Howlor
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| posted on 26/5/06 at 06:56 PM |
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LED's - Damaged by 2 +'s?
Hi,
Just a quick question. I wired an LED into a circuit that is normally a switched negative, but I want to turn it off by switching the neg lead from
earth to positive. The thing is that the LED is now dead. Does it normally dmage LED's with two +'s?
Thanks,
Steve
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DIY Si
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| posted on 26/5/06 at 07:00 PM |
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Why would you switch it off by going to +'ve? Isn't it easier to just switch the connecton off, ie to an open/not connected part of a
switch?
Oh, and no it shouldn't damaged it if both +'s are the same voltage.
[Edited on 26/5/06 by DIY Si]
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Bob C
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| posted on 26/5/06 at 07:13 PM |
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LEDs die easy - too many reverse volts, or no current limit in forward direction. A few microseconds is all it takes....
Bob
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tks
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| posted on 27/5/06 at 07:59 AM |
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mhhh
just put a diode in that V+ line
for example a N4007
then the voltage wound reach the led and it would shut down...
they are really cheaps
a diode is like a one way valve..
it consumes som pressure (0,6 voltage)
but it will block reversing voltages...
Tks
The above comments are always meant to be from the above persons perspective.
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MikeRJ
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| posted on 27/5/06 at 11:34 AM |
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quote:
Originally posted by Howlor
Hi,
Just a quick question. I wired an LED into a circuit that is normally a switched negative, but I want to turn it off by switching the neg lead from
earth to positive. The thing is that the LED is now dead. Does it normally dmage LED's with two +'s?
Thanks,
Steve
The answer is that applying the same potnential to both wires will cause no current flow, and so will not damage the LED. It is the same as
connecting both wires to groud from the LED's perspective.
Putting more than around 5 volts backwards accross an LED will make it break down, and the resulting current flow, if not limited to a low value will
kill it.
Did you use a suitably rated current limiting resistor in series with the LED?
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