Will this one work?

t0m1980 said:

Thanks for the in depth explanation which makes a lot more sense now sort of.....

Until I start looking at resistors, then it becomes a minefield with all the different various types

This one for example looks ok

https://www.ebay.co.uk/itm/25W-75-O...049948&hash=item5923bf57bf:g:5XcAAOSwnvFcm4CY

but do I have to find one with the same ampage as the battery?

So sorry to tie up everybody’s time with this, but if a job is worth doing and all that......

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No, that is a huge resistor, 25W! You just need a little balancing resistor, simple. Try something like this for 10p...

https://uk.rs-online.com/web/p/through-hole-fixed-resistors/6835587/

Shame Maplins is not about any more.

the resistor 100 ohms is to overcome the use of a led normally the 2.2watt lamp is used to excite the alternator ,the resistor is paralleled across the led to do the same job
the diode is I presume to stop the fuel shut of valve staying energised via this resistor and stops a feed being applied to the alternator whilst the engine and alternator are idle
the resistor is 100 ohms at 5% tolerance and would be rated at about 3watt the diode you could possibly use a IN4007

mystery said:

the resistor 100 ohms is to overcome the use of a led normally the 2.2watt lamp is used to excite the alternator ,the resistor is paralleled across the led to do the same job
the diode is I presume to stop the fuel shut of valve staying energised via this resistor and stops a feed being applied to the alternator whilst the engine and alternator are idle
the resistor is 100 ohms at 5% tolerance and would be rated at about 3watt the diode you could possibly use a IN4007

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You are sort of on the correct trail, but not quite. Read my post above.

The resistor resistance needs to be similar to that of the indicator lamp, that is so that the lamp actually lights and was not put there for LED's - these have been going into charge indicator circuits since the dawn of charge indicators! Mine is a 2W lamp and 2W at 12V = 72Ohm - 75Ohm is the fairly standard resistance for these circuits and sold in kit car looms etc. If there was a 3W lamp in you would need even less resistance, i.e 48Ohms. But basically it just needs to be similar.

The diode doesn't do anything for the fuel shut-off and doesn't stop feed going to the alternator (it's polarised the wrong way to do that anyway) when the engine is off, the ignition switch stops feed "to" the alternator - strip the part of the circuit pertaining to alternator wiring out of the larger diagram and what you have is a standard alternator charge light circuit. The diode basically acts as the "alarm" circuit "switch".

think you will find a 2 watt lamp =equals 72 ohms and a 2.2 watt general indicator lamp is generally around 68 ohms taking into account the voltage output can realise 14 + volts
would be 89 0hms so a 100 ohms resistor with a 5% tolerance that could vary to outside temperature conditions I would say is good practice

A massive break through today guys, I've only managed to start the old girl on the key (even if the new ignition is awful, and yes it was a $#!T Part), I even saw the oil light come on and go off when the engine turned over!!! So a massive thanks to all you guys for helping me to get this far, I could never of done it without you and a special thanks to you mystery for simplifying the diagram.

Now back to the diode and resistor I have purchased what has been mentioned and will have to wait now to fit them, but at least I have started to make progress, I will update in due course.

I got both for good measure, but I am leaning towards the one on the right as it looks more robust

mystery said:

think you will find a 2 watt lamp =equals 72 ohms and a 2.2 watt general indicator lamp is generally around 68 ohms taking into account the voltage output can realise 14 + volts
would be 89 0hms so a 100 ohms resistor with a 5% tolerance that could vary to outside temperature conditions I would say is good practiceView attachment 173843 View attachment 173844

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I am a little confused by your post, like you had had some great revelation, read my posts, I had explained it all.

You clearly were not really sure how any of this worked, I explained it to you, then you post pages repeating everything I have just told you. I think you are still confused as to what is going on here based on your comments about diodes stopping fuel solenoids remaining energised and resistors to "overcome the use of a LED".

In response to your last post explaining the resistance of the resistor, I had just explained a 2W lamp is nominal 72Ohm. I wrote: "Mine is a 2W lamp and 2W at 12V = 72Ohm - 75Ohm is the fairly standard resistance for these circuits". Also, now you are all clued up on these matters think for yourself why the next part of your statement is incorrect.

Your assumption regarding the diodes purpose was incorrect and I told you what it was for in post #23, in post #18 also explained what the resistor was for.

discomania said:

No, that is a huge resistor, 25W! You just need a little balancing resistor, simple. Try something like this for 10p...

https://uk.rs-online.com/web/p/through-hole-fixed-resistors/6835587/

Shame Maplins is not about any more.

Click to expand...

So these arrived today,

Do they have to be wired in a certain way, as in, do the colours mean anything?

t0m1980 said:

So these arrived today, View attachment 174120

Do they have to be wired in a certain way, as in, do the colours mean anything?

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They're just resisters. The dun't have a polarity. Can be put into circuit any way round. The colours on them decode to give the resistance value and tolerance %.

Hippo said:

They're just resisters. The dun't have a polarity. Can be put into circuit any way round. The colours on them decode to give the resistance value and tolerance %.

Click to expand...

Excellent thank you very much