69427 
07132012 03:46 AM 
Ignition Physics for regular guys like me.
Ignition Physics
Quote:
Originally Posted by 69427
Over the years I've found myself typing the same stuff over and over again in different ignition related threads, and when asked by members about where to find those posts I can never remember. So, how about an easy to remember thread title that I can refer people to. I thought I'd type up a few items I find interesting, and perhaps a couple others might find the reading useful. First topic: Coils. It's obvious that the coil got its nickname due to the internal construction that involves winding long lengths of wire into a coil shape to concentrate and couple the magnetic field inside it when there is current flow. This prompts a few questions and answers. Q: What is the function of the coil?A: Build up and store energy to later deliver to the spark plugs for ignition. Note that the key word is energy, not voltage. Q: What is the significance of the coil's resistance? A: Essentially nothing. While many advertisers differentiate their coils by mentioning the resistance number, it tells you nothing. A low resistance could mean a) few windings with small wire, or b) a lot of windings with large wire. Both combinations will give you the same resistance value, but potentially very different results when hooked up to your engine. Similarly, a higher resistance coil could be a combination of big or small wire, and long or short lengths in the windings. Result: the customer has no idea what is inside that part. Q: So, what is the correct way to distinguish one coil from another?A: Inductance. Inductance gives you information on how much energy it can store. Think of it as the equivalent of sizing of your shop air compressor tank. Greater cubic feet means a larger amount of air you can store. More inductance means potentially more energy storage capability. Q: Is the energy in the coil of a particular inductance a fixed value?A: No. We know that in an air tank, the higher the pressure the more air that can be stored in that same container volume. A coil has a somewhat similar ability. When the current through it is increased, the stored energy (in the form of the magnetic field) also increases. This all sounds incredibly simple. But, there's a couple details that make coil selection a little tricky for the typical hotrodder, but also makes it interesting for electrical geeks like me.
More to follow.

Quote:
Originally Posted by 69427
Cont'd: We know that when we're trying to fill our portable air tank to take to the track, we just hook up the hose from the garage compressor tank and start to fill up the portable tank. We all know we're going to sit there a while waiting for the pressure in the small tank to finally hit the desired pressure. An ignition coil is a similar deal. Q: Does the coil in my points system run off of 9 volts?A: No. It runs off of 12 volts. Recall in the air tank example that the small tank is fed from the high pressure in the large tank and compressor. The compressor creates the pressure (like an alternator creates voltage) and the large tank stores and smooths the compressed air pulses. A battery does a similar function when it comes to storing and smoothing (filtering) the ripple voltage coming out of the alternator. The compressor/tank is the sole source of pressure that makes the air in the system move, and similarly, the alternator/battery is the only source of electrical pressure (voltage) and volume (amperage) that makes the ignition circuit work. Q: Well then, why do I often hear about the coil running off of 9 volts?A: The wrong tool (voltohmeter) for the job. Back to the air tank analogy. Assume the small tank is empty (0 psi) and the large tank is fully charged (120 psi) and has a modest length (20 feet) air line plugged into the outlet. Just for kicks if you plug your pressure gauge into the air line end, what do you get? 120 psi. Okay, plug the hose into the portable air tank. Quick, what's the pressure in the small tank? If you said essentially nothing (or just a pound or two) go to the head of the class. Why is this? Because the tank is a storage device, and you can't fill it instantaneously. But the air hose had 120 psi in it a minute ago. What happened? We know that once the air actually started to flow we got a pressure drop through the resistance of the hose (120 psi on one end, 0 psi on the other, right at the start of filling). After a minute you take another pressure reading of the tank (30 psi). Another minute (55 psi). Another minute (75 psi). Another minute (90 psi). Notice a trend? At first the air rushes right in, but as the small tank starts to fill (its pressure rising in the process) the air volume rate through the hose keeps slowing down, and the pressure rise rate keeps slowing down. (The rate that the air goes from the big tank to the small tank is determined by the pressure differential, and the flow ability/resistance of the hose.) Note that the pressure reading at the hose end (tank pressure) keeps changing in the process. Obviously there's no one correct number we can say is the pressure at the hose end that's conducting the air into the small tank. Let's say that you accidentally hit the smalltank relief valve (I've done it a couple times on mine). The tank contents will empty extremely quickly, and you're back to zero in the tank. Again, what single pressure number are you going to state for the tank contents? You can't, because it kept changing. Any number you pick was only accurate for the blink of an eye before it was off to another pressure number. An ignition coil is somewhat sim!ilar in that the voltage pressure to charge the coil comes from a main source (alternator/battery), and the charging current goes through the ballast resistance (analogous to the restrictive air hose) to get to the coil. Given that the dwell (and coil charging) is constantly on and off (causing the currents and voltages to keep fluctuating a bunch quicker than the air pressure gauge was moving), a voltmeter is hopelessly too slow to show the fast changing voltages at the coil (C+) terminal. Depending on the ballast resistance, the coil resistance, the coil inductance, the dwell time/angle, and RPM, you can get most any instantaneous number between 5 and 14 volts. Because the voltmeter is so hopelessly outclassed trying to keep up with these voltages, it ends up just averaging the cycling. Another hot rod analogy: Let's say your car runs the quarter mile in 12 seconds flat. You tell your nonhotrod coworker who then does the math and is unimpressed with the thought that your car does 75 miles per hour on the track. You try to explain that your car started at zero miles per hour and had differing acceleration rates and speeds through the gears, and hit 115 miles an hour at the end. He finds that story much too exotic, and prefers to believe that your car was doing 75 miles per hour the whole time. And he would be somewhat correct, in that your car was at exactly 75 mph for perhaps three milliseconds on its trip down the strip.

Quote:
Originally Posted by 69427
Okay, we've been focusing on some general characteristics of the coil when it's in a ballast resistance equipped points system. I want to jump to sophisticated electronic ignition systems (that don't use a ballast) for the moment as the next step in coil characteristics is much simpler to explain this way. (The addition of the ballast resistor adds a bunch of headacheinducing math and performance compromises to the system that are easier to explain after the electronic module discussion.)Recall that the coil is a storage device, and the unit of measurement for an inductor is Henry (similar to labels like Ohm, Volt, Amp, etc). A coil in the 1 Henry (H) range is a darn big inductance, and most common ignition coils are a tiny fraction of that (in the 28 millihenry range). The inductance of the coil has a major influence on how fast the coil will charge. Most of us are familiar with how a car with worn out front shocks reacts to bumps. Depending on the spring rate of the suspension, it will have a characteristic bounce rate/response. And obviously, if you change the spring rate it will change the response. While it's an easy thing to choose a spring that will physically hold up the car (or an inductor to get the energy you want at low RPM), it's an entirely more involved process to pick out a spring that gives good handling at slow and high speeds (or picking a coil/inductor that delivers sufficient energy at low and high engine speeds). Okay, now for some formulas and some simple math. Most ignition systems I've seen seem to deliver between 50 and 150 milliJoules of energy. (A Joule is one watt occurring for one second. 100 mJ (.100 J, the midpoint in the ignition range I mentioned above) would then be equivalent to a 100 watt bulb for one millisecond, or a 1000 watt bulb for 100 microseconds.) We need a way to figure out how to get the energy level that our plugs need to consistently fire no matter what the cylinder pressure load is, or how bad the plugs are expected to get between replacement intervals. The engine and system guys do tests with varying levels of spark energy to see what the engine wants for reliable combustion. Let's say they arrive at an even 100 mJ (so the math will be easier for me after all this typing). On the parts shelf there's an assortment of coils from 1 to 10 mH to choose from. Which one to choose, because we don't want to just close our eyes and grab one of them. The formula that tells us how to determine the coil's energy content is: Energy = (1/2)(L)(I^2). L is the inductance of the coil (in henries), and I is the current (in amps) passing through the primary windings at the moment just prior to the plugs firing. Let's start with the medium inductance coil (6 mH, ie: .006 H) from the shelf. Plugging into the formula: .100J= (1/2)(.006H)(I amps^2)=> 5.8 amps. As long as we run 5.8 amps through the coil we are guaranteed to have 100 millijoules of energy to the plugs (I am ignoring efficiency losses for now.) We're pretty sure this will work at idle (just like the first choice springs holding up the car), but will it work at higher RPM? Remember, available dwell time (not dwell angle) shrinks as the RPM increases.
Next, determining if our initial coil choice will work at high RPM .

Quote:
Originally Posted by 69427
Okay, I changed my mind. There's a topic I want to discuss, and I think I'll let everyone digest the last (dry) post for a while, and do this one now. A forum member asked me about coil overheating issues, and we discussed a few things that I think may be of interest to the collective. Recall that I mentioned that the DC resistance in a coil is pretty much a byproduct of the wire size and length. What happens in a points ignition system if you mix and match coils? Let's assume the alternator voltage is 13.2 volts, the dwell is 30* (out of 45*) and the original ballast resistor (assume it's a removable ceramic resistor rather than a ballast wire for this discussion) is 1.5 ohms, and coil primary winding has a similar 1.5 ohms. At idle or low speed the coil primary current ramp up time is small enough to ignore for this discussion. So, the calculations: The primary current is 13.2v/(1.5+1.5=3 ohms)= 4.4 amps. The ballast and coil resistive wattage numbers (I^2 x R) come out to (4.4^2)(1.5)=29 watts (instantaneous) on each component (because they're the same resistance). The dwell is 67% duty cycle (30/45) making the average power on each device 19.5 watts while the engine is sitting there idling. Let's say the coil accidentally gets damaged, and is replaced by a generic parts store coil. You put your ohmeter on the primary terminals and find that the reading is 3 ohms. You know that the increased resistance in the system will lower the primary current (and energy), so you decide to bypass/shunt the ballast resistor to get back to a total of three ohms in the primary circuit. You hit the starter, and the engine fires right up. Success! Okay, lets see if anything actually changed in the circuit. The current (4.4 amps) is the same as before, and the system resistance (3 ohms) is the same, but now it's all in the coil, as the ballast is shorted out. Let's do the math. The resistive wattage is still (I^2)(R) so (4.4^2)(3)= 58 watts instantaneous, or 58 x (30/45) = 39 watts average. The careful observer will notice that the average power (dissipated in the coil) went from 19.5 watts to 39 watts (doubled!). Can the coil survive this continuous heating? I don't know. It depends on whether the coil can conduct away this heat, which is difficult to do when it is bolted to a surface that is 160 200 degrees. Regardless, it's not a formula for reliability.Third combination: Assume the parts store coil is one (1) ohm. We can do two things. Run it as is, knowing that the current will be higher [13.2/(1.5 + 1)]= 5.28 amps (hopefully the manufacturer has built in some safety margin, but good luck getting him to tell you what that margin actually is). The wattage in the coil will be 27.9 watts instantaneous, and 18.7 watts average. That's slightly lower than the original 19.5 watt baseline, so that should help offset the higher current concern a touch. (The corresponding wattage in the ballast resistor will be higher than the baseline though, and the reader can do the math if interested.) Second option is replace the 1.5 ohm ballast with a 2 ohm model to restore the current level to the original 4.4 amps if points life is a concern. The coil's average wattage in this situation is 13 watts. Directionally correct for increased component reliability. So, two things: While most of the time we can swap coils around and not have the ignition fail on us while we're driving down the road, we are impacting the durability of the part (for better or worse) when the parts specs are changed. And secondly, with this new coil, did we increase the spark energy, or did we lose ground after spending our hard earned money? Unless we can actually get a direct answer from the manufacturer on what the inductance value is (or an honest answer of the calculated energy if you tell him what current level you're running) we won't know. You have to admit none of us would buy a cam or a set of tires if the manufacturer wouldn't publish the technical specs of the part they're selling.

Quote:
Originally Posted by 69427
Let me wander off the electrical reservation for the moment and discuss an item that has always amazed me. For many of us who have had the sheer fun of driving our car at triple digit speeds, it's pretty obvious that it's a whole different experience stopping your car from 120 mph than it is from 60 mph. We agree that it's the brake components' job to stop the vehicle, and that the brakes get darn hot after a high speed stop. We all know that that happens because the brakes are converting the car's (kinetic) energy into heat. The formula for kinetic energy is KE=(1/2)(mass)(velocity^2). It's easy to see that if we increase the car mass (weight) it will be harder to stop. A 10% increase in the car's weight will increase the KE by 10%. The amusing part is what happens when we increase the speed. If we increase the speed by 10%, we increase the KE by the square of the speed (110%^2=121) resulting in 21% more KE, although the speed only increased by 10%. (Or, at 120 mph the kinetic energy is 4 times what it is at 60 mph.)Recall back in post #6 I mentioned the formula for calculating the energy in an inductor or ignition coil. That formula was Energy= (1/2)(inductance)(amperage I^2). Notice the similarity of this electrical energy formula with that of the mechanical kinetic energy formula. Like mass, changing the inductance will change the energy by the same proportional amount (10% more inductance = 10% more energy potential). Now, remember when we changed the speed (velocity) by 10% that the energy increased by 21%, we get a similar response if we increase the coil current by 10% we get 21% more energy. [For those with CD ignition systems the energy stored in the prime storage capacitor is determined by the formula Energy=(1/2)(Capacitance)(Voltage^2). Look familiar?] Note how the energy available in each of these examples is directly and significantly influenced by the factor that is dynamic (more speed, more current, more voltage), and the static parameters (weight, inductance and capacitance) are the lesser influence.I'm sure I'm in the minority, but I'm fascinated by the similarity of these formulas despite the very different appearances of these forms of energy in action.

Quote:
Originally Posted by 69427
Coil required dwell time (Electronic ignition and ideal conditions) Okay, earlier we selected a coil (6 mH) and calculated that we needed 5.8 amps of primary current (supplied by the 13.2 volt alternator setting) to get us to our 100 mJ spark energy target. Because this ignition system example is a high end electronic system (with adaptive dwell and peak current control) we have no need for a ballast resistor or unnecessary winding resistance to prevent the primary circuit from becoming an underhood toaster. The module will control and limit the peak current passing through the coil windings (so for this example we will designate this coil as a higher performance coil with minimal, or zero, winding resistance). The next step is to determine if that coil will actually work in the real world where enthusiasts like to run their engines to redline on occasion. We do this by some simple (for now) calculations, and then certainly by monitoring the coil waveforms while driving the car over the complete RPM range of the engine (in a safe legal location, of course). In order to not waste a bunch of manpower, fuel, and wear and tear on development vehicles we do the calculations and lab stuff first. The formula for coil charge time (or risetime, due to the coil current's characteristic slope appearance as it ramps up from zero amps to the target peak, 5.8 amps in this example) is: V = L (dI/dt). V is the voltage across the coil (13.2 volts), L is the coil inductance (6 mH) and dI/dt is the change (delta) in current (I) per time (t). Plugging in the numbers: 13.2 = (.006)(5.8/dt), my calculator says 2.6 milliseconds. As long as the adaptive dwell circuit in the module turns the coil on (dwell) for 2.6 ms every cylinder cycle, we should have our full energy requirement (100 mJ) available to be dumped into the plugs. As we discussed earlier we can't charge the coil up instantaneously, and we also can't dump it out instantaneously. We electrical types refer to the time interval that it takes to discharge the coil via high voltage and current at the spark plug as "burn time", not to b!e confused with the a/f mixture burn rate/burn time that the engine guys use. To make the math easier for everybody, lets assume a burn time of 400 microseconds (somewhat realistic), resulting in a minimum charge and discharge cycle time requirement of 3 ms (the sum of 2.6ms + 400 microseconds). Let's check the available dwell time at idle, as that's the slowest engine speed and the point where maximum dwell time is available. We're all familiar with points dwell times (typically 30* out of a possible 45* in an 8 cylinder), but remember that these numbers are distributor degrees, not crankshaft degrees. I have to bounce back and forth between crank degrees and distributor degrees to make my points, and I hope it doesn't confuse the issue too much. At idle (assume 600 RPM) a different plug fires every 90 crank degrees, or every 45 distributor degrees (which you all know, but I'm trying to tie timebased dwell, which an HEI type system uses, to anglebased dwell that points use in my explanation) which is 25 millisecond intervals. Because the calculations say we only need 3 milliseconds to charge the coil and fire the plugs, this coil has a lot of dwell safety margin at low RPM. The risetime (2.6 ms) of this coil only needs the equivalent of 5* of distributor dwell time. (A nice thing about this is that the coil dwell is off so much that the coil stays "nice and cool".). Let's rev the engine a bit. At 3600 RPM (near the torque peak and high VE/cylinder pressure) we have an available dwell time of 4.2 ms (45 distributor degrees), but we only need 2.6 ms (28*) to charge the coil, then another .4ms (6*) to discharge the coil. Total we are using up 28 + 6 = 34 degrees of the available angle. Life's still good.Let's rev the engine to 6000 RPM. At this speed the cylinders fire every 2.5 ms. This poses a potential problem because we need 2.6ms to fully charge the coil and .4ms to fully discharge the coil. Something has to give, and it's always the coil charge up time. Assume for simplicity that the 400 microsecond (.4ms) burn time is fixed in the module (that is, it won't turn on dwell again until it's sure the coil is fully discharged), so we're left with 2.5ms cycle time (45 distributor degrees), but subtracting .4ms leaves us with only 2.1 ms (38 distributor degrees) available for charging the coil. Obviously the coil will not be fully charged. Plugging 2.1 ms dwell time into the formula gives us 4.6 amps (down 21% from 5.8). Plugging 4.6 amps into the energy equation gives us 64 millijoules (down 36%!). At first glance this looks like a problem, and a better suited coil would be needed. But fortunately that's not the case. 6000 RPM is significantly above the torque peak, and the VE and cylinder pressure have dropped enough there that it doesn't take as much energy to supply the lower arcover voltage and current to sufficiently light the plugs. (This VE/pressure drop phenomena is a key ingredient in allowing points based systems to still work satisfactorily at higher RPMs.) So, it looks like our first coil choice for the electronic system worked well. (I'm as surprised as you. I'm calculating as I'm typing, and expected the final coil choice to be the second or third calculation.) The choice of a coil for a points based system with ballast resistance (for protection) is a bit more involved. The math gets messy, but I'll do my best to minimize the glassy eyes effect. Side note: Prior to each post I thought I could explain the concept in about 15 lines. I'm amazed how incorrect I was. Hopefully my math calculations are significantly more accurate. I'll try to make the next post topic a bit lighter, and then delve into the points coil selection in the following post.

Quote:
Originally Posted by 69427
Okay, let's see if I can come up with a lighter post with a bit less math hopefully. I've noticed the large number of ignition coil ads in racing catalogs, with some seeming to try to keep trumping their competition. One manufacturer says his coil puts out 45,000 volts, another claims their's puts out 50,000 volts, and the third claims 55,000 volts. It definitely sounds impressive, but, in the real world, what does it mean? A common misunderstanding is that the coil sets the voltage that the plugs see. So, this 55,000 volt coil must be the hot setup, right? Not necessarily. In the real world the plug ionization/breakdown/arcover voltage is what determines what the secondary voltage rises up to and is clamped at. The big numbers in the advertisements are more an indication of the dielectric strength of the coil packaging, rather than how "hot" (as some say) the spark is. Remember, we're interested in a coil producing energy, and the dielectric rating does not tell us what the energy capability of that coil is. Let's go back to the air tank analogy for a moment. Who cares if someone is advertising an air tank that is good for 500 psi when your compressor never gets over 125 psi? Nice insurance, I'm sure, but is it the one I want? If it is a 20 gallon tank, but I can get a 50 gallon tank that is safe to 300 psi for the same price (and hooking to the same 125 psi compressor), I'm probably going to go with the larger capacity tank. Same with coils. The dielectric strength does give an indication of the part's integrity under extreme/fault conditions, but it tells you nothing about how it performs in daily, normal use. Like the larger tank, we would prefer to have a coil that stores a larger amount of energy. So, what is coil energy, in simple terms? Basically, it's Voltage x Current x Time. But remember in the real world there's no free lunch. If you open up the plug gap you increase the required voltage to jump the gap, so necessarily the system will shorten the burn time (the time that current flows through the gap) and possibly the spark current as well. Changing one factor changes the others.Okay, what happens on the coil's primary side if the plugs are arcing over at 30,000 volts, clamping the rise of the secondary voltage? In simple terms the coil will react like a simple transformer. Assume it's a typical coil with a 100:1 ratio of secondary to primary winding count. If the secondary voltage is 30,000 volts, then the primary circuit will see 300 volts (between C and ground, or across the points or switching transistor). Larger gap plugs or high cylinder pressure drives the plug arc voltage up, and this drives the primary voltage up. This is why it's necessary to equip electronic ignitions with a robust high voltage switching transistor, and the need for a quality, high voltage condenser in points systems to prevent damage and failure in the switching circuitry. That's probably enough for now.

Quote:
Originally Posted by 69427
I'm working on the pointsignition calculations. In the mean time, let me pose two questions for your consideration: Assume we're using the coil from above (6 mH) and a total system resistance of 2.3 ohms to keep the current the same as above (5.8 amps, and pretend the points don't burn up at this current level) so that the energy capacity is the same. Assume the coil is 1 ohm, the ballast is 1.3 ohm, and the alternator voltage is 13.2 volts. 1) What is the voltage at the coil C+ terminal 1 degree before dwell starts? 2) What is the voltage at the coil C+ terminal 1 degree after dwell starts? Please feel free to start a thread with your perspective on this. I will hopefully have my next post for this thread up soon. There's a ton of math to do in it, and it's slow going. I will try to minimize the dry stuff where I can.

Quote:
Originally Posted by 69427
Okay, did a crapload of calculations and hopefully I can condense it down into something reasonably readable.Remember the earlier analogy of filling a small air tank from a hose attached to a big tank? Recall that the air rushes in quickly at first, and then slows down as the small tank fills up and the pressure differential between the tanks becomes smaller. A points controlled ignition system works very similar. I posed a couple questions in the previous post. The following bits of math and simple electrical principles will hopefully make the answers clear.Recall that in a sophisticated electronic ignition controller that the math formula was easy when trying to figure out the coil charge time. A simple V= L (dI/dt). There is always system voltage V (1214 volts) across the coil during dwell, so the math is easy. It's different with points. The coil in a points system has a continuously changing voltage across it during dwell due to the charging current causing a changing voltage drop across the ballast resistance. The formula that describes the charging current in a points system is: I=Imax[1exp(tR/L)]. Essentially it shows that the current charges up in an exponential fashion (fast at first, then slowing down just like the airtank example). I is the instantaneous current, and Imax is the maximum current that the circuit will allow due to the ballast and coil winding resistances, using Ohms Law. Note that there is a single R, as the formula (and current) doesn't care if the resistance is in the coil or the ballast. The letter "t" is for the time mark in the coil charge, but because we're car guys I converted the time unit into dwell degrees. Hopefully that will make things a bit more comfortable. Using the formula above, I made a bunch of dwell calculations for idle speed (600 RPM), midrange (3000 RPM), and high RPM (6000) using a 6 mH coil (and 5.8 peak amps) like I mentioned in the prior post. First, idle speed and some current and voltage calculations:
Dwell point (out of 30*)_____Current_____C+ voltage
1* ____________________1.12 A_______11.74 V
5* ____________________3.38 A_______8.82 V
8* ____________________4.73 A_______7.06 V
12*____________________5.35 A_______6.24 V
15*____________________5.56 A_______5.97 V
20*____________________5.72 A_______5.77 V
30*____________________5.79 A_______5.67 V
Note that the primary current charges up quickly at first, but then slows down, and also that the voltage at the coil C+ terminal is not a fixed value. Also, we met the 100 mJ energy target!
Now, 3000 RPM:
1*_____________________.240 A________12.89 V
5*_____________________1.11 A________11.76 V
8*_____________________1.67 A________11.03 V
12*____________________2.32 A________10.18 V
15*____________________2.74 A_________9.64 V
20*____________________3.32 A_________8.88 V
30*____________________4.18 A_________7.77 V
Note that the final current before the points open is down 28%, and the energy therefore drops 48% (from 100 mJ at idle to 52 mJ).
And at 6000 RPM:
1*_____________________.122 A________13.00 V
5*_____________________.584 A________12.40 V
8*_____________________.908 A________12.02 V
12*____________________1.31 A________11.50 V
15*____________________1.58 A________11.14 V
20*____________________2.01 A________10.60 V
30*____________________2.74 A_________9.60 V
The current is down 53%, and the energy has dropped to 23 mJ (down 77%).
This shows a couple things. The ballast resistance, and even the coil winding resistance do nothing good for the system when there is limited time to charge the coil. The resistance, like a long, small ID air hose, prevents a fast flow of the stuff you're interested in (electrons or air molecules). Second, when the dwell is limited to a fixed angle, the system can't compensate when the angle time gets extremely small at higher RPM. (Dual points enabling a couple more dwell degrees would be beneficial at the higher speeds, although it would increase coil heating at idle speeds. No free lunches in this world.) These calculations were done with the coil parameters I discussed earlier. In the real world you would need data from the engine guys to see if the engine would still run with 23 mJ of spark energy at high RPM. If it wouldn't, then additional calculations would be made with other coil inductance/resistance combinations to see what works best. This example used 5.8 amps primary current to help keep the comparisons closer (to the electronic system), but points contact durability would be an issue in the real world. As I mentioned earlier, sophisticated electronic systems don't see the energy level drop as much as a points system due to the electronic system being able to extend the dwell angle out to as much as 4042 degrees (leaving a couple degrees on the table for the coil to discharge at ignition time), and also, because resistance in the system is not needed for peak current control, the coil can be made with less resistance, allowing a faster chargeup ability. But as I mentioned earlier, if the upper RPM VE and cylinder pressure drop sufficiently then the spark energy requirements drop also, and a marginal ignition system can still do the job.
Hope it made sense!

Quote:
Originally Posted by 69427
Cranking effects and the starter shunt circuit. So far most of the discussion has been on engine speeds of idle and up to near redline speeds. Obviously ignition systems have to also work at speeds much below that to get the engine running while being cranked at very low speeds. There's a few things happening while the engine is cranking that makes this condition different from other speeds, and this affects the electrical portion of the engine (ie: the ignition). I'll try to discuss the more familar items. Cold spark plugs. If the engine has been sitting for several hours, or overnight in a typical winter, most of the vehicle components have reached ambient temperature. That includes the plugs. Physics says it's a bit easier to get an arc flowing from a hot surface than a cold surface. (Recall that vacuum tube equipment used a lot of power compared to transistor based circuits, in large part due to the heater filaments that tubes had to promote electron flow from the cathode to the anode. In a similar fashion, traditionally ignition coils were wired so that the high tension/voltage output to the [hotter] plug center electrode was negative polarity.) The net effect is that it takes more kV (voltage) to jump from a cold electrode than from a warm electrode (like after the engine has been running for a while). "High" cylinder pressure: Because the cranking speeds are slow, that gives the cylinders plenty of time to fill with air (and fuel). We've all seen the numbers when we've run compression tests on our engines. If you're trying to clear the engine after it got flooded, the throttle plates were wide open allowing the cylinders to pull in a lot of air. This relatively high pressure requires a relatively high arcover voltage (compared to high manifold vacuum conditions). Spark timing: Because it's such a low RPM during cranking we have no centrifigal advance. If the throttle is open a bit we also have no appreciable manifold vacuum, and no vacuum advance. We are stuck with base timing (perhaps 4* with the factory timing). This places the spark event at just about the most difficult place for the ignition system: near top dead center. This region has the highest compression pressure, and therefore requires more kV to jump the gap, compared to if the spark happened further down the cylinder at 20 or 30* BTDC. Okay, we have a short list here of conditions where the spark plugs are needing a high voltage to break down the gap. (Recall a few posts back when I mentioned that the coil energy is used up as Voltage x Current x Time.) When the gap uses extra voltage, then the current or arc time must get smaller in return. Well, that stinks. Oh, did I mention that while all this extra nuisance stuff was going on that winter morning, the starter was trying to turn over the engine, dragged down by a pan full of molasses? Due to the fact that a starter uses a lot of current to develop the torque/power to spin the engine, and that the battery is not an infinite source of current, the battery voltage can drop significantly from its 12 volt origin. We would all like our engine to start if the starter can still keep the crank rotating, and often that point is down at battery voltages of 68 volts. Do we have enough ignition/coil energy to actually light off the air/fuel mixture? Let's see:It seems that many forum members have indicated that they are seeing an average of about 3 ohms of resistance in the primary circuit. To make the discussion a bit easier let's divide that up evenly between the coil and the ballast resistance for this example. During normal running the system voltage is supplied by the alternator. As we've been doing in several posts, let's assume the regulator is set at 13.2 volts. Assume the engine is idling, so lack of dwell time is not as issue (ie: the coil gets fully charged each time). We can calculate the dwell current peak (13.2v/3 ohms = 4.4 amps). Assuming a 6 mH coil (somewhat typical) we have a spark energy of 58 millijoules while the engine is at idling speed. But, what happens the next frigid morning when you're trying to start the engine? Let's say the engine is cranking over slowly, and your voltmeter in the dash says 8 volts. Do we have enough energy in the coil to light the plugs? Let's do the math. We know we have plenty of dwell time at that sloooow cranking speed, but how high will the dwell current go? At 8 volts battery (because the alternator certainly isn't doing anything productive yet) and 3 ohms (1.5 ohm ballast and 1.5 ohm coil) system resistance we have 2.67 amps. This gives us (with a 6 mH coil) an output of 21 millijoules. The energy level dropped 64%! Not a good thing with cold, fuelsoaked plugs. But a solution was developed by a clever individual to solve this problem. By plugging into the starter solenoid (as this item is active during those times when the problem is occuring), we can implement a useful electrical bandaid. By switching the system voltage (8 volts in this example) from the starter solenoid straight to the coil C+ terminal (rather than through the ballast resistance) we can increase the coil current. Let's see what we have with the shunt circuit in place: 8v/1.5 ohms = 5.33 amps. The coil energy is 85 millijoules! (This is 47% higher than when the engine is idling and the coil is being fed from the 13.2 volt system buss.!) If thea/f ratio is anywhere reasonable, the engine should start. So, what do we have: We know that cold starts are one of the worstcase conditions for the ignition system. We also know that the coil can potentially put out more energy at 8 volts battery than 13.2 volts from the alternator. Why? Current. The coil doesn't care what the voltage source is, or how big it is, it only cares what level of current is allowed through it to create the magnetic field that stores the energy.

Quote:
Originally Posted by 69427
Members, if you encounter any item or items in these posts that you would like to see additional explanation of, or clarification on, please PM me. I am happy to post additional information, or answer via a PM if requested.
Thanks,
Mike
I'm working on a couple additional items to post, probably some HEI and TI stuff. I'm looking through some old files to refresh my memory on some schematic details.

