Neil D. Kenneady (Aus)
There is a cheap and easy to install solution to this annoying problem and it will cost you less than $50.00, if you do it yourself. Before I disclose all about this wonderful solution, allow me to describe the reason for the existence of this starting difficulty. We need to look at each component of the starting system and it's function during the starting operation.
Battery - The battery is a lead type which is about 30% efficient when operating in ideal conditions. As temperature drops so does the battery efficiency. At 0 deg.C the battery is down to 12-15% efficiency. Age of the battery is also another factor, plates sulphate up and current delivery is severely affected. In short on a cold morning with an aged battery you will be very lucky to have 11 volts available from your 12V battery. A good condition battery, at 12 deg.C, should caryy around 12.5V.
Wiring - This is possibly the next most significant area of voltage loss after the battery. When the car was designed there was an almost fanatical obsession to save weight. The wiring did not escape this obsession, with minimum size wiring being used throughout the car. With age this wiring will develop some internal resistance, and combined with the push in type connectors used, we can experience significant voltage drop through the wiring harness - up to 1V. The condition of the earth cable from body to engine block (firewall to back of engine, just below the cylinder head) will also affect the operation of the engine electrics, particularly the performance of the starter motor.
Starter Motor - The Lucas starter motor is a terrible piece of work at the best of times, but when it is worn and tired it is producing a lot of internal mechanical drag which increases the current (amps) it requires to crank the engine. A cold engine with thick oil also increases the work for the starter motor which also increases the current (amps) required. During cranking the starter motor may draw upwards of 200A, depending on conditions.
Ignition Coil - The coil is a transformer with a 12V primary coil winding over a secondary winding. When the current flow through the primary winding is interrupted by the distributor points (opening) a high voltage (upward of 30,000V) spark is produced in the secondary winding and is directed to the appropriate spark plug, via the distributor rotor and cap.
Distributor - This device is nothing more than an engine driven, combination low and high voltage switch, which makes and breaks the circuit to the primary winding of the coil to earth.
At the instant the distributor points open, the magnetic field around the primary winding of the coil collapses and induces a high voltage output at the secondary coil winding, and this high voltage is delivered to the centre of the distributor cap via the coil lead and is directed to the appropriate spark plug (a cylinder which is on the compression stroke) by the distributor rotor. The result, a spark and ignition of the cylinder charge, release of energy and engine rotation.
All this waffle is fine you say, but how does this relate to a starting difficulty?
Let us look at a typical winter morning; the temperature is below 10 deg.C and the car has been sitting outside, unprotected all night. The temperature of the battery is likely to be lower than 10 deg.C and the oil will be about the consistency of thin honey. Battery voltage (dependent on age battery) could be around 11V. You pull the choke and start cranking the engine. The cold, thick oil makes the engine stiff to crank and the starter motor is attempting to extract 200+ amps from the battery (which is at 11V), which results in a further battery voltage drop of 2.5V (or more).
Now let's look at the ignition circuit. We have a 12V transformer (the coil) operating at 8.5V (or less) which has no hope of delivering the intended 30,000V to the spark plugs. We also have a rich air/fuel mixture (the result of using the choke) in the cylinders, which is cold and requires heat to ignite it (a good spark). Not a good set of conditions but with perseverance you usually get the car started.
If the engine starts it will be just as you release the key from the start position, the time when the starter motor load is removed and a battery voltage of near to 12V is available to the coil to produce a good high voltage spark.
But remember, the more you crank the engine, the more drain you apply to the battery and the lower the voltage gets. The only way you will get it started is by restoring the battery voltage, by charging or using jumper leads from another battery.
The solution to this problem is a 'Ballasted Ignition Coil' with it's own starting circuit. (I suspect it may have been the Japanese who introduced this device to the industry when they started copying the English engine designs and improving them.) The ballast resistor is the key to success of this conversion, however it requires some explanation.
We spoke earlier of cranking voltage being around 8.5V. By using an ignition coil designed to operate at 7.5V we have the ability to deliver a full voltage high tension spark at cranking voltage. However once we stop cranking, the engine starts and the charging system comes on line, we have a 14.2 to 14.8 operating voltage - the ignition coil is operating at nearly twice its design voltage. This condition will result in high current (amps) draw which will overheat the coil and induce arcing across the distributor points (burning).
By introducing a ballast resistor into the circuit between the ignition switch and the coil we are able to maintain the design wattage (volts/amps) of the coil while operating at a high system voltage.
Obviously we need to devise a way of switching the ballast resistor in and out of the ignition circuit for it to be of any use. In order to use such a system we require two circuits within the ignition circuit, one for normal running, which includes the ballast, and the second circuit, activated by the key starter switch, which bypasses the ballast and applies full battery voltage direct to the coil.
When the key starter switch is released the system must restore the ballasted ignition circuit. To achieve this a 15A relay is introduced which is activated by the signal from the key starter switch to the the starter motor solenoid switch, which in turn will switch battery voltage direct to the igntion coil. When the solenoid is deactivated the relay contacts open and leave the battery to coil circuit open.
Warning: You cannot take a power supply direct from the starter solenoid switch to the coil, nor can you take a power supply from the starter motor side of the solenoid switch. Either of these acts will results in power being fed to the solenoid switch or starter motor through the ignition circuit. The relay must be used to leave the battery to coil circuit open except when starting.
This modification can be easily made to the car without the need to cut or alter any of the original wiring - use 'LUCAR' type piggy-back connectors on the solenoid and coil. You require:
Disconnect the activating wire (small) from the starter motor solenoid and using a test light, connect the activating wire to the battery earth terminal. Turn the ignition switch to the start position and observe that the test light illuminates.
Disconnect and remove the battery and remove the distributor cap and rotor (leave the plug leads connected and secure the assembly out of the way).
Remove the existing coil and install the new coil in the same location. Mount the ballast resistor near the coil but ensure the body of the resistor is not touching the coil or any wiring (the ballast gets extremely hot during operation). Now mount the relay near the coil, perhaps on the opposite side of the coil, and carry out the wiring as detailed in the wiring diagram (to come! - ed.).
Complete the job by tape wrapping the two wires from the solenoid in a tape of similar colour to the original wiring (silver grey) and secure the wiring to surrounding wires and/or the battery carrier with plastic cable ties - loose wiring can result in broken wires. Replace the rotor and distributor cap and re-install the battery.
Note: The wiring diagram is for the original positive earth generator charging system, using the activating solenoid. If you have reversed the polarity (negative earth) or have installed an alternator, you will need to install the coil with the negative (-) terminal to the distributor.
This is a worthwhile modification to make to any car and if done carefully the only visible additional wires will be the power supply from the battery side of the solenoid and the one from the solenoid activating connector. Good luck.
©1997 Neil D. Kenneady