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Old 29th February 2008, 10:52 PM   (permalink)
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BUMP !! Just get this LED topic relivened

Cheers, Raymond
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Old 29th February 2008, 11:09 PM   (permalink)
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Quote:
Originally Posted by RODALCO
Hero999
It still protects the LED.
The 1N4007 is a 1 kV diode, the LED is in series with it, and can only handle 6 Volts reversed max. If reversed voltage sharing with two diodes is required both diodes have to have same voltage and current rating ideally, IMO.
I agree with you. The reverse leakage of the 1N4007, while very small, could cause the LED to be reversed biased during the negative half of the cycle. When rectifier diodes are placed in series for a higher PIV, then balancing resistors are required:
http://www.eng.uwi.tt/depts/elec/sta.../s2_sconn.html
Since the LED is a diode, the same issues arise and hence the inclusion of the extra diode. You could probably accomplish the same thing with a resistor (10K) across the LED.
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Old 29th February 2008, 11:32 PM   (permalink)
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I doubt it would damage the LED, especially if it's a red LED. When I was a child I hooked up a red LED and a 10k resistor to the mains with no diode in reverse and it didn't damage the LED. White and blue LEDs seem to be more sensitive and are destroyed when subjected to similar abuse.
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Old 29th February 2008, 11:44 PM   (permalink)
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Interesting Hero999, I did the same when i was about 13 years young and it worked with a LED on 220 Volts in The Netherlands.

Somehow the older inefficient LED's stand more abuse. I did the LED with an about 68 k 1/2 watt resisitor which went up in smoke after a while. a 1 watt worked but got very hot but lasted.

In that experiment I used a yellow LED with probably 5 mCad light output, now these days the LED's are 1000 + times brighter.

Ok back to the topic, In case of indicator LED's The cost of a servicecall for a failed indicator far outweights the costs of a 5 cts 1N4148 diode.

You probably get away with it but RELIABILITY is the issue for me.
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