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12 Volt Charging

J

John Popelish

Jan 1, 1970
0
Randy said:
Greetings Group,

This is a hypothetical question concerning an automotive charging
system. I know that the voltage regulator controls battery charging by
regulating the field current in the alternator between 13.5 and 14.5
volts. The question I pose is this:

If the voltage from the alternator (unregulated) was held at a constant
11 volts , that's 11vdc to a peak of 15.55vdc, full ripple, no filter,
would this damage the lead-acid battery?

I don't understand your Dc versus peak voltage numbers. Automotive
alternators are 3 phase generators with full wave rectifiers. The
ripple voltage is very much lower than what you specify (about 5%, I
think). See:
http://www.tpub.com/content/construction/14273/css/14273_49.htm
Theoretically, in this model, the battery works in conjunction with
alternator to supply the load instead of intermittent use. When the
alternator voltage drops below 13.5, the battery starts conducting its
supply to the load. When the alternator voltage exceeds battery voltage,
the battery is charged.

This part makes sense.
 
J

John Popelish

Jan 1, 1970
0
Randy said:
This is a gray area for me. Rectified AC is still AC to me, it's just
all positive, so scratch the dc reference. OK. The AC is rectified and
all positive with no smoothing. From 0 degrees up until the alternator
voltage exceeds that of the battery, the battery supplies the load.

Yes. This is true for a single phase AC sine wave.

But an automotive alternator is a 3 phase source of AC, with 3 sine
waves overlapping and shifted by 120 degrees, each. When full wave
rectified, this produces 6 half sine waves (one arriving every 60
degrees) overlapping per cycle. So there is no time when all of the
half sine waves are zero volts, at the same time. By the time one of
the peaks has gone down by 5%, the next one rises above that, and
takes over. So the peak to peak ripple voltage is only about 5% of
the peak voltage. if the peak is assumed to be about 14.5 volts, the
minimum the alternator puts out is 95% of that or 13.77 volts. And
this does not take into account the flat topping effect of the current
on the not zero impedance of the alternator.
What
concerned me in this example is the voltage at the peak causing damage
to the battery during the degrees of charge. I've read that voltages
over 14.5v can cause the battery to overcharge and lessen its life. To
achieve a 12v rectified output, you need 13.5vrms from the alternator
minus the 1.4v drop across the rectifier. That gives a peak voltage of
19v. Will this damage the battery?

Quite probably. For this reason, single phase battery charging
systems do not supply load current, full time, unless they contain
filtering components to reduce the ripple voltage.
 
R

Randy Gross

Jan 1, 1970
0
Greetings Group,

This is a hypothetical question concerning an automotive charging
system. I know that the voltage regulator controls battery charging by
regulating the field current in the alternator between 13.5 and 14.5
volts. The question I pose is this:

If the voltage from the alternator (unregulated) was held at a constant
11 volts , that's 11vdc to a peak of 15.55vdc, full ripple, no filter,
would this damage the lead-acid battery?

Theoretically, in this model, the battery works in conjunction with
alternator to supply the load instead of intermittent use. When the
alternator voltage drops below 13.5, the battery starts conducting its
supply to the load. When the alternator voltage exceeds battery voltage,
the battery is charged.

Where is the flaw in this line of thinking?

Randy Gross
 
R

Randy Gross

Jan 1, 1970
0
John said:
I don't understand your Dc versus peak voltage numbers. Automotive
alternators are 3 phase generators with full wave rectifiers.


This is a gray area for me. Rectified AC is still AC to me, it's just
all positive, so scratch the dc reference. OK. The AC is rectified and
all positive with no smoothing. From 0 degrees up until the alternator
voltage exceeds that of the battery, the battery supplies the load. What
concerned me in this example is the voltage at the peak causing damage
to the battery during the degrees of charge. I've read that voltages
over 14.5v can cause the battery to overcharge and lessen its life. To
achieve a 12v rectified output, you need 13.5vrms from the alternator
minus the 1.4v drop across the rectifier. That gives a peak voltage of
19v. Will this damage the battery?
 
J

John Popelish

Jan 1, 1970
0
Randy said:
...There is one more question I may as well address while we
are on the subject. The main focus was based on alternators. Assume for
this question that the source of power is 60 HZ,

(single phase, I assume)
rectified and smoothed with a peak of approx. 14.5v.
The load is 12v with a maximum current draw of 15 amps.

If the peak is 14.5 volts, then the load must see 14.5 volta during
those peaks.

Are you saying that the supply averages 12 but peaks at 14.5?
Can I drop the 2.5v without shunting a lot of current
from the main circuit or, would losses make this unnecessary?

I am not sure what you are asking. Are you asking if there is some
way to clip the peaks off so that the load never sees anything higher
than 12 volts? That is what regulators do.

Or are you asking if the increased current that the load draws during
those peaks helps to sag them down a bit? It does, since no power
source is truly a zero resistance source.

Or am I missing your question, entirely?
 
J

John Popelish

Jan 1, 1970
0
Randy said:
I want to use an automotive system in a Electric Vehicle I'm attempting.
I want to use a 12v Lead-acid battery to start a 12v, 10 amp motor. Once
the motor is engaged, I want the alternator to float recharge the
battery and supply the 12v motor.

What is the source of power turning the alternator?
 
J

John Miller

Jan 1, 1970
0
Randy said:
I want to use an automotive system in a Electric Vehicle I'm attempting.
I want to use a 12v Lead-acid battery to start a 12v, 10 amp motor. Once
the motor is engaged, I want the alternator to float recharge the
battery and supply the 12v motor. I'm not even sure it will work as
intended. I have one shot, currently, because of the expense and serious
mistakes will hurt. I'm attempting to take the concept one step further
and eliminate the need to plug the vehicle in every night for a
recharge, charge-as-you-go so to speak.

As written, it sounds as though there is nothing in the system but a 12v
motor, an alternator and an automotive battery. If that is the case,
you're wasting your time, for reasons that countless persons here will be
happy to explain.

If that is not the case, please clarify the configuration, particularly the
external energy source driving the alternator.
--
John Miller
Email address: domain, n4vu.com; username, jsm

Reportedly, in Blythe, California, a city ordinance declares that a person
must own at least two cows before he can wear cowboy boots in public.
 
J

John Miller

Jan 1, 1970
0
Randy said:
That's the reason I'm here John. The question that keeps reoccurring is:
"How do I keep this thing running without draining the battery?" But,
before I go on, is the "no go" from the group based on the battery
supplying current to the field?

No, it's based on your attempt -- whether you realize it or not -- to build
a perpetual motion machine. The energy consumed by the motor must be
replentished by a source external to what you have described so far.
 
P

Peter Bennett

Jan 1, 1970
0
Initially, the battery starts a 12v motor that drives the alternator.
One phase(?) is dedicated to driving the motor with assistance from the
battery when needed. This is the gray area. Theoretically, the
alternator should power the motor once the operating speed is achieved
but, (not being an electrical engineer) I still know there's a flaw
there. So, I included a battery not to drive the motor, but to assist,
which is why I chose single phase. This would allow the battery to
discharge and recharge intermittently and assist the alternator in
driving the motor.

You are attempting to build a perpetual motion machine - it won't
work!! You will end up with a complex battery discharger.

Neither the motor, battery nor alternator are 100% efficient. If, for
example, you put 100 watts into the motor, you may only get 70 watts
(a wild guess) out. Putting that 70 watts into the alternator might
give you 50 watts out - which is not enough to drive the motor, much
less allow the overall system to do any useful work.


--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca
 
R

Randy Day

Jan 1, 1970
0
Randy said:
That's the reason I'm here John. The question that keeps reoccurring is:
"How do I keep this thing running without draining the battery?" But,
before I go on, is the "no go" from the group based on the battery
supplying current to the field?

Just to clarify; you're saying the
alternator runs the motor which does
.... what? Charges the battery? Powers
the alternator?

If you answer 'yes' to either of these
questions, look up 'Law of Entropy' and
'perpetual motion machine' on the
internet.
 
J

John Miller

Jan 1, 1970
0
Randy said:
It's not Perpetual Motion. There are two sources of power here.

Well, there's the error. There two fewer sources of power than you think.

The battery is a device that stores power.
The alternator is a device that converts power.
Neither is a source of power.

--
John Miller
Email address: domain, n4vu.com; username, jsm

In case of fire, stand in the hall and shout "Fire!"
-- The Kidner Report
 
J

John Popelish

Jan 1, 1970
0
Randy said:
Initially, the battery starts a 12v motor that drives the alternator.
One phase(?) is dedicated to driving the motor with assistance from the
battery when needed. This is the gray area. Theoretically, the
alternator should power the motor once the operating speed is achieved
but, (not being an electrical engineer) I still know there's a flaw
there. So, I included a battery not to drive the motor, but to assist,
which is why I chose single phase. This would allow the battery to
discharge and recharge intermittently and assist the alternator in
driving the motor.

You have a big problem. Both alternator and motor are energy
conversion devices. Neither makes energy. It takes more than 1
mechanical watt into the alternator before 1 watt electrical comes out
(it is not 100% efficient). The motor takes more than a watt
electrical in to produce 1 watt mechanical out (it is also not 100%
efficient). All you get by adding the alternator to the motor battery
combination is that the alternator losses have to come out of the
battery power. If the alternator was used only during braking, you
could, theoretically put some of the energy that would have gone into
heat production in the brakes into the battery, but that is the best
you can do.
 
R

Ratch

Jan 1, 1970
0
John Miller said:
Well, there's the error. There two fewer sources of power than you think.

The battery is a device that stores power.
The alternator is a device that converts power.
Neither is a source of power.

You all mean energy, not power, right? Ratch
 
P

Peter Bennett

Jan 1, 1970
0
This is a start Peter. Like other advances in the past, they had to have
a beginning. It may not work properly to begin with but, there are great
minds out there that may be able to find the pieces to the puzzle,
within the law, to make this a reality. It is not Perpetual Motion. It
is applying the law in ways thought not possible. Entropy tells me this
device will try to find a state of equilibrium. OK! Entropy also tells
me I must keep it in "disorder" to work. I haven't circumvented the law.
The law told me what to do.

Yes, the system will reach a state of equilibrium - that state will be
with the battery totally discharged, and the motor and alternator
stationary.

You have no "source of energy" in the system - the motor, alternator,
and battery all _convert_ energy, with less than 100 % efficiency. (I
suppose the battery is initially a source of energy, but it will not
last for long.)

It does not matter how you design the alternator - you can't get more
power out of it than the motor puts into it.

(perhaps you should have a talk with feerguy, who posts about
impossible energy storage on one of the other sci.electronics
groups...)




--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca
 
R

Randy Gross

Jan 1, 1970
0
John Popelish wrote:
Quite probably. For this reason, single phase battery charging
systems do not supply load current, full time, unless they contain
filtering components to reduce the ripple voltage.

Understood! There is one more question I may as well address while we
are on the subject. The main focus was based on alternators. Assume for
this question that the source of power is 60 HZ, rectified and smoothed
with a peak of approx. 14.5v. The load is 12v with a maximum current
draw of 15 amps. Can I drop the 2.5v without shunting a lot of current
from the main circuit or, would losses make this unnecessary?

Randy
 
R

Randy Gross

Jan 1, 1970
0
John said:
(single phase, I assume)

Single phase
If the peak is 14.5 volts, then the load must see 14.5 volta during
those peaks.

Are you saying that the supply averages 12 but peaks at 14.5?


I am not sure what you are asking. Are you asking if there is some
way to clip the peaks off so that the load never sees anything higher
than 12 volts? That is what regulators do.

Or are you asking if the increased current that the load draws during
those peaks helps to sag them down a bit? It does, since no power
source is truly a zero resistance source.

Or am I missing your question, entirely?

I want to use an automotive system in a Electric Vehicle I'm attempting.
I want to use a 12v Lead-acid battery to start a 12v, 10 amp motor. Once
the motor is engaged, I want the alternator to float recharge the
battery and supply the 12v motor. I'm not even sure it will work as
intended. I have one shot, currently, because of the expense and serious
mistakes will hurt. I'm attempting to take the concept one step further
and eliminate the need to plug the vehicle in every night for a
recharge, charge-as-you-go so to speak.

At the present time, I'm experimenting with a 60 hz source, single
phase. If I can get the alternator to drive the motor with minimal
assist from the battery, I will have crossed the first hurdle.
 
R

Randy Gross

Jan 1, 1970
0
What is the source of power turning the alternator?

Initially, the battery starts a 12v motor that drives the alternator.
One phase(?) is dedicated to driving the motor with assistance from the
battery when needed. This is the gray area. Theoretically, the
alternator should power the motor once the operating speed is achieved
but, (not being an electrical engineer) I still know there's a flaw
there. So, I included a battery not to drive the motor, but to assist,
which is why I chose single phase. This would allow the battery to
discharge and recharge intermittently and assist the alternator in
driving the motor.
 
R

Randy Gross

Jan 1, 1970
0
John said:
As written, it sounds as though there is nothing in the system but a 12v
motor, an alternator and an automotive battery. If that is the case,
you're wasting your time, for reasons that countless persons here will be
happy to explain.

That's the reason I'm here John. The question that keeps reoccurring is:
"How do I keep this thing running without draining the battery?" But,
before I go on, is the "no go" from the group based on the battery
supplying current to the field?
 
R

Randy Gross

Jan 1, 1970
0
John said:
No, it's based on your attempt -- whether you realize it or not -- to build
a perpetual motion machine. The energy consumed by the motor must be
replentished by a source external to what you have described so far.

I was thinking that this was the common thought. No. Perpetual motion is
a dream. What I am attempting deals with or remains within the law. I'm
trying to arrange the law to work in this situation. I was so involved
in describing the whole, I neglected to describe the alternator.

This alternator is hand built with two banks of 3 phase windings. The
rotor has 16 permanent magnets separated on a two 8 pole disk. This is
the power source and, along with the battery, are consumable. I am
simply trying to arrange it to where they work together to produce the
result I want and stay within the law. Each one of these devices
produces power, one has to be charged, the other rotated to produce. My
problem is the handshake.
 
R

Randy Gross

Jan 1, 1970
0
You are attempting to build a perpetual motion machine - it won't
work!! You will end up with a complex battery discharger.

Neither the motor, battery nor alternator are 100% efficient. If, for
example, you put 100 watts into the motor, you may only get 70 watts
(a wild guess) out. Putting that 70 watts into the alternator might
give you 50 watts out - which is not enough to drive the motor, much
less allow the overall system to do any useful work.

This is a start Peter. Like other advances in the past, they had to have
a beginning. It may not work properly to begin with but, there are great
minds out there that may be able to find the pieces to the puzzle,
within the law, to make this a reality. It is not Perpetual Motion. It
is applying the law in ways thought not possible. Entropy tells me this
device will try to find a state of equilibrium. OK! Entropy also tells
me I must keep it in "disorder" to work. I haven't circumvented the law.
The law told me what to do.
 
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