Introduction: Making a 120V LED C7 Bulb

Picture of Making a 120V LED C7 Bulb

I wanted a blue LED light bulb for our water dispenser. So I decided to make one.

You need...

3 blue or white LEDs
1 100K 1/2W resistor
1 1N4004 Diode
1 dead (or not) C7 socket light bulb
3 inch of 1/2 " shrink tube

some hot glue
Soldering Iron
Paper tube (paper towel or toilet paper tube)

Step 1: The Circuit

Picture of The Circuit

The circuit is composed of a 1N4004 diode, a 100K 1/2W resistor and 3 blue LEDs

Step 2: Brake a C7 Bulb But Don't Brake It!

Picture of Brake a C7 Bulb But Don't Brake It!

I used a standard C7 type bulb because this is what fits inside the water dispenser.

This will be our base to attach the circuit

You need to brake the bulb, but, you must NOT brake the inside. To do this I inserted the bulb inside a used paper towel tube. and then squeezed the bulb using a vise. When the bulb is broken, it is possible to remove and brake away the rest of the glass, leaving the inside untouched.

Once you have the inside you can remove the tungsten filament, keeping the two leads.

Step 3: Preparing the LEDs

Picture of Preparing the LEDs

In order to make the 3 LED in a perfect circle, I used a pen and taped them up together in a circle.
Make sure you have the long and short leads all in the same direction.

Solder the first long lead to the next short lead, and so on. Leaving you at the end with one long and one short lead not soldered. These two leads will be the negative and positive of the LEDs series.

Step 4: 3 Blue LED in Series

Picture of 3 Blue LED in Series

This is the LED element

Step 5: Solder the Resistor and the Diode to the Two Bulb Wires

Picture of Solder the Resistor and the Diode to the Two Bulb Wires

Solder the resistor and the Diode to the two bulb wires.

One wire has the 100K 1/2W resistor.
The other wire has the 1N4001 diode.

Take note of the diode polarity, the output of the diode will have to go to the short lead of the LED module made before.

The resistor will go on the long lead of the LED module.

Step 6: Tape the Components

Picture of Tape the Components

Tape the components to keep them in place and to isolate the 120V from each other and the rest of the components.

Step 7: Install the LED Module

Picture of Install the LED Module

install the LED module.

Slide it on and solder the diode and the resistor to the long and short leads.

The Diode + should be on the short lead of the LED.
The resistor should be on the long lead of the LED.

Step 8: Add the Head Shrink and Glue

Picture of Add the Head Shrink and Glue

Add the heat shrink tube over the circuit,

leave a bit other the metal socket and shrink it a bit so it make a seal. DO not shrink the hole tube.
Only over the socket.

Fill the tube with hot glue. let it cool down in a refrigerator for 15 minutes. This will make sure the glue is cold. This will give you a chance to heat the shrink tube at the and without melting the glue too much.

Step 9: This Is It

Picture of This Is It

Final product. You can bend the LED to the side a bit then add more hot glue to isolate

Screw this in the socket and there you have it a C7 120V blue LED light.

And yes you can add more LED in series and parallel
You will have to recalculate the resistor

You might think this is a long process for nothing. You are right! you can actually buy C7 120V LED bulb already made... But mine was made with stuff I already had around and it was fun to make...

Step 10: View From the Inside the Dispenser

Picture of View From the Inside the Dispenser

This is the installation. The new LED bulb fits right in where the old lamp was.

Step 11: Second Shop Without the Flash

Picture of Second Shop Without the Flash

you can see the LED is directed to the tab you push to get the water... It's a discrete light that will not light up half the kitchen like the old one...


traisjames (author)2015-01-06

Doesn't the resistor get really hot having to drop the voltage so much?

jeanrenaud (author)traisjames2015-01-11

Blue LEDs have a forward drop voltage of about 3.3 volts. Since we have 3 in series, plus a 0.7 Volt drop on the diode, we have 10.7 volts drop on all of them. 109.3V remains, to be lost in the resistor.

The total power in the resistor will be : V²/R = (109.3)² / 100000 = 0.119 W. Since the resistor can take 0.5 W, the heat should not be a problem, but nothing stops you from using a bigger resistor if you want.

Scurge (author)2008-08-06

This may be a dumb question, but I'm an amature when it comes to circuitry. You are talking about running this bulb on 120v AC 15-20A, like out of a standard U.S. wall socket right? I checked on what a C7 bulb is for and it was saying AC and a C7 is one of the big old Christmas lights. Sorry if this is a dumb question but i thought you had to have a bridge rectifier and a couple of caps to make LEDs work with AC after power correction (or b4). Like I said, I'm new to circuitry, so bear with me please.

TheWelfareWarrior (author)Scurge2008-08-16

The voltage is the potential and the amps is the actual flow of electrons. By putting the resistor in you cut down the amps, and because it is a light emiting DIODE, if you put reverse voltage to it (alternating) it will block it (for the most part) and not light up. What you are actually seeing is it flashing at 60 times a second. Otherwise, yes, you would need a rectifier and caps. Sorry for the spelling, my "check spelling" button is having problems.

Scurge (author)TheWelfareWarrior2008-08-16

ok, so if i wanted to brighten them up a bit, i would have to use a cap to smooth out the signal right? what farad value should i use? (for that matter how does one choose a cap value for any application?) i've got 3 super-bright white leds in series and they're not that bright at all. not to mention that they are pretty directional and i end up with 3 circles on the lampshade. i was gonna use some lighting gel diffusers to try and spread the light out, but it still looks dimmer than they should be.

TheWelfareWarrior (author)Scurge2008-08-17

No, you would use a lower resistance. Be careful if you go too low you will burn out your LEDs sooner than they should. The caps would be to filter out AC noise from the line. You could try putting the LEDs in parellel instead of series.

blargg (author)TheWelfareWarrior2013-09-26

If you put the LEDs in parallel one will tend to draw more current and dim the others. Parallel ensures that they all get the same current. Just lower the resistor if you want more current.

blargg (author)Scurge2013-09-26

You don't need a rectifier to make them brighter, just more current. In another post I calculated under 2mA driving the LEDs, so there is a lot of room for higher current and thus brightness. They're being pulsed so you can drive them higher than their DC current rating.

Lynxspring (author)Scurge2008-08-17

Why do you what to use a cap for? I have no caps and the light is not flashing. LED are diodes, BUT they don't have a very high reverse voltage, this is why the diode is there on the AC. The resistor is for limiting the current. Each blue LED drops about 3V. So the rest of the voltage will be at the resistor and diode...The circuit works, it does not heat, and there is no flicker. This is a K.I.S. circuit... There is no reverse current to the LED because of the diode 1N4001 and the current is fairly low and not stressing the LED. This is going to last a long time. Plus it's so compact, I was able to make it fit within the size of the original bulb.

Kind of an old comment, but If you are only using one diode its a half wave rectifier and therefore only flashes at 30 times a second. If you were to use a bridge rectifier, then it would be 60 times a seconds or hertz.

blargg (author)geeklord2013-09-26

Actually it's 60 flashes a second with the half-wave rectifier; a 60 Hz sine wave has 60 positive peaks and 60 negative peaks per second.

blargg (author)TheWelfareWarrior2013-09-26

You can double the flashing rate to 120Hz without any rectifier or caps (see my other posts).

Lynxspring (author)Scurge2008-08-06

It will work on any 120V circuit. I used a C7 (Christmas lights socket) because this is what's used into my fridge water dispenser. Originally, I left the thing running for a week on my desk before I actually made the final assembly. There was no significant heat. The final assembly has been lighting up my water dispenser for a month now. Don't think it's CSA or UL :')

jules15 (author)2013-04-07

what does the 1 diode do? and how is the led not overvolted??

blargg (author)jules152013-09-26

Oh, and regarding "overvolting", the resistor prevents that. When the LEDs have sufficient voltage running through them, they'll draw pretty much whatever current you feed them, so the resistor will drop the voltage to their forward voltages. Let's say the LED's forward voltage is 3.3V. Three in a row give 9.9V, plus the ~0.6V drop of the 1N4004 diode, so about 10.5V. The 120V (RMS) AC peaks at 170V, so the resistor will have 155V across it and deliver 155V/100K = ~1.6mA @ 3.3V to each of the LEDs during the peak of the cycle.

blargg (author)jules152013-09-26

The diode presumably has a higher reverse-breakdown voltage than the LEDs, so on the negative half of each cycle, it ensures that there's negligible reverse voltage on the LEDs. But a more elegant solution is to just put the LEDs in parallel in opposite directions (see my other post).

mguima (author)2010-03-20

in the darkness, red light is more comfortable to the eyes 

blargg (author)mguima2013-09-26

Blue is also apparently bad for humans to be around at night because it disrupts the sleep cycle.

n0ukf (author)mguima2012-02-22

Yes, Blue light is harsh on the eyes in the dark. Green would be better than blue, but Red is much better yet. Star Gazers use nothing but red to move around their telescope sites to preserve their night vision.

losergeek (author)2010-01-23

Well, it's not my circuit, but I can tell you that the 1N4004 is because LEDs typically don't handle the high reverse bias voltage that will be applied to them (60 volts) on the negative half of the AC power.  So only the positive part of the AC power will be applied to the LEDs.
Not sure what current the LEDs are rated for,

blargg (author)losergeek2013-09-26

Wouldn't it make more sense to use an even number of LEDs and wire them reversed in parallel, such that on each half of the cycle one pair of LEDs is running? Then neither is getting reverse voltage above a couple of volts, and no need for an extra diode (the LEDs themselves are diodes). This also raises the flickering frequency to 120 Hz, making it much less noticeable.

n0ukf (author)2012-02-22

"Brake a C7 bulb but don't brake it!"

Why would you need to brake the bulb? How fast is it moving?
Braking is how you get your car or bike to stop moving (unless you crash it to stop, too hard on the vehicle... and you).

Breaking is what you are doing to the bulb's envelope when you squeeze it in a vise.

grrrrsty (author)2009-06-02

Your circuit looks very simple. I wanted to try something like this for a painting a friend of mine did. I have a couple of questions. 1. Why did you include the 1N404 diode in the circuit? 2. Are the leds the 3.2v at 20mA variety?

bolsocheio (author)2009-05-02

You must use a bridge rectifier to give full bright to the leds and remove the stroboscopic effect.

Lynxspring (author)bolsocheio2009-05-03

I can assure you there it is not flashing at all... I know it should but it's not and it's the perfect brightness. It's much better than the original that use to light up half the kitchen... :-) there's always more than one way to do this... I tried to keep it as small as possible to fit in...

geeklord (author)2009-01-04

Wouldn't it make more sense to get one of those small Bridge rectifiers from radio shack, but put the resistor before it because they are only rated for about 50v?

qs (author)2008-08-19

If you are REALLY using the circuit on a 120v system, you should be aware that the 1N4001 diode is only rated for 50-volts! Each of the LEDs can only take 5-volts, so all 4 in series are only rated for 65-volts!

You may have been lucky that the 1N4001 you got has a higher tolerance, but others using your plans may not be so lucky.

I suggest you use at least a 1N4003 and/or another 1N4001 to block the reverse AC current!

See this for some calculations and ideas.

Lynxspring (author)qs2008-08-20

Actually you are right and a 1N4004 was used. The reason why I did not use a capacitor in this circuit is the size. The circuit must fit inside the door panel of the water dispenser. Also the reason I used an old bulb socket, was to NOT void the warranty of this new fridge. It would have been easy to make a circuit the is larger then just tap on the existing wire. I did not what to take a chance in case something else on the fridge goes wrong. You know how companies can be with warranties.

TheWelfareWarrior (author)2008-08-05

aww! no pictures of it in use. Ah well. Good job though, I made a similar thing with an old hand powered flashlight bulb that was broken, though it produced 6v ac, so I didn't worry too much about it, works great i used candle wax for an insulator.

New Picture of it on line now...

I will try to post a picture of the inside of the water dispenser... The blue light is hard to see...

About This Instructable



More by Lynxspring:Apple Ibook G3 battery repairPool Solar water heaterProtecting Multimeters and other instruments
Add instructable to: