Verification, please?

[canyncarvr]

This question 'fits' both 4-pin and 2-pin ballast setups.

I understand the purpose of the 4-pin ballast, the higher resistance feeding a reduced voltage to a five pin ECU. The coil compensating side, whatever value it may be, is relevant to the question.

The IG2 (start) wire feeds the coil full system voltage, presumably to help a cold engine start. The IG1 (run) side runs the coil through the compensating ballast.

Fine.

What I find curious: During the IG2 event, pin 1 on the ECU is powered through the compensating resistor, reducing the voltage with which the ECU operates.

That seems odd. Why does the ECU's power supply have any connection through/to the ballast resistor (either the 2 or 4-pin). Is it just to get the one wire powered through both states of the ignition switch (IG1 and IG2)?

A BTW, really: I have a 4-pin ballast, don't use the higher ohm side. Don't use the low side, either, but that's a different bale of whick. There are a lot of wires attached to the thing, complicated by modification for an MSD unit I used for years. I'm looking to clean up the mess, having just broken the main 'IG1' run wire right at the spade crimp.

What is a good choice to join together all the wires that need joining...that ARE joined at the ballast? I've found the existing wiring harness wire too old to effectively solder together. Scraping, cleaning, brushing every strand is a PITA...and even with doing that, chances of a decent solder joint are slim.

Separate ring terminals and a barrier strip might work. I've looked for 'This is what I did...' threads, didn't find one; doesn't mean the question hasn't been covered a bazillion times already.

Thanks!!

 

[Mattax]

IIRC, when I was looking at a '73 FSM, the Ignition on Start feeds the ECU pin through the 0.5 Ohm resistor.
I don't know why they used two pins but one possibility is they are joined internally.
In which case, during Start voltage is reduced from roughly 11 to 12 Volts by the 1/2 ohm resistor.
During Run, voltage is reduced from roughly 14 Volts by the 5 ohm resistor.
Just a guess.

I think the best way to join stuff is with double wire in the open barrel terminal. So one methods will be to clip and replace the Packard 58 (or 56) terminals. See what they did on a '70-'72. I think the run to the Voltage regulator and alternator positive fields are splice into the wire. You may not have to touch those. If you do, a splice connector like 67Dart273 just posted about would be a good way to go.

 

[canyncarvr]

IIRC, when I was looking at a '73 FSM, the Ignition on Start feeds the ECU pin through the 0.5 Ohm resistor.
I don't know why they used two pins but one possibility is they are joined internally.
In which case, during Start voltage is reduced from roughly 11 to 12 Volts by the 1/2 ohm resistor.

Almost. During 'start', available voltage (battery, through a switch, through a bunch of wire) is supplied through the 1/2 ohm or compensating resistor.

A basic ignition 2-pin ballast pic is attached. Not what you find on a '73, but the wiring in question is shown.

During Run, voltage is reduced from roughly 14 Volts by the 5 ohm resistor.
Just a guess.

During Run, system voltage to ECU-1 is NOT reduced. I could understand wanting to route the supply V through the resistor on Run because the voltage is higher, but that's not what happens.

There is a tie point to coil+ in the harness on the Start wire.

Italic tags get all messed up in edits. Don't know why. I don't seem to be able to fix 'em.

I think the best way to join stuff is with double wire in the open barrel terminal. So one methods will be to clip and replace the Packard 58 (or 56) terminals. See what they did on a '70-'72. I think the run to the Voltage regulator and alternator positive fields are splice into the wire. You may not have to touch those. If you do, a splice connector like 67Dart273 just posted about would be a good way to go.

Thanks. I recalled the 'Packard' part, but not the number(s).

I ordered some Packard 58s from Mouser. The local NAPA retailer had no idea what I was after, was not familiar with the Packard number or the back clip retainer design.

 

[67Dart273]

Try not to overthink. I think the way it was wired this way is that Ma discovered "it would work" and they didn't have to modify the ignition switch or add relays, etc to supply the ECU

The other issue is, what with harness/ connector deterioration over the years, I would be using a relay triggered by the original output from the bulkhead to feed all underhood loads, anyhow. So with an MSD, just cut all that loose, wire the IGN1/ IGN2 together, and trigger a relay, with a big "ultimate protection" auto reset breaker.

Then feed off the regulator, alternator field, and your MSD trigger off the relay

Most of the reason for this is voltage drop in the harness/ connectors/ switch, which causes over voltage at the battery from the alternator.

 

[Mike69cuda]

Thanks Del, I have always wondered why it was wired the way it is. Wondered briefly if I should fix it. However, in an unusual display of restraint, I did not attempt to fix something that wasn’t broken.

 

[AJ/FormS]

When the resistor is cold, almost no voltage reduction takes place, so it works just right.

 

[Mike69cuda]

Also, since the ecu pulls a lot less current than a coil, the voltage drop would be a lot less, so the ballast resistor would have a lot less effect on the ecu.

 

[Mattax]

Almost. During 'start', available voltage (battery, through a switch, through a bunch of wire) is supplied through the 1/2 ohm or compensating resistor.

Not sure why you say 'almost'. We wrote the exact same thing, unless from what you write here

A basic ignition 2-pin ballast pic is attached. Not what you find on a '73, but the wiring in question is shown.

and

During Run, system voltage to ECU-1 is NOT reduced. I could understand wanting to route the supply V through the resistor on Run because the voltage is higher, but that's not what happens.

I think I misunderstood. You were asking about the single resistor arrangement....and I was describing the dual resistor arrangement, hence my reference to the '73 wiring.

Ths MSD trigger hardly any current. There's a relay in the MSD to draw current from whatever you attach the ravy red wire to. That's one reason the old instructions suggest just attaching it to the original coil feed (which then sits outboard on an insulated tab). That way its easy to switchback to a factory ECU or points if the box failed for any reason.

 

[canyncarvr]

Not sure why you say 'almost'. We wrote the exact same thing, unless from what you write here
and

I think I misunderstood. You were asking about the single resistor arrangement....and I was describing the dual resistor arrangement, hence my reference to the '73 wiring.

I said, 'Almost' not because anything you said is incorrect, but because part of what you wrote does not apply to the question.

Re:

IIRC, when I was looking at a '73 FSM, the Ignition on Start feeds the ECU pin through the 0.5 Ohm resistor.
I don't know why they used two pins but one possibility is they are joined internally.
In which case, during Start voltage is reduced from roughly 11 to 12 Volts by the 1/2 ohm resistor.
During Run, voltage is reduced from roughly 14 Volts by the 5 ohm resistor.
Just a guess.

Yes. On Start, the ECU is fed through the coil compensating ohm resistor.

This part, 'During run, voltage is reduced...by the 5 ohm resistor' is true for the 5-pin ECU, but is not pertinent to my question.

An aside: The lower supply voltage via the 5ohm part of the 4-pin ballast used in the 5-pin ECU is still required in the 4-pin ECUs, but is developed internally. On a Standard LX101 (a real one) you can see a big wire-wound resistor sticking out of the potting. No reason for a large component of that type in an electronic assembly unless it is being used as a voltage divider for some power hungry device.

I posted the 2-pin ballast resistor pic to indicate that the 5-pin ECU isn't relevant. Yes, that happens to be in my car, and I originally had a 5-pin ECU. Not relevant.

I am questioning why they thought it necessary to bypass the compensating resistor for coil+ on Start, but also reduced the voltage to the ECU during Start. I could understand full voltage to the ECU during start (battery voltage is lower), and compensated voltage during Run (alternator voltage is higher) but that is the opposite of what was done.

Usually, what seems odd in a wiring harness does make sense and it was engineered that way for a reason, it's just that I don't understand the whys and wherefores about it.

MSD reference was a past tense sentence. It quit some years ago.The use of the MSD is one reason the wiring in the ballast area is not as clean as it should be. I've been running no ballast and a Chrysler 'style' ECU for years. What coils do or with an uncompensated coil+ isn't the issue.

The main point of this post is about the ECU power feed on pin-1: Why reduce it on Start?

 

[Cuda Al]

I suspect the factory ECU didn't have and internal voltage step down and some part of the electronics needed 12v and some a reduced amount (too lazy to do the math) so they did it with the dual ballast.

The 4 pin ECU will work on a car with the dual ballast.

A 5 pin ECU must have the dual ballast.


Alan

 

[canyncarvr]

This thread got off-track and sidewise. To the extent I obscured the matter or was unclear originally, I'm sorry about that.

Re: 4-pin or 5-pin ECU? Doesn't matter. Both are the same in regard to the question.

Re: 4-pin (dual) or 2-pin (single) ballast? Doesn't matter. Both are the same in regard to the question.

The ECU gets a compensated supply voltage Start. It does not get a compensated voltage on Run. That struck me as odd.

Overthinking it? Guilty as charged!

 

[JoeSBP]

Okay, just to make sure I understand correctly...

During START, the coil will be supplied with a full 12v, while the ECU is supplied with reduced voltage..
During RUN, the coil is supplied with reduced voltage, while the ECU is supplied with a full 12v..

That correct?

 

[JoeSBP]

Having ZERO knowledge what so ever about how the factory ECU's work, just thinking purely about why they would possibly reduce voltage during cranking, one thought comes to mind, and a resulting question..

..It sure would be nice to run max initial timing at idle, but have some simple way to retard timing when cranking to aid in hot starts

..When the factory ECU sees reduced voltage, does that have any effect on the timing?..

When the resistor is cold, almost no voltage reduction takes place, so it works just right.

this also would play into timing retard, as you wouldn't need it on a cold engine

 

[canyncarvr]

Okay, just to make sure I understand correctly...

During START, the coil will be supplied with a full 12v, while the ECU is supplied with reduced voltage..
During RUN, the coil is supplied with reduced voltage, while the ECU is supplied with a full 12v..

That correct?

That is what I see happening, yes. (With a few exceptions: Start voltage is well below static battery (12V) voltage due to starter operation and Run is well over battery voltage due to alternator (14 or so V) output.)

IF...it is considered important to feed a 'full' available system voltage to the coil+ for Start, why feed a reduced voltage to the ECU during Start?

It may well all be a matter of 'overthink'. Say the thinking was...the coil can handle full voltage and works better that way, but it can NOT run that way constantly when in Run mode. OK. Switch it. At the same time, the ECU will run on a reduced voltage during Start and better on the higher voltage of Run, but it works fine either way, so don't worry about it.

In such a case, the indicated wiring is the simplest, least expensive way of doing it.

..and that may be all there is to it. In which case, 67Dart saying, '...it would work..' is just how it was done.

 

[Mattax]

Okay, just to make sure I understand correctly...

During START, the coil will be supplied with a full 12v, while the ECU is supplied with reduced voltage..
During RUN, the coil is supplied with reduced voltage, while the ECU is supplied with a full 12v..

That correct?

Close.

During START, the coil will be supplied with battery voltage, usually around 11 V, while the ECU is supplied with slightly reduced voltage..
During RUN, the coil is supplied with voltage reduced from 14 Volts by a 1/2 ohm resistor, while the 4 pin ECU is supplied with a full 14 V.
While the original factory 5 pin ECUs were supplied at 14 Volts greatly reduced by a 5 Ohm resistor.

Having ZERO knowledge what so ever about how the factory ECU's work, just thinking purely about why they would possibly reduce voltage during cranking, one thought comes to mind, and a resulting question..

Like 67Dart273 said, it was probably just an easy way to wire it up. It really only needed power at 5 Volts to do work.

..It sure would be nice to run max initial timing at idle, but have some simple way to retard timing when cranking to aid in hot starts

On a race engine, sure. That's what DC did in the race distributors.

..When the factory ECU sees reduced voltage, does that have any effect on the timing?..

No. The ECU is just an amplifier. The magnetic pulse from the distributor pickup signals a switch allowing power to flow through the coil.

 

[TrailBeast]

No. The ECU is just an amplifier. The magnetic pulse from the distributor pickup signals a switch allowing power to flow through the coil.

Or more accurately, when the reluctor sends it's signal to the box it drops the power to the coil causing the field in the coil to collapse causing the coil to fire.
But close enough for conversation.

 

[Mattax]

Or more accurately, when the reluctor sends it's signal to the box it drops the power to the coil causing the field in the coil to collapse causing the coil to fire.
But close enough for conversation.

It's good to be accurate.
So if I understand correctly, in the first part you're saying the circuit to the coil is normally complete and the ECU opens it with each pulse.

 

[TrailBeast]

It's good to be accurate.
So if I understand correctly, in the first part you're saying the circuit to the coil is normally complete and the ECU opens it with each pulse.

Correct, and is called "dwell time" for the coil to build a good field.
It's also why the OE coil needs it's voltage dropped so it doesn't overheat.
The OE box is just an electronic set of points if you think about it.
Points are normally closed but open at the set intervals the points cam dictates.
That closed time is dwell.

 

[Mattax]

Yes that's a good way to think about it.

One thing I commonly see is the idea that more voltage through a coil will make a better spark. That's may be true at high rpm when there is not as much dwell time to build a field to high strength. This is the idea behind the external ballast resistor cooling and of course the dual point distributors for racing extends the dwell to give more time for the field to build.
The other part is the actual energy needed to get a good flame going. More can be better, but at some point there are no more gains. The excess we see as ringing on scope.* I don't think that's bad, but it does show that increasing voltage or current through a coil during those conditions is just going to create more heat in the coil.

*Not everyone agrees with that, but I still think its correct.
https://board.moparts.org/ubbthread.../2579668/re-ballast-resistor.html#Post2579668

 

[canyncarvr]

As long as we're almosting and close-ing...

During RUN, the coil is supplied with voltage reduced from 14 Volts by a 1/2 ohm resistor, while the 4 pin ECU is supplied with a full 14 V.
While the original factory 5 pin ECUs were supplied at 14 Volts greatly reduced by a 5 Ohm resistor.

When referring to the 5-pin ECU as having a reduced voltage supplied on Run through the 5-ohm side of a dual ballast, keep in mind that pin-1 on both the 4 and 5-pin ECUs is also/still supplied with system voltage.

Also, in the interest of understanding:

When the resistor is cold, almost no voltage reduction takes place, so it works just right.

Not really. ;)

That sort of thing might happen when there is a poor connection that generates heat; when it's cold it may show less of a voltage drop than when it's hot.

In the case of a wire wound resistor (the ballast), as long as it's operating correctly the resistance it shows, the voltage drop it makes, will be close to the same whether it is hot or cold. A cold 1.25ohms won't be different from a hot 1.25ohms. All of this when within design limits. A resistor designed to handle 50 watts might get real hot and work just fine at 40 watts. Make it try to drop 70 watts....THEN heat is a problem.

One other aside (and having nothing to do with the original question) that may be helpful: In an OEM setup, the Start wire has a tie-point in the harness you don't see that feeds coil+. That's how both Start and Run end up on the single wire that is attached to coil+ while taking two paths around/through the ballast.