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Relay To Turn Lights Off When Starting


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karlloss 
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Posted: January 15, 2011 at 9:06 AM / IP Logged  

I have a Triumph Speed Triple. The headlights are constantly live once the ignition is switched on, even when the engine is cranking under starting. it has twin headlamps, I have found a plug in relay and harness for single lamps called Easy Start Battery Saver which says "Simply plug-in control module into existing headlight wiring, now when ignition is in “on” position, only position light will be turn on while headlight stays off; this will allow more battery power available to start the bike. Once engine’s been started, simply flip high-low beam switch to turn on bike’s headlight and headlight will stay on until ignition is turned off again. This product originally was design for customer to have the option of turning off headlight when starting the bike. It is particularly useful when trying to start the bike in cold climate, when battery does not have much cold cracking power left; it would also allow people to run ECU or ignition test on the bike without having to hook-up to a larger battery because headlight is turned off. Unfortunately the kit for just one head light is Ł30, and I would need two kits for twin lights, therefore I am hoping to construct my own for less cost.

I am presuming this is a latch relay type set up, but I can't figure out the wiring for it, or if I should be using another type of relay, can anyone advise me please?

Karlloss
oldspark 
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Posted: January 15, 2011 at 10:16 AM / IP Logged  
Geez.....
If the headlights are controlled by a relay (der, yes?).
And it's a +ve signal to turn them on.... (depends on vehicle).
Then just ground the relay thru the starter circuit or the starter solenoid connection.
It costs the length of some wire, plus your time & labor.   
Just ensure your headlight relay does not include a spike-suppression diode (resistors and no suppression is fine) - just in case you crank and the headlight controller is not on (energised) (even though that shouldn't happen - but why fit suppression to hat anyhow?)
karlloss 
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Posted: January 15, 2011 at 10:56 AM / IP Logged  

Old Spark,

Ok so I groung the relay through the starter, as the ground or negative is common through the chassis of the bike how is that going to diconnect the lights whilst starting, as all I will be doing is moving the earth/ground for the headlight relay to a different location, the relay will still be energised and the lights on??

Karlloss
oldspark 
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Posted: January 15, 2011 at 3:18 PM / IP Logged  
You don't connect it to the ground, you connect it to the solenoid or cranking +12V signal - but preferably the solenoid if it has one (starter solenoids are low resistance - they typically consume a few Amps; usually 20A-40A for cars).
IE - the grounded end of your lighting relay is moved to the +12V cranking circuit.
karlloss 
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Posted: January 16, 2011 at 10:46 AM / IP Logged  

The bike has a starter solenoid, I'm still confused to how connecting the ground to the positive work, how will the lighting relay ground through the positive starter circuit, as when the the starter button is not pressed then the circuit will be open circuit.  Any chance of providing a diagram, as I'm still not clear on this?

Karlloss
oldspark 
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Posted: January 16, 2011 at 11:46 AM / IP Logged  
So when the +12V end of the starter solenoid is no longer connected to the +12V through the "open" starter switch, what voltage will it be?
karlloss 
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Posted: January 16, 2011 at 12:52 PM / IP Logged  

well it'll be zero volts, but the ground from the lighting relay needs to be -12v to make the circuit for that relay, but the starter circuit when the starter switch is not operated will be zero volts with no continuity. Obviously I'm not understanding something here, do not all circuits need to have continuity to operate, and the component, in this case the lighting relay, needs to have a +12v to the internal electronmagnet and a -12v to the other side of the electromagnet in the relay to operate it, but going through an open starter circuit will not provide the -12v supply ??

Karlloss
oldspark 
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Posted: January 16, 2011 at 3:32 PM / IP Logged  
The ground of the lighting relay needs to be at ground - that is zero volts, NOT -12V.
And no continuity? Are you saying that the +12V end of the starter coil/solenoid is not connected to anything?
Is there not a relatively low resistance coil/solenoid with its other end connected to ground? (Answer, yes. It's just like any other switched load - whether a coil, resistor, light bulb, motor - they are either +12V and operating, else de-energised = NOT switched to +12V - ie, the +12V switch is open.)
FYI - Do not get confused with "ground is negative 12 volts" and -12V.
A supply with +12V and -12V is a "24V" supply {12 - (-12) = 12+12 = 24}.
A supply with +12V and 0V (zero volts, aka ground in "negative chassis vehicles") is a 12V supply {12-0 = 12}.
I remember hating my electronics magazine(s) when they converted from -V to 0V.
EG - +9V & -9V meant the +ve & -ve terminals of a 9V battery. Simple.
But then came split supplies (amplifiers etc) where +9V & -9V literally meant + & - 9V.... ie, 18V; usually with ground at 0V.
Now I cannot understand how we lived with that confusing old convention!
i am an idiot 
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Posted: January 16, 2011 at 8:19 PM / IP Logged  

I know OldSpark's idea seems to go against the flow of everything you know, but it is right and it will work.  A Tyco (formerly Bosch) relay draws only 160 milliamps.  The coil part of the solenoid on the starter draws at least 3 amps of current.  So if you wire both of these in series, the only one that will energize is going to be the relay. 

You can not judge a user by their screenname. The Old part is probably correct, but not so much on the Spark part.

karlloss 
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Posted: January 17, 2011 at 1:17 PM / IP Logged  

Guys, I appreciate the help, but just to make sure I don't fry my loom or relays, I take the ground from the lighting relay and connect it to the +12v feed to the starter solenoid/relay, this feed comes from the starter button. Any chance of a diagram just to be sure?

Cheers

Karl

Karlloss
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