Homemade DIY LED Bulb

Hi Todd,

Your resistors will come out to 1K 1 watt as configured in your schematic.

John

Thanks for the information and correction. I have gathered further research concerning over-voltage and made a modification to the project.

Would this schematic and project work/be safe? Thanks.

Hi Todd,

I am assuming that you are using 120 volts 60 Hz AC power for this LED light. If this is the case your 272 capacitor will have approximately 1 meg of capacitive reactance Xc. This would mean a current of only 0.1 mA. This will not be enough to make your LEDs visible. If they are garden variety indicator style LEDs I would expect something close to 20 mA to be necessary. If they are LEDs suitable for usable lighting the mA will be much higher. I suspect that the configuration of the LEDs in your latest schematic will work if supplied with the proper power but personally I liked the way you had them wired in the first schematic better. I can think of no reason to connect them in the center of the LED array. You can try it but you may find that a 274J 1200V capacitor or a 474J would get you closeer to the current needed for regular indicator LEDs, similar to those used in a night light. 

A word of caution!!!!! Your questions indicate to me that you may not have the experience necessary to mess with the 120 volt mains. The 120 volts in this circuit can; burn you, injure you, and even kill you or someone else. Proper techniques need to be followed to experiment with 120 volts and proper construction techniques must be followed to produce a device that is properly insulated and able to be in an environment where children or others may touch it.

Since I do not know you or your technical experience I just wanted to make sure you are safe.

John

Hi Todd,

I went back and read the questions on your link in the first posting. I understand that you are concerned with over voltage on the LEDs. Though I don't see much difference between 2 LEDs protecting 2 LEDs or having the LEDs paired as in the second schematic either configuration should work. Where did you get the value of 272J for the capacitor. I could not find any reference to this value in the original link. The author mentions substituting a capacitor for a series resistor but I did not see a value. In your last schematic if you use a 472 capacitor there will be approximately 100 volts RMS across the capacitor, 6 volts RMS across the LEDs and 20 volts RMS accross the resistors. If I get the chance this evening I will bread board this circuit and verify my predictions.

John

Mr. Wiltrout,

Thank you for the continued information and for your help. The 272J capacitor came from the inside of the CFL bulb that I dissected. Being a novice, there wasn't any real consideration in the way of calculations as to which of the several capacitors to use. This one just happened to look like it was in a shape that would ultimately fit rather well in the CFL enclosure that I would be recycling. There are also other capcitors I pulled from there:

222J 1200V

102J 1200V

333K 250V

473 250V (no letter after the 473)

There was also a 200V 10uF capacitor, but it was polarized and I didn't think this was suitable for an AC project without adding a diode. So in regards to the 272J capacitor, it was a shape that I felt was appropriate and it was rated for much more volts than would be used. That was the only reason for choosing that one.

In regards to your previous to last post, you are correct in your statements. I will be using good old standard American 120V 60hz AC electricity. The clear LEDs I'll be using are just standard 5mm LEDs I got in a multi-color pack from China. Nothing really special about them. You're also correct in that I have not done a project or really worked with 120V yet. However, I do fully understand the risks involved. Having worked with explosives and heavy weaponry in the Military, I can certainly appreciate the need for safety and safeties for the safety. Part of that safety being lots of research, planning, and seeking assistance from those, such as yourself, with more experience and knowledge. I also try to think through everything that "might" go wrong and plan according backups and contingencies.

So that's pretty much where I stand and my experience. The most I've dealt with is 12V and Arduino/Pi projects to date. I do really appreciate your concern. Given the world we live in today it's nice to run across someone like yourself who's concerned for another person. So in that regard, thank you.

Hi Todd,

I took the time to verify my back of envelope calculations and took some pictures. This first picture is of the circuit using a 272 or .0027 uF capacitor. This is about 0.1 mA and as you can see the LEDs are barely lit.

This next picture is of the same circuit using a 224  0.22uF capacitor. This translates to about 10 mA and is probably the safest level.

This last picture is of your circuit using the 474 or 0.47 uF capacitor which translates to about 20 mA of current.

This last circuit with its 20 mA current is causing the 1K resistor to have about 20 volts across it. This translates to about 0.4 watts of power that needs to be dissipated by the resistor. In this case it is a 1/2 watt resistor. For practical use this is too close and I would probably go with 1 watt just like you have shown in your circuit diagram. I like the way you have decided to use the salvage parts from the CFL bulb. I salvage all my failed CFLs. The 333 or .033 uF cap that you have will probably give you at least some light but you may want to get a value between 0.2 uF 224 and 0.3 uF 334 for good results.

John

Great work John

This last circuit with its 20 mA current is causing the 1K resistor to have about 20 volts across it

I would tend to add some more LED's and use the power that is otherwise being wasted.

Mark