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LED Flasher, Remove 12V(+) When Key Is On


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carrlos 
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Location: Florida, United States
Posted: September 04, 2011 at 8:57 AM / IP Logged  

I have installed a "visual deterrent" on my motorcycle, basically a red LED flasher. The only problem is that it is wired directly to the battery and runs all the time. I only want it to activate when the switch is OFF.

Can someone please help me configure a relay that will remove the 12V when the key is ON so that I don't have a blinking LED in my face at night while riding. In turn, when I turn the bike off, I want the 12V activated again so the LED starts flashing again.

KPierson 
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Joined: April 14, 2005
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Posted: September 04, 2011 at 9:56 AM / IP Logged  
Many ignition systems rest in the ground position. Check the main ignition wire on your bike and see if it rests at ground. If it does, simply connect the (-) side of the LED to the ignition wire.
Alternatively, wire a relay like this:
Pin 85 - ignition power
Pin 86 - ground
Pin 30 - + out to LED
Pin 87 - constant 12vdc
Kevin Pierson
carrlos 
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Location: Florida, United States
Posted: September 04, 2011 at 10:07 AM / IP Logged  

Thanks for the quick reply KPierson. I will check the ignition first as it seems to be the easiest solution.  If that doesn't work, then I'll use the relay.

But I have one question: am I correct in thinking that using the ignition wire option, when the key is on there will be a postiive 12VDC on the negative side of the LED? The reason I ask is because I have a resistor on the positive side of the LED to bring it down to 2.5VDC (it operates between 2.5 and 5VDC).  If so, I would have 12VDC on one end of the LED and 2.5VDC on the other end. Will this harm it in any way?

Maybe the relay option is the best after all.

Thoughts??

oldspark 
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Posted: September 04, 2011 at 10:42 AM / IP Logged  
No relay needed.
Connect the LED+ to battery +12V, and the LED -ve end to IGN +12V.
(Where the "LED" is a 12V LED - ie, 2.5V etc LED plus resistor.)
carrlos 
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Location: Florida, United States
Posted: September 04, 2011 at 11:13 AM / IP Logged  
Thanks oldspark. So I will have +12VDC on the negative end of the LED and +2.5VDC on the positive end when the ignition is on, which should prevent it from activating, yes?
i am an idiot 
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Posted: September 04, 2011 at 1:23 PM / IP Logged  

If you are using a 12volt LED, connect the positive lead to the battery.  Connect the negative lead of the LED to the vehicle's positive Ignition wire.  This wire will have 12volts on it when the key is on.  When you turn the key off, this wire will provide enough of a ground to illuminate the LED.

If you are using a conventional LED, you will need to insert a resistor in series with the LED.  Most LEDs if not 12volt require around 2.5 volts. 

oldspark 
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Posted: September 04, 2011 at 4:34 PM / IP Logged  
carrlos - yes, what you said is correct, though it took me w while to realise that.
i am an idiot is also correct (as usual) but he is describing your set up.
It probably gets confusing because once a LED has whatever resistor is required for 12V, I/we think of the LED+resistor simply as a "12V LED" or load.
In other words, it's not a LED with 2.5 (or 1.7 or 2.4 whatever) Volts across it, but simply a 12V load or circuit or LED.
(And that LED could be 1 or 2 or 3 or more LEDs in series with suitable resistor. Or parallel LEDs etc.)
The good thing is, that with that (current limiting) resistor for 12V, it doesn't matter how you connect is across 12V, you cannot blow the LED.
i am an idiot 
Platinum - Posts: 13,660
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Posted: September 04, 2011 at 6:10 PM / IP Logged  
He threw me off when he said there would be 2.5v on one lead of the LED.  With no ground, there will be 12v everywhere.  Next time I will read deeper into it.
oldspark 
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Posted: September 04, 2011 at 7:51 PM / IP Logged  
Like I said, it took me a while... ie, several reads.
But I thought carrlos is agreeing despite referring to 2.5V... not "+12V at each end" hence 0V hence no LED.
Technically carrlos is incorrect - there is no 2.5V, the lot is ~+12V when the LED off, or there is 0V from one end to the other (resistor & LED)...
But I knew what he meant.
Besides, if not for the zero-current flow (and non-linearity of LEDs), he is correct.
And since there will be current flow since IGN +12V will always be less than the battery +12V (until started and charging), he is correct! (In principle.)

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