This instructable is about How to make a LED Lamp with 121 LEDs with just 4 simple steps And the Light is very effective to lighten a room and only with a Battery of 6v - 9v.

Thing you"ll need -:
1. PCB with Holes
2. 121 LEDs
3. 9v battery
4. 330 ohms
5.wire
6.soldering iron
7.soldering wire

## Step 1:

Take led and connect them in series like shown in the picture.

## Step 2:

Connect each line of leds with next line of leds

## Step 3:

Connect the resistor with a led from first line and then connect to the battery

## Step 4:

After completion of the fixing leds connect the Battery second terminal to the led of last line,if the led do not lighten up then exchange the terminal with each oter and after checking this solder all the leds with each other.

plzz vote.
<p>&quot;Can you include a schematic diagram? I wasn't able to figure out your series/parallel arrangement from the photo.&quot;</p><p>&quot;Sorry but I cant give you the diagram because I have lost it.&quot;</p><p>My &quot;Name&quot; says it all!</p>
That's interesting, I've wanted to do something like that. Can you include a schematic diagram? I wasn't able to figure out your series/parallel arrangement from the photo.
<p>They Are All Parallel Arranged, and Only One Resistor Of 330ohm Is Hooked Up In The First LED.</p>
<p>Hi, I appreciate your efforts of building and coming up with the circuit, but i guess the circuit is not optimized. All battery powered devices have to be designed with maximum power efficiency.</p><p>Each white led's needs atleast 3.3 v DC and max of 20 mA current for best illumination. Thus when the matrix is powered from a 9v battery, the total power dissipated is around 23958mW or 23.958W. Further, the array draws a current of 2420 mA of current or 2.420 Amp. </p><p>Now there are two ways of optimizing the circuit. Either make a matrix of 2x60 +1 led matrix limited by 150 ohm and 330 ohm resistors respectively which dissipates 11.718Watts and draws 1.220 Amps of current, or make a 3x40 + 1 led matrix where limit the 1 led with a 330 ohm limiting resistor. This will make you compromise on brightness marginally, but is way efficient as the configuration draws only 820mA of current and dissipates 7.408 Watts.</p><p>Please feel free to contact me for any assistance.</p><p>Regards</p><p>Shiharan Choudhury </p>
No it can last upto 2-3 Hrs
how much does the battery last? <br>3/5 hours? <br> <br> <br>
No the battery is not getting hot and I have used a normal 330 ohm resistor.
Sorry but I cant give you the diagram because I have lost it.
So this is using just the one resistor? What's the power rating on it? <br> <br>Also, did you notice the battery getting hot?