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the inductive field collapses. Remember, as the field collapses the current continues to flow in the same direction from the the coil, thus reverse charging the capacitor. The inverter must then overcome the reverse charge as it begins to charge for the next spark. Thus no energy is saved from the previous spark to be used for the next spark. More Insight: The coil voltage initially goes negative when the SCR fires. As the capacitor discharges, the coil voltage goes to zero and the coil's field reaches maximum. The coil's field then begins to collapse causing the current to continue in the same direction, reverse charging the capacitor. At which point the SCR turns off and the capacitor is thus reverse charged from what's required for the next spark. |
source = sorcery http://forums.pelicanparts.com/suppo...eys/paddel.gif
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Hi,
I was out yesterday and returned here to find more posts than I expected! Dave is correct in that the cap changes polarity and ends up being fully discharged when the SCR switches off. No charge is 'returned to the capacitor'. wwest, You are right in thinking that more voltage is needed to jump a gap that is wide or if the cylinder pressure is high - you can think of this as increased resistance. Once this resistance gets so big, all that happens is you get no spark at the plug (or it finds another path to ground - your 'flashover' example). Quote:
Here is a scope trace of the primary transformer (coil) winding of a Bosch CDI @ 2000 RPM. The first negative going (-360V) is the initial discharge, then the polarity swing to a positive (+200V). As there is no current flow at the top of the rise, the SCR switches off and the remainder of the charge is dumped (that's the wibbly bit at the end). http://i804.photobucket.com/albums/y...psg7pnp7nd.png Now, Bosch CDI @ 7000 RPM http://i804.photobucket.com/albums/y...psktgjqkw4.png What is interesting is that the primary voltage does reduce at higher RPM. I have seen some boxes fall to around 240V at the higher end. This is a shortfall of the Bosch inverter design. So more RPM ends up with a weaker spark in the Bosch design. As you may know, my team (Classic Retrofit) have redesigned the CDI circuit from the ground up to iron out some of the shortfalls. Firstly our supply delivers a constant 300V at all RPM. This is 20% less than the Bosch design so puts less stress on the original coils. The coil does have to perform 'double duty' in our system as it is dual spark, right up to max RPM. Classic Retrofit CDI+ @ 8000 RPM. Spark interval is 400us. http://i804.photobucket.com/albums/y...pspiyhuzqb.png There is a technical manual at www.classicretrofit.com if you're interested in more info. |
nice.
how does the MSD fair at hi RPM? your website needs some work. pics distorted and the red banner convers the text. in case you dont know. |
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The website is one of the 'new fangled' scrolly ones, just scroll down! What are you viewing on? |
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spark energy (43 milliejoules - 1/2 X (240^2 X 1.5 x 10^-6)) at that voltage. Porsche used the same CDIs as the production CDIs in all of its race cars without any problems over the years. So to imply that the Bosch CDIs are lacking, is a mis-leading. |
you should test the MSD and the permatune. there is always much debate on which of the 3 is better.
i really like the MSD because of the dedicated power to the pulse forrming network/ HV side instead of running all the power thru the ign sw. |
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power source. With regard to the MSD providing better performance, there hasn't been any real test on a running car, e.g. a dyno, to prove the claims, i.e. maybe more "snake oil" sales. |
1. The plug will only fire once the rising HV reaches the point at which the path ionizes.
High compression, 40K volts.., idle, no load, 25K volts. The spark will be sustained only as long as enough energy remains in the capacitor, this is not necessarily, rarely is, a full discharge point. 2. Yes, the polarity of the coil reverses once spark extinguishes, it now becomes a power source instead of a load |
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I agree 240V is adequate but I'd rather know that the spark is the same all the time irrespective of RPM. Like you, I don't subscribe to the "more volts is better". If it is too high, you'll roast the coil and be susceptible to mis/cross firing if your HT insulation isn't 100%. Hence MSD require their own coil and aftermarket leads. Anyway, you seem like a knowledgeable chap and probably know that already!!! :) p.s. Actual dyno results coming soon for our unit. |
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Jonny, your lack of understanding of the theory of operation of a CD ignition surprises me.
For instance, why would you not at least recommend (require?) an upgrade to a E & I core transformer type "coil" when the advantages are so obvious...? And: Perhaps the Bosch CDI starts with such a high initial capacitor charge so that the high RPM drop-off only drops to 300V.. |
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The original Bosch coil is a high quality item which is more than adequate. Many modern ones are poorly made in China, plus they look completely out of place fitted to our older German automobiles. The Bosch CDI produces around 400V at idle and tails off up the RPM, so yes they designed high at low RPM because of that. We designed ours to be constant 300V all the time. Do you have a problem with that? I'm not going to be drawn into a fight with someone who twists my own and their statements around to suit their own needs. I suggest you review what you said and present evidence (like I did) to back it up. As a reminder you stated that the HT voltage rises with RPM and load: "High compression, 40K volts.., idle, no load, 25K volts." My traces show that the primary voltage actually falls slightly with RPM, from 360 to 300v. Given that the transformer is 100:1, the secondary e.g. HT voltages must be 36kV and 30kV respectively. So that is actually falling with RPM. |
Jonny, At the instant the SCR fires, the "coil" inductance resists the input current flow rise in accordance the inductive value, so what we see on the secondary side is a rising voltage slope.
Once that rising voltage is high enough to overcome the "atmosphere" (for wont of a better word) across the plug gap the secondary voltage will rapidly fall to that required to ONLY sustain the arc, ~3-4KV. Think of the way HID ballasts work, ~25KV to strike the arc, then drops to a low level, just enough to sustain the arc. The "coil" magnetic field will not begin to collapse until there is not enough energy being transferred to the secondary to sustain the arc. Once that happens, arc extinguished, field collapsing, the lines of magnetism are now "cutting" the coil wires in the opposite direction so both the coil voltage and current flow will reverse. It is at this point that the SCR turns off due to the reverse voltage, and the remaining energy in the coil's magnetic field is "dumped" into the capacitor. |
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wwest, I despair. You speak in flowery language and your arguments don't make sense. You swerve the issue being discussed (current draw) and insult me for trying to explain in simple terms the operation of the ignition system. You tell me I made a 'major mistake' in explaining how a step up transformer works. Really? What is your agenda here?
To the OP I can only offer an apology for this diversion. It tried to keep it simple but have been baited into a Riddler's lair. For the record, the company we partnered with to design the inverter and switcher for our CDI+ have been designing HV power supplies for military air, land and sea craft since 1956. We have spent two years analysing, re-engineering and improving the original design. I think we probably do understand how it works... So trust me on this, the current does not increase with higher cylinder pressure. No doubt wwest will reply with more riddles but I'm out of here.... Don't forget people, cars are meant to be fun! Off to stick a CDI+ on a race car! |
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"...the company we partnered with..." So, you provided the idea, and your partner supplied the technical expertise/KNOWLEDGE. :rolleyes: Have your partner's project design engineer respond to this post series.. You might find that very enlightening... |
But again, I assume that you do not question the fact that input current will rise with RPM.
My point is that the coil secondary voltage will only rise as high as needed to initiate an arc across the spark plug, and then drop to a much lower voltage that sustains the arc until the power available is not high enough to sustain the arc. That MIGHT mean ~40KV for an engine under LOAD, extreme compression level, or for simple idling, ~25KV. Numbers are not meant to be taken as factual. I will assume that you would agree that the inductance of the coil will limit the rate of rise of the current flow in the primary, and thus the rate of rise of the secondary voltage. So, at what point will the primary current flow be at the 25KV firing time vs 40KV... Understand..?? |
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Here we go.............. Quote:
Patience. I am not able to technically comment but I am able to see a pattern after six years. Willy, here's a dude that knows a lot. Why do you need to do this? Just bark and make your point. |
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The power draw difference between firing a plug in a low pressure environment vs a high pressure environment might be in the 10% range but denying it exists is like an Ostrich. |
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