pre lit
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| posted on 23/12/14 at 03:43 PM |
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Coil output
The ignition system on my crossflow is putting out nearly 19 v thro the negative terminal
Is this right the reason I Ask is my shift lights don't work correctly they seem to sense too many revs
Even h and h ignitions don't know what's causing it
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gremlin1234
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| posted on 23/12/14 at 04:02 PM |
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whats the battery voltage? (at its terminals), also check the voltage at the alternator with and without the engine running
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pre lit
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| posted on 23/12/14 at 04:22 PM |
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Battery voltage is 13.1 alternator puts in 14v back into battery
I was just wondering why the output is that high thro the coil as my shift light is connected to the neg terminal of the coil
Everything is working as it should except the shift light I have had more than one type of shift light on and they all do the same
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Dick
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| posted on 23/12/14 at 07:01 PM |
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Think i would check your using a 12 volt ignition coil and not a 6 volt one if it was me
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theprisioner
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| posted on 23/12/14 at 07:03 PM |
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The negative terminal of the coil can have 100V or so on it as it is generated by the back emf of the magnetic core of the coil. You should not
connect it to anything electronic. I am amazed it is still operating. You need to capacitive couple it at the very least (most). And possibly connect
a diode to kill of negative spikes.
http://sylvabuild.blogspot.com/
http://austin7special.blogspot.co.uk/
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pre lit
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| posted on 23/12/14 at 07:20 PM |
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So why does my Rev counter feed and shift lights all be recommended to be connected to the coil negative side
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AdrianH
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| posted on 23/12/14 at 08:22 PM |
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On the old Ford cars they all connected things like rev counters there, but, they also went through a small inductive coil. I think to stop some of
the high volatges going back to the instrument or possibly to stop a type of ringing in the circuit to the rev counter.
There is also a suppression cap on my coil, but can not remenber what side of the coil it is on, possibly the +Ve.
Adrian
Why do I have to make the tools to finish the job? More time then money.
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gremlin1234
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| posted on 24/12/14 at 03:22 PM |
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are you using points in a distributor?
if so worth changing the 'suppressor' to see if that helps.
or changing to an 'optical' or 'hall effect' electronic setup.
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pre lit
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| posted on 24/12/14 at 03:37 PM |
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It's a full electronic system from h and h ignitions
No points just high power electronic
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theprisioner
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| posted on 24/12/14 at 04:00 PM |
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Same problem
http://sylvabuild.blogspot.com/
http://austin7special.blogspot.co.uk/
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MikeRJ
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| posted on 24/12/14 at 08:30 PM |
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quote:
Originally posted by theprisioner
The negative terminal of the coil can have 100V or so on it as it is generated by the back emf of the magnetic core of the coil.
It can be a fair bit higher than as well, especially if you pull HT leads off to diagnose ignition problems (never a good idea). A meter will average
the short duration high voltage spike across the spark periods, so it will read much lower.
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pre lit
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| posted on 24/12/14 at 08:34 PM |
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Does any one have any suggestions how to stabilise the voltage to around a constant 9v I think the changes in the voltage are confusing the shift
lights
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AdrianH
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| posted on 24/12/14 at 08:44 PM |
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Resister and 9 Volt Zenner diode, something like a 10 K Ohmn couple of watt resister and a 3 watt Zenner diode.
The feed to the shift lights at the junction of the resister and diode and the other end resister to the -ve terminal, bottom end of diode to
ground.
Adrian
Why do I have to make the tools to finish the job? More time then money.
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pre lit
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| posted on 24/12/14 at 08:54 PM |
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I tried a resistor from from the neg coil output but this reduced voltage to only 5 volts so the shift lights would not light if I could just fits a
resistor of correct rating to supply a constant 9v would this be ok
With the multimeter whatever's the input voltage it was only 5v thro the end of the resistor
What size resistor would I need to give a constant out of the voltage i require 9v ish
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theprisioner
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| posted on 24/12/14 at 09:10 PM |
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You could try resistor from negative terminal (of coil) to shift lights (I/P) say 1Kohm (high voltage resistor) and diode from +12V (cathode) to
shift lights I/P (anode). The resistor will isolate you from the high voltage and the diode will clamp the pulses to +12V max.
You may need a fiddle about with the value of the resistor otherwise you may reduce the spark.
If I were attempting it I would use an oscilloscope to verify design.
http://sylvabuild.blogspot.com/
http://austin7special.blogspot.co.uk/
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gremlin1234
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| posted on 24/12/14 at 09:27 PM |
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having re-read the thread a couple of times,
I think you need a 'flyback diode' across the coil. 1n4001 should do, but I would go for a higher voltage rating, probably 1n4004 or 5
this wikipeadia article also suggests a very low value resistor in the circuit, (protects the diode!) - but most circuits don't bother. (hence
the use of the higher spec diode... )
http://en.wikipedia.org/wiki/Flyback_diode
edit:
actually rather than 1n4005, just go for a 3 amp, 600 volt one. 1n5406
http://uk.rs-online.com/web/p/rectifier-schottky-diodes/7743335/
or even the 1000 volt one 1n5408
http://uk.rs-online.com/web/p/rectifier-schottky-diodes/7743344/
(postage will cost more than the parts!!!)
[Edited on 24/12/14 by gremlin1234]
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theprisioner
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| posted on 24/12/14 at 09:57 PM |
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Sorry this is not correct. The flyback or ringing is the energy coming out of the core of the coil producing a back emf because there is no load. If
you clamp this you will get no spark. The high voltage on the primary is proportional to the high voltage on the secondary hence the spark.
http://sylvabuild.blogspot.com/
http://austin7special.blogspot.co.uk/
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pre lit
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| posted on 24/12/14 at 10:42 PM |
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All above my head I'm afraid
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theprisioner
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| posted on 25/12/14 at 09:42 AM |
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Does this help?
http://sylvabuild.blogspot.com/
http://austin7special.blogspot.co.uk/
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