5 Watt LED Bulb





Introduction: 5 Watt LED Bulb

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.

Dear Friends , this Led Bulb has light output equal to 12 watt CFL. Saving of electricity is approx. 80 Units which saves INR 500 / $ 8 per year on basis of 24 hours usage.

Step 1: Components & Their Purpose

1) Metallized Polyster Capacitor

1 uf/250 Volt(105K) ,
.1uf/250 Volt(104K)
Together they are capable of providing approx. 85mA current to circuit.
These are used in parallel for acquiring 5 Watt output.

2) Noise Suppression Capacitor

103 M or 103 K X1 ,Y2 (.01uf)
Voltage rating should be more than 250 Volt.Here 2 KV rating is used.
The purpose of using this component is to lowering down the High frequency signals generated by the circuit itself and eliminates the buzzing sound in metallized polyster capacitor.

3) Negative Temperature Coefficient (NTC) Surge protection Thermistor.

10D-9 Ohm Ntc is used as ICL(inrush current limiting).
It is energy efficient in comparison of bleeder resistance.
It's resistance decrease with increase in temperature.So in operation it initially provides some resistance and after that negligible resistance is offered by NTC.

4) Electrolytic Capacitor

Alternatively use 100uf/100 V capacitor.
Due to local availability, two 470 uf/63 V capacitor are conneced in series ,so that their voltage rating becomes doubles and filteration capacity just halfed according to series rule of capacitor.
This series combination is equivalent to 235 uf/126 V.

5) Discharging resistor

a.)47 kilo ohm / 1 Watt or 1/2 Watt resistance is responsible for quickly absorption of charge present in the electrolytic capacitor when the switch is turned off.
Discharging Cap Value is choosen from R-C time constant formula.

b.) 1 Mega Ohm / 1/4 watt is used for metallized polyster cap.

6) Series Resistance

1 Ohm/2 Watt
For avoiding the Short Circuiting of LED's when the surge is bypassed to load , this 2 watt resistance drops surge across it before going to the load.So it ensures safety of load in such type of conditions.

7) 27 x 1 watt Led

Use adhesive on the back side of led and place directly on aluminum sheet of around 24 gauge in thickness .

8) General purpose diode

1N4007 is the most common diode(1000Volt/1 Amp) which forms full wave bridge rectifier and provides 80% efficiency in compare of half bridge rectifier.

Step 2: Circuit and Heatsink Building

1.)Heatsink Building

We Require 2mm MDF or higher for making heatsink.
For placing 27 led's we require a suitable heatsink.
With 2.5 inch Hole cutter ,cut four circles and two aluminium sheet
or alternatively use scissors for cutting MDF board and aluminum sheet.
Join them as shown in picture i.e layer by layer.
After that take LED's and bend their anode and cathode terminal to atmost 90° for avoid shorting with heatsink and place them such a way that they form a circle and also their positive and negative terminal comes around the hole of heatsink. Apply solder paste and solder all the led in series fashion if they are far from each other then place the cutting pins of components for making connection. Avoid any solder contact with heatsink.

2.) Circuit Building

Follow the circuit and picture carefully for reference .
In this circuit a jumper wire is placed between Capacitor neagtive terminal and bridge rectifier for simplicity.

Don't touch circuit in ON Condition ,this gives you a lethal shock.

Step 3: Testing and Measurements

Individual Led measurement

1. Current- 57mA
2. Voltage- 2.9 VDc

Individual Wattage
= Voltage x Current
= 2.9 x .057
= 0.16587Watt
Total Wattage
= total no. of led x wattage of individual led
= 27 x 0.16587 watt
= 4.48 Watt

Power monitor readings at 248 Volt A.C

=5.11 watt / 0.250 Power factor

So power lossed is
= 5.11 Watt - 4.48 Watt
=0.63 Watt

This loss is due to

1.) Poor Power factor of unregulated capacitive power supply.
2.) High ESR of Electrolytic capacitor.
3.) Stray Capacitance of Solder Tracks.

Scope of Improvement

1.) If S.M.P.S is used it will deliver good power factor i.e low wattage loss.(Buck,Buck-Boost,Flyback topologies)

2.) Attain a large potential drop from these supplies which is equal to rectified volatge i.e 300 Volt or higher for power factor greater than 0.50 which is dangerous and not possible for bulb enclosure.

Challenges with Switch mode power supply

1.) High Component Count
2.) High Complexity
3.) Smps transformer is not readily available.



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    29 Discussions


    1 year ago

    This one is even better than last one. keep it up bro.

    I want to replace a 40w florescent tube light in my kitchen with more or same amount of light as the tube rod. kindly enlighten my way.

    1 reply

    Bro, 40 watt tubelight is easily replaced by 18 to 24 watt led tube,for same amount of light output(1800 lumen) a 22 watt led tube is sufficient. Use 4 feet,3mm in thickness aluminium strip and place the led and heatsink with the help of adhesive. In driver, you can use inductor from damaged cfl circuit.Please refer to the diagram. Thanks


    thanks for idea ...i have too many 1watt LEDs in my stock ...

    If my power source is 127V AC, what I need to change?

    1 reply

    Then,you have to change the value of 1 uf metallized polyester capacitor to 2.2 uf.

    Nice project!
    You can use 5360 SMD LED they are cheaper and emit much lumens than the LED you used.

    5 replies

    Thanks buddy, smd led often requires metal core pcb for better thermal management and of course they have more lumens/watt and also these are originally made for automatic pick n place machines.

    That's right!

    But I got an Idea of using two sided copper boards, on the one side we can mount the led and the metal plate on the other side can be used for heat control.

    Nice Idea!
    This will be great.
    It will require pcb etching and from both side it will be etched until the bottom side is coated with marker or other substance.
    Waiting for your execution.

    That's right!

    But I got an Idea of using two sided copper boards, on the one side we can mount the led and the metal plate on the other side can be used for heat control.

    24 gauge aluminum is very thin, severely limiting its ability to spread heat flow to the sides. Since you are not operating the 1W LEDs at their rated current (or even close to it), there is some tolerance for heat buildup in the LED chip. Also, thermally conductive self-adhesive backings are not very good at transferring heat compared to other methods (as with many things, compromises made for convenience). So what I see here is a design where the temperature conditions within the LED package are potentially limiting their useful lifetime - so very little surface area in the heat sink (although it won't help to extend the Al sheet unless it is made much thicker or replaced with a real heat sink design). If this bulb were to be completely enclosed from outside airflow, the heat situation becomes dramatically worse (including heat losses from the power supply).

    Note that the voltage drop of the LEDs is specified at a certain operating (chip) temperature and current if you go by the manufacturers data sheet.

    It seems very common for LED light experimenters to make assumptions about heat management for the LEDs, usually being too optimistic about how much heat is being drawn from the LED chip inside its package, or how well the bulb will work out in the long term (because it's been working for them for a while without failing yet), or if people put them in different environments. A good thermocouple, certain specifications from the manufacturer, and some simple math will confirm the actual LED chip temperature under various operating conditions. I suspect many people would be surprised at the results of doing this, and to pay much more attention to how they are removing heat from their LEDs in the future.

    1 reply

    Dear pal, your explanation is true.
    Here 24 gauge aluminium is used for spreading heat to it's bottom layer and four 2.3 mm MDF are used for absorbing that heat and that heat is transfered via air through Six 6 mm holes from bottom side of bulb housing.
    This is a cheap yet efficient way for thermal management. Bulb has been tested for a long run of 24 hour usage for 5 days,no problem was detected with it's light output.
    Led's are driven below 20% of it's full capacity for ensuring durability. If we want to enhance light output from this bulb, we should use several 24 gauge aluminum sheets or can use 3mm aluminium.
    If the bulb is to be completely enclosed for a given output or higher output then we choose another design which uses thermal conductive paste, around 18 gauge aluminum sheet and for absorbing heat, bulb housing should be made from thermocouple material.

    These Led's were bought from local market and due to unknown company, datasheet couldn't be found which clears more about variation of forward voltage with thermal resistance of led. Although reference datasheet has been taken.

    What will happen with 470 uf/63 V capacitors if LED chain is burned out?

    1 reply

    If Led Chain is burned out,the capacitor finds the shortcircuit path which dramatically increases the ESR of cap at very high level causing swollen capacitor's vent which then turned into non electrical conduction and capacitor act likes an open circuit.
    But this situation won't come until the led's doesn't able to take off it's generated heat or the heatsink is improperly designed.

    dont u think this transformer less supply cause more losses or can say loss the current not in use.?

    1 reply

    Kavish, by using capacitive power supply ,circuit generates more reactive power (VAR) in comparison of smps and power factor can't be acheive above 0.6 . So it make power losses, harmonic distortion, and the current is not fully utilized by load which is then returned to neutral line.

    looks great. But these lamps are for sale for just a few dollars.
    nevertheless, well done

    Looks great. I am looking for a transformer less power supply for TBA810 7W audio amplifier. Is it Possible?