Lotus Elan and Lotus Elan Plus 2 Forums

Yes stu it is working properly again. Unfortunately my brain is "wired" for mechanical work and not electrical work, which is handy when all the problems I've been having are electrical

This has been a useful post; one item I seem to be missing is if you have a 12 v ie internal ballast coil, with the usual two terminals, then I presume a relay output doesn't do anything as cannot bypass the internal ballast? (See below I didn't measure a diff. between coil voltage and solenoid output terminal).
I note the lotus manual shows what looks like a ballast resistor mounted just beside the input terminal on the coil with the two connections ballast and not ballast, however I do not have such.
I measure about 12 v at both battery and coil wire (disconnected from coil) and about 9.5 v while cranking, (again disconnected). I measured the output from the solenoid take off cranking and it too was 9.5 ( note a geared small starter with the solenoid on the starter). My coil measures about 1.8 ohms so should be changed to a. ~ 3.2 ohm. By the by, would an overheating coil cause some missing or just fail completely?
Jack

Many people seem to misunderstand the ballasted coil system.

Let me try explain in a nutshell.

1. The ballasted coil is DESIGNED to work at about 9 volts.
2. During starting the coil is connected to the battery supply which would usually be around 12v but because the starter motor is taking a lot of current, the main battery supply has dropped to about 9 volts.
3. So during starting the coil is working at its designed voltage and gives a healthy spark.
4. during normal running the main battery supply rises to its normal 12 volts approx.
5. continuous running at 12 volts would damage the 9 volt coil, so during normal running it is fed through a ballast resistor, which effectively drops the voltage to about 9 volts so again it is working at its design rating.

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It is tricky to get the full battery voltage feed to the coil, (bypassing the ballast resistor) during starting. First thoughts might simply be to connect it to the fat terminal on the solenoid that drives the starter motor. The snag with that is that then during normal running there is a back circuit through the ballast resistor, then backwards down your bypass wire to the fat terminal and thence to the starter motor. The motor is low resistonce, (it will try to turn, but not with full uumph), but the ballast resistor will overheat and burn out.

You could put a diode in the bypass wire to prevent the back flow, but that defeats the purpose, because diodes always lose about 1.5 volts across them, so the starting voltage on your coil would be 9 volts approx, less 1,5 volts i.e around 7.5 volts, so you don't get full sparks.

So the only practical method is that the wire that bypasses the ballast resistor has to be unconnected during normal running. So it could be fed from an extra contact on the starter switch or an extra contact on the solenoid or if you haven't got either of those on your car, you could arrange a relay with its coli wired in parallel with the starter solenoid and its one pair of contacts arranged to connect the battery supply direct to the ignition coil, bypassing the ballast resistor.

Thanks Bill; ok, but if no external ballast (12 v coil) then there is no separate resistor to bypass, regardless of source?
The solenoid aux. coil output is the 9v on cranking, but then once running, the normal wiring would yield 12 v to the coil.

I think many if not all ballast coil ?lans have the ballast resistor hidden within the wiring loom in the form of a long resistive wire, on Fords it used to run a continous uninterrupted loop up to the o?l pressure switch and back, presumably to dissipate the heat generated.

Fitment of a non ballasted coil to such a vehicle will result in a vastly reduced HT voltage unless either the resistive wire is removed or a seperate 12V feed is made to the coil.