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Default Alternator regulator on mower, G G L C WE terminals, meaning?

In article ,
Chris Green wrote:
Roger Hayter wrote:
Rod Speed wrote:

"Roger Hayter" wrote in message
...
The Natural Philosopher wrote:

On 24/04/18 09:42, Chris Green wrote:

OK, yes, I agree. 12 volts across 82 ohms is around 160mA.
No, its around 146mA.

What's the voltage of a '12V' lead acid battery?

Usually taken as 13.8 but it obviously varys a bit.


That was my point: 13.8/82=0.168

That's the 'on charge' voltage. A car battery which is not being
charged won't provide 13.8 volts, we're just used to car batteries
with alternators which mean that whenever the engine is running it's
being charged.


Take a look at:-


http://scubaengineer.com/documents/l...ing_graphs.pdf


Even a fully charged battery will only show just over 12.5 volts when not
being charged.


The point is when calculating near anything car electrics wise is you have
to make allowances for the voltage when running being rather higher than
the nominal 12v. As very little is only ever used with the engine
stopped. And when starting the engine, you also need to allow for the
voltage dropping considerably due to the starter load.

--
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Default Alternator regulator on mower, G G L C WE terminals, meaning?

"Dave Plowman (News)" wrote:
In article ,
Chris Green wrote:
Roger Hayter wrote:
Rod Speed wrote:

"Roger Hayter" wrote in message
...
The Natural Philosopher wrote:

On 24/04/18 09:42, Chris Green wrote:

OK, yes, I agree. 12 volts across 82 ohms is around 160mA.
No, its around 146mA.

What's the voltage of a '12V' lead acid battery?

Usually taken as 13.8 but it obviously varys a bit.

That was my point: 13.8/82=0.168

That's the 'on charge' voltage. A car battery which is not being
charged won't provide 13.8 volts, we're just used to car batteries
with alternators which mean that whenever the engine is running it's
being charged.


Take a look at:-


http://scubaengineer.com/documents/l...ing_graphs.pdf


Even a fully charged battery will only show just over 12.5 volts when not
being charged.


The point is when calculating near anything car electrics wise is you have
to make allowances for the voltage when running being rather higher than
the nominal 12v. As very little is only ever used with the engine
stopped. And when starting the engine, you also need to allow for the
voltage dropping considerably due to the starter load.

But the current we were debating is the one that flows through the
'lamp' often when the engine hasn't yet been started so it will
probably be when the battery voltage is around 12.5 volts.

It's not a critical current (as in its value isn't critical) anyway so
its exact value is unimportant anyway.

--
Chris Green
·
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Default Alternator regulator on mower, G G L C WE terminals, meaning?

On Thu, 26 Apr 2018 10:40:45 +0100, Dave Plowman (News) wrote:

In article ,
Chris Green wrote:
Roger Hayter wrote:
Rod Speed wrote:

"Roger Hayter" wrote in message
...
The Natural Philosopher wrote:

On 24/04/18 09:42, Chris Green wrote:

OK, yes, I agree. 12 volts across 82 ohms is around 160mA.
No, its around 146mA.

What's the voltage of a '12V' lead acid battery?

Usually taken as 13.8 but it obviously varys a bit.

That was my point: 13.8/82=0.168

That's the 'on charge' voltage. A car battery which is not being
charged won't provide 13.8 volts, we're just used to car batteries with
alternators which mean that whenever the engine is running it's being
charged.


Take a look at:-


http://scubaengineer.com/documents/

lead_acid_battery_charging_graphs.pdf

Even a fully charged battery will only show just over 12.5 volts when
not being charged.


The point is when calculating near anything car electrics wise is you
have to make allowances for the voltage when running being rather higher
than the nominal 12v. As very little is only ever used with the engine
stopped. And when starting the engine, you also need to allow for the
voltage dropping considerably due to the starter load.


This where the positive coefficient of the 2.2W ignition warning lamp's
filament resistance might prove useful[1] since even a good condition 12v
car battery's terminal voltage will fall to around the 8 to 10 volt mark
from the starter cranking load (once the motor has accelerated to
cranking speed - the voltage can easily dip to less than 6 volts on the
initial starting surge of 300 odd amps or more before it settles down to
the 100 or so amps once up to cranking speed).

[1] Thinking about that, it might be better to avoid bootstrapping the
excitation process prematurely and wait till the engine is actually
ticking over under its own power rather than whilst it's still being
cranked by the starter. :-)

Relying on the much lower 10 to 20 mA of an LED ignition warning lamp
alone might prove to be a more optimal choice of initial excitation
current from the point of view of electrically starting an ICE. Even with
an open circuit ignition lamp, the alternator could usually be
bootstrapped into excitation from the weak residual magnetism of the
rotor by blipping the engine revs well above tickover (say 3000rpm or so
for a car engine).

In this case, the evidence that the alternator had started producing
output could be seen either by observing the suddenly increased
brightness of other dashboard lights or the varying brightness of the
headlights when accelerating the engine speed a modest amount above
tickover. It could also be detected by the increased vibration and slight
drop of engine tickover speed when flashing the headlights on and/or
switching on a heavy electrical load such as a screen heater.

The secret to the alternator's ability to provide useful output at
tickover speed compared to that of the older dynamo charging system was
simply down to the much stronger construction of its rotor assembly
compared to that of a dynamo armature assembly which meant it could be
spun much faster before flying apart from centripetal forces, allowing it
to be geared up to a sufficient speed to produce close to full output
even at tickover speed without risk of it flying apart at the engine's
maximum rev limit.

Elimination of the high friction commutator brush assembly, as used by
the classic dynamo, by use of a silicon rectifier bridge, not only
improved efficiency by eliminating mechanical friction, it also allowed
much higher output currents to be delivered than the typical 22A rating
of most car dynamos, a rating arrived at as a compromise between minimum
requirements and the frictional losses incurred by the commutator/brush
assembly which would otherwise incur even greater frictional loss with
the heavier duty brush gear needed to handle greater levels of output
current.

The frictional losses of the alternator's slip ring brush assembly were
only a tiny fraction of that of a segmented commutator further reduced by
the much lower contact pressure needed to handle a mere 3 or 4 amps of
excitation current. The slip ring brushes, as a consequence of the lower
pressures and an absence of gaps in the slip rings, enjoyed much less
wear and had a much longer service life, further enhanced by using much
longer 'carbons' and springs in the slip ring brush holder assembly.

Service life ratings for a motor vehicle alternator's slip ring brushes
were typically some 75000 miles or so the last time I saw any figures on
this. Dynamo brushes had to be routinely replaced during the expected
service life of the dynamo (every 10000 miles or so?) with the commutator
needing maintenance at about every 2nd or 3rd brush renewal cycle. Little
wonder then that the alternator took over from the dynamo the moment that
cost effective silicon diode rectifier packs became commercially
available.

--
Johnny B Good
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Default Alternator regulator on mower, G G L C WE terminals, meaning?

In article ,
Johnny B Good wrote:
The point is when calculating near anything car electrics wise is you
have to make allowances for the voltage when running being rather
higher than the nominal 12v. As very little is only ever used with
the engine stopped. And when starting the engine, you also need to
allow for the voltage dropping considerably due to the starter load.


This where the positive coefficient of the 2.2W ignition warning lamp's
filament resistance might prove useful[1] since even a good condition 12v
car battery's terminal voltage will fall to around the 8 to 10 volt mark
from the starter cranking load (once the motor has accelerated to
cranking speed - the voltage can easily dip to less than 6 volts on the
initial starting surge of 300 odd amps or more before it settles down to
the 100 or so amps once up to cranking speed).


Doesn't really matter, since it would be silly for the alternator to be
attempting to charge the battery while the starter is operating.

--
*If you try to fail, and succeed, which have you done?

Dave Plowman London SW
To e-mail, change noise into sound.
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Default Alternator regulator on mower, G G L C WE terminals, meaning?

On Thu, 26 Apr 2018 23:58:17 +0100, Dave Plowman (News) wrote:

In article ,
Johnny B Good wrote:
The point is when calculating near anything car electrics wise is you
have to make allowances for the voltage when running being rather
higher than the nominal 12v. As very little is only ever used with
the engine stopped. And when starting the engine, you also need to
allow for the voltage dropping considerably due to the starter load.


This where the positive coefficient of the 2.2W ignition warning
lamp's
filament resistance might prove useful[1] since even a good condition
12v car battery's terminal voltage will fall to around the 8 to 10 volt
mark from the starter cranking load (once the motor has accelerated to
cranking speed - the voltage can easily dip to less than 6 volts on the
initial starting surge of 300 odd amps or more before it settles down
to the 100 or so amps once up to cranking speed).


Doesn't really matter, since it would be silly for the alternator to be
attempting to charge the battery while the starter is operating.


Hence note[1]. :-)

--
Johnny B Good


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Default Alternator regulator on mower, G G L C WE terminals, meaning?

Johnny B Good Wrote in message:
On Thu, 26 Apr 2018 23:58:17 +0100, Dave Plowman (News) wrote:

In article ,
Johnny B Good wrote:
The point is when calculating near anything car electrics wise is you
have to make allowances for the voltage when running being rather
higher than the nominal 12v. As very little is only ever used with
the engine stopped. And when starting the engine, you also need to
allow for the voltage dropping considerably due to the starter load.


This where the positive coefficient of the 2.2W ignition warning
lamp's
filament resistance might prove useful[1] since even a good condition
12v car battery's terminal voltage will fall to around the 8 to 10 volt
mark from the starter cranking load (once the motor has accelerated to
cranking speed - the voltage can easily dip to less than 6 volts on the
initial starting surge of 300 odd amps or more before it settles down
to the 100 or so amps once up to cranking speed).


Doesn't really matter, since it would be silly for the alternator to be
attempting to charge the battery while the starter is operating.


Hence note[1]. :-)


Appendix shurely?
--
Jim K
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