starter issue

[PetesPonies]

The issue about the starter sticking on is most often a case of the starter relay contacts welding together. It is easy enough to check if it is the ignition switch by disconnecting the small wire to the relay. If the starter stops turning, it's the ignition switch, if it continues to turn, the solenoid contacts are stuck.

The reason the starter contacts weld together is excess current draw. The typical reason for excess draw is low voltage. Yes, I know Ohm's law does not work that way, but a series wound motor which must put out a certain amount of work will compensate by increasing the current draw. Becomes a lot of calculation including inductance which I'll not even attempt here. Most often, this is caused by a poorly charged, or defective battery.

So, it sounds like you have already tackled the obvious issues, so what next?

I'd like to know which style of starter relay you are using.

The old style was used with the series wound starter motors. It has the terminals out the side of the cylinder.

The new style solenoid is used with the PM starters. It has the terminals on the face of the cylinder.

I don't have specs, but the PM starters use a second solenoid on the motor itself and the fender mounted solenoid passes much less current. I suspect using the new style solenoid with a series wound motor may exceed the current ratings of the solenoid which would promote sticking.

Just a theory of mine, may be worth looking into.

I don't know why you think Ohm's Law is not represented here? I'm being totally truthful, no kidding. Ohm's law is always the case. A motor is measured in Watts. It draws XXX amount of Watts. How are Watts derived? Voltage and Amperage. If you lower one, the other must be higher. You don't generate the amount of Watts needed, the motor won't run.

[NotEnoughTrucks]

I don't know why you think Ohm's Law is not represented here? I'm being totally truthful, no kidding. Ohm's law is always the case. A motor is measured in Watts. It draws XXX amount of Watts. How are Watts derived? Voltage and Amperage. If you lower one, the other must be higher. You don't generate the amount of Watts needed, the motor won't run.

Watts are a measure of work performed. The physical load on a starter motor will remain relatively constant, so a certain amount of work is performed. Ohms law is used to predict this, but the lower voltage would limit the wattage given a fixed resistance. This is where inductance comes into play and again, the factors involved in exactly why the electric motor behaves as it does are quite complex, but in the simplest terms, the motor will require more current as the load increases because the inductance lowers the effective resistance. For example, 1 horsepower is approximately 745 watts and also is defined as a force of 550 foot pounds per second. At 12 volts, 62 amps of current will be required to produce 744 watts. At 6 volts, 124 amps will be required to produce the same power, (approximately 1 horsepower). Ohm's law tells us the R value at 12 volts would be slightly less than 0.2 ohms. At 6 volts, this becomes less than 0.05 ohms. The actual resistance of the copper used in the motor windings will likely be something entirely different, but the effective resistance will change as a result of the effect of inductance. How this inductance works falls into a study of magnetic fields.

The DC resistance of the copper windings will be the lowest resistance you will encounter. Should you be able to maintain 12 volts into a stalled starter motor, I can pretty much guarantee that you will be making smoke. As the motor spins, the inductance of the windings increase, reducing the current. The physical forces of turning the engine over represent a load. Ultimately, energy is neither created or destroyed. Either physical work is done, (motion and force), or heat is produced.

Hope this helps!

[Gary Lewis]

Watts are a measure of work performed. The physical load on a starter motor will remain relatively constant, so a certain amount of work is performed. Ohms law is used to predict this, but the lower voltage would limit the wattage given a fixed resistance. This is where inductance comes into play and again, the factors involved in exactly why the electric motor behaves as it does are quite complex, but in the simplest terms, the motor will require more current as the load increases because the inductance lowers the effective resistance. For example, 1 horsepower is approximately 745 watts and also is defined as a force of 550 foot pounds per second. At 12 volts, 62 amps of current will be required to produce 744 watts. At 6 volts, 124 amps will be required to produce the same power, (approximately 1 horsepower). Ohm's law tells us the R value at 12 volts would be slightly less than 0.2 ohms. At 6 volts, this becomes less than 0.05 ohms. The actual resistance of the copper used in the motor windings will likely be something entirely different, but the effective resistance will change as a result of the effect of inductance. How this inductance works falls into a study of magnetic fields.

The DC resistance of the copper windings will be the lowest resistance you will encounter. Should you be able to maintain 12 volts into a stalled starter motor, I can pretty much guarantee that you will be making smoke. As the motor spins, the inductance of the windings increase, reducing the current. The physical forces of turning the engine over represent a load. Ultimately, energy is neither created or destroyed. Either physical work is done, (motion and force), or heat is produced.

Hope this helps!

Would you be willing to re-write, or completely start over, on my page re Electricity 099? Given what you've said mine is wrong. Too simplistic.

[NotEnoughTrucks]

Would you be willing to re-write, or completely start over, on my page re Electricity 099? Given what you've said mine is wrong. Too simplistic.

Gary, your page is not wrong at all. What's going on here is something called Faraday's Law. It's a pretty intensive read. If I said I understood it completely, I'd be lying!

If that previous post is at all helpful, feel free to cut and paste and add to your page.

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

[Gary Lewis]

Gary, your page is not wrong at all. What's going on here is something called Faraday's Law. It's a pretty intensive read. If I said I understood it completely, I'd be lying!

If that previous post is at all helpful, feel free to cut and paste and add to your page.

https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction

Ray - Your statement of "As the motor spins, the inductance of the windings increase, reducing the current" is the key. I think we can safely assume that a starter relay/solenoid is sized correctly to handle the in-rush current to the starter, but that the designers of the solenoid assumed that the starter will spin at a certain rate, thereby reducing the current. However, if the starter doesn't spin at the assumed rate then the continuous current draw will be more than they designed for, and even contacts made with good material will weld together.

So, a dragging starter would cause this problem, and a dragging starter is one where the bearings have worn to allow the armature to rub against the stator. Even a tiny rub causes heat, which makes the armature grow, which causes more friction, which causes more heat, which.....

And even a poor connection can cause the problem if it keeps the starter from spinning fast enough to reduce the current substantially and, therefore, stay in the high-current mode.

Does that make sense?

[NotEnoughTrucks]

Ray - Your statement of "As the motor spins, the inductance of the windings increase, reducing the current" is the key. I think we can safely assume that a starter relay/solenoid is sized correctly to handle the in-rush current to the starter, but that the designers of the solenoid assumed that the starter will spin at a certain rate, thereby reducing the current. However, if the starter doesn't spin at the assumed rate then the continuous current draw will be more than they designed for, and even contacts made with good material will weld together.

So, a dragging starter would cause this problem, and a dragging starter is one where the bearings have worn to allow the armature to rub against the stator. Even a tiny rub causes heat, which makes the armature grow, which causes more friction, which causes more heat, which.....

And even a poor connection can cause the problem if it keeps the starter from spinning fast enough to reduce the current substantially and, therefore, stay in the high-current mode.

Does that make sense?

Yes it does.

I think I would be careful to maintain that the increased load is the cause of the higher current. Not so much directly by friction, but by induction changes due to speed.

You could demonstrate this effect with a small motor connected to a battery with an ammeter measuring current. Slowing the motor by increasing the load will increase the current.

The thing about the starter motor scenario is that presumably, you have reduced the speed by lowering the voltage. Certainly something similar is going on in your electricity 99 description of the blower motor? What is happening in that scenario is that you have limited the current by means of the series resistor. Your Ohm's law power calculation kicks in and you can no longer deliver the same horsepower, frictional losses kick in and the motor speed slows. With the starter motor and no resistive losses, the battery is still able to provide full current. The voltage drop comes from internal resistance within the battery and the poor starter motor just wants to continue to crank the engine. So what does it do? It slows down, inductance decreases, current flow increases and the power calculations maintain the required force to crank the engine. Of course, there are practical limits to this effect. The DC resistance internally of both the motor and the battery would define those limits, but for all practical purposes, we will experience starter solenoid failure due to overcurrent.

One last thing to consider is the type of work we are asking the motor to do. All mechanical losses can be considered frictional losses, but we need to keep in mind that work is a product of force, motion and time. In the case of a blower motor, the frictional losses are largely associated with airflow and are reduced as fan speed is reduced. Not so with the starter motor which operates at relatively slow speeds, but faces a constant demand for torque, (force). If the starter motor was simply freewheeling, the current increase due to voltage reduction would be much smaller. Under mechanical load, it skyrockets.

[Gary Lewis]

Yes it does.

I think I would be careful to maintain that the increased load is the cause of the higher current. Not so much directly by friction, but by induction changes due to speed.

You could demonstrate this effect with a small motor connected to a battery with an ammeter measuring current. Slowing the motor by increasing the load will increase the current.

The thing about the starter motor scenario is that presumably, you have reduced the speed by lowering the voltage. Certainly something similar is going on in your electricity 99 description of the blower motor? What is happening in that scenario is that you have limited the current by means of the series resistor. Your Ohm's law power calculation kicks in and you can no longer deliver the same horsepower, frictional losses kick in and the motor speed slows. With the starter motor and no resistive losses, the battery is still able to provide full current. The voltage drop comes from internal resistance within the battery and the poor starter motor just wants to continue to crank the engine. So what does it do? It slows down, inductance decreases, current flow increases and the power calculations maintain the required force to crank the engine. Of course, there are practical limits to this effect. The DC resistance internally of both the motor and the battery would define those limits, but for all practical purposes, we will experience starter solenoid failure due to overcurrent.

One last thing to consider is the type of work we are asking the motor to do. All mechanical losses can be considered frictional losses, but we need to keep in mind that work is a product of force, motion and time. In the case of a blower motor, the frictional losses are largely associated with airflow and are reduced as fan speed is reduced. Not so with the starter motor which operates at relatively slow speeds, but faces a constant demand for torque, (force). If the starter motor was simply freewheeling, the current increase due to voltage reduction would be much smaller. Under mechanical load, it skyrockets.

You are beyond me. I understand what you are saying if I stop and concentrate on it, and it makes sense. However, the following causes me some wonder:

I think I would be careful to maintain that the increased load is the cause of the higher current. Not so much directly by friction, but by induction changes due to speed.

Isn't the speed change in my scenario of a dragging starter due to friction? Yes, it isn't the friction itself that is causing the higher current, but it is causing a slow down in RPM for the starter and, therefore, keeping the current high. Right?

Ditto the poor connection, which reduces voltage and, therefore, power to the motor. And that reduces RPM. If you have enough of that then you could conceivably keep the motor in high-current mode.

 

[NotEnoughTrucks]

You are beyond me. I understand what you are saying if I stop and concentrate on it, and it makes sense. However, the following causes me some wonder:

I think I would be careful to maintain that the increased load is the cause of the higher current. Not so much directly by friction, but by induction changes due to speed.

Isn't the speed change in my scenario of a dragging starter due to friction? Yes, it isn't the friction itself that is causing the higher current, but it is causing a slow down in RPM for the starter and, therefore, keeping the current high. Right?

Ditto the poor connection, which reduces voltage and, therefore, power to the motor. And that reduces RPM. If you have enough of that then you could conceivably keep the motor in high-current mode.

It's one of those cases where I'm careful how I word it.

Yes, friction slowed down the motor and yes, slow speeds change induction and increase current. Thus the use of the word directly.

On the poor connection, yes that will slow the motor down and yes, the loaded motor will try to compensate, but that poor connection will also limit current. Possibly not as destructive to the solenoid, but the net effect will be to slow down the starter.

I confuse myself sometimes as well! Lots of things to consider.

[PetesPonies]

Ray - Your statement of "As the motor spins, the inductance of the windings increase, reducing the current" is the key. I think we can safely assume that a starter relay/solenoid is sized correctly to handle the in-rush current to the starter, but that the designers of the solenoid assumed that the starter will spin at a certain rate, thereby reducing the current. However, if the starter doesn't spin at the assumed rate then the continuous current draw will be more than they designed for, and even contacts made with good material will weld together.

So, a dragging starter would cause this problem, and a dragging starter is one where the bearings have worn to allow the armature to rub against the stator. Even a tiny rub causes heat, which makes the armature grow, which causes more friction, which causes more heat, which.....

And even a poor connection can cause the problem if it keeps the starter from spinning fast enough to reduce the current substantially and, therefore, stay in the high-current mode.

Does that make sense?

Absolutely makes sense Gary. Anything that impedes the starter from turning easily ( engine, heat, inductance, etc ) will cause the starter to draw more amperage. If you had a sample starter and connected it to a 4 cylinder engine, you would expect it to draw less current than the same starter connected to a large 12 cylinder engine. In this case, for nothing more than the mechanical resistance of the larger engine. So when that current draw increases, so does the heat in the wiring and all contacts. So yes, this could be a reason why the solenoids have issues with the contacts welding together.

[Jcris]

Absolutely makes sense Gary. Anything that impedes the starter from turning easily ( engine, heat, inductance, etc ) will cause the starter to draw more amperage. If you had a sample starter and connected it to a 4 cylinder engine, you would expect it to draw less current than the same starter connected to a large 12 cylinder engine. In this case, for nothing more than the mechanical resistance of the larger engine. So when that current draw increases, so does the heat in the wiring and all contacts. So yes, this could be a reason why the solenoids have issues with the contacts welding together.

Just a follow up here regarding the starter issue. After several start-ups the starter issue has not resurfaced. My brother-in-law replaced the ignition switch and apparently that has resolved the issue. I'm a skeptic and so will keep an eye on it as I think this will happen again. Money has been an issue or I'd just do the PMGR starter now. We'll see how this goes. Thanks for all the help

Jcris