A common automotive ignition coil can generate an output voltage on the order of 30,000 volts. This is a sufficient voltage to produce arcs which jump an air gap of an inch or more. Driven properly they can also be used to create a Jacobs Ladder display
Ignition coils are inexpensive and readily available, and constructing a circuit to drive the coil is straightforward.
This article provides an overview of ignition coil operation, as well as a drive circuit design and system capable of driving the coil to produce arcs for your entertainment and experimentation.
Of couse, as with any high voltage project, safety warnings are in order. Excecise caution and good sense when working with high voltage. Remove power from the system before making adjustements.
Don't touch the arcs. Be mindful that the potentials generated can jump a significant gap, and that insulation on tools like regular pliers may be inadequate to prevent you from getting a shock.
Keep in mind that some parts of the system can get hot. The arcs can be hot enough to to ignite paper and plastic, so operate it in a safe location.
High voltage discharges generate Ozone gas, which can cause irritation if breathed in. High voltage discharges also generate some ultraviolet radiation, so limit your exposure and don't stare directly at the arcs.
First, enjoy some video!
Ignition coils are inexpensive and readily available, and constructing a circuit to drive the coil is straightforward.
This article provides an overview of ignition coil operation, as well as a drive circuit design and system capable of driving the coil to produce arcs for your entertainment and experimentation.
Of couse, as with any high voltage project, safety warnings are in order. Excecise caution and good sense when working with high voltage. Remove power from the system before making adjustements.
Don't touch the arcs. Be mindful that the potentials generated can jump a significant gap, and that insulation on tools like regular pliers may be inadequate to prevent you from getting a shock.
Keep in mind that some parts of the system can get hot. The arcs can be hot enough to to ignite paper and plastic, so operate it in a safe location.
High voltage discharges generate Ozone gas, which can cause irritation if breathed in. High voltage discharges also generate some ultraviolet radiation, so limit your exposure and don't stare directly at the arcs.
First, enjoy some video!
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An ignition coil produces a high voltage output on its secondary when the current flowing in the primary winding is interrupted. The first step in generating a high voltage from an ignition coil is to store energy in the inductance of the primary winding. That energy is then released, generating the high voltage output.
Energy is stored in the primary of the coil when current is flowing through it. This energy is provided by a DC power supply, usually the 12 volt battery of a vehicle’s electrical system.
The primary circuit is driven by closing a switch to ground, which allows current to flow from the power supply through the primary. When the switch is initially closed, the current in the coil remains zero, as the inductance of the primary does not allow the current to change instantaneously. The current in the primary then increases exponentially until it reaches its steady state value.
The steady state current is the maximum value the current will have. It is determined by the voltage of the power supply and the total series resistance of the primary circuit.
The steady state current in the primary is determined by:
I_steady_state = V_source/R_primary
Rprimary consists of the total resistance of the wire in the primary coil, as well as the resistance of the wires and other connections in the primary circuit.
The length of time needed for the current in the primary to reach its steady state value is determined by the time constant. The time constant is a parameter which is equal to the primary inductance (in Henries) divided by the primary resistance (in ohms), and so it has the units of seconds. In equations it is commonly represented with the Greek letter tau. A smaller time constant means that the current will increase more quickly.
The equation for the current in the inductor is:
i(t)=I_steady_state*(1-e^(-t/time_constant)
The graph shows how the current in the coil increases after the switch is closed. The graph expresses the current as the percentage of steady state value. Note that after 4 time constants have passed, the current is at about 98% of its steady state value, and the after 5 time constants, the current is at more than 99% of its steady state value.
The relevant equations for the primary circuit are summarized in a chart below.
Energy in the Primary Winding
The energy stored in an inductor is a function of its inductance and the current flowing through it. The equation for the energy stored in an inductor is:
Energy= ½ * L * I^2.
Refer again to the graph that shows the current and energy in the primary coil versus the time elapsed. Note that the longer the switch is closed, the less additional energy is stored in the primary inductance. After the first time constant, the energy stored is 38% of the maximum. After two time constants have elapsed, the total energy stored is 73%. After three time constants, the total energy stored is 90%. After four time constants, the total energy is 96%, and so on. After five time constants, the energy stored is essentially at 100% of the maximum possible value for the power supply voltage and primary resistance in the circuit.
Once the primary current has reached its steady state value, no additional energy can be stored in the primary of the coil. If the switch remains closed after this point, the energy from the source will simply be dissipated as heat in the resistance of the primary circuit. It therefore makes no sense to keep the switch closed longer than 4 or 5 time constants, as past that point the energy will simply be wasted.
Energy is stored in the primary of the coil when current is flowing through it. This energy is provided by a DC power supply, usually the 12 volt battery of a vehicle’s electrical system.
The primary circuit is driven by closing a switch to ground, which allows current to flow from the power supply through the primary. When the switch is initially closed, the current in the coil remains zero, as the inductance of the primary does not allow the current to change instantaneously. The current in the primary then increases exponentially until it reaches its steady state value.
The steady state current is the maximum value the current will have. It is determined by the voltage of the power supply and the total series resistance of the primary circuit.
The steady state current in the primary is determined by:
I_steady_state = V_source/R_primary
Rprimary consists of the total resistance of the wire in the primary coil, as well as the resistance of the wires and other connections in the primary circuit.
The length of time needed for the current in the primary to reach its steady state value is determined by the time constant. The time constant is a parameter which is equal to the primary inductance (in Henries) divided by the primary resistance (in ohms), and so it has the units of seconds. In equations it is commonly represented with the Greek letter tau. A smaller time constant means that the current will increase more quickly.
The equation for the current in the inductor is:
i(t)=I_steady_state*(1-e^(-t/time_constant)
The graph shows how the current in the coil increases after the switch is closed. The graph expresses the current as the percentage of steady state value. Note that after 4 time constants have passed, the current is at about 98% of its steady state value, and the after 5 time constants, the current is at more than 99% of its steady state value.
The relevant equations for the primary circuit are summarized in a chart below.
Energy in the Primary Winding
The energy stored in an inductor is a function of its inductance and the current flowing through it. The equation for the energy stored in an inductor is:
Energy= ½ * L * I^2.
Refer again to the graph that shows the current and energy in the primary coil versus the time elapsed. Note that the longer the switch is closed, the less additional energy is stored in the primary inductance. After the first time constant, the energy stored is 38% of the maximum. After two time constants have elapsed, the total energy stored is 73%. After three time constants, the total energy stored is 90%. After four time constants, the total energy is 96%, and so on. After five time constants, the energy stored is essentially at 100% of the maximum possible value for the power supply voltage and primary resistance in the circuit.
Once the primary current has reached its steady state value, no additional energy can be stored in the primary of the coil. If the switch remains closed after this point, the energy from the source will simply be dissipated as heat in the resistance of the primary circuit. It therefore makes no sense to keep the switch closed longer than 4 or 5 time constants, as past that point the energy will simply be wasted.
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If I use no driving circuit and just apply approx 13v DC to the coil and have 1 wire out of where a lead to a distributor would go, and 1 wire to the negative ( for the sparks, are these called anodes and cathodes?), then have the positive isolated with a switch.
Is it true that I will get just 1 quick spark just a split second after I turn switch off?.
What i am trying to understand is if the driving circuit just switches the supply power on & off very quickly!?!?!? and a spark is discharged when supply is interrupted. If this is true, does allowing an earth have the same effect as switching off current, except current is still on.
In other words will the mere act of presenting an earth then trigger a spark while un-driven power is applied? Or is it a case that absolutely nothing will happen if constant uninterrupted power is applied?
I would love to try your PC based circuit but I currently have no way of programming a PIC and have only Macintosh laptops at my disposal. I suppose I could invest in a PIC programmer and borrow a PC to just load the files on to the PIC. There are other projects such as the Aurora 9x18 RGB LED art (PIC24F08KA101) that also use a PIC from Microchip. Could I just buy the Microchip's PICkit 3 In-Circuit Debugger/Programmer? Would this work with the files you provide here?? Or could you sell me a few programmed PICs? I could also buy a few PICs and fedex to you for programming, if you would be willing to help me.
Also, could you send me a BOM for the parts list?
OK to email direct: kuriloff@nyhni.org
There is a misconception among some that an ignition coil operates like a Tesla coil. An igntion coil is actually an iron core transformer, with a turns ratio on the order of 75:1. Switching off the primary current causes the primary voltage to rise to several hundred volts, which is in turn stepped up by the turns ratio.
What is important is to have an idea of the inductance and the resistance of the primary, as that lets you determine the time constant. From there you can configure your pulse generator to turn the switch on for an appropriate length of time.
I'm not familiar with the PICkit 3 In-Circuit Debugger/Programmer. If it is a product made by Microchip for general in circuit programming then it should be able to be used. I have actually never programmed any parts using in circuit programming before, but I know that there is an app note on Microchips web site explaining what you need to do regarding connections to the processor.
The files I have attached should work regardless of the programmer you are using. You will have to use the 12F683 with the files supplied, unless you know how to update the source code for a different PIC and then assemble to get a hex file for programming the different PIC. That isn't terribly difficult if you know a bit about the assembly language.
I don't have a BOM of the circuit per se, but the schematic has manufacturers PNs for all the more critical parts, like the processor and the IGBT. The other stuff is just common resistors and caps. 1/4 Watt resistors work fine. The caps you select must be rated to handle whatever maximum voltage you plan on applying to them. For example, the ones on the input to the +5V regulator will see the power supply voltage, so select them depending on the max power supply voltage you intend to use.
Dan
The inductance of the secondary was too high to measure with my meter. The secondary resistance was more than 10k ohms if I recall correctly.
I've seen projects where some people have driven two coils, with opposite polarity output, so that the potential difference between the two is double. If this is what you have in mind, I would suggest that you use two of the exact same coils. The reason being is that the discharge on the output is fairly brief, and the timing is determined by the inductances and resistances and turns ratios of the coils. If two different coils were used, the output voltage peaks may occur at different times between the two coils, and the output may no be as great as you hope. Again, I haven't tried this type of dual coil arrangement, so I am sort of speculating here.
The primary of an ignition coil has a fairly high inductance, and so the current can't be driven with a high enough "carrier frequency" which could then be modulated to produce any kind of reasonable quality audio. Flyback transformers are usually what is used for plasma speakers.
At any rate, the drive to the switch needs to be a digital signal to turn the switch either on or off, and so you wouldn't want to apply an analog signal directly to the gate/base of the switch you are using.
I asked because it is easier for me to get for free a ignition coil than a flyback.
One more question. What type of material should be used for the electrodes, so they would not melt after a couple of minutes of arc production?
The wires may get a little warm after producing arcs continously for a while.
I've only run the setup here for about 3 minutes at a time. If you are planning on making something to run for much longer, its best to check it out and see just how hot it gets. An of course, don't leave it unattended.
The method I have used in the pasts is to drive che primary by discharging a 1 /uFcapacitor charged at 400V with a thyristor. With hi-voltage coils one can reach 60,000-70,000V without any damage to the coil.
The system you describe, where a cap is charged to 400V and then discharged through a coil via thyristor is called capacitve discharge igniton (CDI). I've never experimented with such a system, but I belive they use a different type of coil. Anyway, the older inductive discharge "Kettering" type system that my design is easier to implement, as CDI needs a circuit that can generate the ~400V needed to first charge the cap.
To achieve the 400V in the capacitive discharge method, I used the common voltage multiplicator circuit (Villard circuit and similar) to go from the typical household AC voltage line to the desired voltage, making the circuit quite simple..
I'd like to try a capacitve discharge type circuit, but I think I would need to use a coil intended for a CDI system. The coils I used in the project here were all intended for inductive disharge systems, and have an inductance of about 3 to 5 mH.
For a confirmation of the purpose of the capacitor please read the following explanation from:
http://en.wikipedia.org/wiki/Ignition_system
(quote):....
At the same time, current exits the coil's primary winding and begins to charge up the capacitor ("condenser") that lies across the now-open breaker points. This capacitor and the coil’s primary windings form an oscillating LC circuit. This LC circuit produces a damped, oscillating current which bounces energy between the capacitor’s electric field and the ignition coil’s magnetic field. The oscillating current in the coil’s primary, which produces an oscillating magnetic field in the coil, extends the high voltage pulse at the output of the secondary windings. This high voltage thus continues beyond the time of the initial field collapse pulse. The oscillation continues until the circuit’s energy is consumed.
(end)
I personally designed, build and sold SCR ignition units in the early seventies.
One of the various occupations that help me to pay for my MSc EE.
The standard coil had no problem holding the voltage, because the spark at the sparkplugs acts as a voltage limiter. On the other hand, if one bench drives a standard coil with a large airgap, it is 100% certain that the standard coil will be damaged. As I said with a hi-voltage coil and a SCR unit, it is easy to achieve 60,000-70,000Volts
I am afraid there are no misconceptions. In any case job well done.
From: http://en.wikipedia.org/wiki/Ignition_system
(Quote)
At the same time, current exits the coil's primary winding and begins to charge up the capacitor ("condenser") that lies across the now-open breaker points. This capacitor and the coil’s primary windings form an oscillating LC circuit. This LC circuit produces a damped, oscillating current which bounces energy between the capacitor’s electric field and the ignition coil’s magnetic field. The oscillating current in the coil’s primary, which produces an oscillating magnetic field in the coil, extends the high voltage pulse at the output of the secondary windings. This high voltage thus continues beyond the time of the initial field collapse pulse. The oscillation continues until the circuit’s energy is consumed.
(End of quote)