Ballast resistor and primary current on Jubilee

Geo-TH,In · Jul 27, 2015

There has been a lot of chatter about ballast resistors, burnt points and primary current so I decided to measure primary current and coil voltage.

Keep in mind when I bought my Jubilee it had already been converted to 12v. My battery is a 2007. The battery voltage is a little weak, 11.96v, tractor RUNS JUST FINE and STARTS JUST FINE TOO.

I found something very interesting. When the points are closed and I turn the key on, the coil voltage starts out at 6v. It steadily decreases to 4v in less than a minute. Battery voltage is holding at 11.96v.

The primary current starts out at 4 amps, like JohnT and JMOR said it would. However the current steadily decreases to 2.82 amps as the coil voltage decreases. BTW, I'm still using the original points that came on the tractor. I've put over 250 hours on it. Again, the tractor runs just fine. Spark plugs are very clean, no carbon.

The decreasing voltage and current can only be explained one way. Before I post my theory as to why, I'll like the experts opinion on what's going on.

Now when tractor is running, the alternator is putting out a little over 14v, so coil will get slightly more volts and slightly more current. How would you measure the amps when tractor is running? There will only be primary current when points are closed and no current when points are open. So is my tractor operating with only 2.82 amps going to the coil when the points are closed?

In one of the pics, you may see a diode with the anode pointing at the coil. This is my way of by passing the ballast when I hit the start button. I'm taking the starter voltage, the cranking voltage, and applying it via the diode to the coil. Yes, I know the diode is using .6v, so not all the cranking voltage is applied to the coi. I get a hotter spark that way. By passing ballast is the old school way my 1965 barracuda was wired.

JMOR · Jul 27, 2015

(quoted from post at 17:31:25 07/27/15) There has been a lot of chatter about ballast resistors, burnt points and primary current so I decided to measure primary current and coil voltage.

Keep in mind when I bought my Jubilee it had already been converted to 12v. My battery is a 2007. The battery voltage is a little weak, 11.96v, tractor RUNS JUST FINE and STARTS JUST FINE TOO.

I found something very interesting. When the points are closed and I turn the key on, the coil voltage starts out at 6v. It steadily decreases to 4v in less than a minute. Battery voltage is holding at 11.96v.

The primary current starts out at 4 amps, like JohnT and JMOR said it would. However the current steadily decreases to 2.82 amps as the coil voltage decreases. BTW, I'm still using the original points that came on the tractor. I've put over 250 hours on it. Again, the tractor runs just fine. Spark plugs are very clean, no carbon.

The decreasing voltage and current can only be explained one way. Before I post my theory as to why, I'll like the experts opinion on what's going on.

Now when tractor is running, the alternator is putting out a little over 14v, so coil will get slightly more volts and slightly more current. How would you measure the amps when tractor is running? There will only be primary current when points are closed and no current when points are open. So is my tractor operating with only 2.82 amps going to the coil when the points are closed?

In one of the pics, you may see a diode with the anode pointing at the coil. This is my way of by passing the ballast when I hit the start button. I'm taking the starter voltage, the cranking voltage, and applying it via the diode to the coil. Yes, I know the diode is using .6v, so not all the cranking voltage is applied to the coi. I get a hotter spark that way. By passing ballast is the old school way my 1965 barracuda was wired.

ecreasing current & voltage: Many ballast resistors (including the original 12250 on N tractors) increase resistance as they heat up.
How to measure amperes when running? A current probe on an oscilloscope. It looks like this (lower trace):

Is my tractor operating at 2.82A when running? No, it will be somewhat higher at 'peak', because the average current is lower due to both the exponential rise (see scope current trace) and the fact that points are closed only for the dwell percentage of the cycle. Since the average current is lower, the ballast resistor will cool down relative to steady ON situation, thereby dropping less voltage and giving you a somewhat higher than 4v coil input & thus higher than 2.82A peak.

Geo-TH,In · Jul 27, 2015

That's what I thought, the resistance of ballast increases as temp increases. I guess
I'll have to measure resistor temp after it sets with key on for 1 minute while not
running and then measure the temp when it's running.

Ballasts resistors are cermanic resistors and are rather hot when in use, right?

Which scope pattern represents the voltage across the ballast, the bottom one? If so,
what is the vertical ampilifier set on?

JMOR · Jul 27, 2015

(quoted from post at 18:31:05 07/27/15) That's what I thought, the resistance of ballast increases as temp increases. I guess
I'll have to measure resistor temp after it sets with key on for 1 minute while not
running and then measure the temp when it's running.

Ballasts resistors are cermanic resistors and are rather hot when in use, right?

Which scope pattern represents the voltage across the ballast, the bottom one? If so,
what is the vertical ampilifier set on?

es, the resistors typically get very hot. Using the open (not encased) N tractor resistor (12250) connected to variable power supply, it is only warm at at an amp or two & resistance of a couple tenths of an Ohm, but increase that to 4 amperes & leave it powered for 1 to 2 minutes at 4A, & it becomes red (~800F) & will scorch paper touching it. Resistance goes up to over an Ohm.
Lower scope trace is current, upper is coil primary voltage.
To obtain ballast voltage, multiply current times value of resistor. Here, for example, the peak current was 3A & if the resistor was 1 Ohm, then the product would yield 3 Volt peak resistor voltage.

John T · Jul 27, 2015

The gang has it, resistance increases with a temperature rise is what's taking place. For a typical Kettering coil distributor ignition around 4 amps is the max current points by design (sure they could be designed for 1 or 10 amps but around 4 is typical) have to carry and switch. Since the points are opening and closing rapidly there's a duty cycle and dwell time involved so one way to measure current is like JMOR showed using an oscillloscope. Some cars instead of using a fixed ballast resistor used a temperature compensating resistor (Thermistor I believe) such that the coil carried more current during start up, but as it warmed up resistance increased and voltage on the coil and current passing through it decreased. NOT rocket science but still fun sparky chat.

John T

mj · Jul 27, 2015

(quoted from post at 16:00:52 07/27/15)

(quoted from post at 17:31:25 07/27/15) There has been a lot of chatter about ballast resistors, burnt points and primary current so I decided to measure primary current and coil voltage.

Keep in mind when I bought my Jubilee it had already been converted to 12v. My battery is a 2007. The battery voltage is a little weak, 11.96v, tractor RUNS JUST FINE and STARTS JUST FINE TOO.

I found something very interesting. When the points are closed and I turn the key on, the coil voltage starts out at 6v. It steadily decreases to 4v in less than a minute. Battery voltage is holding at 11.96v.

The primary current starts out at 4 amps, like JohnT and JMOR said it would. However the current steadily decreases to 2.82 amps as the coil voltage decreases. BTW, I'm still using the original points that came on the tractor. I've put over 250 hours on it. Again, the tractor runs just fine. Spark plugs are very clean, no carbon.

The decreasing voltage and current can only be explained one way. Before I post my theory as to why, I'll like the experts opinion on what's going on.

Now when tractor is running, the alternator is putting out a little over 14v, so coil will get slightly more volts and slightly more current. How would you measure the amps when tractor is running? There will only be primary current when points are closed and no current when points are open. So is my tractor operating with only 2.82 amps going to the coil when the points are closed?

In one of the pics, you may see a diode with the anode pointing at the coil. This is my way of by passing the ballast when I hit the start button. I'm taking the starter voltage, the cranking voltage, and applying it via the diode to the coil. Yes, I know the diode is using .6v, so not all the cranking voltage is applied to the coi. I get a hotter spark that way. By passing ballast is the old school way my 1965 barracuda was wired.

Click to expand...

ecreasing current & voltage: Many ballast resistors (including the original 12250 on N tractors) increase resistance as they heat up.
How to measure amperes when running? A current probe on an oscilloscope. It looks like this (lower trace):
Is my tractor operating at 2.82A when running? No, it will be somewhat higher at 'peak', because the average current is lower due to both the exponential rise (see scope current trace) and the fact that points are closed only for the dwell percentage of the cycle. Since the average current is lower, the ballast resistor will cool down relative to steady ON situation, thereby dropping less voltage and giving you a somewhat higher than 4v coil input & thus higher than 2.82A peak.

Right! ........ Gotcha! :shock:

Geo-TH,In · Jul 27, 2015

JohnT,
It may not be rocket science. I thought it was interesting that the ballast increases resistance with temp. It must get very warm. Someday when I get bored, I'll measure the actual temp while running and while key is on and not running.

I think in my case, a good way to say there is 4 amps in primary is for the coil to have 6v across it.

Not only is resistance increasing, so is the battery voltage when charged by the alternator.

Do you think it would be safe to conclude that the proper sized ballast can be determined with an ammeter, instead of measuring it's resistance, 4 amps when it's powered up?

JMOR · Jul 27, 2015

(quoted from post at 21:33:22 07/27/15) JohnT,
It may not be rocket science. I thought it was interesting that the ballast increases resistance with temp. It must get very warm. Someday when I get bored, I'll measure the actual temp while running and while key is on and not running.

I think in my case, a good way to say there is 4 amps in primary is for the coil to have 6v across it.

Not only is resistance increasing, so is the battery voltage when charged by the alternator.

Do you think it would be safe to conclude that the proper sized ballast can be determined with an ammeter, instead of measuring it's resistance, 4 amps when it's powered up?

ot to butt into John's response, but measuring current is what I have been preaching for years, but too many don't know how & without care , inserting the ammeter may introduce additional resistance. Much better current readings than very low resistance readings are obtainable by the average Joe.

John T · Jul 27, 2015

Hey Hoosier Neighbor

"Do you think it would be safe to conclude that the proper sized ballast can be determined with an ammeter, instead of measuring it's resistance, 4 amps when it's powered up? "

I don't have a problem with the use of a good analog ammeter myself. An "ideal" (none such made however) has zero resistance so as to NOT affect the reading. The critical limiting factor (to prevent premature points failure) in a typical old tractor coil ignition is how much current the points can pass and switch so if you get that in the 4 amp range and provided 4 amps is a sufficient coil current I say compute the ballast resistance by that method makes sense. That's where they get the typical 3 ohms vicinity for 12 volt coils and 1.5 for 6 volt coils as that yields the 4 amps of points current. NOTE these are not all exact figures, I've seen stock 12 volt coils from 2.5 to 3.5 ohms.

Fun chat for sparkies at least

John T

old · Jul 27, 2015

I have not seen any body that has a clue what or how to use an o-scope in well over 30 years now. I have one that is Navy surplus sitting right here and have used a good many of them over the years but then again I was a Navy Electronics teck

Hobo,NC · Jul 28, 2015

Google ignition coil raping are Low amp probe

https://www.google.com/search?q=ign...p;ved=0CDMQsARqFQoTCIWJqMbP_MYCFUQnHgodk5IL-A

You can do allot with a DVOM even more with a high are low amp clamp.

For those that think they know it all but forgot it DSO's are common place I have 5 and use them all the time... Not that I know what I am doing but google is my friend...

I have a saying tho if you have to drag out a lab scope to chase a intermediate problem problem you are in deep DODO.... I spent 3 hr. with one last Friday chasing down why a speed-o would quit working...

Texasmark1 · Jul 28, 2015

I'm surprised the copper in the primary of the "coil" didn't change more than the tungsten wire in the ballast resistor. On scopes, I left the factory Jan1, 2005. Scopes were so complex to operate then it took "Norman New Guy" right out of school to operate them. Can't imagine how they work today. I guess the knobs have disappeared and you just swipe your finger across the face to select vertical and horizontal scales and all that.

MF uses 5 amps as their break current. That's the only time I saw a number in a service manual. I have 5 Ford manuals back to the XXX engines and don't recall seeing the number they use. Course my tractors are all diesel so I have little need for such.

Nice picture JMOR. I doubt anybody has ever done what you did in showing the interaction of the inductance of the "coil" with the other parts of ignition firing. While you have everything hooked up, do everybody a favor, if you would, and put your scope V input on the black wire going to the distributor and snap a picture of the points opening with the condenser installed and without. Might give everybody an idea as to just what it does to protect the points from pitting.....Thanks.

Hey Old, I have an old Tektrinix 545 dual trace with the plug in vertical amplifiers I bought from the company surplus store 30 years ago. Don't know if it still works or not.

Hey Geo. My Yellow HF $5 cheapie works better than your red! Grin

JMOR · Jul 28, 2015

(quoted from post at 08:21:45 07/28/15) I'm surprised the copper in the primary of the "coil" didn't change more than the tungsten wire in the ballast resistor. On scopes, I left the factory Jan1, 2005. Scopes were so complex to operate then it took "Norman New Guy" right out of school to operate them. Can't imagine how they work today. I guess the knobs have disappeared and you just swipe your finger across the face to select vertical and horizontal scales and all that.

MF uses 5 amps as their break current. That's the only time I saw a number in a service manual. I have 5 Ford manuals back to the XXX engines and don't recall seeing the number they use. Course my tractors are all diesel so I have little need for such.

Nice picture JMOR. I doubt anybody has ever done what you did in showing the interaction of the inductance of the "coil" with the other parts of ignition firing. While you have everything hooked up, do everybody a favor, if you would, and put your scope V input on the black wire going to the distributor and snap a picture of the points opening with the condenser installed and without. Might give everybody an idea as to just what it does to protect the points from pitting.....Thanks.

Hey Old, I have an old Tektrinix 545 dual trace with the plug in vertical amplifiers I bought from the company surplus store 30 years ago. Don't know if it still works or not.

Hey Geo. My Yellow HF $5 cheapie works better than your red! Grin

" While you have everything hooked up, do everybody a favor, if you would, and put your scope V input on the black wire going to the distributor and snap a picture of the points opening with the condenser installed and without. Might give everybody an idea as to just what it does to protect the points from pitting.....Thanks. " I would do that, but I did all that back in the winter of 2007 & now on this July in Texas day, it is just too darned hot to get out there. In my notes that I made at the time, however, I see that there was little difference in the voltage appearing across the points with caps from 0.01 to 0.4 microfarads. What is clearly visible is the amount of arcing at the points. The least with a "factory" value of 0.2-0.3 & increasing with larger value & increasing a large amount with much smaller values until the greatest with no cap at all. With no cap at all, the HV spark is weak enough to cause missing, but the ring up voltage spike remained more less the same voltage. I believe, because, voltage of 'inductor' will still rise to whatever is necessary to establish an arc at plug. Clearly evident however, is the loss of energy that is being burned up/dissipated in points arcing and this is seen as greatly reduced duration of plug arc time (from 2.5milliseconds to 1 millisecond with no cap). As a side note, on of our 8Ns will not run with no cap.........must be marginal?

Geo-TH,In · Jul 28, 2015

JMOR,
I thought of a way to determine the resistance of a hot ballast. When I applied power to
my ballast, the coil used 6 v and the ballast also used 6v. Therefore, the resistance of
the coil is equal to the resistance of the cold ballast. So, after running the tractor
the ballast will warmed up and it's resistance will increase and it will use more than
6v and the coil will get less than 6v. If that's the case, then a coil can still produce
a hot spark with less than 4 amps.

When I turned my key on, the coil used 6v and the current was 4 amps, like you said.

JMOR · Jul 28, 2015

(quoted from post at 17:20:44 07/28/15) JMOR,
I thought of a way to determine the resistance of a hot ballast. When I applied power to
my ballast, the coil used 6 v and the ballast also used 6v. Therefore, the resistance of
the coil is equal to the resistance of the cold ballast. So, after running the tractor
the ballast will warmed up and it's resistance will increase and it will use more than
6v and the coil will get less than 6v. If that's the case, then a coil can still produce
a hot spark with less than 4 amps.

When I turned my key on, the coil used 6v and the current was 4 amps, like you said.

hat is all in the definition of a "hot spark". Certainly it will be enough spark to run. Is it the same as it was cold? Very unlikely. The ballast should reach its maximum temperature when ign on, points closed (stalled engine). It should cool down from that once started.

Geo-TH,In · Jul 28, 2015

I'll post back with my findings the next time I mow the yard, about 35 minutes of run
time.

JMOR · Jul 28, 2015

(quoted from post at 18:42:46 07/28/15) I'll post back with my findings the next time I mow the yard, about 35 minutes of run
time.

hen you finish mowing, grab the coil for a 'feel'. It will likely be to hot to hold onto very long, so that means its winding resistance will be up by 20-25%.

Texasmark1 · Jul 28, 2015

Hot Spark!!!!!! The energy that is dispersed in the spark across the plug gap is stored in the inductor (coil) as 1/2 L [inductance of the primary in this case (7 mh for a number) x actual current flowing at the instant the points open squared].

So for a 7 mh coil and 4 amps you have 1/2 (7 exp-3) x 4 exp 2 for 56 millijoules....(mj defined as a volt x amp x second and is a measure of energy. Get it going via pulses per unit of time and you are consuming energy at some rate which is work, measured in watts or horse power, 746 w/hp)

But for the same coil and 5 amps you have 88 mj, some 60% more energy.

The voltage across the coil both primary and secondary per the turns ratio will rise to infinity theoretically when the current attempts to stop due to the points opening, in an attempt to keep it moving.....nature of the beast. The actual breakdown value of the plug depends on the pressure on the plug from it's environment and the gap. The "heat" of the spark is the energy in the arc (volt-amp-period) across the gap and that comes from the coil. So low current, exponential loss of energy. Question is, how much does it take to fire the plug in a given engine at a given time that contains enough energy to light off the available fuel air mixture present at the gap?

On the resistor getting hot, that's the reason it's built the way it is. It's in a high wattage package with provisions for mounting to a heat sink...a piece of sheet metal on the tractor somewhere.....sorry heat sink but better than nothing.

Talking about coils getting hot from running, what always got me was the mounting of the coil right on the side of the engine block. Convenient, yes, but there are better places that are cooler.....course the fan blows across it and I guess that makes it ok.

Mr. JMOR sir, got time for those pictures of the points with and without the capacitor (condenser)?

Thanks,
Mark

JMOR · Jul 28, 2015

I guess you didn't read my response to that question, above?

Geo-TH,In · Jul 29, 2015

Jessie, Good point. Guess I'll just have to measure the current and voltage when it's
hot. Oh well, 6v on coil and 4 amps when cold is a good starting point. Or simply choose
a ballast to apply 6v to a 6v coil.

Don't recall, How many what's the voltage to the old 12v coils of the 60's? I want to think, they operated on 8v and ballast used 4v

JMOR · Jul 29, 2015

(quoted from post at 06:45:20 07/29/15) Jessie, Good point. Guess I'll just have to measure the current and voltage when it's
hot. Oh well, 6v on coil and 4 amps when cold is a good starting point. Or simply choose
a ballast to apply 6v to a 6v coil.

Don't recall, How many what's the voltage to the old 12v coils of the 60's? I want to think, they operated on 8v and ballast used 4v

pile of GM stuff I have indicated the GM 12v systems were around 1.8 ballast & 1.5-1.6 coil until 1968, when they went to 1.35 ballast & 1.77-2.05 coil until the transistorized igns came about & they went to lower resistance coils. Using those numbers & 13.5 volts, will show about 6.2v ballast/7.3v coil on early and about 5.5v ballast/8v coil for post 1967 (I used 1968 Nova and 1970 Camaro).

Texasmark1 · Jul 29, 2015

I looked 3 times. No where did I see a comment about a condenser.

JMOR · Jul 29, 2015

(quoted from post at 14:12:11 07/29/15) I looked 3 times. No where did I see a comment about a condenser.

ell, it is back there sometime yesterday, at least in Modern View, but here it is again:
" While you have everything hooked up, do everybody a favor, if you would, and put your scope V input on the black wire going to the distributor and snap a picture of the points opening with the condenser installed and without. Might give everybody an idea as to just what it does to protect the points from pitting.....Thanks. " I would do that, but I did all that back in the winter of 2007 & now on this July in Texas day, it is just too darned hot to get out there. In my notes that I made at the time, however, I see that there was little difference in the voltage appearing across the points with caps from 0.01 to 0.4 microfarads. What is clearly visible is the amount of arcing at the points. The least with a "factory" value of 0.2-0.3 & increasing with larger value & increasing a large amount with much smaller values until the greatest with no cap at all. With no cap at all, the HV spark is weak enough to cause missing, but the ring up voltage spike remained more less the same voltage. I believe, because, voltage of 'inductor' will still rise to whatever is necessary to establish an arc at plug. Clearly evident however, is the loss of energy that is being burned up/dissipated in points arcing and this is seen as greatly reduced duration of plug arc time (from 2.5milliseconds to 1 millisecond with no cap). As a side note, one of our 8Ns will not run with no cap.........must be marginal?

Texasmark1 · Jul 29, 2015

Appreciate your reply.

In the time difference of 2.5 down to 1, that says something right there about the voltage in the spike. With the joule being VAT, if T
shrinks by a factor of 2.5 then V or A....gotta be V cause the coil wouldn't let A change immediately, must have gone up 2.5x.
What's the bandwidth of the vertical input of your scope? Possible that the spike jumped up there and the scope didn't catch it;
maybe a little tweak of the V sink would have caught it.

I spent a lot of time in industry with pulse modulated circuits, high voltage, arcing, and all that and know that when you attempt to
interrupt the current in an inductor, you had better do one of two things or both: limit the actual voltage with something like a
zener diode or VDResistor, or limit the rate of rise to something less than what would produce the arc. The C is there to do that.
Surprised you went through the range you did with no results. Course since you were increasing the C from what I learned was THE
value, you already had the arc under control. Always heard that .01 was the value of caps used across breaker points. Don't know
as I never measured one nor saw it stamped on the case.

I've been retired 10 years and forget and forgot a lot. I enjoy these brain tweaks as they keep what little brain I have left, active and I
hear that helps to ward off Alzheimer's, speaking of which my lil'sis was here for a couple of days with her hubby who is in the
preliminary stages. Really sad, sad, sad. He's just like a little kid. Drives her crazy but she has a good spirit and loves him. The
visit did her good as she got to see her son home from overseas and his family plus a couple of weeks of breaking with the routine.

JMOR · Jul 29, 2015

(quoted from post at 21:44:03 07/29/15) Appreciate your reply.

In the time difference of 2.5 down to 1, that says something right there about the voltage in the spike. With the joule being VAT, if T
shrinks by a factor of 2.5 then V or A....gotta be V cause the coil wouldn't let A change immediately, must have gone up 2.5x.
What's the bandwidth of the vertical input of your scope? Possible that the spike jumped up there and the scope didn't catch it;
maybe a little tweak of the V sink would have caught it.

I spent a lot of time in industry with pulse modulated circuits, high voltage, arcing, and all that and know that when you attempt to
interrupt the current in an inductor, you had better do one of two things or both: limit the actual voltage with something like a
zener diode or VDResistor, or limit the rate of rise to something less than what would produce the arc. The C is there to do that.
Surprised you went through the range you did with no results. Course since you were increasing the C from what I learned was THE
value, you already had the arc under control. Always heard that .01 was the value of caps used across breaker points. Don't know
as I never measured one nor saw it stamped on the case.

I've been retired 10 years and forget and forgot a lot. I enjoy these brain tweaks as they keep what little brain I have left, active and I
hear that helps to ward off Alzheimer's, speaking of which my lil'sis was here for a couple of days with her hubby who is in the
preliminary stages. Really sad, sad, sad. He's just like a little kid. Drives her crazy but she has a good spirit and loves him. The
visit did her good as she got to see her son home from overseas and his family plus a couple of weeks of breaking with the routine.

hat Alzheimer's is truly a sad, sad, thing to witness. Mark Tomek mean anything to you?

Texasmark1 · Jul 29, 2015

Ballast resistor and primary current on Jubilee

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