Generating a short duration spark

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A system is being developed to generate a short-duration spark, approximately 20ms, to ignite a methane jet for studying gas concentration fluctuations. The proposed method involves using a car ignition coil connected to a battery, with the spark duration controlled by the circuit's inductance, capacitance, and resistance. To automate the sparking every few seconds, a NE555 timer can be employed, but the switch must handle high voltage and current. An alternative approach is to utilize an old capacitor discharge ignition circuit, which can also be triggered by the NE555. Additional guidance and references on the subject are requested for further understanding.
percy1
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Hi

I'm trying to build a system to generate a spark with sufficient energy to ignite a freely expanding jet of methane, but with a quite short duration, ca. 20ms (if that is classed as short?) The purpose being to examine concentration fluctuations of gas (in a jet) above and below the flammability limits due to turbulence. I have pretty limited experience of electronics so am I hoping for some guidance, primarily so I have confidence in what I am doing, and am not going to do my self an injury.

From some internet research i think the way to go is to use a car ignition coil connected to a car battery, then the high voltage wire coming off the coil will spark with a suitable ground when the low voltage circuit is completed and then broken. Does this sound right? Are there any other components that i should have?

With a system like this how do you control the spark duration? Does varying the spark gap does this? (To keep the expt’s relevant I need to have a gap below 2/3mm), And also how can I determine the spark duration?

Any guidance or references to textbooks on this subject would be much appreciated

percy
 
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Here is a traditional (before transistors) automobile ignition circuit:

https://www.physicsforums.com/attachment.php?attachmentid=20126&d=1250393723

A 2 mH ignition coil (with 12-ohm internal resistance) is in series with a 0.02 uF capacitor in parallel with a switch to ground. When the switch opens, a 300-volt pulse is generated across the coil primary. With a 1:100 turns ratio, the coil secondary develops a 30 kV spark a fraction of a millisecond long. Note: the 300-volt pulse is also generated across the capacitor and the switch.

Bob S
 
Bob

thanks for your help. one or two more question if that is ok. How can i vary the spark duration, does the switching play a role in that? And also if i wanted to automate the circuit to spark every couple of seconds what kind of components would achieve that?

ta
percy
 
The high voltage oscillation frequency in the circuit

https://www.physicsforums.com/attachment.php?attachmentid=20126&d=1250393723

is proportional to sqrt(1/LC), where L and C are the inductance and capacitance in the circuit. The pulse duration is proportional to exp[-Rt/L], where t= time in seconds, and R is the total circuit resistance. Once a spark occurs (usually on the first half cycle), the oscillation is quenched. To fire the circuit every couple of seconds, a NE555 timer could be used as a timing source, but the switch would have to be able to open (switch 1 amp) quickly, and the open-switch hold-off voltage would have to be over the coil primary voltage (~300 volts). (Note in above circuit that the high voltage occurs when the switch opens).

An alternative is to get an old CD (capacitor discharge) ignition circuit, which uses a dc-dc converter to charge a capacitor to ~300 volts, and uses an SCR to discharge it into an ignition coil. These could be triggered directly by an NE555. Some old CD ignition modules are available on eBay (they were popular 1965 to '75) (Mark Ten is a good brand), and new ones are available in automotive supply stores (very $$$).

Bob S
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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