Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne Ignition Coil
Name: Paul
Status: student
Age: N/A
Location: N/A
Country: N/A
Date: N/A


Question:
When a car ignition coil increases the voltage from 12v DC to 10,000v DC, what happens to the value of the current? Voltage and current have a direct relationship. Does the current increase also?



Replies:
Nope. As voltage is increased, current DEcreases. What stays the same is power, which is voltage times current.

Richard Barrans Jr., Ph.D., M.Ed.
Department of Physics and Astronomy
University of Wyoming


No, Paul, it is the inverse, for transformers. what is conserved is power. So when voltage goes _up_ by a factor of N, the current goes _down_ by a factor of N.

But I should tell you that is for an efficient AC transformer, which transfers energy continuously, and is never suddenly interrupted. Ignition coils waste some power and are also calculated differently.

They store and discharge energy. While the 12v is applied and the primary-coil current ramps up, energy is being stored. The energy stored is E = (1/2) * L * I^2, where L is the inductance of the primary coil, and I is the current. (It is exactly like the formulas for energy on a charged capacitor: E = (1/2) * Capacitance * Voltage^2 or kinetic energy of a moving mass: E = (1/2) * Mass * velocity^2. )

After the current is ramped up, all further time running high current adds no more magnetic stored energy, it is just wasted in the resistance of the wire, making the coil get hot. (Many gimmicks on market to minimize that. Capacitor-discharge ignition is one of them.)

Then a switch (the distributor points) opens and stops the primary current. That forces a corresponding current to flow in the secondary, about 1000x less current, but capable of charging the spark cable to 1000x higher voltage. More, actually, if the cable's capacitance is small. That stored energy insists on getting out somewhere, and merely charging the cable to high voltage is not quite enough. At some point the spark-plug reaches its breakdown voltage, and all the stored energy is discharged into that spark.

End of story until next time the points close, re-starting the 12v charge-up.

It is hard to get a good value of L of an ignition coil, but E = (1/2) * Voltage * ChargeTime * FinalCurrent also works. Charge-time is the time mentioned above, the time during which the primary coil current is linearly increasing. (if it's not linear, rather an exponential decay to he final current, extrapolate the initial ramp rate with a straight line to where it would reach the final current, and read that time interval.) It's some milliseconds; you can see it on an oscilloscope when hooked up to measure current in the primary side of the coil.

And I think the actual turns ratio of wire-windings inside an ignition coil may be less than 1000:1, maybe it is around 200:1. The top 5x or 10x of voltage increase happens because of the inductive voltage surge that always happens when an inductor with current is suddenly open-circuited. The peak voltage the ignition coil can reach must always be higher than what the spark-plug requires to spark over. Maybe 2x higher.

All that being said, thinking of it as a transformer still kind of works, including the inverse relationship of voltage and current.

Jim Swenson


Yes they have a direct relationship, but not in the manner you see with typical transformers in steady-state applications. In a step-up transformer, output voltage is greater, but at a lower current than the input current (conservation of energy). On the 12V side of an ignition coil, current ramps up over time (inductors resist changes in current) until it reaches a few amps, slowly building up energy in the magnetic field. When it gets disconnected, the magnetic field collapses and creates "back EMF" at a much higher voltage on the secondary side because of the turns ratio. There is no current until the voltage reaches the dielectric strength of the air-fuel mixture at the spark plug, at which point the gasses ionize and form nearly a short circuit. At this point there is a very brief spike of current (around 10-100 microseconds long) that spikes and decays. The current can be quite high (nearly a tenth of an amp), but being relatively brief, does not violate conservation of energy.

Paul Bridges


Yes, voltage is directly related to Current times Resistance. (Ohm's Law) When you have a step up voltage coil what happens is the voltage increases But the current decreases.

In physics you have the law of the conservation of energy (Joules). Power is rate of energy consumption (Joules/sec) Electrical Power (Watts) is the product of current (amps) times voltage (volts).

So if you increase the voltage, the current decreases unless you have what is called an "active device" to add power, Like a battery, alternator/generator, or Power company adding power to the system, somewhere.

"Passive devices" just consume power, "active devices" add power to a system. (but they really don't add power they just convert power from one form (chemical power in batteries) to electrical power.

Sincere regards,

Mike Stewart



Click here to return to the Physics Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (help@newton.dep.anl.gov), or at Argonne's Educational Programs

NEWTON AND ASK A SCIENTIST
Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory