70bigblockdodge
Old Man with a Hat
Or someone buys a car "fixed" by someone else and we work harder to fix it.
Summit,Jegs mention 2 times in one post. Banner ads are coming Fast n Furious.
ballast resistors
- Thread starter 70Tom
- Start date
Summit,Jegs mention 2 times in one post. Banner ads are coming Fast n Furious.
Ballast resistors are designed to limit the amount of current to the coil. Voltage will drop depending on the amount of resistance in that ballast resistor and the amount of current draw in the coil.
Think of the ballast resistor as a kink in a garden hose.
FWIW, A single ballast resistor is 1.2 ohms resistance and a dual ballast resistor has 5 ohms and 1.2 ohms resistance.
Think of the ballast resistor as a kink in a garden hose.
FWIW, A single ballast resistor is 1.2 ohms resistance and a dual ballast resistor has 5 ohms and 1.2 ohms resistance.
70bigblockdodge
Old Man with a Hat
Dammit, quit explaining my ramblings with correctness and logic.
You bring up a good point.
Change to a .5 ohm ballast resistor and you will go faster, and your points will be happier, and your wife will love you more, no don't do that.
traintech55
Senior Member
(How do we bring up the laughing emoji).Dammit, quit explaining my ramblings with correctness and logic.
You bring up a good point.
Change to a .5 ohm ballast resistor and you will go faster, and your points will be happier, and your wife will love you more, no don't do that.
70bigblockdodge
Old Man with a Hat
Next to picture icon on bottom of toolbar.
There is a sentence that was gobally-guk 15 years ago.
There is a sentence that was gobally-guk 15 years ago.
70Tom
Senior Member
So the moral of the story is...? Keep the ballast no matter what?
Carmine
Old Man with a Hat
I actually agree with both you and Big John and add one more item... the voltage drop was also to decrease arcing across the points, another feature rendered obsolete as time marched along.
Dave... 6 volt cars charge around 8-9 volts when running
You may go remove your red X from here at anytime
E=IxR, I=E/R, R=E/I
You are right Big_John... but when the resistance is fixed, the relationship between volts and current put simply... as voltage increases, so does the amperage...
Don't stop... I love watching you bust Dave's balls when he picks on my posts
That was kinda a big miss... thanks for the correction.
That's true, but you are saying it a little backwards. The less amperage draw at the coil will mean higher voltage after the resistor.E=IxR, I=E/R, R=E/I
You are right Big_John... but when the resistance is fixed, the relationship between volts and current put simply... as voltage increases, so does the amperage...
Don't stop... I love watching you bust Dave's balls when he picks on my posts
Things to remember about voltage and current... Voltage won't kill you... Static electricity from scuffing your feet across a carpet can hit 4000 volts really easily. Current is what will kill you. Ten amps of current through your body will make you very dead.
I didn't post anything in this thread to bust Dave's (or anyone else's) balls. I like to make people think about how stuff works.
traintech55
Senior Member
Remember the golden rule when working with electricity, (A/C or D/C), treat it like it is your father, give it all the respect it is due and you will be fine, mess with it and it WILL kick your ***.
Ok... Gotta break out the High Voltage training story.
Back when I worked for GE, there was a mandate that everyone in our division got "high voltage" training. It really didn't apply to us, but it was a safety thing aimed at a lot of their people in the division that did.
So... we all sat through the class and it wasn't too bad. The instructor kept it interesting etc. It lasted two days.
At the end of the class, the teacher started going around the room to ask questions about what we had learned. He started with me and gave me a scenario where I was to disconnect some high voltage transformers at a power plant. I said "Honestly, I wouldn't feel comfortable doing it and would call for someone that knew what they were doing". The teacher got a little pissed and said "Didn't you learn anything in this class?". I said "yep... I learned to that it would be too dangerous for me to do the job as I do not have the experience".
He growled something under his breath and went to the next guy... who happened to be my boss... and asked the same question. My boss said, "I would do the same thing John would do, call someone that knew what they were doing". The teacher got a little more pissed and went to the next guy who said "Yep, calling someone that can do this safely is the thing to do". The questions ended at that point and we all got our training certificates and went back to work.
70bigblockdodge
Old Man with a Hat
Those numbers may be true. When a 6volt system is starting it is maybe 5 at best so in the dead of winter you want that coil to light a fuel air mixture at barely half of its designed voltage?Dave... 6 volt cars charge around 8-9 volts when running
You may go remove your red X from here at anytime
The ballast resistor is bypassed while cranking sending full battery voltage to coil which when cranking on a 12 volts system which usually drops below 10 volts right in the designed voltage for the coil.
@Big John I should not have said Dammit stop, should have said ahhh yes, no I did not think you were breaking balls just sorry I did not think to help explain correctly.
commando1
Old Man with a Hat
Please, everybody, keep the discussion going about ballast resistors. To me, they are finally no more little blocks of white ceramic and getting confused why some have 2 wires and some have 4. 
I thought that was hilarious!
Pretty funny! Love this forum!
OK, no problem...Please, everybody, keep the discussion going about ballast resistors. To me, they are finally no more little blocks of white ceramic and getting confused why some have 2 wires and some have 4.
I'm assuming that you mean the electronic ignition, so here's what I understand.
The first production electronic ignitions had a double ballast resistor.... actually two resistors housed in the same white ceramic. One resistor had 1.2 ohms resistance and the other 5 ohms.
When you twisted the key to start, the current flowed through the 1.2 ohm resistor. After the car started, and you released the key, the current flowed through the 5 ohm side.
This was the earlier version and is known as the "five pin" because of the five connections on the ignition box.
I'm not sure when or where it happened, but a "four pin" version of the ignition box came into play later on. This box used a single 1.2 ohm ballast resistor and the start position bypassed the resistor. Release the key back to "run" and the current flows through the ballast resistor.
I have no idea why the early one was designed like that just as I have no idea what (other than less pins) was changed in the later box. We're beyond my knowledge.
FWIW, the whole system seems pretty flexible. My car was wired by a PO for a five pin box with a dual ballast resistor, and the box was a four pin. It still ran fine.... I switched some wires around and it's wired correctly now.
I've also seen a lot of cars wired incorrectly using the four pin box with both positions (start and run) of the ignition switches running through the 1.2 resistor. They seem to start and run OK, although they are also used when the weather is nice etc. There's a few incorrect wiring diagrams floating around on the internet that have brought this about.
Then I've seen cars that are wired using the four pin that have bypassed the resistor all together. The resistor is still there... bolted to the firewall and there are wires going to it... but it's not wired right and they seem to have issues with failing coils that just can't be explained. If you ever see a coil mounted to a fenderwell because "it gets too hot and fails when it's mounted to the engine", chances are strong it's wired this way.
A small refinement to your thinking.
The ballast resistor and points and coil are all in series. This means if you add the resistance of the two devices and divide that result into the input voltage you calculate the current (amperage). When the points are closed, current flows, when the points are open no current flows and the voltage across the points is the full input voltage. The is no voltage drop across either the ballast resistor or the coil since there is zero current. So what is the purpose of the ballast resistor? You might also wonder what the condenser is supposed to do.
The coil is in fact a transformer, a coil wound inside a coil such that an alternating voltage on one coil, the primary, will induce a voltage on the other, secondary, coil. If the primary and secondary coils have the same number of windings, the input and out voltages will be the same. If the the primary has say 50 wraps and the secondary has say 1000 wraps, the ratio is 1:20. So 12V AC in would produce 240V out. (actually a little less due to transformer efficency) Notice all the voltages must be alternating or rapidly switched off and on. Notice also no mention of current because current does pass through a transformer.
So how does this transformer do it's thing? A current passing through a wire creates a mild magnetic field, wind the around a piece of iron to make coil and you have an electromagnet. Now wind a second wire on top of your first coil and you have a transformer. Why, because current flowing through a wire creates a magnetic field and as that field builds the field cuts through the secondary coil and induces a voltage. This secondary voltage works across the secondary load to create current, ie I=V/R.
Note also, power = V times I and since you can't "create" power, the power output of the transformer will always be equal or less than the input power.
Anyways, a transformer is an inductor and it "backfires". When the voltage in is rising, the magnetic field rises and the secondary voltage rises. When the input voltage drops to zero, the magnetic field collapses, but the collapsing secondary voltage induces a reverse magnetic field and this induces a negative voltage spike in the primary. This voltage is what causes the points to arc and produce a negative voltage ripple backwards into your input power supply.
To control this negative spike a capacitor, another non-linear impedance, will short this pulse to ground, but block non-alternating voltage, ie your battery DC. This in fact why a few capacitors in your wiring harness will greatly reduce radio static.
So why the ballast resistor? The coil is an oil filled transformer which has difference inductive characteristics when it is running and at operating temperature. To make starting easier with a "hot" spark, the starter switch bypasses the Ballast resistor. Once running, you don't need as hot a spark and you don't need to drive the coil quite as hard, so the ballast resistor is added to split the input voltage across the resistor and the coil in a ratio equal to their individual resistance.
All of this is just the basic story, the real story is a lot more complicated because we have a chopped, electrically noisy input voltage working in a resistive, inductive and capacitive circuit. In electronics speak this is an tuned oscillator with a temperature and load variable inductor. So don't muck about without careful thought. Clean properly gapped points are critical, the capacitor is critical and the ballast resistor must be sized to the coil being used. Ideally the ballast resistor should be variable such that you could adjust it according to how the coil is heating.
Does this help or is it too much tech speak?
The ballast resistor and points and coil are all in series. This means if you add the resistance of the two devices and divide that result into the input voltage you calculate the current (amperage). When the points are closed, current flows, when the points are open no current flows and the voltage across the points is the full input voltage. The is no voltage drop across either the ballast resistor or the coil since there is zero current. So what is the purpose of the ballast resistor? You might also wonder what the condenser is supposed to do.
The coil is in fact a transformer, a coil wound inside a coil such that an alternating voltage on one coil, the primary, will induce a voltage on the other, secondary, coil. If the primary and secondary coils have the same number of windings, the input and out voltages will be the same. If the the primary has say 50 wraps and the secondary has say 1000 wraps, the ratio is 1:20. So 12V AC in would produce 240V out. (actually a little less due to transformer efficency) Notice all the voltages must be alternating or rapidly switched off and on. Notice also no mention of current because current does pass through a transformer.
So how does this transformer do it's thing? A current passing through a wire creates a mild magnetic field, wind the around a piece of iron to make coil and you have an electromagnet. Now wind a second wire on top of your first coil and you have a transformer. Why, because current flowing through a wire creates a magnetic field and as that field builds the field cuts through the secondary coil and induces a voltage. This secondary voltage works across the secondary load to create current, ie I=V/R.
Note also, power = V times I and since you can't "create" power, the power output of the transformer will always be equal or less than the input power.
Anyways, a transformer is an inductor and it "backfires". When the voltage in is rising, the magnetic field rises and the secondary voltage rises. When the input voltage drops to zero, the magnetic field collapses, but the collapsing secondary voltage induces a reverse magnetic field and this induces a negative voltage spike in the primary. This voltage is what causes the points to arc and produce a negative voltage ripple backwards into your input power supply.
To control this negative spike a capacitor, another non-linear impedance, will short this pulse to ground, but block non-alternating voltage, ie your battery DC. This in fact why a few capacitors in your wiring harness will greatly reduce radio static.
So why the ballast resistor? The coil is an oil filled transformer which has difference inductive characteristics when it is running and at operating temperature. To make starting easier with a "hot" spark, the starter switch bypasses the Ballast resistor. Once running, you don't need as hot a spark and you don't need to drive the coil quite as hard, so the ballast resistor is added to split the input voltage across the resistor and the coil in a ratio equal to their individual resistance.
All of this is just the basic story, the real story is a lot more complicated because we have a chopped, electrically noisy input voltage working in a resistive, inductive and capacitive circuit. In electronics speak this is an tuned oscillator with a temperature and load variable inductor. So don't muck about without careful thought. Clean properly gapped points are critical, the capacitor is critical and the ballast resistor must be sized to the coil being used. Ideally the ballast resistor should be variable such that you could adjust it according to how the coil is heating.
Does this help or is it too much tech speak?
70bigblockdodge
Old Man with a Hat
T
Thanks.
That is awesome. I am far from a expert but I think you used commas properly.
Thanks.
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