# Running LEDs from an AC supply

Discussion in 'General Electronics Chat' started by aravindhoysala, Nov 24, 2008.

1. ### aravindhoysala Thread Starter Active Member

Aug 6, 2008
31
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hi all,
is there any circuit for Running LEDs from an AC supply(230v)?

Mar 24, 2008
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3. ### markm Member

Nov 11, 2008
16
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Bill, the circuit diagram in your article is good, but some of the other information is rather dated. White LEDs for illumination are no longer under development - I've got a customer that's building white LED fixtures as fast as they can put them together. The LED's run at 700mA, Vf = 3.5 to 3.9V, and produce about 80 lumen. Four will light a hotel bathroom, fifty replace a big street light.

They also generate a fair amount of waste heat. They're surface-mount, heatsinking to the board through solder joints, but unless you can spread them out inches apart you're probably going to need to bolt the board to a heatsink. Otherwise, just a few degrees too warm and you're taking years off the LED lifetime. I expect that a lot of lighting companies with little understanding of electronics are going to fail as they move into this area because they don't comprehend the cooling requirements.

Also, you get a lot better power efficiency if you use a switching regulator in a constant current configuration to run a series string of LED's with no series resistor except a fractional-ohm current sense resistor. There are IC's on the market designed specifically for this usage. These circuits cost considerably more than a limiting resistor, but the energy savings ought to pay for them in the long run. Go for as long a string of LEDs and as high a total series voltage as you can (the limit is either how many LEDs you need, the supply voltage if on batteries, or the max voltage of the ICs). E.g., for the 50 LED fixture, we've got 5 strings of 10 each, Vf = 35 to 39 on each string, 48VDC power, and 5 switching regulator IC's in a buck configuration. The switchers turn most of the energy difference between 48V and Vf into more current for the LED's, so nearly all the power actually reaches an LED.

I think white LED's are inherently less efficient than the single-colored ones. They use phosphors to change a single color to a range of colors, so you've got two rather lossy energy conversions (electricity to light and blue light to green/red) instead of just one. But lumens aren't just the emitted power, they're the emitted power times the human eye response. Most colored lights are rated only in candelas (power), so it's hard to compare the efficacy of white and colored light.

4. ### aravindhoysala Thread Starter Active Member

Aug 6, 2008
31
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thanks for the link Bill_Marsden, but its all about diode, i need schematic diagram because i want to use general purpose led's(i will use many to get enough light,in India high flux led are not available. )

5. ### Wendy Moderator

Mar 24, 2008
21,838
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Not my article, it is part of the online All About Circuits book. I have submitted an update however, on this thread. White LEDs are alive and well, and getting brighter, check out the Rebel series mention in the update, it is modern. If you go to the manufacturers website you can get complete specs, easy as pie.

6. ### Wendy Moderator

Mar 24, 2008
21,838
3,047
If you check there were schematics, and how to calculate for your application. Did you read the entire article on LEDs?

I'll help, but we'll need more info. What is the AC power standards of India in voltage and frequency. How many LEDs do you plan on stringing together? If you have the typical Vf on them this would also help.

Last edited: Nov 26, 2008
7. ### aravindhoysala Thread Starter Active Member

Aug 6, 2008
31
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its 240v 50Hz(some times it goes to 270v also,here voltage fluctuation very much)

i am trying to get high flux LED from abroad ,
if its not possible i have to go for locally available LED's which are usually 2to3.5V and up to 30ma max and i want to string 50 such LED's (which i will confirm within couple of days).

thanks for the link , i will go through it .

8. ### aravindhoysala Thread Starter Active Member

Aug 6, 2008
31
0
Hi Bill_Marsden,
sorry for the delay in replaying .
as it will take time to get high flux LED's ,i want to try general purpose LED's first.
the ratings are :3V 30ma (max)
and i want to use 50 such LED's OR suggest me how much i should use.
thank you

9. ### mindmapper Active Member

Aug 17, 2008
34
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First, full wave rectification. Help you get rid of high reverse voltage that othervise vill kill the LED's. After rectification you have a peak voltage of 240V * SQR2. We forget the voltagedrop of the rectifierdiods.

From that peak voltage you subtract the voltagedrop of the LED's wich is 50 * 3V. (50 LED's in series.)
Whats left is the voltage drop you will have in the resistor limiting the current through the LED's. Ohms law will give you the size of the resistor.

Using the peak voltage in the calculation will limit the current to be within the maximum limit (30mA).

Don't forget to calculate the power dissipated in the resistor to select a resistor with a proper rating!

Last edited: Dec 2, 2008

Aug 6, 2008
31
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what is SQR2

11. ### leftyretro Active Member

Nov 25, 2008
394
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Square root of 2 = 1.414, it is the value X line voltage that the caps will charge to, about 340vdc in your example.

Lefty

12. ### aravindhoysala Thread Starter Active Member

Aug 6, 2008
31
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doubt 1:what if input increased ?
doubt 2:is there any constant current circuit?
because in india voltage fluctuation is common thing and some time we are getting spike of up to 1800v hence it is better to build a constant current with spike protector

(I lost many LED's in th previous project before joining All About Circuits
sorry..! if i am asking silly doubts .

13. ### nomurphy AAC Fanatic!

Aug 8, 2005
567
13
You'll need a 600V (minimum), 1 amp, rectifier diode to protect the LEDs, a 6.8K 5W ballast resistor, and 50 LED's connected as follows:

250VAC HOT --> a-Diode-c --> 6.8K RES --> a-LED1-c --> a-LED2-c ...LED50 --> 250VAC NEUT

where diode/LED: a = Anode, c = Cathode

If you want to stay below 30mA, use a 10K, 5W resistor.

Last edited: Dec 2, 2008
14. ### aravindhoysala Thread Starter Active Member

Aug 6, 2008
31
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a single diode is enough ? or i need 4(for bridge),
can you give me diode NUMBER?

15. ### leftyretro Active Member

Nov 25, 2008
394
5
Use four, bridge, 1N4007 will work well

16. ### aravindhoysala Thread Starter Active Member

Aug 6, 2008
31
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is this safe enough?

can you provide me schematic

is this normal resistor?

thanks for the immediate response nomurphy

17. ### nomurphy AAC Fanatic!

Aug 8, 2005
567
13
I wouldn't use a bridge, it's just more power to dissipate. The single diode half-wave will give you 25Hz, which usually is not noticable.

6.8K, 5W, resistors are a standard value that should be readily available (same with 10K, 5W).

18. ### mindmapper Active Member

Aug 17, 2008
34
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1: Use the maximum.

2: Sounds terrible, is it possible to use computers in India? Instead of constant current circuits, use transorbs or VDR. I don't know whats availible to you.

One option is to take the protection from an old damaged computer power supply. Depending of availible protection put in before the full wave rectifier or/and split the resistor in two. Use two resistors in series and put the protection in the midpoint of these two and - of the rectifier (cathod of the last LED.

A half wave rectifier is not enough, (cause of reverse voltage) and will give you 50Hz (60Hz).

19. ### Wendy Moderator

Mar 24, 2008
21,838
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The chapter I mentioned earlier covered that, don't worry about it. Either put diodes in parallel the other way or regular diodes, you don't need to rectify the voltage (it just complicates the issue).

All these issues and more were covered in the article.

To the other people trying to help, please read the AAC book chapter I recommended before trying to help. You are confusing the issue without researching the background.

Last edited: Dec 2, 2008
20. ### Wendy Moderator

Mar 24, 2008
21,838
3,047
OK, figuring 25 pairs (back to back), you would drop around 75 volts (3 X 25). We'll use 240VAC at 50Hz at 20ma as a base line. 240 - 75 = 165.

165V/.02A=8250Ω reactance.

F=1/2πfC or C=1/(6.28XFX8250)=0.39μF

Don't forget the 1.2KΩ resistor to prevent surges.

OR...

If you use regular diodes in parallel you will drop 150V. C=0.71μF

Using 20ma gives you some flex room for surges.

Last edited: Dec 2, 2008