Low resistance spark plug wires on a 6.0L engine with Marelli ignition

I had posted earlier (7 years ago) on this subject, but had not gotten a conclusive answer.
Here is the old thread, for posterity:

This time, I am (or at least I think) better prepared, having read the write-up on the magnecor website about the the “truth”:

https://www.magnecor.com/magnecor1/main.htm

and having studied the specs for spark plugs wires in the Jaguar manual:

My question is – is EMF interference the only concern with running low resistance spark plug wires on this engine?

I’ve completely given up on the idea to make the idle speed silky smooth and the reason for the question is of mostly academic interest in the subject.

Thank you for all responses.

The coil stores an amount of energy, and when the spark jumps the plug gap that energy is released. If there is no resistance in either the wires or the plugs themselves, the current skyrockets and all that energy is released instantly – which, apparently, is not good for ignition. By having resistance in the circuit, the current is limited, and it takes longer for the energy from the coil to discharge. So the spark lasts longer. And supposedly that’s better for ignition.

Also, the high currents are probably not good for the plugs, the cap and rotor, etc.

Magnecor wires limit the spark current via inductance rather than resistance.

This is all understood, and I am running BR7EF plugs, which are resistor plugs.
Plus, the aftermarket wires in question also have built-in resistance, albeit lower than specified.
The question is how exactly the extra 2-3 kOhms in the wires “help” with interference?

Also, the Benzes with V12s form the same vintage also had a few kOhms in the wires, but oddly enough, call for NON-resistor plugs.

The problem with RFI and EMI is spark current flow. More current flow, more RFI and more EMI.

Plugs and plug wires usually include an intended, engineered-in resistance that adds to the natural resistance of the ionized, semi-conductive plug gap, rotor gap, and the various connections the “spark pulse” has to travel through to complete it’s circuit from and back to the coil.

As said, current can be limited by inductance as well as by resistance. In electrical terms, this is known as a “choke” as well as “inductor”.

Doesn’t matter HOW the spark current is limited. As long as the current is limited, RFI and EMI are reduced. More resistance = less current flow. More inductance = less current flow.

Measuring the “ohms” means little if current is also limited by inductance.

Strictly speaking, an inductor attempts to stop the rate of change of current - i.e. it tries to keep current flow constant.

Prior to the advent of fuel injection, the only substantive piece of electrical equipment was the radio and so long as the aerial didn’t pick up electrical interference, all was good. Now you will have a computer network, fuel injectors, electronic control units and many sensors operating at low voltages, so a small amount of electrical noise can swamp genuine control signals. The various techniques, such as adding inline resistance, limit the initially huge inrush of current as these electrical switches connect and disconnect so their footprint or shadow isn’t imprinted onto the weaker signals.

The MB cars probably had the bonnet well earthed so as to screen the rest of the car from ignition noise. I seem to recall that in earlier and far less sophisticated days, the Fiat 126 used the bonnet as the aerial for its radio.

kind regards
Marek

That is an interesting question, and I think it is a lot more complex than people think.

As I understand it, there are 3 phases to the spark.

  1. Breakover. In this phase the voltage builds up across the parasitic capacitance within the spark plug, across the electrodes. The electric field ionizes the molecules and creates streamers creating plasma channels. While the conduction builds up, the impedance decreases dramatically. The result is several hundred AMPS of current. i = C dv/dt. So even with 40pF of capacitance (tiny), with 10kV and 2ns you have 200A of current flowing during breakover (but only for 2 nanoseconds - only 4 mJ)

  2. Arc Phase. During the arc, the voltage across the electrodes rises to about 100V and about 100mA of current flows. The energy for the arc comes from the parasitic capacitances and inductances in the coil and cable and the characteristics are controlled by the external impedances (resistance of your cable is one). (Perhaps this is where the marketing claims about spiral wound cables etc come from?) The arc phase is sustained by electrons emitted from hot spots on the negative electrode (which is where some erosion might occur and talk of spark polarity is probably relevant).

  3. Glow Phase. When the current drops further (<100mA) the voltage rises to 300-500V for a few milliseconds. This energy is delivered by the coil and limited by the cable resistance. Usually 10-100mJ of energy.

So different parts of the system are critical to different parts of the process.

To get more energy into the spark, you need more efficiency in the ignition circuit. Which means less resistance. So less resistance is good in this respect. (More energy goes into the spark, less into heat in the cables etc)

However, EMF or EMI is related to di/dt and dv/dt. That is the rate of change of current (or voltage) with respect to time. Therefore, in another sense, the lower the current the better, which results from higher resistance in the system.

But … the current in the secondary can be modelled with an LR circuit, where there is an exponential component e raised to the power of -Rs/Ls*t, where Rs is the combined series resistance of the coil secondary, the cable and the plug and Ls is the secondary inductance of the coil. Therefore, increasing anyone of those resistance results in a FASTER change in current, but lower overall current (and lower overall energy). Higher resistance results in shorter spark (glow time), while lower resistance makes for longer spark duration. To complicate matters, the effective resistance of the spark gap is an exponential function, with high resistance (>50k) at low current (10mA) and low resistance (<1k) and higher current (>100mA). So the spark is effectively changing resistance during the event.

The solution is a compromise.

The whole system is interrelated, with the coil secondary inductance (and resistance), the cable resistance and the plug resistance. You can bet Mercedes knew what they were doing using low(er) resistance wires. I would guess they were controlling spark duration with a certain coil to make it all work at a rated rpm on a V12.

If it were me, I’d try the wires you want and put a scope on the primary to “observe” the spark duration. This is also how you can find weak coils when dealing with COP systems (short duration sparks).

I think there are also other issues at play, but this has gone on long enough! Sorry.

Anyway, the real answer is to use Coil On Plug. Hopefully someone out there is trying that. They still tend to use resistor plugs (or have resistors in the secondary path). There is a Ford coil DG508 that is common on Crown Vic V8’s. Cheap and cheerful. This seems to have the perfect angles to fit the V12 with the fabrication of a small bracket.

Thank you for the informative replies.
Form the above and from reading more on various other forums (there are literally WARS on the subject), two main observations can be made:

  1. OE leads with carbon-core are inherently high-resistance and on purpose - to suppress interference
  2. Aftermarket leads that are advertise as high-performance = low resistance still have resistance built into them, although much lower than OE (my case). They have different mechanism to suppress interference
  3. Very low resistance solid metal wires are for racers only

The basic operation principle of the ignition is this:
The battery voltage is applied to the primary winding of the ignition coil. While the voltage is applied, the current ramps up until either all of the battery voltage is dropped accross the winding reistance and there is no voltgae left to increase current in the inductance, or the primary current gets interrupted. When the primary current gets interrupted, the current in the inductor commutates to the secondary winding. The voltage accorss the spark plug air gap rises quickly, because before the spark is fired, the current can only be driven into (parasitic) capacitance present in the system. Once the spark fires, the current flows through the air gap and ramps down to zero while the inductance gets discharged. The ramp is determined by the inductance of the coil and the burn voltage of the spark. If a resistance is introduced in series with the spark plug, there is an additional voltage drop accross that reistance which adds to the burn voltage. And that means, the spark actually burns for a SHROTER TIME with the resistance than it does without. Some of the energy stored in the coil is discharged into the resistance, and therefore less is discharged into the spark.

What the resistance does is, it dampens the ringing between the coil inductance and the parasitic capcitance(s). This ringing results in voltage spikes which stress the coil and also cause EMI.

The resistance does not limit the current though. The current is determined by the inductor current. The inductor is a current source rather than a voltage source.

Cheers,
Harald

PS: I have tried to explain this in simple words, but inductors aren’t that simple, so I have omitted a couple of things.

I just saw an ad for fancy new spark plugs. These $15 plugs contain a capacitor to “store up the spark from the coil and release it all rapidly”.

I chuckled. A $180 dollar experiment I won’t be undertaking.

A capacitor rated at 50KV?

Pulstar

Looks like the plug body is a big capacitor.