240 Volts 100 Hertz 29 LED lamp

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If everything is done right an LED can beat a CFL in lumens per watt. This is a rare case however. LED's will get a better lifetime. The quoted of hours of life of CFL are usually way over rated and assume you never turn off the CFL. Turn off - turn on's are where most of wear and tear occurs on CFL's.
 
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i was confused with the intensity of illumination of a led light bulb in that project because he doesnt tel the lumenous effect produced by that led..it must nice if RODALCO says for that.
 
The led lamp is not as bright as a CFL lamp. I mostly use these as nightlights in the house or in the porch and at our local fire station.

Considering about 11mA going through the leds versus 100 mA through a CFL it is obvious that a CFL lamp is a lot brighter.
 
thank you so much RODALCO...by the way i have another question for you..if i used 6 volt LED with 50pcs instead of 3 volts drop across per each LED,what value of resistors and capacitors may i use for it??.is just the same with the 3volts led?.if not,can you show me how to get those value??...the problem here is just i want to compensate the 20W incandescent bulb...
 
You might want to add a small smoothing capacitor to reduce visible flicker. Visible flicker at 100 Hz comes when you move what you are looking at, when the light leaves a trace with gaps in it on your persistence of vision. So if you can remove the gaps, it doesn't matter the capacitor is quite small so the LEDs dim quite a lot in between the mains peaks. The LEDs continue to give out some light even at tiny currents.
 
Calfox , I'm a bit confused here 6 Volt LED, Do you mean to run the leds of 6 Volts supply?

6 Volts case.

Most white and blue leds run at 3.0 to 3.6 volts, so to put two in series may not give you full brightness at 6 volts supply.
In your case with 50 LED's you require a series resistor for each LED. (e.g. 3.3 Volts Fw U drop per led)
and leds are connected in parrallel.
U=I*R therefore (6-3.3)= 20mA*R
R = 135 ohms. I would take 150 ohms in that case to keep led current below 20mA.

240 Volts case

50 leds * 3.3 = 165 Volts total drop across leds. (leds in series)
U=I*R therefore (240-165) = 20 mA*R
R = 3750 ohms I would go for 4 resistors (1watt) of 1 k.ohm in series.
current with 4 k.ohm is 19 mA for the leds.
from P=I²R therefore .019²*4000= 1.44 Watts so with a 4 watt rating you are within a good safety margin.
 
Diver300

A small smoothing capacitor may be added across the bridge rectifier if flicker is of concern.
To be honest I did not notice it at 100 Hz, check my YouTube video.

Bear in mind that a capacitor, depending upon its size, will increase the voltage across the leds by a value of sqrrt2 x U led string , and resistor values need to be recalculated.

That capacitor has to be rated for at least 400 Volts.
 
God's grace..

thank you so much REDALCO..I mean,i use it in 240 mains supply..i;m so thankful for your reply for further explanation...
 
"goodmorning RODALCO"

By the way i have some question for you for the assurance of this project.

"I want" to used a 50 LED's and i wanna let to divide in two section to be connected in each section in parallel form..first section is composed of 25 LED's and the second section is the same with the first..In this case,"what value of resistors i am going to connect in the 240 volts that can protect the LED's???...
It is because i just want to assure for my 50 LED's that can be operate safely...please..please help me "RODALCO"...

In my idea:
i would used a 4(2.2K-1 watt)in series with the bridge and by each section i would rather to series a 7K-ohm(1 watt) for the further protection of my LED's...
is this right??please guide me "RODALCO"..THAnKS ADVANCE FOR THE REPLY..God bless you always...
 
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It comes to my mind that it much nice to devide into two section in that way if one of led's to be dimming then it doesn't distinguish in all circuit..for the long reliability..if i am right.!..
 
Hi calfox by having 2 strings of 25 white leds in parrallel the resistor design has to be as follows.

25 x 3.3 = 82.5 Volts
240 V supply hence 157.5 V to be dissipated.
I led < 20 mA
157.5 / 0.02 = 7875 ohms (use 8k2) 157.5 / 8200 = 0.0192 A
I²R then 3 Watt is dissipated in the R's

Because you have two strings of 25 leds in parrallel the resistors have to rated for at least 6 Watts!
In this case you end up with a lot of heat in a small place so I would personally not take this option and use a single string of 50 leds in series.

My experience with leds is that they fail in a continuous conducting state and not emitting any light when faulty, so the defective led can be easily traced and replaced if need be.

Again with white leds run them at about 15 mA's or even less to ensure long term reliability.
 
rodalco---

why is it the 50pcs of led lights i was made is not bright than cfl??? it is dimmed in the eyes..i used a white LED's and even i used a reflector it's not clear to see things project by the light.why?????operated in 19mA..if i increase the current to 23mA i think i adds brightness right?
 
if i combined a yellow,green and blue led's it is possible to produced a light brighter than white LED's?
 
Calfox

Current type 5 mm Ø of LED's are not up to the job of house lighting, exept the perhaps expensive luxeon LED's.
As said in the beginning of this thread, these led lamps can be used as hall or porchlights to find your way around.
For reading lights use the normal incandescent lamps or CFL lamps, cheaper and more light output.

You can try different colours led's but doubt you get more light.

If you increase current to over 20 mA, you will shorten the life of the leds (more heat) and will not get much more noticeable light output.
 
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