GOVTMOD's dumb electrical question of the day; solenoid "I" post as power source?
#16
11-10-2012, 01:08 PM
I don't know if this would work or not but what about hooking it up to the "S" (stator) terminal on the alternator?
This is the terminal that powers the electric assist for the automatic choke and it only puts out power while the engine is running. Its AC and measures around 6 volts.
Like I said, I don't know if this is a good idea or not, just throwin' it out there.
This is the terminal that powers the electric assist for the automatic choke and it only puts out power while the engine is running. Its AC and measures around 6 volts.
Like I said, I don't know if this is a good idea or not, just throwin' it out there.
#17
11-10-2012, 02:33 PM
I don't know if this would work or not but what about hooking it up to the "S" (stator) terminal on the alternator?
This is the terminal that powers the electric assist for the automatic choke and it only puts out power while the engine is running. Its AC and measures around 6 volts.
Like I said, I don't know if this is a good idea or not, just throwin' it out there.
This is the terminal that powers the electric assist for the automatic choke and it only puts out power while the engine is running. Its AC and measures around 6 volts.
Like I said, I don't know if this is a good idea or not, just throwin' it out there.
BTW, here's my truck about an hr ago before I moved it to avoid a ticket.
Does this quailify as a full load?
#18
11-11-2012, 07:31 AM
ATTENTION FOLKS
Do NOT load the 'I' terminal of the starter solenoid. Remember that the 'I' post of the solenoid and the coil ballast resistor are hard-tied at the coil positive. Therefore power at the 'I' terminal back-feeds through the coil ballast resistor with the key in RUN. Once you start pulling extra current through this circuit in RUN, it can cook the wiring harness because of the extra power dissipation through the ballast. It's also unfused, so there's nothing to stop it from happening.
A relay coil does not draw much current and probably wouldn't cook the wiring harness in this case, but any extra current through the ballast that is NOT due to the coil can increase the IR drop across the ballast and lower the voltage available to the coil. So there are actually two reasons not to do this; the latter being the more important issue here.
Nothing tied to the ignition system is meant to carry any more load than exactly that of the coil primary winding. Nothing on this circuit is fused either, first because this system can't risk failing due to a fuse, and second because the factory did not intend to make it accessible for any kind of modification. In other words, don't touch the ignition system.
Looking back through the thread it looks like Mike already caught this, so nice work Mike
Your request is very simple and there's a much better way to do it. See the diagram below. For fused switched power to the relay coil, you can use the idle-stop solenoid power which comes from a two-connector pigtail along the firewall. Let me know if you need help finding it.
Also, this is not quite correct:
The coil itself does not have any different requirements while the engine is in START. The whole point of a ballast resistor is to compensate for the fact that the battery voltage drops significantly during cranking. When the engine is cranking, the battery can drop below 10 volts, so the coil (and ignition system) have to be designed to work below 10 volts. But as soon as you let off the key, without a ballast, the voltage would jump back up to 12 to 14. That's why the coil is connected straight to the battery in START, but through a ballast in RUN - it keeps the voltage to the coil "somewhat" close between the two modes of operation so that the ignition system can be designed for a single operating range.
Finally, to answer Mike's question:
That's an interesting idea, but would not work reliably. Most automotive relays are rated for 12 volts DC and the RMS of the stator waveform may not be high enough to close the relay. The 6 to 7 volts people see is just a byproduct of looking at it with a cheap meter reading DC. It happens to work fine for the electric assist choke because it's just a "dumb" heating element that just needs to get hot. If someone tried it and it happened to work, I still wouldn't trust it across all conditions.
Do NOT load the 'I' terminal of the starter solenoid. Remember that the 'I' post of the solenoid and the coil ballast resistor are hard-tied at the coil positive. Therefore power at the 'I' terminal back-feeds through the coil ballast resistor with the key in RUN. Once you start pulling extra current through this circuit in RUN, it can cook the wiring harness because of the extra power dissipation through the ballast. It's also unfused, so there's nothing to stop it from happening.
A relay coil does not draw much current and probably wouldn't cook the wiring harness in this case, but any extra current through the ballast that is NOT due to the coil can increase the IR drop across the ballast and lower the voltage available to the coil. So there are actually two reasons not to do this; the latter being the more important issue here.
Nothing tied to the ignition system is meant to carry any more load than exactly that of the coil primary winding. Nothing on this circuit is fused either, first because this system can't risk failing due to a fuse, and second because the factory did not intend to make it accessible for any kind of modification. In other words, don't touch the ignition system.
Looking back through the thread it looks like Mike already caught this, so nice work Mike
Your request is very simple and there's a much better way to do it. See the diagram below. For fused switched power to the relay coil, you can use the idle-stop solenoid power which comes from a two-connector pigtail along the firewall. Let me know if you need help finding it.
Also, this is not quite correct:
Finally, to answer Mike's question:
I don't know if this would work or not but what about hooking it up to the "S" (stator) terminal on the alternator?
This is the terminal that powers the electric assist for the automatic choke and it only puts out power while the engine is running. Its AC and measures around 6 volts.
Like I said, I don't know if this is a good idea or not, just throwin' it out there.
This is the terminal that powers the electric assist for the automatic choke and it only puts out power while the engine is running. Its AC and measures around 6 volts.
Like I said, I don't know if this is a good idea or not, just throwin' it out there.
#19
11-11-2012, 09:07 AM
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Hello LMC400,
Given the casual context to which my comments relate, my comment is correct enough !
Had the context required more than a layman's answer, I would have given more detail, like this :
The whole point of a ballast resistor is certainly not to compensate for the fact that the battery voltage drops significantly during cranking, but rather to keep the circuit resistance, measured in ohms, in line with what is required so as not to destroy the points or ignitor in the dizzy, depending on what is fitted.
The coil itself will have an ohms measurement from which one can determine if in fact a ballast resistor is required in the ignition circuit.
For those who are curious, no ohms reduction (via ballast resistor or whatever) in the ignition circuit renders the starter solenoid ''I'' terminal redundant.
The wiring diagram is almost perfect though.
It just needs to state what amps the relay should be, and what amps the fuse should be, for each item GOVTMOD is going to fit.
The best place for the relay fitment wouldn't go amiss either.
Back to basics though, a starter solenoid is exactly that, and is for the starter circuit only.
Just like the alternator is for the charging/regulating circuit. (Layman's overview there!
)
Cheerz4now,
Nick
Given the casual context to which my comments relate, my comment is correct enough !
Had the context required more than a layman's answer, I would have given more detail, like this :
The whole point of a ballast resistor is certainly not to compensate for the fact that the battery voltage drops significantly during cranking, but rather to keep the circuit resistance, measured in ohms, in line with what is required so as not to destroy the points or ignitor in the dizzy, depending on what is fitted.
The coil itself will have an ohms measurement from which one can determine if in fact a ballast resistor is required in the ignition circuit.
For those who are curious, no ohms reduction (via ballast resistor or whatever) in the ignition circuit renders the starter solenoid ''I'' terminal redundant.
The wiring diagram is almost perfect though.
It just needs to state what amps the relay should be, and what amps the fuse should be, for each item GOVTMOD is going to fit.
The best place for the relay fitment wouldn't go amiss either.
Back to basics though, a starter solenoid is exactly that, and is for the starter circuit only.
Just like the alternator is for the charging/regulating circuit. (Layman's overview there!
)Cheerz4now,
Nick
#20
11-11-2012, 10:00 AM
Mike and FMC400 are absolutely correct. I have the same set up but I made my own power tap. I get my IGN B+ from the fuse block. Now that I just replaced my fuse block with a Painless Fuse block the connection is even easier (and it's fused). Be sure to fuse everything from the power source to protect the wiring and the devices on it.
#21
11-11-2012, 12:08 PM
The whole point of a ballast resistor is certainly not to compensate for the fact that the battery voltage drops significantly during cranking, but rather to keep the circuit resistance, measured in ohms, in line with what is required so as not to destroy the points or ignitor in the dizzy, depending on what is fitted.
You are completely correct that a ballast resistor limits the current into the coil (and therefore the current through the points, or the power transistor in the ignition module). And why is this necessary? Because the system is designed to run below 10 volts. And why is it designed to run at this lower voltage? Because the battery voltage drops below 10 volts during cranking. Because the system is designed to operate at that lower voltage, it cannot operate at the 12 to 14 volt range - and therefore a ballast resistor is used to limit the current when the key is turned to RUN.
Therefore, the voltage at the coil stays relatively constant: in START, the battery is lower, but connected straight to the coil. In RUN, the battery is higher, but this "extra" voltage drops across the ballast resistor. This is why it gets the nickname "ballast." This is explained further in this Wikipedia article under "fixed resistors" - it even mentions automotive ignition systems as an example: http://en.wikipedia.org/wiki/Electrical_ballast
The purpose of my comment was to point out that your original statement made it sound like an ignition coil must see a different voltage based on what the engine is doing, which is not the case. However, per your clarification (thank you) I now see that this was not what you were trying to say.
#24
11-11-2012, 03:44 PM
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Hi fmc400,
I just noticed that I got your name wrong earlier. Not intended. Sorry! Don't I feel like a dick now!
(I browse the 'LMC catalogue' for reference so often that I subconsciously put lmc instead of fmc.)
Quote : ''I don't think you realize that you and I are actually saying the same thing.''
LOL, you're right !! I wasn't looking at the ignition system from the starter, or coil, point of view.
What's funny is that if we were having a chat over coffee, it would take a matter of seconds for us both to be on the same page !
Quote : ''a ballast resistor limits the current into the coil (and therefore the current through the points, or the power transistor in the ignition module). And why is this necessary? Because the system is designed to run below 10 volts. And why is it designed to run at this lower voltage? Because the battery voltage drops below 10 volts during cranking.''
In case I'm coming across as argumentative, it's also not intended !!
That said, LOL, I disagree with the above comment.
The ballast resistor, or resistor wire or whatever, is 'ignored' by the starting circuit at starting!
The resistance is necessary to prevent damage to ignition components during normal running, not at starting(cranking).
Here's my detailed summary of the thread so far, and I hope it puts us on the same page.
The starter solenoid, acting in principle similar to that of a relay, receives a 'low' current to the 'S' terminal via a 'thin' wire from the 'start' position of the ignition key switch, and only from the 'start' position.
This 'low current' activates a switch inside the solenoid to transfer a 'high' current from the battery to the starter via 'thicker' heavier duty wires (cables).
The 'I' terminal also receives a 'low' current when the key is in the 'start' position regardless if a wire is connected or not.
The 'I' terminal should only have a wire connected to the coil for starting purposes, period.
And that's only if it's required !
When would it be required ?
If the ignition system has a resistor of some sort in it, that's when.
A battery voltage with the engine NOT starting or running will be +/-12v.
The same battery will reflect, assuming in good condition, a voltage of +/-13.5 to +/-14.5 when running.
However, if there's a resistor of some sort, it's for a reason, and here's an example which hopefully clarifies why.
An engine has an MSD Blaster coil and a Pertronix ignitor (electronic ignition which replaces points).
The coil has 0.7ohms resistance, but the ignitor requires a current with a minimum resistance of 1.5 ohms.
Resistance is a reduction of a current, so if 0.7 ohms is fed to the ignitor, the resistance will not be enough, and the ignitor will receive an 'overload' thus making it rest in peace pretty quickly.
In order to increase the resistance to at least 1.5 ohms, a ballast resistor or resistor wire etc, with at least 0.8 ohms (0.7 + 0.8 = 1.5) needs to be installed in the ignition circuit.
That will then safeguard the ignition component during normal engine running.
However, when cranking the engine, thanks to the ballast resistor (or whatever), the ignition system has less power than that required to fire up the engine.
The solution is to connect a wire from the starter solenoid 'I' terminal to the coil (+ terminal) to restore the power needed for starting.
The starter itself draws so much power that the minimum resistance required by the ignitor is not reduced excessively.
If however, a Pertronix coil is fitted, not an MSD blaster, the resistance is actually 1.5 ohms and so a resistor would not be required to be installed, and the wire from the 'I' terminal to the coil would not be needed, thus making the 'I' terminal redundant.
Say an engine has no need for a resistor, and one is fitted, difficult starting, and a loss of engine power whilst driving, would be experienced.
Without a resistor, the ignition circuit will measure +/- 13.5 to +/-14.5 volts (with the engine running), but less with a resistor connected. (I mentioned 9 volts in an earlier post, but I've forgotten the actual voltage. It's probably higher than 9.)
So what does GOVTMOD do now if he can't use the 'I' terminal ?
He uses the diagram above and creates a new circuit !
The amperage of the fans, for example, will determine what amp relay and therefore what amp fuse will be required, and the relay should be fitted as close as possible to the fans.
The shorter the 'high load' wires (cables) the better.
Volts is one measurement of the electrical circuit, but amps is the actual power running through the circuit. A periodic check of the amps being generated is a good idea.
I hope I didn't
too much.
And I hope we're at this stage fmc400 :
.
Cheerz4now,
Nick
P.S. The pictures of the truck in the snow are awesome !
I just noticed that I got your name wrong earlier. Not intended. Sorry! Don't I feel like a dick now!
(I browse the 'LMC catalogue' for reference so often that I subconsciously put lmc instead of fmc.)
Quote : ''I don't think you realize that you and I are actually saying the same thing.''
LOL, you're right !! I wasn't looking at the ignition system from the starter, or coil, point of view.
What's funny is that if we were having a chat over coffee, it would take a matter of seconds for us both to be on the same page !
Quote : ''a ballast resistor limits the current into the coil (and therefore the current through the points, or the power transistor in the ignition module). And why is this necessary? Because the system is designed to run below 10 volts. And why is it designed to run at this lower voltage? Because the battery voltage drops below 10 volts during cranking.''
In case I'm coming across as argumentative, it's also not intended !!
That said, LOL, I disagree with the above comment.
The ballast resistor, or resistor wire or whatever, is 'ignored' by the starting circuit at starting!
The resistance is necessary to prevent damage to ignition components during normal running, not at starting(cranking).
Here's my detailed summary of the thread so far, and I hope it puts us on the same page.
The starter solenoid, acting in principle similar to that of a relay, receives a 'low' current to the 'S' terminal via a 'thin' wire from the 'start' position of the ignition key switch, and only from the 'start' position.
This 'low current' activates a switch inside the solenoid to transfer a 'high' current from the battery to the starter via 'thicker' heavier duty wires (cables).
The 'I' terminal also receives a 'low' current when the key is in the 'start' position regardless if a wire is connected or not.
The 'I' terminal should only have a wire connected to the coil for starting purposes, period.
And that's only if it's required !
When would it be required ?
If the ignition system has a resistor of some sort in it, that's when.
A battery voltage with the engine NOT starting or running will be +/-12v.
The same battery will reflect, assuming in good condition, a voltage of +/-13.5 to +/-14.5 when running.
However, if there's a resistor of some sort, it's for a reason, and here's an example which hopefully clarifies why.
An engine has an MSD Blaster coil and a Pertronix ignitor (electronic ignition which replaces points).
The coil has 0.7ohms resistance, but the ignitor requires a current with a minimum resistance of 1.5 ohms.
Resistance is a reduction of a current, so if 0.7 ohms is fed to the ignitor, the resistance will not be enough, and the ignitor will receive an 'overload' thus making it rest in peace pretty quickly.
In order to increase the resistance to at least 1.5 ohms, a ballast resistor or resistor wire etc, with at least 0.8 ohms (0.7 + 0.8 = 1.5) needs to be installed in the ignition circuit.
That will then safeguard the ignition component during normal engine running.
However, when cranking the engine, thanks to the ballast resistor (or whatever), the ignition system has less power than that required to fire up the engine.
The solution is to connect a wire from the starter solenoid 'I' terminal to the coil (+ terminal) to restore the power needed for starting.
The starter itself draws so much power that the minimum resistance required by the ignitor is not reduced excessively.
If however, a Pertronix coil is fitted, not an MSD blaster, the resistance is actually 1.5 ohms and so a resistor would not be required to be installed, and the wire from the 'I' terminal to the coil would not be needed, thus making the 'I' terminal redundant.
Say an engine has no need for a resistor, and one is fitted, difficult starting, and a loss of engine power whilst driving, would be experienced.
Without a resistor, the ignition circuit will measure +/- 13.5 to +/-14.5 volts (with the engine running), but less with a resistor connected. (I mentioned 9 volts in an earlier post, but I've forgotten the actual voltage. It's probably higher than 9.)
So what does GOVTMOD do now if he can't use the 'I' terminal ?
He uses the diagram above and creates a new circuit !
The amperage of the fans, for example, will determine what amp relay and therefore what amp fuse will be required, and the relay should be fitted as close as possible to the fans.
The shorter the 'high load' wires (cables) the better.
Volts is one measurement of the electrical circuit, but amps is the actual power running through the circuit. A periodic check of the amps being generated is a good idea.
I hope I didn't
too much.And I hope we're at this stage fmc400 :
.Cheerz4now,
Nick
P.S. The pictures of the truck in the snow are awesome !
#25
11-11-2012, 04:27 PM
Thanks guys! You all are a great help. OK leave the solanoid alone, got it. Told you it was a dumb question.
For the record, this relay's sole pupose is to interupt the ground of a thermal switch and headlight relays so they can't come on with the key off and drain the battery. Any of the circuits involved will be already fused at one point or another.
Here's a the thread on my Taurus power box: https://www.ford-trucks.com/forums/1...tall-help.html
And my Mark VIII fan set up: https://www.ford-trucks.com/forums/1...r-dumbies.html
fmc400, is the plug you are talking about over by the ICM? I haven't had time to look but those are the only plugs that come to mind currently. The diagram is pretty simple, Is the relay illustrated "my" relay or one that is already present under the hood somewhere?
For the record, this relay's sole pupose is to interupt the ground of a thermal switch and headlight relays so they can't come on with the key off and drain the battery. Any of the circuits involved will be already fused at one point or another.
Here's a the thread on my Taurus power box: https://www.ford-trucks.com/forums/1...tall-help.html
And my Mark VIII fan set up: https://www.ford-trucks.com/forums/1...r-dumbies.html
fmc400, is the plug you are talking about over by the ICM? I haven't had time to look but those are the only plugs that come to mind currently. The diagram is pretty simple, Is the relay illustrated "my" relay or one that is already present under the hood somewhere?
#26
11-11-2012, 06:17 PM
Again, no worries, and the same goes for me
The starter solenoid, acting in principle similar to that of a relay, receives a 'low' current to the 'S' terminal via a 'thin' wire from the 'start' position of the ignition key switch, and only from the 'start' position.
This 'low current' activates a switch inside the solenoid to transfer a 'high' current from the battery to the starter via 'thicker' heavier duty wires (cables).
This 'low current' activates a switch inside the solenoid to transfer a 'high' current from the battery to the starter via 'thicker' heavier duty wires (cables).
I guess I'm still not seeing why; to me our two explanations are equivalent. There is not much to this decades-old circuit. Agreed though, sometimes it's hard to get on the same page on the internet.
#27
11-11-2012, 06:23 PM
No such thing as stupid questions!
No, it's closer to the passenger side. Here's a shot of mine, currently being used for an aftermarket electric choke. It's the black pigtail hanging vertically with the blue butt splice (please excuse the nasty wiring - I have since cleaned it up). The WHITE with BLACK stripe wire is the alternator stator signal for the factory electric choke assist; the BLUE wire is fused hot-in-RUN originally routed to the idle stop solenoid on the carburetor.
The relay is one that you add yourself; a standard Bosch-style relay will suffice.
The relay is one that you add yourself; a standard Bosch-style relay will suffice.
#28
11-11-2012, 08:48 PM
OK I think I know which one you are talking about. At least it's one of 2 that come to mind. I either have the standard electric hoke plugged into one side of it or it is just next to it about 4" to the center with nothing at all plugged into it. I don't think my 2150 had the solenoid that you are talking about. I'll check both for key on power when I get the box ready to wire up.
I'm just using the standard Ford FOAB 14B-192-AA relay... I took the Bosch type block out of the power box and put in one for another 2 FOABs.
I'm just using the standard Ford FOAB 14B-192-AA relay... I took the Bosch type block out of the power box and put in one for another 2 FOABs.
#29
11-12-2012, 04:06 AM
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Hi fmc400,
I like this thread, very informative !
I'm sorry I didn't find one as detailed and conclusive 3 years ago when I needed it !
Hi GOVTMOD,
Your links led to interesting threads !
You asked in one of them if you should change the 195* sensor / 195* t/stat setup to a 195* sensor / 180* t/stat, but I didn't see any replies to that.
Changing the t/stat to a 180* would be preferred !
Cheerz gents.
I like this thread, very informative !
I'm sorry I didn't find one as detailed and conclusive 3 years ago when I needed it !
Hi GOVTMOD,
Your links led to interesting threads !
You asked in one of them if you should change the 195* sensor / 195* t/stat setup to a 195* sensor / 180* t/stat, but I didn't see any replies to that.
Changing the t/stat to a 180* would be preferred !
Cheerz gents.
#30
11-12-2012, 07:34 AM
Hi fmc400,
I like this thread, very informative !
I'm sorry I didn't find one as detailed and conclusive 3 years ago when I needed it !
Hi GOVTMOD,
Your links led to interesting threads !
You asked in one of them if you should change the 195* sensor / 195* t/stat setup to a 195* sensor / 180* t/stat, but I didn't see any replies to that.
Changing the t/stat to a 180* would be preferred !
Cheerz gents.
I like this thread, very informative !
I'm sorry I didn't find one as detailed and conclusive 3 years ago when I needed it !
Hi GOVTMOD,
Your links led to interesting threads !
You asked in one of them if you should change the 195* sensor / 195* t/stat setup to a 195* sensor / 180* t/stat, but I didn't see any replies to that.
Changing the t/stat to a 180* would be preferred !
Cheerz gents.
Yeah I have 180 ready to install. All my components are tested and ready I just need to get to installing everything.