Troubleshooting: LED Lighting Tube with 240VAC Input - Fuse Burning Issue

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Homework Statement



Im currently doing a LED lighting Tube. I am using 240VAC 50Hz input. The connection will be the Input Source connected through a Rectifier,then a resistor and 68pcs of LED connection. When we try to testing the circuit, all the LEDs did not light up but the 13A fuse inside the 3-pin plug burn off immediately after we turn on the switch. We connect the Life and Neutral wire to the positive and negative of our circuit.Left our the Ground wire.



Homework Equations





The Attempt at a Solution



So what is the problem actually that the fuse burn off immediately after we turn on the switch?Overcurrent?
 
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Do you have something like this?

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

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 knowledge and experience to do it safely.
 
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My connection is exactly what you draw in the Figure. The rectifier i used is the round head rectifier which having 4 legs. The rectifier is something like in http://www.nedis.com/Articles/Diotec/B250C1500R.php

Im not so sure about the specification. So what is the specification of rectifier i need in that circuit?

We connect the rectifier with a type G plug from the power source which is 240ACV 50Hz.
The type G plug is something like in http://www.world-electric-guide.com/plug_g.gif .

The type G plug have 3 different kind of wire,which is Life, Neutral and Ground. How am i suppose to connect the 3 wire of my type G plug to my rectifier and resistor?

Based on the knowledge we have we think that our connection is correct but end up the 13A fuse burn immediately as we turn on. All the LEDs did not flash.

Why the 13A fuse will burn?Is this telling us the supply current is too big?

Thanks and i appreciate your help so much =)
 
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Yes the connection is exactly the same with your diagram. The rectifier we used is the round head rectifier with 4 legs. The rectifier is connected to type G plug with 240VAC 50Hz power supply.The type G plug have 3 wire, Life, Neutral and Ground. We only connect the Life and Neutral wire of the plug to the rectifier and resistor.
Based on your knowledge we think that the connection should be alright but the 13A fuse inside the plug burn off immediately as we turn on the switch.
Why is this happen?is it overcurrent of the power supply or we accidentally short circuited it because the wrong connection?
Thanks and we appreciate your help a lot =)
 
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.
 
We are trying to use small value of resistor as we want to minimized the power dissipated in the circuit.At first we use 50ohm resistor, then we try 10kohm but still the fuse get burn. We'll try to test it again and reconnect the connection of rectifier using single diode and see whether it works.
Thanks =)
 
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.
 
The LED's we using is 5mm superbright daylight colour. The specification can be found at http://www.ledshoppe.com/Product/led/LE1002.htm
 
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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.
 
The calculation we had make is
Source Voltage = 240V
3.5V for each LED X 68 column = 238V
Voltage consumed in each cycle for the full-wave rectifier = 1.4 V
Remaining voltage = Resistor voltage = 0.6V
Current of Resistor = 0.6V/50ohm = 12mA
I am curious how the peak current will go to 2.6A if we using 50ohm?Is our calculation wrong?
 
syee10 said:
The LED's we using is 5mm superbright daylight colour. The specification can be found at http://www.ledshoppe.com/Product/led/LE1002.htm

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.
 
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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.
 
syee10 said:
The calculation we had make is
Source Voltage = 240V
3.5V for each LED X 68 column = 238V
Voltage consumed in each cycle for the full-wave rectifier = 1.4 V
Remaining voltage = Resistor voltage = 0.6V
Current of Resistor = 0.6V/50ohm = 12mA
I am curious how the peak current will go to 2.6A if we using 50ohm?Is our calculation wrong?
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.
 
vk6kro said:
The connection will be the Input Source connected through a Rectifier,then a resistor and 68pcs of LED connection.

Yeah, well, it didn't appear to be a homework problem, at the time. It is titled "LED saving energy light," after all.
 
May i know what is the model of the thermistor can i use? What is the specification of the thermistor i need to consider? The supporting voltage and forward current? How about the resistance of the thermistor?
 
Did you find out why your setup was blowing fuses?

Which components were faulty?
 
yup because of the rectifier can't afford the voltage supply and it get burn at the 1st time we test the circuit.We replace it with 4 single diodes and now the circuit function well but weird thing is the voltage across the rectifier diode that we build is 100V++.Based on the calculation it shudnt takes so much voltage for the rectifier.
 
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.
 
Ok..Just want to make sure again..is my calculation correct this time?
Supply voltage = 340V
Voltage consumed by 68 LEDs in series = 3.5V X 68 = 238V
Voltage consumed by diode rectifier (4 pcs of 1N4007) = 1.4V
Voltage consumed by resistor = 340 - 238 - 1.4 = 100.6V
Resistance set it to 4.5k, Hence the current is 100.6/4.5k = 22.4mA
The power of Resistance we need is at least 100.6 X 22.4m = 2.25W
 
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.
 
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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

If you have room, you could think about adding a capacitor across the output of the rectifier.
This would hold the 350 volts between 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.
 
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Is the connection of capacitor and resistor in the jpeg file correct?Is that what you mean?
Hopefully the practical current will be smoothed after adding the capacitor inside the circuit.
 

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The capacitor would be connected like this:

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

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.
 
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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.