Making an eternal light bulb

Homemade cars 218 490 25

Anyway, let's get started :)
In order to start fleshing out this topic we will need:

- “Super Moment” glue or any other second glue.
- Light-emitting diode
- light bulb housing
- resistor*

* the resistance of the resistor for different LEDs is needed differently, I have seen it so far LEDs at 3.5 and 1.5 volts or with an integrated 12 V resistor.
Here is the required resistance:
1.5 V - 1.5-1.8 kOhm
3.5 V - 1.1-1.2 kOhm
----------------------

So let's get started:
1) To one leg LED We solder the resistor, and we try to solder it as close to the LED as possible so that the structure is as short as possible, otherwise it may not fit in the light bulb body. Next, we solder 3-4 “hairs” of thin copper wire 2 centimeters long to the second leg and insulate the resistor using a thermal tube.

2) We insert the resulting structure into the body of the light bulb, insert the resistor into the “navel”, and stick out the wiring soldered to the other leg LED, and glue it all to the body of the light bulb. After the glue has dried, solder it.

3) Result:

In general, of course, everything is clear here from the photographs, but still....here the most difficult part is to glue everything together correctly, the easiest and most effective way is to pour a little glue into the light bulb body and onto the bottom part LED, they will hold on iron-clad.

come back

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Comments (25)

#1 tyt Guests December 24, 2010 11:42
2

But the author did not describe that it is necessary to determine in advance where LED "+" and where is "-"
(the easiest way to achieve this is experimentally)
Yes, and I also think they come across more often LEDs at 3 and 4.5 volts... I personally have not seen 1.5!!
- Answer
1. #2 Edward Guests 27 August 2017 19:35
  3
  
  polarity LEDs can be determined by the length of the leads, the positive lead is LEDs longer. Can be even longer Light-emitting diode look, the short pin inside it is +. And the wide pin on which the crystal is located is “-”
  
  Answer
  1. #3 Edward Guests December 4, 2017 05:55
    1
    
    As a rule, craftsmen use BU LEDs, you can’t do without the poking method. And so, everything is very correct
    
    Answer
    1. #4 Guest Valery Guests 19 August 2019 15:44
      2
      
      Usually, at LEDs Two legs are visible inside the lens body. The big leg is "-", and the small leg is "+"
      
      Answer
2. #5 Victor Guests 11 February 2018 21:31
  1
  
  on the rim LED "-" has a flat.
  
  Answer
#6 vitaliy Guests 30 January 2011 14:45
0

in the photograph you can see that Light-emitting diode new!(Long leg +)
- Answer
#7 IvanGovnov Guests 30 January 2011 20:01
0

thing! They bought a mot from me because of these homemade products
they don't last forever
- Answer
#8 Nemecida Guests 29 October 2011 18:19
2

Eternal!!! If used correctly!!
- Answer
1. #9 Edward Guests 27 August 2017 19:42
  9
  
  If you use the LED correctly, then Light-emitting diode will become eternal. Cause of burnout LEDs for Chinese-made cars and motorcycles in their simple connection diagram, and a resistor with an underrated value, that is, when checking at the market or in a store, sellers show that LED lamps shine brightly, but hide what LEDs they operate at the maximum current for themselves, and when the generator on a car or motorcycle is running, the voltage in the network is already 14.5 volts instead of 12, as a result, LED lamps quickly burn out. In such lamps I integrate a 78L05 stabilizer chip in a TO-92 housing, with a stabilized voltage 5 volts and soldered to the output of the stabilizer through a 56 Ohm resistor Light-emitting diode.all this fits in the light bulb base
  
  Answer
#10 bodzik Guests 5 January 2012 17:30
3

If you look into the diode, there is a thin part +, and a thick part - .
- Answer
#11 Veent Guests 19 July 2012 21:48
6

pattern,
Edison turned over in his grave...
- Answer
#12 Dmitry Vladimirovich Guests 8 January 2013 13:20
8

Powering an LED through a resistor is almost like shooting sparrows from a cannon.
Let's do some math. Plastic white diodes consume 20 mA with a voltage drop of approximately 3.5 V, therefore, when powering one ice from 12 V, the power consumed by it is 0.02 x 3.5 = 0.07 W, and ballast resistors uselessly dissipate 0.02 x (12-3.5) =0.17W. Total power efficiency of this misunderstanding: 0.07/(0.07+0.17)*100%=29.17%.It is more profitable and optimal to use three 3mm ice in series with such a base.
Conclusion: if you are already using economical appliances, then take care of their economical power supply, otherwise, to put it mildly, the result is not serious.
- Answer
#13 Dmitry Vladimirovich Guests 8 January 2013 13:22
1

Veent,
Edison has raked up thousands of things like this, so he is no stranger to tossing and turning. :lol:
- Answer
#14 Kirill Guests 14 June 2013 19:44
1

The LED has two contacts, long this + short this -
- Answer
#15 Eugene Guests December 27, 2013 14:00
1

Quote: Dmitry Vladimirovich
Powering an LED through a resistor is almost like shooting sparrows from a cannon.
Let's do some math. Plastic white diodes consume 20 mA with a voltage drop of approximately 3.5 V, therefore, when powering one ice from 12 V, the power consumed by it is 0.02 x 3.5 = 0.07 W, and ballast resistors uselessly dissipate 0.02 x (12-3.5) =0.17W. Total power efficiency of this misunderstanding: 0.07/(0.07+0.17)*100%=29.17%. It is more profitable and optimal to use three 3mm ice in series with such a base.
Conclusion: if you are already using economical appliances, then take care of their economical power supply, otherwise, to put it mildly, the result is not serious.

You are not talking seriously, although this is out of ignorance. It has long been known that LEDs they must be powered through a current-limiting resistor, otherwise they will burn out in just a couple of minutes. Stick it in Light-emitting diode at 3.5V to a 3V source without a resistor and note its operating time)))
- Answer