The Stovebolt Site Forums The Shop Area The Electrical Bay Ignition coil and resistor question

Hello.
Today I was at the parts store looking to purchase a spare coil.
Usual questions- what year, make, modle, engine size etc.
Which made me wonder, is there that big of a difference.

As I understand it, there are two posible coils for pre 80s american autos.
6 v coil till mid 5os. (same coil- ford, chevy, dodge)
12 v coil mid 50s to early 80s (same coil- ford, chevy, dodge)

Am I correct? Or will a Chevy coil only work on a Chevy?

Also do all resistors reduce the voltage to the coil from 12v down to 6-7 volts?

Can you eliminate the ceramic resistor if you use a coil with internal resistor?

Looking forward to learning more.

With the points closed I like to see around 8 volts. Points open should be battery voltage.


Yes, but you will not get as quick of a start as you would with a bypassed ballast resistor and a coil without an internal resistor.

"12V" GM coils are actually made with a primary (low voltage) winding that works right at around 9-9.5 volts. That's approximately the voltage a 12V battery has available during cranking, due to the load the starter puts on the electrical system. The ballast resistor is there to drop the voltage available to the coil low enough to prevent point damage when the charging system is operating against a fully charged battery- - - - -around 14.5 volts. During starting, the resistor is bypassed to allow for a hotter spark with the low battery voltage that's available. GM coils from about 1955 to the early 1970's (before HEI) should be interchangeable, regardless of the vehicle they happen to be listed to fit. Ford and MOPAR "12V" coils usually don't have the right primary winding resistance to work well in GM applications.
Jerry

Bill and Jerry.

Thank you for answering my questions.

I am by no means an expert or even a novice but had some similar questions on my 6 volt system.
I was told by others on here that no 6 volt system should use a resister - internal or external.
So, I replaced my internal resisted coil to non-resisted. I also do not have an external resistor.
Maybe some others on here can give a better explanation of why - it has to do with voltage drop at starting.

The reason ignition systems run a resistor in the primary circuit is for current control. Mechanical points can switch approximately 5 amps of current with a reasonable amount of reliability, so the current flow in the primary windings needs to remain below that threshold. Higher current flow results in very short point life. The problem is, that current flow isn't constant- - - - -at 2000 RPM, the points of a 6 cylinder engine are opening and closing 100 times per second- - - - -and the primary current is starting and stopping at the same rate- - - - -going from no current flow at all to maximum, and back to zero- - - - -repeatedly. The resistance of the primary coil winding is one factor to consider, plus another little detail caused by current flow in the primary windings called "inductive reactance" caused by the coil windings, and also voltage spikes in the primary caused by the magnetic field inducing up to 100 volts where there used to be anywhere from 4.5 to 7.5 volts, which is also happening 100 times per second. The engineering involved in even a simple point-switched ignition circuit is pretty amazing!
Jerry

Jerry, a really good explanation. I know that when microcontrollers switch relays or motors that have inductive coils there is a usually a flyback diode in parallel with the coil to prevent voltage spikes from damaging the controller. Are diodes ever used to prevent high back current from the collapsing magnetic field in automotive coils?

Ignition coils predate diodes about like Pterodactyls VS the space shuttle. The closest thing to induction dampening in a point type ignition system is a pretty primitive capacitor that tries to prevent flashover across the point contacts when the magnetic field collapses across the primary windings. It shunts the voltage spike to ground long enough for the points to get a few thousandths of gap established before the condenser charges fully. Then the condenser must discharge and recharge several times to dissipate the voltage before the points close again.
Jerry

Well wow. That's a lot. At the moment I am suffering from information overload. I am unfamiliar with the terms and concepts used in the explanations. I'm going to have to look them up to to get a better understanding. it's like cutting open a can and finding a whole universe inside.

Bear with me, I'm trying to understand.
When you start the engin there is a bypass wire that bypasses the resistor. Full 12 V for start up.
Once running 12v flowes through the resistor dropping to around 6 v through the coil primary winding to the points.

am I correct?

also what is the role of the secondary winding and the condensor?

The secondary winding is made up of thousands of turns of hair-fine wire, wrapped around a laminated steel core. As the magnetism generated by the current flowing through the primary winding collapses as the points open, it cuts across the secondary and induces voltage into each turn of wire. Several thousand volts are created, enough to jump the gap of the spark plugs. This process is repeated each time the points close, then open- - - - -one for each compression stroke of the engine. That's why the distributor must be timed correctly, so the spark happens just as each piston reaches the top of the compression stroke, with the right mixture of fuel and air in the combustion chamber.

The condenser acts as a shunt to minimize damage to the points created by the magnetic induction in the primary winding- - - -which momentarily spikes the primary voltage from 6 or 12 volts to a hundred or more.
Jerry

Jerry , fascinating explanation, in plain language.
I’m feel like I might be beginning to understand ignition!
So to follow through with this, I would love to know exactly what the effect is when the capacitor fails.
And the symptoms, a very important bit of knowledge I don’t have!
Thanks for all your help here on the ‘Bolt.
You are a font of knowledge, and a fount too.

Jerry didn't answer "am I correct " about the bypass.

Yes, you are correct.

Condensers can fail in two ways- - - - -open circuit, or shorted. A short, which is extremely rare, results in an immediate engine stop- - - - -just like the points staying closed all the time and preventing the coil from producing a spark. An open circuit, where the condenser fails to absorb some, or all of the voltage spike in the primary as the coil fires, results in a high current arc across the points every time the coil fires. The points burn, and the engine stops- - - -just not as quickly as with a short. Ideally, a "12 volt" coil with the proper size resistor in series with it will see around 7.5 to 9 volts applied to the upstream primary terminal when the resistor is in the circuit. The remainder of the charging system voltage (up to about 14.5 against a fully charged battery) will be dissipated by the resistor.
Jerry

To expand on this a little more, starting around 1960 GM had the bypass wire coming off the starter solenoid. There was a terminal mark " S " and one marked " I " on the solenoid. The " I " was for IGNITION, it supplied full battery voltage to the coil when the solenoid was pulled in while the starter was cranking. Once the engine started and the key was returned to RUN, battery voltage dropped off the bypass wire, and the coil used the resistor wire for voltage supplied by the key switch.

When electronic ignition came out in 1973, GM used the key switch to send power to the coil in the start mode doing away with the bypass and resistor wire.

The 1954 version used an upstream resistor, and a single wire to the coil with a second wire spliced in right after the resistor. In start mode, the key would apply battery voltage to the coil wire via the spliced in wire after the resistor, once it started, the battery voltage dropped off the spliced in wire and the voltage came through the resistor and out to the coil by way of the key switch run position. 1953 and prior used a single wire to the coil.

I went through the wire diagrams here, https://chevy.oldcarmanualproject.com/electrical/wiring/ it looks like 1954 was when they started using the bypass wires and resistors with variations of how they switched between them. Prior to '54, they were all wired using battery voltage from key switch to what must have been a coil with resistors built into them. By 1960 they settled on the starter solenoid / key switch method and stayed with it till '73.

If I got something wrong, feel free to point it out, I will change or delete so its accurate.

Chevrolet (and GMC with a positive ground) small trucks used 6 volt electrics until second series 1955. GM started their change from 6 volt to 12 volt cars with Cadillac, Buick and Olds in 1953. Pontiac and Chevy cars changed to 12 volt in 1955.

The 6 volt systems did not use any resistors in the ignition system, internal or external to the coil.

"6 volt" Delco-Remy coils were designed to work properly with a system voltage between about 4.5 (cold cranking) to around 7.5 during full charged battery operation at cruise speed, without a voltage control resistor.
Jerry

Thanks everyone for the information, I think I'm getting the basics but the higher level stuff is still foggy. Im going to have to think about it some more.

So this is what I am understanding so far.

Voltage to coil during start is 12-14v, during run the reducer drops the voltage to around 6-9v.

at the coil, power goes two seperate paths.

1. Primary path. Power goes into the (+) through the primary windings where the current is increased, then out through (‐) to/through the condenser where the excess volts are
stored so only 6-9 v go through the points, where when open they trigger the colaps of the electromagnet in the secondary system of the coil.


2. Secondary path. Through the secondary the voltage is increased via an electro magnet, which collapses when the points open. The voltage then travels through the cap to the rotar where it is distributed to the correct spark plug igniting the fuel.

Some questions.

Am I wrong with the above?

As I understand it, to test the condenser you test it's ability to absorb, hold and discharge energy.
Does the condenser absorb the extra volts then store them when the ponts are open and discharge when the points are closed?

If I where to test the Voltage at the coil +
during start it should be 12-14 V and when in run, it should show 6-9v. Is this correct?

Im still a little vague on what happens in the coils primary and what exactly the condenser does.

The primary purpose of the condenser is to prevent arcing (burning) of the points as they open.

The coil is a "step up" transformer. The voltage and current flowing through the primary winding is induced into the secondary winding (connects to the big wire going to the distributor) producing a lot more voltage (enough to jump the gap of a spark plug) and a lot less current.

Maybe a drawing will help.

Bill,
Thank you for the picture. I drew a similar one when I was trying to visualize the flow of current to the plugs and the points. Once I get home I will take a Pic and post it so it can be scrutinized for errors or mis conceptions.

So how does the condenser prevent the arcing. Does it absorb the excess voltage? If so what happens to the excess. And by current do you mean amps?

When I asked my original question, I didn't expect the amount of detailed information I would get. Which increased my curiosity.

Thanks to everyone for assisting in my learning and answering my questions.

Not the best rendering but it's what I used to visualize the path of the Volts.
Though it is not noted, I understand that in Start the volts are 12-14 v depending on your Battery and that the resistor drops the volts to 6-9 V during run. Just not captured on the diagram.

Looks to me the resistor drops the volts to the coil, the coil increases the volts and then the condensor drops the volts again before the points.

What am I missing? Is it a misunderstanding Volts v Amp?

If you had an oscilloscope attached to the negative side of the coil (or the wire lead on the condenser which is electrically the same place) you would see a picture that looks something like the attachment below. It shows one ignition cycle (one spark plug fires) with voltage and amps at the - terminal of the coil. Voltages would be less on a 6 volt system, but that spike when the points first open would probably still be in excess of 200 volts.

The equivalent of the condenser in your home plumbing would be an anti-hammer valve.

Close, but no cigar. Points and condenser are in parallel to each other as shown below.

Thanks Bill.
It's starting to make sense.

Bill,

The part which has me confused is that the voltage reducer drops the volts to "protect" the points, then there is the condensor which also "protects" the points. it's like there is a guard at the door and a body guard in the room.

Is it that, if the volts stayed a constant 12+, the volts coming from the coil would be increased to such an extent, that the condensor would not be able to handle the excess and the spill over would be to much for the points to handle over an extended period. ?

"Full" battery voltage may be down around 10 volts while the starter is engaged. When you are starting the engine, full battery voltage is switched to coil+ to provide a strong spark.

Once the engine starts and you release the key the voltage through the resistor is applied to coil+. The resistor reduces the DC battery voltage to around 8 volts at the coil+ when the points are closed. This reduces the current (amps) going through the coil primary winding extending coil and points life.

When the points first open the magnetic field in the coil collapses providing lots of volts from the secondary winding in the coil, out through the big wire to the distributor and on to one of the spark plugs. Meanwhile, back at the points there is a very short electrical ringing (see previous post's attachment) that arcs across the barely open points. That arcing over many miles is what causes "burning" of the points. When the points "burn" enough over time, the resistance through the points increase which reduces the amount of voltage drop and current through the coil primary, which reduces the amount of spark at the spark plug. The condenser's job is to short as much of that high voltage ringing to ground while still maintaining the 8 volt or so of DC voltage at the connection between coil- and the open moveable point.

Bill,
Thank you for the time and effort you have put into helping me understand. I believe I finaly get what's going on and how the separate components work together to provide ample spark without burning out.

Electricity and water follow the same laws of physics. Path of least resistance and once flowing it doesn't want to stop. The condenser on a point switched system is much like the expansion tank on a water heater or well pump.

When I went through USAF basic electronics classes in 1966 one of the instructors would use plumbing parts to help explain how electrics work. He said he could make a hydraulic radio, except he couldn't figure out how to get the antenna to work.

Almost the same conversation when I have a new mechanic and he's unsure or scared about hydraulic work. First I'll ask if they know basic electrical and have confidence in the subject. Then I let them know hydraulics is the same principle, it's all about flow. How it changes direction or creates an action is pretty darn close, too. Just different kinds of "switches." And like a 12v system, the fluid starts at the tank and ends at the tank.

Something like that.