©By: John T. Blair (WA4OHZ)
1133 Chatmoss Dr., Va. Beach, Va. 23464; (757) 495-8229
Originally written: circa 1996
The ignition subsystem of a car is probably one of the least understood items by
most hobbyists. So, in this article, we'll take a look at the parts comprise this
subsystem, and how they work.
This discussion will assume that the reader has a basic knowledge of electricity
theory. If not, it is suggested that you read the article - Electricity primer.
When the ignition switch is closed, the electrons can flow along the wire, into and out of the coil, and to the points. If the points are open, then again the circuit is broken (opened) and no current can flow. When the points close, the current can pass across the points to ground, via the metal parts of the distributor, engine block, ground straps, and back to the battery. Simple!
The secondary side is a little harder to understand. Notice there is no battery! So how does it get any electricity? This is where the coil comes into play. The term coil is actually a misnomer here, it is really a voltage step up transformer. When an alternating current (AC) flows in the primary of a coil, a voltage is induced, created, in the secondary. This secondary voltage is very high, starting at about 20,000V.
To get this high voltage to the spark plugs another switch is needed, enter the distributor cap and rotor button. The high voltage enters the distributor via the thick wire from the coil to the center of the distributor cap. Inside the cap, there is a small rod that is spring loaded and will rest on the top center of the rotor button. The rotor button is keyed so that it will stay in place on the distributor shaft. As the engine turns, so does the distributor shaft, and since the rotor button is physically locked to this shaft, the button turns also. As the tip of the rotor button goes around inside the distributor cap, it almost touches the numerous contacts that the spark plug wires are attached to. If there happens to be any hi voltage present in the wires, as the rotor passes a contact, the current will flow to the contact and the attached spark plug wire, in turn.
Now that we have an understanding of the basics lets start looking at some of the
parts more closely and get a better understanding of the theory behind how this all
It should be noted here that there are both 6V and 12V coils. The main difference between them is the amount of amplification they performs on the input voltage. For example: The output of a given coil is to be 120V. If 6V were to be applied to the coil, it would have to be multiplied 20 times. If 12V were applied to a coil, it would have to be multiplied only 10 times. This amplification factor and many other parameters go into the actual design of the coil.
Remember that a plain piece of wire has very low resistance, and the primary of the
coil is just a long piece of wire to a DC voltage. If the ignition switch were in
the "on" position, the engine not running, and the points just happen to be closed;
there would be a complete circuit through the coil to ground allowing for very high
current to flow. Consequently, the coil's primary or the wiring harness would burn
up. Enter the ballast resistor.
A resistor is used to control electron flow and in so doing will cause the voltage to decrease across the resistor. (Remember Ohms Law, E=I*R; - from the electrical primer.) The 12V, from the battery, is applied to the input of the resistor. The resistor changes some of that energy to heat and the output voltage of the resistor becomes less than 12V - 7V for example. So the actual voltage at the input terminal of the coil will depend on whether there is an external ballast resistor or not. If the voltage is less than 12V there is an external ballast resistor somewhere. This resistor may be a resistor or special resistive wire.
On some ignitions systems the external ballast resistor is automatically switched out
of the circuit when the car is being started. The rational is that the voltage applied
to the coil will be higher and since the coil was designed for a specific multiplication
factor, the output will be higher. For example: normally 7v applied to the coil. If
there is a multiplication factor of 10 then the output of the coil is 70V. When the
input voltage is allowed to be 12V, with the same x10 factor, the output is 120V or
50v higher. This can be very helpful in cold weather starting.Most of the British cars of the 50s and 60s had their ballast resistor built into the
coil. I hadn't seen the external resistors, on any of my British cars, until I
started playing with my 77 Spit.
Once the ignition switch is closed, the primary circuit has voltage, 12V, applied all
the time. The coil's primary return wire goes to the points and condenser. When the
points are open there is NO ground on the coil's primary side and the condenser is
allowed to charge. When the points are closed, it **shorts** out the primary's
voltage to ground and the condenser is allowed to discharge.
There is a "Charge time" (time constant) associated with the a condenser. The charge time must be designed to match the time between required pulses to the spark plugs.
The condenser has 3 basic functions.
The distributor has jobs to perform in both the primary and secondary ignition circuits and is the most complicated item in the entire ignition system.
In the primary circuit, the lobes on the distributor shaft open and close the points, helping to turn the DC from the battery into pulsating DC. There will be 1 lobe on the distributor shaft for each spark plug. In the secondary circuit, the distributor shaft and rotor button form a large rotating mechanical switch to distribute the high voltage to each of the spark plugs at the required time. The distributor cap is the static part of the switch and routes the spark to each plug.
In addition to distributing the spark, the distributor with its mechanical and vacuum advance mechanisms can change the timing of when the spark is delivered to the plugs.
The distributor cap can collect moisture and or carbon dirt. When this happens, the high voltage is allowed to travel through the water or carbon dirt to the metal base and "short" out the distributor. The engine will miss. This can be readily fixed by removing the distributor cap and wipping it with a clean paper towel.
I have seen in a few rare cases, the rotor button short out. The high voltage line comes into the center of the distributor to a spring loaded contact. This contact presses on the rotor button, which sits on the distributor shaft. If the material the rotor button is made of breaks down, the high voltage is allowed to go through the rotor button and to ground via the distributor's drive shaft.
To test the low voltage side of the distributor and coil, remove the high voltage wire from the distributor cap. Hold it near a ground, any metal on the engine block, and have someone try to start the car. If there is a nice spark everything including the coil is functioning correctly.
To test the high voltage side of the distributor, simply remove a spark plug wire from
one of the spark plugs and insert a Phillips screwdriver into the connector at the
end of the plug wire. Hold the shaft of the screwdriver near a ground, any metal on
the engine block, while someone tries to start the car. If there is a nice spark
everything is functioning correctly.
This ends the basics of the ignition system. I hope this has helped take some of the mistery out of this system. A more indepth information on transformer theory follows for those that are interested.
Enjoy your Morgan
The following is a more indepth look at some of theory behind the operation of a Coil.
Scientists found that moving a magnet across a coil of wire would induce (make) a current flow in the wire. This is the bases of how a transformer works.
To make a transformer a core (something to wind the wire around) is needed. The core is usually made of soft iron and have any shape. However, the material and shape will determine how it works in an AC circuit (beyond the scope).
In a DC circuit. When power is applied to the transformer primary a magnetic field will be established. This building of the field will induce 1 spike across a transformer. To induce a continual voltage on the output (secondary of a transformer) the magnetic field must continually build and collapse, build an collapse. This can be achieved by using AC source or in automotive use - pulsating DC.
In an AC circuit, as the positive half of an AC signal flows in the primary it sets up a magnetic field around the secondary windings. This will produce 1 and only spike (its size, duration, and ringing are based on things like the ratio of primary to secondary windings, RC time constants). When the incoming AC signal starts to flow in reverse, the initial magnetic field collapses across the secondary and again induces a spike. Therefore, the output frequency of the coil is the same as the input frequency. Or if you put a coil (transformer) on a typical 60Hz household circuit, the output will be 60Hz also.
Since a coil is nothing more than 2 very long pieces of wire, and the primary wire is
relatively large, it has only a few Ohms resistance to DC (from the battery).
To paraphrase the conservation of energy, You can't get something for nothing. The best you can hope for is for the input and output to be almost equal. Therefore, in a perfect environment, power into a device would be equal to the power out of a device, Pin=Pout. We can substitute the P=I*E into the conservation of energy equation and get: Iin*Ein = Iout*Eout. And this can be rearranged to:
Ein Iout ----- = ------ Eout Iin
This equation, shows that the Voltage and Current across a transformer are inversely related. That is to say that if the output voltage is increased across a transformer, the output current is decreased by the same factor. This is why you don't get killed when you pull a spark plug wire off a running engine. There is enough voltage to cause your muscles to spasms but there isn't enough current to kill you.
A transformer is nothing more than 2 coils of wire, a primary and a secondary. For a voltage step up transformer the secondary will have more turns of wire than the primary. So if you want to multiply the input voltage by a factor of 2, then the secondary will have to have twice as many windings as the primary. The equation for this becomes:
Vi = Vo --- --- Np Ns
where Vi is the voltage applied to the primary,
Vo is the voltage from the secondary
Np is the number of turns (coils of wire) in the primary
Ns is the number of turns (coils of wire) in the secondary
To make a step up transformer, we will need 3 items:
Return to the Index of Tech. articles
To email me with comments or questions.