Homework Help: LED saving energy light

Do you have something like this?

http://dl.dropbox.com/u/4222062/LED%20string.PNG [Broken]

If so, maybe you have the rectifier connected the wrong way around. It is possible to get almost a short circuit if you do this.

The best way would be to test the components and see what is blown up. You didn't mention a bright flash, so maybe the LEDs are OK.

Resistor R should be about 5000 ohms and about 10 watts.

The bridge rectifier is connected to the mains, so you need a big safety margin. So, the bridge should be a 1000 V piv type.

This is a public Forum, so I should warn anyone against making direct connections to the mains unless they have the necessary knowlege and experience to do it safely.

 

You will have to do some measurements.

Measure the input resistance of the bridge rectifier (the resistance between the points where the A and N connect). If it is a short circuit, it was probably connected wrongly or maybe it did not have enough voltage rating.

You can replace it with individual diodes, but use the highest voltage rating ones you can find. 1000 volt diodes are cheap and you can easily use them.

Check every LED to see if it is the right way around and it still works.

What is the size of your resistor?

There should not be any surge currents, but it is important to find out what caused the short circuit before you apply power again.

 

What LED's are you using? They can vary widely in forward voltage drop. It would help if you could supply a link to a datasheet for the LEDs in question.

 

When you used a 50 ohm resistor, the peak current would have been about 2.6 amps which would have blown up some or all of your LEDs.

If you have a multimeter, you really need to go through and test everything before you apply power again.

 

I didn't see anywhere where you said how many LEDs you have in each string. It makes a difference.

Other than that, there is a better way to drive leds off the AC lines. An associate and I came up with an eligant and energy efficient solution, with little power dumping, except at start-up time. There are no very lossive resistive elements required. It involves a capacitor and a bridge rectifier and a soft-start thermistor and the diode string. I don't know if it's patent protected. Does it matter?

Connect one of the AC lines to a negative coefficient thermistor, then to a capacitor. This is a high voltage capacitor. Connect the capacitor and the other AC line to a bridge rectifier. Connect the DC outputs of the bridge to your LED string. The arrangement constitutes a somewhat-constant current source, which is what your want. Size the capacitor to develop the current given in the LED specs (<30 ma rms) for nominal line voltage. The capacitor needs to be nonpolar. The current developed through the string is dependent upon the AC line voltage and the capacitor tolerance, with lesser dependence upon the voltage drop across the LEDs when the LED string voltage is, say, <1/3rd of 340 volts, in your case. But if you have a lot of LEDs like 60 or 70, it will still work OK.

You could model it in spice if it helps.

The thermistor is required to prevent surge currents from blowing out the string. Otherwise the full rectified line voltage of 340V could appear across the LED string at turn-on time. Size the thermistor so that it's cold resistance current doesn't exceed the 80 ma current rating of the LEDs.

However if your design requirments are to turn the LEDs on within a few seconds of turning them off this design solution will not work. The thermistor requires time to cool.

 

The connection will be the Input Source connected through a Rectifier,then a resistor and 68pcs of LED connection.

They have 68 LEDs all in one series string.

Somehow there is a 13 amp fuse blowing instantly.
Hopefully it is just a bridge rectifier or an accidental short circuit and not that expensive string of LEDs.

 

The peak value of the input voltage is 1.414 times 250 or 353 volts.

The LEDs add up to 238 volts so there is 353 minus 238 or 115 volts across your 50 ohm resistor.

So the current will be 115 volts / 50 ohms or 2.3 amps.

 

Yeah, well, it didn't appear to be a homework problem, at the time. It is titled "LED saving energy light," after all.

 

Did you find out why your setup was blowing fuses?

Which components were faulty?

 

No. That's OK.

You can't measure it with a meter. You would then be measuring the reverse voltage as well which is much bigger.

You would need a CRO (Oscilloscope) to do it, but it would be very dangerous doing this when connected to the mains.
Try it at low voltage some time.

 

Yes, that looks OK. However that is only peak current.
The average current will be about 10 mA. So the LEDs may not be very bright. They are rated at 30 mA, I think.

Try this with just the rectifiers first to make sure there are no fireworks.

 

The average current will be pretty low because the current only flows if the input voltage is greater than 238 volts.

http://dl.dropbox.com/u/4222062/LED%20current.PNG [Broken]

If you have room, you could think about adding a capacitor across the output of the rectifier.
This would hold the 350 volts betwen input pulses and allow a steady current to flow in the LEDs.

The current is quite small, so about 10 uF should be enough. 450 volt rating or higher.
It would need a suitable resistor across it to discharge it when the power is turned off.
About 100 K, 2 watts.

 

The capacitor would be connected like this:

http://dl.dropbox.com/u/4222062/LED%20string%20with%20C.PNG [Broken]

This is called a filter capacitor and it converts pulsing DC into smooth DC.

The 100 K resistor across the capacitor has to be there because otherwise the capacitor would stay charged up to about 238 volts and be dangerous even after the power was removed.
The resistor removes the charge in a few seconds.

I think I mentioned before that this is very dangerous stuff you are doing.
Do you have someone who can check your setup and advise on its safety?
That capacitor and the two resistors and the diodes should be in a sealed plastic box, at least. The 5 K resistor is going to get quite hot, so the box needs ventillation holes.

 

y d resistor 5k need 10 W?how u calculate it?

 

That 5 K should be 4.7 K.

The 4.7 K resistor will dissipate about 2.7 watts. (350 - 238) * 0.0238 A.

So, you could use a 5 watt resistor.