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led lightbulb
by：Demo
2020-09-10
This is just a lark that itches in the back of my head for a while.
This is an LED bulb powered by power.
This project does not need much.
Incandescent bulb. A .
47 microfarad 200 V capacitor.
1/4 W 1 kg ohm resistor for pairing of LEDs and other items.
You need to start with cleaning up the bulb and there are a lot of guidance on this step and I will give up here.
The circuit consists of two reverse connected LED, I ground the LED to a place not far from the chip and stick them together to make a single double LED.
Twist the legs of the LED together and weld the capacitor on one side and resistance on the other. Simple.
Here, I fixed the circuit to the socket with my clothes, which of course is the recommended test procedure. ; -)
Stick the circuit to the bulb, fix it in place with some hot glue, be sure to have some cold water and you will burn your fingers.
Try putting the LED in the center of the bulb.
Once the circuit is in place, bend one of the leads on the base and fix it with an aluminum strip.
The second lead is twisted around the brass screw inserted into the hot glue.
Look at the shorts. you should be able to go.
First of all, this project needs a non-polarized capacitor, muy importante, to solve the supervision problem.
Now, how does this work?
We all know that in order to run the LED from a power supply above the rated voltage, we limit the current with a resistor.
In fact, in this case, we can limit the current with a resistor of about value. 6.
8 k Ohms, but a few watts of resistance is required! ! !
Not a good thing.
Since we are using an AC power supply, we can take advantage of the properties of the capacitor that is subjected to AC action, called resistance.
We can equate resistance with resistance.
It is a simple formula to calculate the reactor = 1 /(2*Pi*Freq*C)
Solving this problem for C will provide us with the size capacitor needed to limit the LED current.
So, why do we have resistors?
When the power supply is on, there is a current peak value, and the 1 k ohm resistance can limit the current peak value of the peak value.
Finally, why two LED?
The LED is a diode, because we are dealing with the AC here, so we need to connect the two LEDs so that the waveform can complete its cycle.
Basically, each LED flashes at 60 hz, but the phase is the opposite.
This is an LED bulb powered by power.
This project does not need much.
Incandescent bulb. A .
47 microfarad 200 V capacitor.
1/4 W 1 kg ohm resistor for pairing of LEDs and other items.
You need to start with cleaning up the bulb and there are a lot of guidance on this step and I will give up here.
The circuit consists of two reverse connected LED, I ground the LED to a place not far from the chip and stick them together to make a single double LED.
Twist the legs of the LED together and weld the capacitor on one side and resistance on the other. Simple.
Here, I fixed the circuit to the socket with my clothes, which of course is the recommended test procedure. ; -)
Stick the circuit to the bulb, fix it in place with some hot glue, be sure to have some cold water and you will burn your fingers.
Try putting the LED in the center of the bulb.
Once the circuit is in place, bend one of the leads on the base and fix it with an aluminum strip.
The second lead is twisted around the brass screw inserted into the hot glue.
Look at the shorts. you should be able to go.
First of all, this project needs a non-polarized capacitor, muy importante, to solve the supervision problem.
Now, how does this work?
We all know that in order to run the LED from a power supply above the rated voltage, we limit the current with a resistor.
In fact, in this case, we can limit the current with a resistor of about value. 6.
8 k Ohms, but a few watts of resistance is required! ! !
Not a good thing.
Since we are using an AC power supply, we can take advantage of the properties of the capacitor that is subjected to AC action, called resistance.
We can equate resistance with resistance.
It is a simple formula to calculate the reactor = 1 /(2*Pi*Freq*C)
Solving this problem for C will provide us with the size capacitor needed to limit the LED current.
So, why do we have resistors?
When the power supply is on, there is a current peak value, and the 1 k ohm resistance can limit the current peak value of the peak value.
Finally, why two LED?
The LED is a diode, because we are dealing with the AC here, so we need to connect the two LEDs so that the waveform can complete its cycle.
Basically, each LED flashes at 60 hz, but the phase is the opposite.
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