to make the array since there is
two separate lamps, and I don't know how to put the array together
The way you recommend. (

P.S. I plan to make a cable that connect the two lamps together with a
headphone cable/Jack, or something like this.

 

You have a working schematic on Post #20. The difference between 100Ω and 91Ω is very small, but you have what you need. All you need to do is build it.

 

Ok guys !!!
Thanks a lot
I will bring you photos of the
system completed !!!
Again !
THANKS A LOT, to all of you guys !!