Homemade CDI ignition

[BrownOut]

Hi again, it occured to me that just because your module burned out, that doesn't mean your coil is bad. Chances are it's still good and can be re-used. I should stand corrected in my comment that you probably don't have CDI. Guess all my 2strokes are too old, and I'm not up on the newer technology. I'm glad to hear these old PWC's are still running, because I'm looking for a good deal on one myself

 

[DerStrom8]

I doubt you have a 200-300 VAC alternator on your jetski.

This is probably true, but it is not uncommon for the voltage across, for example, the primary of a regular car ignition coil, to spike at 100-200 volts. When 12 volts is applied, and then quickly removed, the voltage in the primary often spikes to a couple hundred volts. This may be what the OP was talking about.

 

[BrownOut]

This is probably true, but it is not uncommon for the voltage across, for example, the primary of a regular car ignition coil, to spike at 100-200 volts. When 12 volts is applied, and then quickly removed, the voltage in the primary often spikes to a couple hundred volts. This may be what the OP was talking about.

Hi Der Storm8, I was refering to the site that was linked that shows 200-300 volts are required for CDI ignition. Do you know why such high voltage is requried ( I ask because of your high-voltage experiments ) But on further refilection, I take back what I said ( see my post just above ) There are more small engins that are using CDI than I realized. This is an interesting subject for me, as I own and operate a number of small marine motors.

PS: Thanks buju357, you validated what I thought was right,but wasn't quite sure.

 

[crutschow]

Hi Der Storm8, I was referring to the site that was linked that shows 200-300 volts are required for CDI ignition. Do you know why such high voltage is required ( I ask because of your high-voltage experiments )

The storage capacitor is charged to 200-300Vdc. The SCR then fires and applies the cap voltage across the coil primary. The 100:1 or so coil turns-ratio generates the 20kV-30kV at the secondary that fires the plug(s).

This is unlike a Kettering type ignition which uses the coil inductance in a flyback type circuit. With the points closed the current builds up in the coil primary. When the points open the inductive energy from this current generates the couple hundred volt spike on the primary which is then stepped up by the coil secondary turns ratio. The CDI uses the coil only as a transformer, the coil inductance is not used for energy storage.

 

[SimonTHK]

I have worked extensively with 2 stoke engines.
all 2 stoke twins fire at the same time , when one piston is on the compression and the other on the exhaust. they have 2 pickups 180 degrees opposed on the stator housing. 1 coil splits into 2 leads , each going to a cylinder. The reason they program cdi's is to get better low and top end power by varying the timing. effectively if your 2 stoke revs to say 8000 rpm its going to be sparking at 8000 x 2 = 16000 rpm.
a small car coil WILL work , its one of the tricks to get a higher voltage spark at the pliugs or if the original coil burns out.

btw - I have a 1982 yamaha RD 350 LC with 45k km's on the clock and the coil is still working , coils seldom burn out.

thank you, this was very usefull.

My motor is 2 stroke, 2 cylinders place in a line.

So basically when I say that 1 spark plug was working and the other didnt (and then that the last one burned a little later), it cannot happen unless a wire is cut to the spark plugs, cause they simply just spark at the same time. From same coil.

How can it give more top end power by decreasing the times it spark?

 

[SimonTHK]

Hi again, it occured to me that just because your module burned out, that doesn't mean your coil is bad. Chances are it's still good and can be re-used. I should stand corrected in my comment that you probably don't have CDI. Guess all my 2strokes are too old, and I'm not up on the newer technology. I'm glad to hear these old PWC's are still running, because I'm looking for a good deal on one myself

If you are gonna get one, then look for either a very well tuned 650 kubik (it aint cheap to tune it yourself, but it is cheap to buy a tuned one) or a 750 kubik, else they are not powerfull enough for the fun, believe me.

The current coil is very well placed inside a box filled with glue to keep it away from moist, I cannot get it out in pieces, I tried with my old one, but I will try since this model is abit different.

 

[DerStrom8]

Hi Der Storm8, I was refering to the site that was linked that shows 200-300 volts are required for CDI ignition. Do you know why such high voltage is requried ( I ask because of your high-voltage experiments ) But on further refilection, I take back what I said ( see my post just above ) There are more small engins that are using CDI than I realized. This is an interesting subject for me, as I own and operate a number of small marine motors.

PS: Thanks buju357, you validated what I thought was right,but wasn't quite sure.

Thanks crutschow for answering this. To be honest, I don't know much about ignition, and I guess I misunderstood what you were saying, BrownOut. All I know is what I have read, and I have not had much experience with CDI. I guess I should have stayed out of this one

The storage capacitor is charged to 200-300Vdc. The SCR then fires and applies the cap voltage across the coil primary. The 100:1 or so coil turns-ratio generates the 20kV-30kV at the secondary that fires the plug(s).

This is unlike a Kettering type ignition which uses the coil inductance in a flyback type circuit. With the points closed the current builds up in the coil primary. When the points open the inductive energy from this current generates the couple hundred volt spike on the primary which is then stepped up by the coil secondary turns ratio. The CDI uses the coil only as a transformer, the coil inductance is not used for energy storage.

 

[SimonTHK]

This is probably true, but it is not uncommon for the voltage across, for example, the primary of a regular car ignition coil, to spike at 100-200 volts. When 12 volts is applied, and then quickly removed, the voltage in the primary often spikes to a couple hundred volts. This may be what the OP was talking about.

I cannot tell myself right now, but how else should it work?

 

[SimonTHK]

The storage capacitor is charged to 200-300Vdc. The SCR then fires and applies the cap voltage across the coil primary. The 100:1 or so coil turns-ratio generates the 20kV-30kV at the secondary that fires the plug(s).

This is unlike a Kettering type ignition which uses the coil inductance in a flyback type circuit. With the points closed the current builds up in the coil primary. When the points open the inductive energy from this current generates the couple hundred volt spike on the primary which is then stepped up by the coil secondary turns ratio. The CDI uses the coil only as a transformer, the coil inductance is not used for energy storage.

Very nice So you say that it is the coils that charge the 400v capacitor and not the alternator which just charges the 12v battery on the ski?

 

[DerStrom8]

I cannot tell myself right now, but how else should it work?

I am still not sure how it works with these coils. I have a constant current running through my primary, with what voltage? And then when I discharge the capacitor, it wil create a spark?

Here, read this post from crutschow. It seems to explain it fairly well

The storage capacitor is charged to 200-300Vdc. The SCR then fires and applies the cap voltage across the coil primary. The 100:1 or so coil turns-ratio generates the 20kV-30kV at the secondary that fires the plug(s).

This is unlike a Kettering type ignition which uses the coil inductance in a flyback type circuit. With the points closed the current builds up in the coil primary. When the points open the inductive energy from this current generates the couple hundred volt spike on the primary which is then stepped up by the coil secondary turns ratio. The CDI uses the coil only as a transformer, the coil inductance is not used for energy storage.

 

[BrownOut]

Thanks crutschow for answering this. To be honest, I don't know much about ignition, and I guess I misunderstood what you were saying, BrownOut. All I know is what I have read, and I have not had much experience with CDI. I guess I should have stayed out of this one

I think your praticipation is fine and appreciated. If I suggested otherwise, I apologize. The advantage of using high voltage, as I understand, is that is allows faster charging of the ignition system, thus allowing higher engine RPM. The high voltage required for a spark comes from rapidly changing the current in the seconday coil, and in accordance with the equation VL=L*dIL/dT. So, the higher the change in current ( the dIL/dT ) the higher the output voltage. There are two ways to accomplish a rapid change in current. The more common, as mentioned, is to use a low voltage primary in a flyback configuration. A low voltage is applied to the primary and the primary current slowly builds up, creating an expanding magnetic field. Then, the current is suddenly interrupted, and the field collapses. The secondary voltage is given by V=Ns*dθ/dt, where θ is a product of the magnetic field created by the primary current. when the primary current is quickly interrupted, the dθ/dt term is very high, creating the high dIL/dt in the secondary. Note that it takes some time for the magnetic field to build up in this scheme.

Another way to get a rapidly changing magnetic field is to apply a high voltage to the primary. Thus, the equation IP=LP∫VP*dt, where IP is the primary current, LP is the primary inductance, and VP is the voltate applied to the primary, this current rises quickly, due to the high voltage being applied. This created a rapidly expanding magnetic field that induces the high voltage in the secondary given by the equation above. So, the high voltage is created upon the application of the primary voltage, rather than on the interruption of the primary current. The advantage is that the high voltag doesn't need as much time to charge as the primary current did in the first case, and so higher PRM's can be achieved. This is important in small 2-stroke engines, because for one thing, they develop their power at high RMP, and also because the ignition has to fire at twice the rate as a 4 stroke would.

You might use these principles in your high voltage experiments.

Hope that explanation also answers this question:

How can it give more top end power by decreasing the times it spark?

It doesn't. The CDI allows for higher RPM as described above.

 

[SimonTHK]

Here, read this post from crutschow. It seems to explain it fairly well

ya sry, english aint my first language, I can have difficulties in understanding sometimes. I know I have a coil that works as a transformer from 300v to 30kv, but what charges the primary coil with current? I dont know.

I figured out now that I can buy two different coils placed inside the engine (not being the 30kv transformer coil) the one is called charge coil, the other is called exciter coil. I am not able to look up what exciter means, but I guess that it is another word for start or spark kinda. Sadly I cant find any good info about the electrical in the ski and I aint going to tare it apart when summer is over us.
The two coils can be seen here **broken link removed**
Here is a picture of the two coils **broken link removed**
I dunno if that is called an alternator or just a stator, I dunno the difference.

 

[KeepItSimpleStupid]

Best guesses:

Charge coil = This coil voltage that will be rectified to charge the battery
Exciter coil = This coil voltage will be rectified to delivers the 300 VDC that the CDI requires.

Stator basically means the non-moving part. Rotors are normally electrically magnetized to create the magnetic field.

 

[crutschow]

Another way to get a rapidly changing magnetic field is to apply a high voltage to the primary. Thus, the equation IP=LP∫VP*dt, where IP is the primary current, LP is the primary inductance, and VP is the voltate applied to the primary, this current rises quickly, due to the high voltage being applied. This created a rapidly expanding magnetic field that induces the high voltage in the secondary given by the equation above. So, the high voltage is created upon the application of the primary voltage, rather than on the interruption of the primary current.

That's not quite right. When you apply a sudden high voltage to the primary from the capacitor, the high primary inductance prevents the inductive current IP, from significantly increasing in the short time the voltage is applied (a few microseconds). The majority of the primary current that flows is simply due to the transformer action between the primary and the secondary spark current.

Transformer action does not use the primary inductance as a medium to transfer the energy. It uses the change in magnetic flux which is proportional to the change in voltage, not current.

 

[BrownOut]

Magnetic flux is always proportional to current, per Ampere's law, and not voltage. The only way to get a rapid change is magnetic flux is to rapidly change the primary current. In the low voltage case, that can only be accomplished by suddenly interrupting the current flow in the primary, becuase the primary inductance prevents the current from rising quickly. But in the high voltage case, the primary current rises proportionally higher, to the higher voltage. The governing equation is: IP=∫VP*dt. If the applied voltage is say 300/12=25 times higher, then the result of the equation will be 25 times higher, thus the primary current rises that much higher.

Note: I'm not an expert in CDI, this is only my theory of how it operates. I've never worked on CDI, but I have worked on high voltage capacitive discharge circuits for hemetric sealing. This is how that system worked.

 

[crutschow]

Magnetic flux is always proportional to current, per Ampere's law, and not voltage. The only way to get a rapid change is magnetic flux is to rapidly change the primary current. In the low voltage case, that can only be accomplished by suddenly interrupting the current flow in the primary, becuase the primary inductance prevents the current from rising quickly. But in the high voltage case, the primary current rises proportionally higher, to the higher voltage. The governing equation is: IP=∫VP*dt. If the applied voltage is say 300/12=25 times higher, then the result of the equation will be 25 times higher, thus the primary current rises that much higher.

You are still somewhat confusing the small magnetizing current of a normal operating transformer with the large inductive current of a flyback transformer.

In a flyback transformer, the primary inductance stores energy when the primary is conducting for transfer to the secondary when the primary current is interrupted. The stored primary energy is transferred to the secondary at that time. It is a two-step process -- Storage of energy in the primary in the first step. Transfer of this energy to the secondary in the second step.

In a normal operating transformer the primary current is just sufficient to establish the required magnetic flux based upon the applied voltage and the frequency. Even though the primary voltage is much higher with a CD ignition, the generated primary inductive (magnetizing) current is much smaller due to the short duration (high frequency) of the applied voltage. The energy transfer is through the flux connecting the primary and secondary, not by the energy stored in any inductive current. (For example, the magnetizing current of a power transformer is a constant based upon the voltage and frequency and is largely independent of the amount of power or current being transferred through the transformer.) It is a single-step process with the capacitor energy being (nearly) instantly transferred to the secondary by the transformer process.

 

[BrownOut]

You are still somewhat confusing the small magnetizing current of a normal operating transformer with the large inductive current of a flyback transformer.

No. My post covered both the CDI and the flyback modes of operation. I never said the CDI system uses flyback current.

In a flyback transformer, the primary inductance stores energy when the primary is conducting for transfer to the secondary when the primary current is interrupted. The stored primary energy is transferred to the secondary at that time. It is a two-step process -- Storage of energy in the primary in the first step. Transfer of this energy to the secondary in the second step.

I covered that in what I called the "low voltage" case.

Even though the primary voltage is much higher with a CD ignition, the generated primary inductive (magnetizing) current is much smaller due to the short duration (high frequency) of the applied voltage.

Primay current in the CDI system can be very high, and have a short diration. The voltage in the primary has to rise >10KV/uS. I don't see this operation as being comparable to a 'normally' operating transformer.

The energy transfer is through the flux connecting the primary and secondary, not by the energy stored in any inductive current

I described two different and seperate cases. I think you've somehow combined the two cases into one.

 

[crutschow]

Primay current in the CDI system can be very high, and have a short diration. The voltage in the primary has to rise >10KV/uS. I don't see this operation as being comparable to a 'normally' operating transformer.

I described two different and seperate cases. I think you've somehow combined the two cases into one.

The rate of voltage rise has nothing to do with whether it is operating as a "normal" transformer. For the CD ignition the transformer is indeed operating in a normal transformer mode. By normal I mean it is simply converting the voltage on the primary to a voltage on the secondary, and the primary and secondary voltages and currents are proportional to the turns ratio.

No, I did not combine the two cases into one. I was actually attempting to differentiate between them to show that the CD circuit is not somehow emulating the operation of a flyback circuit. The primary inductive current is not the main factor in the operation of a CD circuit, whereas it is fundamental to a flyback circuit.

But perhaps I'm beating a dead horse at this point.

 

[BrownOut]

Well, it has something to do with how the transformer operates. The voltage has to overcome the primary inductance, however small. Even in 'normal' operation, there is an inrush current 5 to 10 times the steady-state magnetizing current. Ignition coils are mostly in the inrush mode. Guess if you had to think about it as being normal, you'd have to be constantly applying the starting current. Anyway, I understand a little of what you're saying. I think we've gone off topic, and I would welcome taking this discussion off line, or starting a new thread.

Ps: one last thing, ignition discussions are always popular, sometimes contintious. But always fun, becuase making sparks is a pure joy!

 

[KeepItSimpleStupid]

My fun time when I was 16 or so is to power an auto ignition coil with a vibrator used for tube radios. I got some pretty long cool jagged blue sparks. I had this thing about HV then. Did an experiment on the effect of 3 kV DC on the growth of radishes. When the home built power supply crapped out, I just used the secondary anode voltage of a portable TV. I got an honorable mention in a regional Science Fair.