jabbahutt
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 posted on 5/10/09 at 10:14 AM |
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small job to keep me sane
Morning all
Been a while since I've been here due to a spell in hospital with kidney problems. all well and good now apart from being banned from the
garage in case I do anything too strenuous and hurt myself.
so sat here twiddling my thumbs and looking at fitting a starter button, no reason it's needed just fancy one.
So I'm looking at fitting a 15A switch from CBS and just want to clarify the wiring.
Is it simply disconnecting the wiring from the solenoid to the igntion and instead connecting to switch and then connecting other side of switch to
the second key position on the ignition (when all the warning lights come on)?
And if the above is correct how would i wire it so the switch lights up before the engine starts and is off when it's started? I was thinking of
taking the power off the ignition warning light as that goes out when the engine is started.
Any help for an electrical fool who can't do much is appreciated as at least I can sit on the sofa mucking around with wire etc as the wife will
aloow that!!
Thanks all as usual
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Mr Whippy
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| posted on 5/10/09 at 10:18 AM |
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As you say, the simplest way would be to use the charge indicator light. Do make sure the button is only live with the ignition switch is turned. It’s
also quite handy if you can have another switch that turns off the coil so you can crank the engine over without starting it up, say if the engines
not been used for a while and you want to circulate some oil round it first.
Fame is when your old car is plastered all over the internet
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blakep82
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| posted on 5/10/09 at 10:43 AM |
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thats how i plan to do mine 
________________________
IVA manual link http://www.businesslink.gov.uk/bdotg/action/detail?type=RESOURCES&itemId=1081997083
don't write OT on a new thread title, you're creating the topic, everything you write is very much ON topic!
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BenB
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| posted on 5/10/09 at 10:47 AM |
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Yup, just splice the supply to the keyswitch and feed the starter button with this, take the other side of the starter button to the starter solenoid.
Job done. Using the ignition light does indeed seem a nice trick, and 99% of the time it will work. The current flowing through the light pre-excites
the alternator doobries (technical term!) so it needs to be a "proper" bulb in the switch. If it's an LED you might well not get the
proper mojo flowing through the alternator packs.
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jabbahutt
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| posted on 5/10/09 at 10:56 AM |
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good point with the LED. If I keep the original ignition light to the alternator in the circuit and wire the LED in series with it then both the igin
tion light and switch light should both work?
Many thanks for the prompt responses, just trying to keep my mind ticking over
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A1
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| posted on 5/10/09 at 11:18 AM |
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ive just fitted a longacre one from demontweeks, it took about 10 mins to get it working. it was just the 3 wires from the ignition barrel wired to
the switch with the third ignition click wired to the button. it came with a wee diagram.
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David Jenkins
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| posted on 5/10/09 at 11:57 AM |
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I thought about doing all sorts of clever circuitry to make sure that the starter button wouldn't be active when the engine's
running...
In the end, I just put the button where only I could reach it, far away from all the usual controls.
Simples!
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02GF74
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| posted on 5/10/09 at 12:41 PM |
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quote:
Originally posted by jabbahutt
good point with the LED. If I keep the original ignition light to the alternator in the circuit and wire the LED in series with it then both the igin
tion light and switch light should both work?
no. the lamp is about 2 W so will draw around 200 mA - that is way too much for a LED - a 5 mm LED lamp would draw 20 mA and needs a drop
resistor.
If you wire the LED with the dorp resitor, unless it has one alread, in parallel i.e. both lamps connectoto 12 V and then to the alternator
contact, all will be well.
If you plan to do away with the ignition lamp, you will need to put a larger resistor in parallel with the LED lamp to provide enough currentto excite
the doobries of the alternator. Note that at low rpm you may see t he LED flash but this is because a LED repsonds far much quicker than filament
bump - it is nothing to worry about.
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jabbahutt
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| posted on 5/10/09 at 01:44 PM |
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thanks for the in depth details about resistors etc. Must admit it doesn't make much sense to me.
You mention using a resistor, in laymans speak couldn't I put the LED in series after the standard ignition bulb then the bulb would get
it's 2W and then wouldn't the voltage drop for the LED?
I already know that theres a reason why you can't do this just typing what pops in my head, plus I like to try and understand why I need to do
something a certain way rather than just doing it.
Many thanks as always with the patience with my electrical voodoo ingnorance.
If not what size resistor will I need to put in parallel with the normal bulb and will the resistor be in series with the led?
Nigel
[Edited on 5/10/09 by jabbahutt]
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neilj37
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| posted on 5/10/09 at 03:47 PM |
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I just wired a starter button in myself. I have included a 4 point relay in mine. I wired one side of the button to a permanent postive. The other
side ran to the relay. From the relay a wire went to earth and the other 2 connections went to the IG and ST wires on the key barrel. Think this
makes sense.
It was done similar to this but only 2 wires from the button as it is not illuminated.
http://pivotjp.com/download/img/pdf-e/ES-e.pdf
This is the kit i bought off ebay as i am an electrics numpty:
Starter button kit
[Edited on 5/10/09 by neilj37]
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02GF74
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| posted on 7/10/09 at 08:35 PM |
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quote:
Originally posted by jabbahutt
You mention using a resistor, in laymans speak couldn't I put the LED in series after the standard ignition bulb then the bulb would get
it's 2W and then wouldn't the voltage drop for the LED?
the bulb can be considered a resistor but its resistance changes since a metal's resistance increases when it heats up, in other
words, the bulb's resistance will be lower when cold.
You can fit bulb in series with the LED.
A LED needs about 1.5 Volts across it before it emits light and this occurs for a given current through it, as given by the spec. sheet (it is non
linear).
Thus the current through it needs to be limited - too much it blows - to about 20 mA.
now if the power supply is constant e.g. 12 V, you can do the sums to work out what resistor value is needed.
Now the bulb can be part of this resistor, albeit of low value so you would still need
a second resitor in series to limit the current.
Ok, so the LED lights up but the current will be so low, maybe 1/10th without the LED and resistor in the curcuit, hence the bulb with be very dim;
you probably won't see that the bulb is on.
Since the bulb isn't getting very hot, its resistance will be not much changed from when it is cold.
The LED will be on, but not the bulb.
If not what size resistor will I need to put in parallel with the normal bulb and will the resistor be in series with the led?
you would run the bulb as it is now and fit the led/resistor in parallel.
the bulb needs no resistor.
depending onthe LED you use, assuming as 5 mm LED, max current 20 mA, Vf 1.5 V,
sums (12 - 1.5)/20 mA is .... 500 ohm.
I hve used 12 as bttery voltage but in reality it will be nearer to 12.5 V when engine is not running.
Note tht Maplin sell 12 V LEDs - in effect these have the drop resistor built in - I ahve them in my Land Rover and have not blown one., and they run
at 13 + volts.
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