Jenko
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| posted on 20/11/13 at 09:16 PM |
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LED Flashers - how do they work?
So, 3 pin led flashers.....they simply need 12v on the supply pin, and the live pin goes on and off....yes fine....but................
.......this means the flasher is in a permanent state of flashing (irrespective of the indicator or hazard switch)... As the the hazard lights need to
be operated with no ignition....won't the flasher draw current and flatten the battery?.
MY BLOG - http://westfieldv8.blogspot.co.uk/
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splitrivet
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| posted on 20/11/13 at 11:40 PM |
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No the circuit is only completed when the switch is thrown, thats what makes it flash.
Cheers,
Bob
I used to be a Werewolf but I'm alright nowwoooooooooooooo
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Jenko
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| posted on 21/11/13 at 07:17 AM |
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You would think....but, this is not the case...with 12v applied to the supply pin, a DVD shows that the live pin is flashing even with no
load!.strange I know.
MY BLOG - http://westfieldv8.blogspot.co.uk/
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Slimy38
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| posted on 21/11/13 at 08:04 AM |
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Considering LED's consume a fraction of normal bulbs, and my car has two interior LED's that permanently flash as part of the alarm
circuit, I think you should be ok.
Also thinking about it from a connection point of view, yes the device might be oscillating, but without anything connected to it, there isn't
any current being drawn.
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peter030371
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| posted on 21/11/13 at 08:41 AM |
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Never played with one but if the flasher circuit is permanently powered then I assume they take naff all current (that's the technical term!)
and the biggest load is only when the lights are connected and flashing. In the case of LED lights though the 'biggest' load is still
relatively small compared to filament bulbs.
I suspect that a great big car/ bike battery will have more self discharge than the drain of a LED flasher 
I also assume that with a permanently powered flasher the indicators also work with the ignition off, is that correct?
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avagolen
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| posted on 21/11/13 at 09:01 AM |
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If it is only a 2 pin unit, the earth will be supplied through the indicators.
This should only occur when they are selected via the hazzard switch or the indicator stalk.
No current drain when not selected. (I hope) !
The Answer for everything, but never the last word....
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Jenko
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| posted on 21/11/13 at 09:04 AM |
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Good point....and it makes sense that the current draw is 'naff all'...so, I shall stop worrying.
As for the indicators...again, you are correct, Unless I can work out another way around this, the indicators will work with ignition off....something
I really don't want!.....I need to start the head scratching now!.
MY BLOG - http://westfieldv8.blogspot.co.uk/
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peter030371
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| posted on 21/11/13 at 10:26 AM |
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A battery isolater can be used to totally disable all systems when you are parked up though
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Not Anumber
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| posted on 21/11/13 at 10:59 AM |
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Why dont you just power it from the live feed to the ignition ?
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Jenko
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| posted on 21/11/13 at 12:14 PM |
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quote:
Originally posted by Not Anumber
Why dont you just power it from the live feed to the ignition ?
Because the hazards need to work outside of the ignition.
MY BLOG - http://westfieldv8.blogspot.co.uk/
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Not Anumber
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| posted on 21/11/13 at 12:26 PM |
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you are not using a seperate relay for the hazards, ok.
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rachaeljf
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| posted on 21/11/13 at 12:29 PM |
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The usual thing is that the hazard switch is a double pole switch arranged to provide either permanent live or ignition live to the flasher unit, i.e.
"hazards off "-> ignition live fed to flasher, "hazards on" -> permanent live fed to flasher. The other side of the
hazard switch, when in the "hazards on" position, connects all the left and right indicators to the flasher so they all work together,
overriding the indicator stalk.
Thus, in the "hazards off" position, the flasher unit only gets a feed when your ignition is on, so you shouldn't have any concerns
about running your battery down.
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