I'm in the process of building an ignition enhancer for the basic Kettering ignition system. I've got the main parts built and have tested the system and it works extremely well. However, it's not very reliable; after extended use I am burning up components. Here is a diagram.

The diodes and capacitors are working swimmingly, not having any trouble. However, I believe the inverter is getting power surges from the secondary side of the system; after extended use the inverters quit working.

I realize that the filtering on the output of the inverter needs to be much more robust, and this is where my questions start.

In addition to the .001uF capacitors across the output, where/how can I include inductors to more effectively absorb/shunt the high voltage surges that I assume are getting back to the inverter??

 

Hi Pinhead,

I'm curious / puzzled - what is the main idea in your circuit design?

 

Hi Pinhead,

I'm curious / puzzled - what is the main idea in your circuit design?

When the original ignition system fires the spark plug, the spark acts as a short circuit through which current can flow. The 300 vdc is able to source much needed current to intensify the spark.

This video shows just how much of a difference the circuit makes.

http://www.facebook.com/v/111293731973

 

delete -- Sorry, misread the schematic.

John

 

I've made a modification to the above schematic that will more-than-likely increase the life of the spark plugs while not taking away much spark intensity. I found an old radio shack "noise suppressor" coil that was designed to run in-line with the DC power for my CB radios. It looks like a transformer with only a primary side (no secondary). The coil was used to reduce EMI that would enter the radio through the power wire.

This video demonstrates the difference. My video phone is pretty low quality; you'll have to listen closely to hear the difference in the spark. The only way I can think to describe the sound, is the spark doesn't sound as "sharp."

http://www.facebook.com/v/112827451973

 

Lets see if I understand this circuit correctly.

You create the initial spark from the ignition coil, but try to substain the spark from the invertor (220VDC)?

You do realize the spark is going to be substained as long as it is feed voltage? While it is happening you effectively have a closed switch on the invertor, voltage doubler, and the rest. I suspect you are overloading all the circuitry (with an effective short) and burning it out. Perhaps you need to break the circuit at some point, and let the 10µF 400V cap take the brunt of the load (circuit designations are your friend in this case).

 

i think Bill hit nail on head.

perhaps try to trigger a HV fet for a pulse period. but as the rpm's increase the frequency of the inverter dumping HV will increase, and if the freq gets too high it will start to look like a DC short.... and, have you done any in-lab experiments with this? you might be surprised, starting the short with HV (spark) and dumping low voltage (12v on large cap) in a field of compressed fuel mixture might yield the results you are looking for. this looks similar to getting a strobe (flash) tube to fire with energy, etc.

my advice, buy a MSD 6AL box.

 

Lets see if I understand this circuit correctly.

You create the initial spark from the ignition coil, but try to substain the spark from the inverter (220VDC)?

You do realize the spark is going to be substained as long as it is fed voltage? While it is happening you effectively have a closed switch on the inverter, voltage doubler, and the rest. I suspect you are overloading all the circuitry (with an effective short) and burning it out. Perhaps you need to break the circuit at some point, and let the 10µF 400V cap take the brunt of the load (circuit designations are your friend in this case).

Before I built the circuit I would have agreed with you. However, I've found that the 300 VDC is unable to sustain the spark on its own (as both videos demonstrate). Even while running on mains voltage the spark duration is the same with and without the 300 VDC applied to the circuit. I know the main power isn't shutting down, verified by voltage readings taken at the AC side of the voltage doubler.

 

Before I built the circuit I would have agreed with you. However, I've found that the 300 VDC is unable to sustain the spark on its own (as both videos demonstrate). Even while running on mains voltage the spark duration is the same with and without the 300 VDC applied to the circuit. I know the main power isn't shutting down, verified by voltage readings taken at the AC side of the voltage doubler.

hmmm, the spark fall time (on to off) should be very fast. what type of output stage does the inverter have? is it sensitive to sudden current transient (on to off)?

 

Before I built the circuit I would have agreed with you. However, I've found that the 300 VDC is unable to sustain the spark on its own (as both videos demonstrate). Even while running on mains voltage the spark duration is the same with and without the 300 VDC applied to the circuit. I know the main power isn't shutting down, verified by voltage readings taken at the AC side of the voltage doubler.

The point I'm making is you mention it fails under sustained use. That loading is only going to increase with acceleration. I didn't say the main power is shutting down, I'm saying the invertor is being forced to feed too much juice to the spark, which is not it's purpose. A arc is an effective dead short, way down there on the ohmage scale. Charge the cap, separate the cap from the invertor, then dump that charge from the cap.

Just my 2¢ worth.

 

Once the flame front has burned away from the vicinity of the spark plug, what purpose is served by continuing the arc? There is nothing left to ignite.

 

hmmm, the spark fall time (on to off) should be very fast. what type of output stage does the inverter have? is it sensitive to sudden current transient (on to off)?

Wouldn't a large capacitor/inductor filter help to buffer the sharp "pulses" that the inverter sees? This was my original question.

Once the flame front has burned away from the vicinity of the spark plug, what purpose is served by continuing the arc? There is nothing left to ignite.

I could write an entire article answering this question but I'll just say it's much more complicated than it seems; it has to do with local a/f ratio vs overall a/f ratio, how well the mixture is homogenized, how much turbulence is in the chamber, amount of engine load, humidity, etc, etc... In other words, I'll just say it is important.

THIS TOPIC at GoFastNews.com could shed some light on the subject.

The point I'm making is you mention it fails under sustained use. That loading is only going to increase with acceleration. I didn't say the main power is shutting down, I'm saying the invertor is being forced to feed too much juice to the spark, which is not it's purpose. A arc is an effective dead short, way down there on the ohmage scale. Charge the cap, separate the cap from the invertor, then dump that charge from the cap.

Just my 2¢ worth.

The problem I see with doing this, is complexity. Plus the fact that the motor that this system will be going on spins up to 10,000 rpm in a waste spark configuration (firing twice per revolution). Without completely rebuilding the entire ignition system, I don't see how this can be a possibility. If I wanted to go that route, I'd just spend the bajillion dollars on an aftermarket CDI conversion and be done with it. :/

If you haven't guessed yet, I'm a tightwad.

 

Originally Posted by DC_Kid View Post
hmmm, the spark fall time (on to off) should be very fast. what type of output stage does the inverter have? is it sensitive to sudden current transient (on to off)?

Wouldn't a large capacitor/inductor filter help to buffer the sharp "pulses" that the inverter sees? This was my original question.

no, i dont think by adding a series inductor with a dc bypass cap will help things. what i'm saying is maybe the output stage of the inverter is being damaged by the suddenly stop of current flow.

what type of inverter is it? is it a 12vdc to 120ac and then you are doing full wave rectification to get your ~200Vdc ??

what you are wanting to do is exactly what my TIG machine does. it super imposes hv (hv frequency) over the dc to help start the dc flow. perhaps you can look into how a TIG inverter works vs. the inverter you have. my TIG inverter can run 100% duty cycle at 160A (but i have to back off because my air cooled torch gets too hot to hold). the difference i see right away is that my TIG has current limiting, whereas your setup is basically shorting out the inverter output.

perhaps incorporate a current limiter like this (http://www.radio-electronics.com/in...rent_limiter/power_supply_current_limiter.php). perhaps design it so it limits to 50% of inverter rated output.

...and can you explain how you are isolating the hv of the spark from the inverter?

 

A 12vdc to 120vac inverter feeds into a simple capacitor/diode voltage doubler circuit using two 330 μF capacitors, the output of which is smoothed by a 330 μF 400v capacitor. I've also got a few ceramic-disc capacitors across the output to help short out some of the spiking. This is all shown in the schematic in the first post.

 

looks like you are trying to use a simple Greinacher voltage doubler like this

looks like your D1 is tied to ground, so the output of the inverter sees 25ohm + diode directly to ground??

seems to me you would want to double the AC and then run it through a full wave rectifier to get DC, or, use a Villard voltage doubler like this:

and how does your 50kV PIV diode stop the spark voltage from entering the inverter?

 

The link is interesting, but mostly from lack of hard data.

Looking at the spark duration issue: At 2500 RPM, the engine is turning at 41.667 RPS. That is 24 msec per cycle. Since it's a 4 stroke engine, we divide by 4 to get the actual duration of the power stroke - .006 sec. Running a spark longer than about 1/4 of that duration is simply wishful thinking. Is there actual data on better combustion resulting from increased spark duration? If so, how much of an increase is significant?

Notice that the duration is proportional to RPM as set out above. If you want a longer spark, use a charged capacitor bank. Something like the capacitive discharge ignition I used back in the 1970's.

Modern spark ignition uses significantly higher voltages than back in the period up to the '80's. More like 40KV, whereas it used to be 25KV. The spark plug gap is significantly wider, for a larger, hotter spark from that alone.

Back in WWII, Charles Lindberg did a lot of work with aero engines and running a very lean cruise. He got surprisingly good results, considering 1940's technology. The very lean cruise regime allowed the P-38's to fly far enough to intercept Yamamoto in 1942.

But an aero engine at cruise never varies the output, which is utterly unlike an auto engine which is continually varying its output to adjust for widn, road grade, and so on. A major improvement in the ECU and better sensor placement could probably lead to better fuel economy through a tighter control loop. When you have to obey emissions standards and so on, it's never quite as good as it might be.

On the other hand, I have a '07 Honda Fit. 109 HP and automatic tranny. I get an average of 38.2 MPG driving in the country. 44 if only highway driving. Shedding weight and body shape can do a lot for mileage.

 

so why not use www.pulstar.com (junk if you ask me, but worth a try on the dyno)

also, its a lot easier to upgrade the spark system and widen the spark gap to get more spark energy into the burn kernel.

with todays latest ECM controls i do not see any significant advancements in spark via higher output spark coils or by silly things like pulstar spark plugs - meaning the results show little gains (if any) on the dyno.

 

looks like your D1 is tied to ground, so the output of the inverter sees 25ohm + diode directly to ground??

and how does your 50kV PIV diode stop the spark voltage from entering the inverter?

The positive output IS tied to ground. However, the VAC is internally isolated from ground inside the inverter; the only steady state current that the inverter sees is the current that flows through the secondary of the ignition coil (about 1 milliamp).

The Kettering ignition uses a collapsing magnetic field in the primary to induce a voltage in the secondary. Due to this collapse, the spark voltage is negative with respect to ground. That is why the diodes look "backwards" compared to what one would initially expect and is how the diodes are able to block the spark from shorting into the inverter.

With regards to the discussion at GoFastNews, I can assure you I'm not basing all of my plans and goals on that thread... I've been working on combustion enhancement for years. I'm much more advanced with combustion and mechanical systems than I am with circuitry, which is why I posted this thread.

Edit: I've looked extensively into Pulstar plugs and while they do "work" they're terribly unreliable (at least they were when I investigated them a year ago). And they're EXPENSIVE.

 

So no suggestions about filtering this circuit?

 

I've found that adding more resistance between the coil and the voltage doubler has eliminated the "filtering" problem and the circuit no longer destroys the inverters. However, I've come across another problem.

Adding resistors in-line with the circuit has proven to be problematic, as the resistors keep burning up. Anything below 420 or so ohms would work for a while without heating whatsoever (not warm to the touch), but then suddenly burn up. I wired a 60w light bulb in-line with the circuit and discovered why the resistors would suddenly fry. For some reason, something sets up a continuous arc instead of a quick spark. Here's a video...

http://www.facebook.com/v/119016156973