# 72 LEDs in series with 220v AC

Discussion in 'The Projects Forum' started by sudhalok, Jun 1, 2009.

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1. ### sudhalok Thread Starter New Member

Jun 1, 2009
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Hi, this might be a really stupid or easy question for u all out there.

The thing is, I have very little idea about all this. I have made four boards with 72 white LEDs connected in series in each of them. The AC supply in my place is about 220v. Now, if I connect any of these boards to my AC source, they light up but flicker(obviously). So, I guess I need to use a rectifier? Is it necessary to use a capacitor and filter the output? I want to use one rectifier and connect two boards in parallel with it. That is, total of four LED boards with 72 white LEDs each in them and two rectifiers.

So, from the power supply circuit, I guess I will be needing output of 40mA. Would be really great if you can help out and guide me with the power supply circuit design.

I am planning to install the LED circuits in a place where it will be really difficult to replace them. Hence, I need such a solution which ensures that the life of the LEDs is maximum. The rectifier circuit will however be placed away from the LED circuit so that the rectifier can be changed easily.

Thanks.

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2. ### Wendy Moderator

Mar 24, 2008
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Read the AAC eBook here (look for the section on LEDs)...

Now for the synopsis, you MUST have either an LED or a diode back to back with every one of the other LEDs. This is because while LEDs are diodes, they no tolerance for reverse voltage, so you have to get around this somehow. BTW, this spec is called PIV (peak inverse voltage). My personal preference is to use a second LED, to match the current characteristics. By putting a diode across the LED facing the other way the max PIV the LED will see is the dropping voltage of the LED.

The other thing the article mentions is using a capacitor as the current limiting component. This works, but you have to be cautious. High voltage AC can kill rather quickly, and over estimate the size of the capacitor and some very expensive smoke escapes.

Jun 1, 2009
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4. ### beenthere Retired Moderator

Apr 20, 2004
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You should be aware that the circuit will always present a lethal shock hazard, and there is no way to make it safe. Using a transformer is the only way to get isolation from the line voltage.

5. ### Wendy Moderator

Mar 24, 2008
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I got this as a PM...

You can use a full wave bridge. We keep talking about the danger involved, it is real, so be careful.

I get the feeling you aren't too familar with electronics or LEDs.

The capacitor will act like a resistor with one important difference, it doesn't disapate heat, the reactance restricts the flow of electricity via other means. It is part of what makes a flourescent light so efficient.

So you use a capacitor to help resistrict the flow. You're still going to need a resistor. More than that, you will need a separate resistor per leg. It will help distribute the current evenly, since every LED has a slightly different voltage drop.

If you aren't familar with LEDs you really need to read this...

LEDs, 555s, Flashers, and Light Chasers

OK, the math. Figure the RMS value 220VAC works out to 308VDC. As a bullcorn estimate 72 white LEDs will drop around 260VDC. If each leg pulls 20ma, and you have 4 legs, so that is 80ma. Total resistance is around 48V/0.08A, or 600Ω. You've never mentioned the line frequency voltage in your country, so I'll assume 50Hz. The capacitor is around 4.7µF 440Volts. The schematic looks something like this...

I'm shooting from the hip on some of these values, but they are in the ballpark. I was figuring on 10µF 250V polarized capacitors, but if you can find 4.7µF non-polarized 500VAC caps they will work (but are harder to find).

We keep saying this, but every single wire on this project can kill you. Don't make the mistake of thinking any part of it is safe. You can tweek the current by adjusting the 470Ω resistors. You can also measure the current though the legs by measuring the voltage across the 470Ω resistors and calculating the current from there.

Last edited: Jun 2, 2009
6. ### sudhalok Thread Starter New Member

Jun 1, 2009
5
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I am learning a lot........

Actually, I want one power supply to power two legs.

So, I will be making two power supplies. And yes, the frequency is 50 hz in my place.

I will definitely be VERY cautious.

BTW, if I increase the no. of LEDs to say 90, do I need to use all this?

If I have 90 LEDs, the p.d. -> 90*3.5 = 315v.

Then, I can just use the diode bridge for rectification. Will The LEDs regulate the current automatically?

Thanks.

Last edited: Jun 2, 2009
Anand n likes this.
7. ### Wendy Moderator

Mar 24, 2008
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You are going the wrong way with this, I'll try to explain.

LEDs require a minimum voltage per LED. This is called Vf. You have white LEDs, which can run between 3.5 to 3.8V, I'll pick 3.6 as the average voltage. 72 X 3.6 = 259.2 Volts. It gets worse with 90 LEDs, 90 X 3.6 = 324 Volts. No Good. Very Bad.

Instead, lets drop them down to 6 LEDs per chain, and use a 24V Wall Wart. Use 15 legs and you have 90 LEDs, and a safe voltage. The power supply provides 0.3 Amps, a very doable figure. If an LED burns out (rare, but it happens) only 6 LEDs go dark, not a chain of 72.

Read the article I provided. The are no circumstances that an LED can regulate current, their function is light, not current regulation. Resistors however, are cheap. They cost around 2 cents each. Try to skimp on resistors and you'll burn out or damage your LEDs

Doing it the way I suggested makes for a completely safe project, and increases reliability at the same time. The moderators and I strongly recommend doing it this way.

I just got through with a thread where I couldn’t get the point about current limiting resistors across to the poster. It was frustrating to say the least. There are other ways to control current through many chains using transistors, but resistors are the cheapest way to do it.

I tried to post this around 12 hours ago, but the AAC site died.

8. ### DC_Kid AAC Fanatic!

Feb 25, 2008
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dont use rectifier, to avoid the flicker for each "leg" add another in the opposite direction so that they are always "on". each will light 50Hz @ 50% duty cycle.

for safety perhaps you could encapsulate the LED array. there are all types of clear epoxies that provide insulating properties.

certainly is not the right SAFE way to drive a LED array, but if you have to with just 220AC and LED's.......

now, if you want to exploit LED ability you might think about using PWM 100Hz @ 10-20% duty cycle and over drive them some..... a 555 and a FET works well.

and btw, even with full bridge rectifier you will have flicker @ 50Hz. 60Hz is recognizeable by the human eye, and is one reasons why i hate cheap ballast florescent light fixtures....

Last edited: Jun 3, 2009
9. ### Wendy Moderator

Mar 24, 2008
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Actually 30Hz is the lower end of persistance of vision, which is why 30 Hz frame rate is used for NTSC (old style american television). Unless it is moving you can't see 30Hz flicker. It also sets up a minimum time you can see a LED flash clearly (30ms), which I've used to good result in some of my 555 experiments. This rate is pretty close to what movies use too.

This isn't the first request for HV LED drivers, people want to treat them like light bulbs, and their not.

A full wave rectifier on a 50hz system will flash the LEDs 100Hz, as each peak gets involved. Also, I put a cap in the original to go for peak values. However, as I and other people have stated, this is a terrible idea for design because of the safety issues.

10. ### DC_Kid AAC Fanatic!

Feb 25, 2008
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yep, and i wanted to correct myself......

however, two segments close together, one facing the + way and the other the - way should nearly eliminate the flicker when viewed from a distance. interleave the whole bunch and it should look ok.

i can see 60Hz! anything that flickers at 60Hz really bugs me, headaches and all. i dont use fluorescent fixtures (unless they have upconvert on their drives), and my PC monitor runs well above 60Hz.

11. ### sudhalok Thread Starter New Member

Jun 1, 2009
5
3

Now I think I will increase the total no. of LEDs to 77, then make 11 legs of 7 LEDs each, just to keep the voltage on each LED a bit down, 7 x 3.5= 24.5. That I can power with a 24v adapter. That way I guess it will be safe and last longer, wont it?

However, one thing, I hope that adapters do not break down frequently.

12. ### Søren Senior Member

Sep 2, 2006
472
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Hi,

It will certainly be much safer!
If you place the adapters at a reachable spot, they would be easy to service/change, but don't buy the cheapest stuff anyway.

You NEED a resistor in each chain of LEDs! as the voltage will probably fluctuate and when you turn on the power, there will most likely be a spike of a higher voltage.

You can use 6 LEDs and a single resistor of 100 Ohm (for ~30 mA) each string, since the resistor needs a couple of volts to do its job properly.
You could drive all the 288 LEDs that you mentioned originally, in 48 strings of 30 mA each, from a single 1.5A (24V) supply.

With a DC adapter you won't get any flicker (just make sure it's not a so called "pulsating DC" adapter).

Don't mount the adapter in a hot spot.
(288 LEDs in a closed enclosure WILL be a hot spot and will need some kind of heat sinking or ventilation.)

Btw. Have you got a datasheet for the LEDs?

13. ### sudhalok Thread Starter New Member

Jun 1, 2009
5
3
No, I dont have a datasheet for the LEDs.

I checked out the market here and they dont have small size DC adapters of 24v. All 24v DC adapters are about 3" X 3" X 3" which is too big for my place. Only 12V DC adapters are there. If I put up 2 of those in series, will it work?

Thanks..

14. ### Wendy Moderator

Mar 24, 2008
21,778
3,023
7 LEDs per leg is too many for 24V. Back to the old formula, 3.6V/led, so 3.6V X qty7 is 25.2V. This voltage is where the LED turns on, like a switch, the voltage over that is used to set the current.

Like I said, for 24VDC you need 6 LEDs per leg, which work out to 21.6V. If the LEDs are 3.8V then it would be 22.8V total, which is still doable.

If the LEDs are 3.6V Vf, then 120Ω will give you 20ma (0.02A), the resistor will drop 2.4V, and the wattage is .048W. If the LEDs are 3.8V Vf, then the resistor will drop 1.2V, and the current will be 10ma. Some experimentation is in order.

When I mentioned the transistors they have ways to compensate, so you get a firm current no matter what the LEDs drop (to a point).

Points that you can not change:

Most of this is explained that that article I pointed you to.

You can not eliminate the resistors, they are absolutely necessary.

The excess voltage between the power supply and the LEDs drop is what the resistor uses to set the current.

Each chain should have its own resistor.

Questions?

Last edited: Jun 3, 2009
15. ### mouly244 New Member

Oct 18, 2009
1
0
thanks a lot but i dont understand about 470 ohm resistor in series with the 72 leds .please tell me wattage of the resistor and circuit information

Nov 29, 2005
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17. ### zmint New Member

Dec 29, 2009
2
0
Sir, Please put some lite on using polarized caps (C1, C2) instead of AC caps. for limiting Voltage & current. What are the points to be kept in mind while working with, this scenario. Say for e.g instead of a .22 uF 400 or 600V AC non pol. cap. (224J, 400V), I want to use polz. cap's, what will be the values of C1 & C2 polz. caps. How do we calculate the values ? Obviously, I will keep in the mind, the line in RED.

Thanks

Apr 5, 2008
19,689
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Hello,

We do not allow circuits directly connected to the mains.
A transformer MUST be used for safety reasons.

Greetings,
Bertus