12 Volt A.C LED Lamp




About: Hey Everybody, I am very thankful to this site owner who gives us a new platform to share the things we love to make.Thanks a lot.

LED's consumes very less power and also eco- friendly.They have great lifetime and also looks awesome.Go Green...!!!

For the original source of inspiration ,check this link https://www.instructables.com/id/AC-LED-Night-Lamp/

I have just designed it my own way and done some calculation about power consumption and every thing remains same about circuit.

Previously I believed that for Led to be "Turned On" ,conversion from A.C to D.C is necessary.

But after knowing this fact is not true,i have developed interest in this project.This LED Lamp is constructed under INR 150 (less than $2.50).

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Step 1: Components Required

1. Step Down Transformer- 12 volts, 500 mA [ INR 35, $ 0.58]

To Lower down the 240 Volt A.C to 12 Volt A.C. This is also known as 12-0-12 transformer,in the case you have 3 wires at output of transformer just pick up the 2 wires from either corner among 3 wires.

2. Light Emitting Diode- 40 Nos, 5 mm [ INR 40, $0.67 ]

30 white and 10 blue led's are required.Generally these works in between 15 mA to 30 mA and consumes voltage in between 2 V to 3.6 V

3. Single Pole Dual Throw Switch (SPDT) - One SPDT switch [ INR 20, $0.33 ]

Also known as TWO WAY Switch.This is used for switching between LED's

4. Perf Board [ INR 10, $0.17 ]

Simply to place the LED's

5. Switch Board[INR 25, $0.42 ]

To place the transformer and SPDT switch in it.

6. 2-Pin Plug and Wire [ INR 30, $0.50 ]

Scrap book paper for decoration purpose ,you can avoid them also.That's it...:D

Step 2: Circuit & Designing

The Circuit diagram is build around the positive and negative half cycles of alternating current.

The 12 volt output from transformer is connected to the middle terminal of SPDT switch while other two terminals are to be connected with the Blue and White Led's wire.

The step down Transformer is a 240 volts to 12 volts. AC.Blue and White LED's working voltage is 3.3 volts and current is about 20mA i.e 0.02 Ampere

Since Voltage from transformer output is RMS Voltage , so the effective voltage from the transformer is 1.4 times higher than the RMS Voltage.( According to the rule)

We divide the single LED forward voltage by the total voltage so as to find the number of Led's in an Array.

So, Total Effective voltage 12v X 1.4 = 16.8 divided by 3.3 = 5.090 so 5 LED's have taken per row.

and Power consumption = Voltage(in Volts) x Current (in Ampere) x Power Factor(70% to 85% in case of A.C)

For Blue Led's

Power Consumption= 16.8 x current in two arrays of blue led's x 0.7(Minimum) or 0.85(Maximum)

= 16.8 x 0.02 x 2 x 0.7 or 0.85

= 0.4704 Watt(Minimum) or 0.5712 Watt(Maximum)

For White Led's

Power Consumption= 16.8 x 0.02 x 4 x 0.7(Minimum) or 0.85(Maximum)

= 0.9408 Watt(Minimum) or 1.1424 Watt(Maximum)

Note- 1. These Led's are connected in series,so forward current remains same in their respective array.For total current of circuit, we multiply the current by number of array.

2. Power factor is assumed between 0.7 and 0.85

Step 3: Final View

Actually i forgot to took a snap while making it since i have no idea about the uploading.

For you reference , please see the picture next to it which is uploaded by Dipankar. Thanks

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    18 Discussions


    4 years ago

    Awesome project sir...thankx
    Some queries....
    1.I have to connect the leds to the 12 and 0v or to both 12v outlets???
    2. I have a 6V 200mA transformer and want to connect it to a single led.. what should i do
    Thank you

    1 reply
    Hemant Jainrshacker86

    Reply 4 years ago

    1.Connect leds to 12V and 0V only,omit another 12 V output or you can use single 0-12 volt transformer instead of 12-0-12.

    2.You need to drop voltage of transformer using resistance.So,
    Resistance= Source Voltage - Led Voltage divided by Current(in Amp) required by Led.
    so,for one white led..
    Source voltage=6 Volts
    Required Voltage=3 Volts
    Current Drawn by Led=20mA(.02Amp)
    Put these values,you will get resistance =150 Ohm
    Power Rating of Resistance
    P=V x I
    =3V x .02Amp
    =60 mW
    So 150 ohm/ quarter watt(1/4 watt or 250 mW) is sufficient.
    Put this resistance in series with one led.


    4 years ago

    Sir can u upload the videos of construction for this project and send it to my email pungathapa@gmail.com

    1 reply
    Hemant Jainpunga

    Reply 4 years ago

    Sorry, i didn't made video for this project.You can simply read the steps of making.


    4 years ago

    good work. check out mine.

    DIY LED-night lamp runs from AC mains - Check out this awesome Instructable.


    1 reply

    4 years ago on Introduction

    if u not use transformer then you use transformer less power supply

    short place & weight less i send u a circuit diagram soon

    1 reply
    Hemant Jaingemgroup

    Reply 4 years ago

    i know that method ,a metallized polyster capacitor say 105k(1uf) and output can be controlled via voltage regulator (7812) or with 12volt/1watt zener diode.

    Hemant Jaingemgroup

    Reply 4 years ago

    Okay,but the idea behind this project was to lit led's directly from stepped down A.C source.


    Hi, Hemant !

    Good ! Literally,you have explained very well this small project in simple steps. This is what DIY means for all whoever wants to do / build such items.

    1 reply

    5 years ago

    Or you use a bridge rectifier, a capacitor and resistors

    1 reply