When I was a kid Sears sold Model T Ford spark coils for about $1. I was about 11 or 12 when I bought my first one to make a Jacob's Ladder. Soon bored with it I tried to take it apart but was stymied by the tar-like potting. Then, about 16 years old, I bought one and made a Tesla coil. It worked, but poorly, probably because I used rolled up newspaper for the secondary coil support.

The other day I googled that coil and learned some interesting things; Primary 212 turns, secondary 16,100 turns, frequency about 200 Hz.

The turns ratio is about 78:1 so, if the output is 7,800 volts (I'm making the math here easy) the input must be about 100 volts. I realize now (but didn’t as a kid) that this voltage comes from the collapsing magnetic field when the current through the primary is interrupted.

A 200 Hz frequency gives a time of only 5 milliseconds for each cycle.

I can't work out how a 6 volt battery can charge a coil/capacitor circuit enough in roughly half of 5 milliseconds to create 100 volts.

 

It all depends on how much charge is used by each spark. If charge is small, then it doesn't take long for the battery to supply it in each cycle.

However, I think this is not germane in this case since, AFAIK, the Model T used a magneto system for feeding the primary side of the trembler coils on each cylinder. I think a battery (for ignition) was used only for starting and even that was optional and not recommended by Ford.

 

I can't work out how a 6 volt battery can charge a coil/capacitor circuit enough in roughly half of 5 milliseconds to create 100 volts.

It's actually a flyback circuit.

• The current builds up in the primary inductance during the time the contacts are closed, storing magnetic energy as ½I²L.
• The inductance is low enough that the current can build to several amps in 5ms.
• When the magnetic field from this current gets high enough it pulls the contacts open.
• This generates an inductive kick in the primary across the points which is absorbed by a small capacitor across the points to prevent the energy being wasted in a spark across the points.
• This primary kickback voltage is converted by the turns ratio to a higher voltage in the secondary.
• If the secondary generates a spark, then the inductive energy that was transferred to the capacitor, now is delivered as spark energy.

Make sense?

 

Well, I am not interested in what Ford recommended. I'm asking about a well known coil. This is not homework—I'm just an octogenarian trying to return to a youthful hobby.

If I understand the formula, and have not messed up the decimal places AND have made reasonable assumptions about the primary coil then:

So let's guess .1 ohms (I just guessed at 10 feet of 20 gauge wire). Let's guess at 10 mH (wild guess).

Time constant = L/R = .00001/.1 = .0001 sec = 0.1 millisecond.

So, assuming a duty cycle of 50% (probably close) the battery would have just under 3 time constants
to charge the coil.

I don't have a clue as to how to calculate the resulting voltage if the battery is 6 volts. Is it likely to be 100?

 

Hello,

Here is a page from the wiki about the trembler coil that is used in de model T ford:
https://en.wikipedia.org/wiki/Trembler_coil

Bertus

 

Well, I am not interested in what Ford recommended. I'm asking about a well known coil. This is not homework—I'm just an octogenarian trying to return to a youthful hobby.

If I understand the formula, and have not messed up the decimal places AND have made reasonable assumptions about the primary coil then:

So let's guess .1 ohms (I just guessed at 10 feet of 20 gauge wire). Let's guess at 10 mH (wild guess).

Time constant = L/R = .00001/.1 = .0001 sec = 0.1 millisecond.

So, assuming a duty cycle of 50% (probably close) the battery would have just under 3 time constants
to charge the coil.

I don't have a clue as to how to calculate the resulting voltage if the battery is 6 volts. Is it likely to be 100?

It's not a matter of what they recommended (or whether you are interested in what they recommended). You said you didn't understand how something was possible and I'm pointing out that you seem to be asking about something that wasn't happening in the first place. It's like me saying that I'm trying to figure out how the Wright Brothers got their first plane off the ground using a diesel engine and someone pointing out that they didn't use a diesel engine.

The Model-T ignition system didn't use a 6 V battery, it used a flywheel driven magneto. When the battery WAS used in later models it was used for starting only.

Leaving that aside, I don't see how using fanciful made-up numbers that have no discernable relation to any real values (i.e., just wild guesses) is useful for figuring out whether the actual voltage in an actual system is close to a particular value or not. Let's say the answer is Yes, it is likely to be 100 V. Does that mean that the voltage in the actual system is likely to be 100 V? No, because the parameter values you have here may not bear any relation to the actual values. Let's say that, instead, the answer is No, it can't be anywhere near 100 V. Does that mean that the voltage in the actual system can't be near 100 V? No, because the parameter values you have here may not bear any relation to the actual values. So regardless of the answer, you won't have learned anything that relates to the question you are trying to answer.

 

I don't have a clue as to how to calculate the resulting voltage if the battery is 6 volts.

The battery voltage affects the charging time of the coil, and thus only indirectly the output voltage.
The output voltage depends upon the primary peak current, the primary inductance, and the capacitor value.
Only when you know all three can you determine the input flyback voltage.
The peak capacitor voltage occurs when all the ½LI² inductor energy has been transferred to the ½CV² capacitor energy during the flyback.

Below is the LTspice simulation of the basic ignition primary circuit.
For the simulation values shown, the primary inductor current builds up to 2.4A in 5ms.
When the switch opens, there is a 350V negative flyback voltage generated across the capacitor.
Notice the peak capacitor voltage occurs when the inductor current has returned to zero (all inductive energy transferred to the capacitor).
If this energy is not dissipated with a secondary spark (which is not simulated here), then the primary will continue to oscillate as an LC resonant circuit until the energy is dissipated in the primary resistance.

 

Thank you @crutschow! Your simulation shows that at 2.5 milliseconds there could be about 200 volts.

And thanks to the other responders; I'm sorry that I did not make myself clear--I have no interest in the history of the Ford coil, or the Ford car or man. I am interested in the physics that make the coil work. Also, as crutschow's simulation shows, my guesses were not too far off.