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Old 22nd July 2006, 08:42 AM   (permalink)
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We used a 470 Ohms ¼ Watt resistor in series with the Capacitor, this one sometimes ruptured by the inrush current. Didn't try the 1 Watt Resistor which may perhaps work ok.

here is the schema, photo taken of hand drawn schema
where

U is 230 Volts 50 Hz
R1 and R2 are 33 or 39 k Ohms 1 Watt resistors
D1 is 1N4007 (1kV 1Amp diode)
D2 is 1N914, 1N4448, any rectifier diode of 50 mA or more will do here.
D3 is high efficiency LED ( 3000 to 5000 mcad or more ) any colour.

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Old 22nd July 2006, 03:21 PM   (permalink)
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thanks dude.
i'll definetly try this one....
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Old 23rd July 2006, 07:26 AM   (permalink)
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last one..
can i keep this in a closed space?
imean if it get(resistor) heatup forming fumes,then it might create problem
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Old 23rd July 2006, 10:40 AM   (permalink)
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Heat up of series resistors is negligible as long you stick to the values quoted in diagramme.
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Old 24th July 2006, 09:55 PM   (permalink)
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D2 doesn't do anything, because D1 is already blocking the negitive cycles.
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Old 25th July 2006, 11:05 AM   (permalink)
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Hero999
It still protects the LED.
The 1 N4007 is a 1 kV diode, the LED is in series with it, and can only handle 6 Volts reversed max. so a nasty spike while the LED is not conducting and revere biased still could kill the LED, for the sake of a 5 cts diode and 20 seconds extra soldering time a extremely reliable potential indicator for 230 Volts is built.
I know the two series resistors will limit the available voltage across the LED anyway, but better reliable than popped LED and repair.
If reversed voltage sharing with two diodes is required both diodes have to have same voltage and current rating ideally, IMO.
In the 1500 or so meters I have installed 3 LED's each in makes 4500 * 230 Volts LED's with NO faillures. Try to beat that. !!
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Old 25th July 2006, 03:14 PM   (permalink)
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Quote:
Originally Posted by RODALCO
In the 1500 or so meters I have installed 3 LED's each in makes 4500 * 230 Volts LED's with NO faillures. Try to beat that. !!
This is a correct argument that the circuit and its element are sufficient to obtain the reliability mentioned.

I can also claim the same reliability if I add two more 1N4007s in series with the original 1N4007 but that does not implies that my two 1N4007 is "necessary" to order to obtain the same reliability.

The point is, in my opinion and maybe others also, same reliability will result if Diode D2 is omitted. This remains an opinion only as I don't have any data to prove it.
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Old 25th July 2006, 03:30 PM   (permalink)
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The 1N4007 is darn slow to turn off and allows a good portion of the reverse voltage to zap the LED.
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Old 25th July 2006, 03:51 PM   (permalink)
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Assume turn off time of 1N4007 at 3mA forward current or less is 20us.

The 50Hz AC waveform timing is 20ms per cycle and so the max reverse voltage that fed through the 1N4007 diode after zero crossing is.....a few volts ????
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Old 25th July 2006, 04:03 PM   (permalink)
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Quote:
Originally Posted by eblc1388
... so the max reverse voltage that fed through the 1N4007 diode after zero crossing is.....a few volts ????
I don't know exactly how slow a 1N4007 rectifier turns off.
The absolute max reverse voltage for an LED is only a few volts.
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Old 25th July 2006, 04:26 PM   (permalink)
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Rodalco designed this indicator circuit for use in an industrial environment.
He wanted good reliability.
The cost of the extra 1N4148 diode is nothing compared with the cost of replacing a faulty indicator unit.

If you were designing these things to be built by the million in China, and sell on the domestic market as cheaply as possible, you may well consider missing out the 1N4148.
If the LED fails in 2 months time, what the hell, you can sell them another one!

Horses for coarses.

JimB
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Old 25th July 2006, 05:31 PM   (permalink)
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well thanks RODALCO.
i built it for a desk crystal ball,and its awesome.
but i want that LED(i am using green) to be little more bright.ofcourse iam using ultra bright one only.so what shld be modified to get little more brightness by keeping the LED safe.
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Old 25th July 2006, 10:50 PM   (permalink)
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fever

Use a higher efficiency LED e.g 15000 Millicandela instead of 5000, or put 2 or 3 LED's in series with their own antiparallel diode each.

We have an interesting discussion going here about these indicators, great.
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Old 26th July 2006, 01:21 AM   (permalink)
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Quote:
Originally Posted by JimB
Rodalco designed this indicator circuit for use in an industrial environment.
He wanted good reliability.
The cost of the extra 1N4148 diode is nothing compared with the cost of replacing a faulty indicator unit.
Yes, this is really an very interesting discussion.

Of course I would not advocate to Rodalco to remove the diode in all his existing circuits.

The argument is based purely on the function of the diode as an diode in circuit theory. But there is still no solid proof, to me at least, via circuit simulation or calculation or otherwise, that the diode D2 is doing its share to enhance the reliability of the circuit. If that is the case, one cannot simply says that the diode is there for good reliability as I can also safely say two 1N4007s in series will also enhance reliability.

I remembered seeing diodes placed in series to semiconductor base to prevent breakdown of the semiconductor junction so these circuits might share some similarity here. There is no anti-parallel diode connected also in those cases.

And, sorry Rodalco I would disagree again with your suggestion that separate anti-parallel diodes are needed instead of using just one across the LED chain.
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Old 29th February 2008, 10:51 PM   (permalink)
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BUMP !! Just get this
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