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Starter SERVO amperage


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Thanks!   The collegial nature of this and many other such discussions is heart-warming, as well as educational!

Pete, you will be familiar with the "pre-engaged" starter design, as opposed to the post-engaged.    The second uses a Bendix drive, and the inertia of the drive sprocket, to throw the sprocket into engagement with the starter ring, as the motor starts to turn.   It uses a helical screw on the back of thr pinion and a fixed nut:

Bendix drive - used - Series Forever

A Pre-engaged starter employs a solenoid (!) that is energised by the starter button, or ignition key switch, to throw the pinion into engagement.     Once there, and not before, it also closes a switch that energises the starter motor itself, thus engaging the starter pre-viously.   Here's a sectional diagram:

 Pre-engaged Starter Motor (Automobile)

 

I had such a starter motor and robbed it of its solenoid.     For this application, it has the advantage of pulling, when (I discover) an O/d solenoid pushes.

 

Alec, thank you for that explanation.  Unreasonably and ignorantly, I assumed the opposite.  Coils of wire, I have, but hang on - vague memories of inductance in a coil, so lay it out loosely on the floor?   And if I only have a multimeter, will that give a wild figure anyway?

Nick, Yes, option 2 is attractive, and I think I may have a momentary switch in the drawer that could do for a test - or else just disconnect the test battery once in reverse.    A "central locking solenoid" - what that??  Until light dawned and I realised that it's a solenoid used in a central locking system. 

DoorAnimation1.gif Now THAT's a good idea!

I fly with this for a while - I'm that far off completion that if this part crashes and burns it's no disaster.

John

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Thank you John, I see now. Was getting confused between the title, and talk of starter motor solenoids.

I also wonder if you are confusing 2.5piman (Alec) for DeTRacted (not Alec!) in your last 2 posts..? 

Anyway, this bit: 

22 hours ago, JohnD said:

Alec, thank you for that explanation.  Unreasonably and ignorantly, I assumed the opposite.  Coils of wire, I have, but hang on - vague memories of inductance in a coil, so lay it out loosely on the floor?   And if I only have a multimeter, will that give a wild figure anyway?

I think you're referring to this post:

Quote

John - as you say if the series resistance you are using is equal to or greater than the resistance of the solenoid, it is that resistor which is mostly determining the current so the reading isn't much use.  To do this you need to use a much lower resistance ideally less than a tenth of the solenoid resistance which is why I suggested using a length of mains cable in my original post, since that will only be about 0.01 Ohms and will not influence the result to any degree.  You do need a meter which will read down to millivolts though. 

If you are certain about the DC resistance of the solenoid I think you are safe to use Ohms law anyway and your 21.3 Amps is correct. The resistance of copper increases as it gets hot so the current can only be lower than that. 

The solenoid is an inductor so opposes applied voltage. That means the current rises from zero to 21.3A as the back-emf dies, so there is no inrush current peak to be worried about either.

You needn't worry about unspooling loads of cable for your measurement, nor for anything much when it's a DC circuit. Your vague memory is good but applies to AC, where the coil increases circuit impedance. But the point of using a known length of cable, with a known cross-section, is that you don't have to measure its resistance - the ohms-per-metre is a known physical characteristic. You just have to measure the millivolts resulting from your current, which a multimeter is fine for.

I would gently dispute a couple of things in the above quote from DeTRacted though; you can't rely on heating the copper to reduce current to a safe level - if the solenoid isn't rated to take whatever current continuously, the eventual effect is a burnt-out coil. This is thermal overload, which I've just spent quite a long time calculating for a tram network, incidentally.

The other thing is, an inductive load does exhibit inrush current when you energise it. A solenoid is quite interesting, as you'd have a large inrush but as soon as the solenoid started to move it would produce a back-EMF and suppress the current to some degree. Then when the solenoid is fully engaged the back-EMF disappears and current will rise again (unless you switch it off or change the circuit resistance). 

Anyway I'm getting carried away now. 

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50 minutes ago, PeteStupps said:

The other thing is, an inductive load does exhibit inrush current when you energise it. A solenoid is quite interesting, as you'd have a large inrush but as soon as the solenoid started to move it would produce a back-EMF and suppress the current to some degree. Then when the solenoid is fully engaged the back-EMF disappears and current will rise again (unless you switch it off or change the circuit resistance

I'm going to partially retract this bit, I am confusing AC & DC. It wouldn't have an inrush, although I would expect the solenoid movement to produce the back-EMF effect. 

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16 hours ago, PeteStupps said:

you can't rely on heating the copper to reduce current to a safe level

I wasn't suggesting you could nor that it would be safe - I was just stating that the current isn't going to go any higher than the initial value even when the coil gets hot.  

15 hours ago, PeteStupps said:

The other thing is, an inductive load does exhibit inrush current when you energise it......I'm going to partially retract this bit, I am confusing AC & DC.

Just for the record, an inductor always opposes any change of current (remember CIVIL - in an inductor the voltage leads the current.) When voltage is applied the start of flow of current into any inductor produces an instantaneous change of magnetic flux in the core and it is this change of flux which induces the back-emf to limit the current.  There is no initial spike, the current rises from zero asymptotically to the final DC value.  This has nothing to do with physical movement of the armature - it happens before the armature starts to move.

Perhaps you are thinking of motors where the back-emf rises with motor speed so they do take more current when stalled. Motor current when running is much less than the DC value due to armature winding resistance alone. 

Rob

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This is remarkably reminiscent of real common room discussions / department meetings, where an innocent question leads to a viscious (not that the above is viscious  in any way!) argument between eminent experts!

 

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23 minutes ago, DeTRacted said:

Just for the record, an inductor always opposes any change of current (remember CIVIL - in an inductor the voltage leads the current.) When voltage is applied the start of flow of current into any inductor produces an instantaneous change of magnetic flux in the core and it is this change of flux which induces the back-emf to limit the current.  There is no initial spike, the current rises from zero asymptotically to the final DC value.  This has nothing to do with physical movement of the armature - it happens before the armature starts to move.

Perhaps you are thinking of motors where the back-emf rises with motor speed so they do take more current when stalled. Motor current when running is much less than the DC value due to armature winding resistance alone. 

Rob

Rob I do agree that an inductor opposes change in current, but made the foolish error of proclaiming things without giving them much thought first. Coming from a power engineering viewpoint, I was thinking of transformers as much as motors. Transformers exhibit magnetizing inrush current but I shouldn't blithely extrapolate that out to a DC solenoid. 

I would like to see the transient current waveform for a solenoid though; I reckon there would be some ripple associated with the initial magnetization, movement & re-magnetization of the armature. But not remotely significant for John's application! 

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Hello Peter,

 

as there is a back emf it follows that the instant of energising must have a higher current than therafter, simply the impedance of the winding being the controlling factor. It is of an extremely short duration that it is not easily detected but it must be there.

 

John,

are you still going with the startersolenoid?

As you say it is only used to select reverse, therafter, if the mechanical linkage is arranged that the solenoid is not needed until the next time I can't see any problem with overheating. After all consider how much cranking a PI needs if the fuel system is disconnected.

Alec

 

 

 

 

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Hello John,

 

PS. An overdrive (A type at least) pulls not pushes, otherwise how cou;ld the primary winding disconnect switch work as it is in the base of the solenoid. If you tried one with the polarity reversed it may push but don't quote me on that?

 

Alec

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10 hours ago, 2.5piman said:

as there is a back emf it follows that the instant of energising must have a higher current than therafter, simply the impedance of the winding being the controlling factor.

Back to the common-room. 

No -  sorry Alec but there really is no initial current spike.

If you wish to consider the impedance as the controlling factor rather than back-emf, remember the impedance of a coil increases with frequency (X=2Pi fL). A step change in applied voltage is high frequency (f=1/t where t is the rise-time) so if the step change could be  instantaneous, the impedance would be infinite preventing any initial current flow. In practice that is not possible of course but the initial impedance is still very large because the voltage is applied quickly, so any initial current flow is necessarily very small and can only increase with time.  

The current can never exceed the value determined by the DC resistance of the coil and will reach that level slowly as the effect of the impedance term decreases over time.  The current curve is asymptotic, equivalent to the voltage curve in charging a capacitor.

 

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