Toshiba TV29C90 problem; Image fades to black...

[John Fields]

On Sun, 12 Jun 2005 14:00:23 -0800, floyd@barrow.com (Floyd L.
Davidson) wrote:

Put an ammeter there and it says zero. That's zero. Electrons bouncing
around in the conductor have an average net displacement, over time, of 0.

Is this an AC ammeter, or a DC ammeter? (And isn't that just a voltmeter
anyway, in most actual cases????) Hmmm...

---
No. With the exception of electroscopes and such like, all voltmeters
are really ammmeters and what was being referred to was really a
galvanometer
---

--
John Fields
Professional Circuit Designer

 

[Floyd L. Davidson]

"--" <dehoberg@comcast.com> wrote:

"Floyd L. Davidson" <floyd@barrow.com> wrote:
John Fields <jfields@austininstruments.com> wrote:
On Sun, 12 Jun 2005 17:15:06 -0700, Don Lancaster <don@tinaja.com
wrote:

Sum a 1 volt peak sinewave with a 0.6 volt dc term and you have a
waveform whose polarity continuously changes but whose average value is
continuous.

---
No, you have a waveform with a polarity which changes _periodically_,
making it an AC signal. Do the electrons traversing the circuit
change direction? Yes. Do the electrons in a DC circuit ever change
direction? No.

Ergo, because of the periodic polarity reversals what you're looking
at is AC.

And, according to what you've said in other posts, if that were a
0.6 volt peak sinewave with 1.0 volt dc, it wouldn't be.

But your definition of AC is faulty, because in fact they are the
same thing, and *both* of them contain an AC component and a DC
component, even if the general direction of electrons is always the
same.

No, both do not - only one of the 1 volt/.6 volt examples given has an
_alternating_ direction component - both examples do have a _variation_ in
their magnitude component.
( This is not a new discussion - and all of the dozen or so engineering
and physics texts and training manuals I have researched on the matter
adhere to the "alternating is reversing" definition of AC. It has been
custom and practice for at least 40 years.)

1) the 1 volt dc with the .6 sine variation does not alternate its
direction of flow. Its flow only varies in the magnitude of the charge
flowing always in one direction.
It has no alternating current ( i.e, it has no regularly reversing, i.e.
_alternating_, charge flow direction)

2) the 1 volt sinewave with the .6 volt dc does reverse charge flow
direction. It is alternating in its flow direction.
It also varies in its magnitude.

The direction of the description vector must alternate in order to have
Alternating Current. If it does not change direction but only varies in
magnitude, the descriptive vector is not alternating, it is merely varying
in magnitude.

3) Impedance laws apply equally to varying DC and to AC.

Item 3 is correct. That is because "varying DC" *is* AC.

It is AC even if the axis is shifted far enough to avoid
polarity reversals relative only to some specifically defined 0
current.

The reversals are relative... to the steady state condition,
not to some magical 0 current where supposedly no electrons are
flowing.

Otherwise, instead of two types, you are dividing circuit analysis
into three types, two of which are identical in all significant
respects other than an arbitrary definition that is meaningless.

It makes no sense to say that "Impedance laws apply equally" and
then claim that the two are not identical.

--
Floyd L. Davidson <http://web.newsguy.com/floyd_davidson>
Ukpeagvik (Barrow, Alaska) floyd@barrow.com

 

[Choreboy]

NSM wrote:

"Choreboy" <choreboyREMOVE@localnet.com> wrote in message
news:42ABDF7C.3A0374F4@localnet.com...

To call a waveform an AC sine wave implies that there is no DC, but this
thread is the first time I've read the claim that all sine waves are AC
sine waves.

FWIW, most waveforms can be created as the sum of sine waves. I wrote an
interesting computer demo once that showed how a sine and it's harmonics
could be added graphically to form a better and better approximation of a
square wave, running through what looked like Butterworth etc. responses.

N

With high frequency and amplitude, a sine wave could be very steep at 0
and 180 degrees. It could also turn sharply at 90 and 270, like the
corner of a square wave. You would need low frequency and amplitude for
a sine wave to approximate the flat peaks of a square wave.

That part is simple enough for me, but I don't understand harmonics. If
you overdrive an amplifier with a sine wave, the output will resemble a
square wave. I know the output can be broken down into the input
frequency and its odd multiples. I'll have to accept it on faith.

 

[Do Little2]

"Dave Plowman (News)" <dave@davenoise.co.uk> wrote in message
news:4d7a4a81d8dave@davenoise.co.uk...

In article <IJ5re.2791$EP2.13759@newscontent-01.sprint.ca>,
Do Little2 <listed@space.com> wrote:
So your system is an upgrade ?? Hmmmm ...

True, serving about 330 million people,
compared to a few on a little island is
a real big upgrade...

Did you know there are approximately 10 times more fires per capita in the
US than the UK caused by poor or inadequate house wiring?

I was not aware of that. So, even with unfortunate fires, chances
are that the US upgrades approximately 10 times more than the UK.

Do Little2

 

[Do Little2]

"Arfa Daily" <monitech@ntlworld.com> wrote in message
news:vGbre.7834$q46.5504@newsfe1-win.ntli.net...

Well, actually, our " little island ", as you so quaintly put it, is
actually connected to many millions of other users in mainland Europe, to
form a hedged power distribution system

On this side of the pond "many millions" is not even close to 330 million +.

which doesn't suffer from cascade
failure when the weather gets a bit cold, or major outages when the sun
flares a bit ... ; -)

True, but think about the enormous length of the electricity network in
the US and Canada compared to the UK. The UK (and that part of
Europe that the UK could possible support) would easily fit about
10 times in Ontario alone....

Just out of curiosity: How many Megawatts of electricity is
the UK capable of exporting before its own network collapses?

Do Little2

 

[Dave Plowman (News)]

In article <SHgre.2830$EP2.14167@newscontent-01.sprint.ca>,
Do Little2 <listed@space.com> wrote:

Did you know there are approximately 10 times more fires per capita in
the US than the UK caused by poor or inadequate house wiring?

I was not aware of that. So, even with unfortunate fires, chances are
that the US upgrades approximately 10 times more than the UK.

It would seem quite the contrary. House wiring more than 40 years old is
uncommon in the UK. Seems not so in many parts of the US with older houses.

--
*It's o.k. to laugh during sexŒ.Œ.just don't point!

Dave Plowman dave@davenoise.co.uk London SW
To e-mail, change noise into sound.

 

[Dave Plowman (News)]

In article <THgre.2831$EP2.14167@newscontent-01.sprint.ca>,
Do Little2 <listed@space.com> wrote:

Well, actually, our " little island ", as you so quaintly put it, is
actually connected to many millions of other users in mainland Europe,
to form a hedged power distribution system

On this side of the pond "many millions" is not even close to 330
million +.

So you think the population of Europe is tiny compared to the US?

which doesn't suffer from cascade
failure when the weather gets a bit cold, or major outages when the sun
flares a bit ... ; -)

True, but think about the enormous length of the electricity network in
the US and Canada compared to the UK. The UK (and that part of
Europe that the UK could possible support) would easily fit about
10 times in Ontario alone....

It's an exchange scheme. Like all proper grids with multiple power
stations feeding them.

Just out of curiosity: How many Megawatts of electricity is
the UK capable of exporting before its own network collapses?

Dunno - it's never happened. The whole idea of a grid is that you share
power, so any fault is likely to be local. And in event of a grid failure,
the power may be routed in a different way.

FWIW, I can't remember an outage here in this part of London. Nor do I
have or need an UPS etc. Of course there are rural parts that still may
have problems due to overhead lines etc.

--
*Why does the sun lighten our hair, but darken our skin?

Dave Plowman dave@davenoise.co.uk London SW
To e-mail, change noise into sound.

 

[Ethic]

Surgeon General's Warnings :

Religion Causes Hatred Cancer,
Xenophobic Disease and Complicates Life.

Quitting Religion Now Greatly Reduces
Serious Risks to Your Mental Health.

* Ethic *


According to Karl Marx, RELIGION IS USED
BY OPPRESSORS TO MAKE PEOPLE FEEL BETTER ...
This is the origin of his comment that :
" RELIGION IS THE OPIUM OF THE MASSES "

( But, Karl Marx is probably censured in USA )

* Ethic *
integrityethics@NOhotmailSPAM.com

"shred" <shred_monger@hotmail.com> nicely replied :
news:1118500257.667619.124790@g47g2000cwa.googlegroups.com...
if theres one thing i can't stand it's born again christians....if
theres someting else i can't stand it's born again christians in the
wrong place,,...keep it for church

 

[John Fields]

On Sun, 12 Jun 2005 16:35:48 -0800, floyd@barrow.com (Floyd L.
Davidson) wrote:

John Fields <jfields@austininstruments.com> wrote:
On Sun, 12 Jun 2005 14:00:23 -0800, floyd@barrow.com (Floyd L.
Davidson) wrote:

My point still stands, that if the current is changing, it is by
definition AC, and current not changing is DC. Trying to look
at it as DC is all in one direction and anything else is AC,
doesn't work.

---
Your point is flawed. Alternating Current, by definition, causes
electrons to move in one direction for a time, and then to reverse
direction for a time.

That isn't true.

---
Yes, it is. If you have proof, instead of just a statement to the
effect that it isn't, I'd love to see it.
---

The sinusoidally varying unipolar voltage under consideration _always_
forces electrons to move in one direction only.

A non-sequitor.

---
"Non sequitur." No. A non-sequitur is an inference or a conclusion
that does not follow from the premises, or a comment that is unrelated
to a preceding one. My error was the omission of a reference, Mr.
Lancaster's: "1 volt peak sinewave with a 0.6 volt dc term"
---

Since the voltage varies, the current will also, but the _direction_
in which the electrons are travelling will never change.

If it varies, it's AC.

---
No, it isn't. What's necessary is the polarity reversal before it can
be considered AC.
---

That means that the signal is DC. A varying DC, but DC nonetheless.

If there is such a think as "varying DC", connect a load to
it... through a capacitor. Now, how do you describe the effect
that load has on your "varying DC". The load see's *only* AC,
even according to your definition. That AC came from somewhere,
and it certainly was not generated by the capacitor.

---
It most certainly was!

Consider a DC coupled audio amplifier running from a single 12V
supply, with its output set to Vcc/2 and feeding an 8 ohm load with a
4VPP sinusoidal signal. Like this:

+12
|
+--o--+
| AMP |---+-->Vout
+--o--+ |
| [4R]
| |
GND GND

Now, with phi being equal to the phase angle of the signal and zero
degrees corresponding the voltage halfway between the most positive
and least positive output voltage, the output voltage excursions will
look like this:

phi Vout
-----+------
0° 6V
90° 8V
180° 6V
270° 4V
360° 6V

Now, connect that magical capacitor between the amp and the load, as
shown below, and watch what happens:

+12
|
+--o--+
| AMP |---[cap]--+-->Vout
+--o--+ |
| [4R]
| |
GND GND


phi Vout
-----+------
0° 0V
90° +2V
180° 0V
270° -2V
360° 0V

Why?

Well,for starters, consider that under quiescent conditions the
left-hand side of the cap will be charged to 6V and the right hand
side will be at zero volts since there is no galvanic path to the
output of the amp through the cap.

Now, imagine that the voltage at the amp's outout starts to go
positive. What will happen is that the amp will start sucking
electrons out of the cap, generating a potential difference across the
cap's plates which causes electrons to flow through the resistor,
making the top of the resistor more positive than the bottom.

Continuing in time, a point will be reached where the output of the
amp will start forcing electrons _into_ the resistor, at which point
the direction of travel of the electrons will be reversed. This
periodic reversal will cause the polarity of the signal into the
resistor to alternate. This alternating voltage will then give rise
to an _alternating current_ in the resistor.
---

That's because AC is *not* defined by any change in direction,
but only by a rate of movement change.

---
Poppycock. It's precisely the alternations in the direction of charge
flow which cause it to be called "Alternating Current".

Your way would have it be called AC by assigning some arbitrary rate
of change, irrespective of direction as the delineation point, which
makes no sense at all. That is, what would you specify as the rate of
change which would delineate between between AC and DC? 0.5A/s?
0.001A/s? 0.1V/s?

--
John Fields
Professional Circuit Designer

 

[--]

Strictly speaking, I believe the reactance (part of impedance)
equations apply to any variation in current magnitude. Their appropriate
application does not in any way require reversing the charge.

1) I think one needs to define the term "alternating current" by its
phenomena rather than define it by what applies to "AC". In other words,
define AC as alternating current -rather than defining AC as "anything
requiring an impedance calculation because of its magnitude variation".
( OK, all scientific definitions require definitions in terms of other
defined concepts; thus voltage and charge are defined in terms of force.
And yes, any phenomena in its purest defined form uses the fewest of the
core units, and only the core units, of the measuring system. And yes,
since, unlike in the British ft-sec-lb system, force is not a core unit of
the metric kg-sec-m system, one cannot be as "pure" in the metric system
with many definitions as one can be in the British system, "decile"
convenience notwithstanding)

2) There are two phenomena and two descriptive words if one uses the
mathematical description of the changes associated with current: changes in
current _direction_ and changes in current _magnitude_.

There are three (or more) phenomena if one uses only the two descriptive
terms _AC_ and _DC_, well evidenced in this thread: changes in direction and
magnitude, changes in magnitude only, or no changes in either magnitude or
direction. Three phenomena defined using only two words for those three
cannot be specific and exclusive enough for a rigorous definition. The
middle condition, the overlap as it were, ends up wanting.

3) In the definition approach to a phenomena, one deals with the
descriptive term and the phenomena itself and ignores the present attached
effects. Once the definition is had, then the phenomena's interaction with
other phenomena can be determined. Yes, having such rigor in a definition
can be more complicated in its application.

In the application approach to defining a phenomena, one defines by
addressing what equations, etc., apply to the condition. In this approach,
you end up in circular arguments, chasing your tail. Something always will
not fit. Like changes in magnitude without changes in direction.


"Floyd L. Davidson" <floyd@barrow.com> wrote in message
news:87slzmv066.fld@barrow.com...

"--" <dehoberg@comcast.com> wrote:
"Floyd L. Davidson" <floyd@barrow.com> wrote:
John Fields <jfields@austininstruments.com> wrote:
On Sun, 12 Jun 2005 17:15:06 -0700, Don Lancaster <don@tinaja.com
wrote:

Sum a 1 volt peak sinewave with a 0.6 volt dc term and you have a
waveform whose polarity continuously changes but whose average value
is
continuous.

---
No, you have a waveform with a polarity which changes _periodically_,
making it an AC signal. Do the electrons traversing the circuit
change direction? Yes. Do the electrons in a DC circuit ever change
direction? No.

Ergo, because of the periodic polarity reversals what you're looking
at is AC.

And, according to what you've said in other posts, if that were a
0.6 volt peak sinewave with 1.0 volt dc, it wouldn't be.

But your definition of AC is faulty, because in fact they are the
same thing, and *both* of them contain an AC component and a DC
component, even if the general direction of electrons is always the
same.

No, both do not - only one of the 1 volt/.6 volt examples given has an
_alternating_ direction component - both examples do have a _variation_
in
their magnitude component.
( This is not a new discussion - and all of the dozen or so engineering
and physics texts and training manuals I have researched on the matter
adhere to the "alternating is reversing" definition of AC. It has been
custom and practice for at least 40 years.)

1) the 1 volt dc with the .6 sine variation does not alternate its
direction of flow. Its flow only varies in the magnitude of the charge
flowing always in one direction.
It has no alternating current ( i.e, it has no regularly reversing,
i.e.
_alternating_, charge flow direction)

2) the 1 volt sinewave with the .6 volt dc does reverse charge flow
direction. It is alternating in its flow direction.
It also varies in its magnitude.

The direction of the description vector must alternate in order to have
Alternating Current. If it does not change direction but only varies in
magnitude, the descriptive vector is not alternating, it is merely
varying
in magnitude.

3) Impedance laws apply equally to varying DC and to AC.

Item 3 is correct. That is because "varying DC" *is* AC.

It is AC even if the axis is shifted far enough to avoid
polarity reversals relative only to some specifically defined 0
current.

The reversals are relative... to the steady state condition,
not to some magical 0 current where supposedly no electrons are
flowing.

Otherwise, instead of two types, you are dividing circuit analysis
into three types, two of which are identical in all significant
respects other than an arbitrary definition that is meaningless.

It makes no sense to say that "Impedance laws apply equally" and
then claim that the two are not identical.

--
Floyd L. Davidson <http://web.newsguy.com/floyd_davidson
Ukpeagvik (Barrow, Alaska) floyd@barrow.com

 

[John Fields]

On Sun, 12 Jun 2005 18:28:16 -0800, floyd@barrow.com (Floyd L.
Davidson) wrote:

"operator jay" <none@none.none> wrote:

It is not changing polarity. I would hesitate to call it alternating
current. On the "dc sine wave" issue, I wouldn't even get into that debate.
To me the terms involved are open to too many interpretations. As evidenced
in this thread, I suppose.

Where *do* you get this requirement for changing polarity? We
don't call it "Alternating Polarity", we call it "Alternating
Current". If the current is being altered, it's AC.

---
No, if the direction of charge flow alternates between two states,
then it's Alternating Current.
---

You keep talking about AP, and it isn't the same.

---
Yes, it is. In order for the current in a load to alternate, the
polarity of the generator's output voltage must alternate as well.

--
John Fields
Professional Circuit Designer

 

[Mike Berger]

The freezing point and boiling point of water are both have
a clearly defined intrinsic meaning to a chemist. "DC sine wave"
is a non-sequiter. It's a made-up self-contradictory term
that won't have an unambiguous meaning to anybody who knows
electronics, no matter how clear it is to you.

jackbruce9999@yahoo.com wrote:

Go back to the original few posts to see how it got started....despite
being explicit about the specs of the wave, someone childishly objected
to my casual usage of "DC sine wave".....would it have been
objectionable had I used "a fully DC-offset sine wave"?......again,
I've never claimed that I was using "official" or
conventionally-correct teminology or nomenclature....I just really
object that anyone would object to what I was saying, when its meaning
was explicitly stated (using actual numbers) and the phrase "fully DC
sine wave", although conventionly queer, is not at all cryptic or hard
to figure out......if I were a chemist and someone said "200 degrees
above the freezing point of water", I wouldn't mock them, just
respectfully point out that it's more common to say "20 degrees above
the boiling point of water".....I would consider the person ignorant of
the conventional terminology, but I would consider the person dead-on
if he were talking about 232 degrees F.

 

[Mike Berger]

It's a different remote. There are at least two variations,
and I've never been able to get a Pioneer shop to sell me
one (and Pioneer won't sell it direct). If you want to maintain
your old LaserDisc players, which also require the service
remote, you're out of luck.

William R. Walsh wrote:

Hi!

You might have all you need and not even know it. If you can do so, try
taking apart your existing remote. You may find unused contact pads with no
(or reduced) button "lumps" above them. Pressing one might give you the
service function you are looking for.

I don't know what the likelihood of this being true is like, but I'd be
surprised if Pioneer made a totally different PCB and controller layout for
a "service remote".

Just be careful if you do find that your remote can do this--going into
service modes can really screw things up and cause what may be fatal damage
to the unit.

William

 

[Floyd L. Davidson]

John Fields <jfields@austininstruments.com> wrote:

On Sun, 12 Jun 2005 16:35:48 -0800, floyd@barrow.com (Floyd L.
Davidson) wrote:
That means that the signal is DC. A varying DC, but DC nonetheless.

If there is such a think as "varying DC", connect a load to
it... through a capacitor. Now, how do you describe the effect
that load has on your "varying DC". The load see's *only* AC,
even according to your definition. That AC came from somewhere,
and it certainly was not generated by the capacitor.

---
It most certainly was!

Do a reality check on what you are saying! Capacitors do *not*
generate AC, and when the rest of your theory depends on the
idea that they do, you've made a mistake.

Consider a DC coupled audio amplifier running from a single 12V
supply, with its output set to Vcc/2 and feeding an 8 ohm load with a
4VPP sinusoidal signal. Like this:

+12
|
+--o--+
| AMP |---+-->Vout
+--o--+ |
| [4R]
| |
GND GND

You haven't drawn the schematic of an amplifier. There is no
input. Call it what you like, but it isn't an amplifier.

Add the input, and then we know where the AC originated...

Now, with phi being equal to the phase angle of the signal and zero
degrees corresponding the voltage halfway between the most positive
and least positive output voltage, the output voltage excursions will
look like this:

phi Vout
-----+------
0° 6V
90° 8V
180° 6V
270° 4V
360° 6V

Clearly AC. (And if you don't treat it as AC, your circuit analysis
will be flawed.)

Now, connect that magical capacitor between the amp and the load, as
shown below, and watch what happens:

+12
|
+--o--+
| AMP |---[cap]--+-->Vout
+--o--+ |
| [4R]
| |
GND GND

phi Vout
-----+------
0° 0V
90° +2V
180° 0V
270° -2V
360° 0V

Why?

Because you feed an AC signal to the capacitor, and hence you
see an AC signal on the other side.

What's your point? Capacitors pass AC and block DC. All you've
done is *prove* that there was AC coming out of the AMP (as well
as DC).

Well,for starters, consider that under quiescent conditions the
left-hand side of the cap will be charged to 6V and the right hand
side will be at zero volts since there is no galvanic path to the
output of the amp through the cap.

Now, imagine that the voltage at the amp's outout starts to go
positive. What will happen is that the amp will start sucking
electrons out of the cap, generating a potential difference across the
cap's plates which causes electrons to flow through the resistor,
making the top of the resistor more positive than the bottom.

And clearly you have an alternating voltage on both sides of the
capacitor, and an AC current passing through it. Not generated
by it, but passing through it.

Continuing in time, a point will be reached where the output of the
amp will start forcing electrons _into_ the resistor, at which point
the direction of travel of the electrons will be reversed. This
periodic reversal will cause the polarity of the signal into the
resistor to alternate. This alternating voltage will then give rise
to an _alternating current_ in the resistor.
---

That's because AC is *not* defined by any change in direction,
but only by a rate of movement change.

---
Poppycock. It's precisely the alternations in the direction of charge
flow which cause it to be called "Alternating Current".

It is defined by a differential (which necessarily will have a
sign reversal), not "polarity" reversals.

Your way would have it be called AC by assigning some arbitrary rate
of change, irrespective of direction as the delineation point, which
makes no sense at all. That is, what would you specify as the rate of
change which would delineate between between AC and DC? 0.5A/s?
0.001A/s? 0.1V/s?

*Any* rate of change (differential) that you can detect, means
you have detected AC.

There simply is no way to do circuit analysis with any other
definition.

Or do we really want three states:

A) DC

B) Varying DC[1]

C) AC[2]


[1] Varying DC is exactly like AC and all functions are
identical.

[2] AC is exactly like Varying DC and all functions are
identical.

That is 3 states in your mind, and only 2 in fact.

Not very reasonable from a logical point of view, but that is
exactly what we do have because of the historical baggage that
we carry along.

Remember when every electrical engineer would tell you that
current flows from the positive terminal of a battery to the
negative terminal... and every electronics engineer would tell
you that when the B battery is connected to a vacuum tube
circuit the current flows from the cathode to the anode. Of
course the positive battery terminal is connected to the anode,
so they can't both be correct.

Of course, then solid state electronics came along, and it became
clear that current wasn't even necessarily the movement of electrons,
but could also be the movement of a lack of electrons! How does
*that* fit your "polarity" requirements?

You are telling me the positive terminal supplies the current, and
the return path is to the negative terminal. I'm telling you that
electrons flow from the cathode to the anode, and I don't care how
many reference books you cite saying that current comes from the
positive terminal on that battery.

Same sort of historical baggage.


(And can the spelling flames. If you haven't got any better
manners than you do logic, you have no place complaining that I
forgot to run the spell check on that article. Your claim that
the referenced statement was not the non-sequitur that I pointed
out it was didn't hold water according to the very definition
*you* supplied!)

--
Floyd L. Davidson <http://web.newsguy.com/floyd_davidson>
Ukpeagvik (Barrow, Alaska) floyd@barrow.com

 

[Floyd L. Davidson]

John Fields <jfields@austininstruments.com> wrote:

On Sun, 12 Jun 2005 18:28:16 -0800, floyd@barrow.com (Floyd L.
Davidson) wrote:

"operator jay" <none@none.none> wrote:

It is not changing polarity. I would hesitate to call it alternating
current. On the "dc sine wave" issue, I wouldn't even get into that debate.
To me the terms involved are open to too many interpretations. As evidenced
in this thread, I suppose.

Where *do* you get this requirement for changing polarity? We
don't call it "Alternating Polarity", we call it "Alternating
Current". If the current is being altered, it's AC.

---
No, if the direction of charge flow alternates between two states,
then it's Alternating Current.

That fits my definition, but not yours! Are you changing your definition
or is that just a momentary bit of logical thought?

The states do *not* have to be plus and minus polarity. Just different
current levels...

---

You keep talking about AP, and it isn't the same.

---
Yes, it is. In order for the current in a load to alternate, the
polarity of the generator's output voltage must alternate as well.

Sure. But it doesn't need to change polarity. All it needs to do
is change level.

--
Floyd L. Davidson <http://web.newsguy.com/floyd_davidson>
Ukpeagvik (Barrow, Alaska) floyd@barrow.com

 

[John Popelish]

Choreboy wrote:

With high frequency and amplitude, a sine wave could be very steep at 0
and 180 degrees. It could also turn sharply at 90 and 270, like the
corner of a square wave. You would need low frequency and amplitude for
a sine wave to approximate the flat peaks of a square wave.

That part is simple enough for me, but I don't understand harmonics. If
you overdrive an amplifier with a sine wave, the output will resemble a
square wave. I know the output can be broken down into the input
frequency and its odd multiples. I'll have to accept it on faith.

You might want to look into the basis of Fourier analysis. It all
falls out of a very simple mathematical property of the sine wave.

If you take any periodic waveform, and multiply its value at every
point in time with the value of any frequency of sine wave at the same
points in time, over all time and add up (integrate) all the products
and divide by the total time (an infinite amount of time), only sine
waves that fit an integral number of cycles within the period of the
waveform will produce nonzero results (infinite integral divided by
infinite time). In fact, it can be shown that you get the same
quotient for harmonics if you use any integral number of periods of
the waveform, including one period. Testing an infinite number of
waves is only necessary to show that non harmonics always produce a
zero contribution. For instance, if you test a sine wave that fits
1.000001 cycles into a cycle of the waveform, you don't reach the
first zero result till you include a million periods of the waveform
(and you get more zeros at every integer multiple of a million cycles,
with a smaller and smaller cycle of results between those millions as
the number of cycles increases because you are dividing by larger and
larger times).

Harmonics (sine waves that fit an integral number of cycles within the
waveform) will produce a finite result representing that frequencies
contribution to the waveform. (Actually you have to test both the
sine and cosine against the waveform to cover all possible phase
shifted versions of the sine. Any phase shifted sine can be broken
sown into sine and cosine components. Another nice property of sine
waves.) Since only harmonics contribute to the total wave shape, you
can skip all the other frequencies, and just evaluate the part each
harmonic contributes to making the total waveform.

That is Fourier analysis.

The rest is about making the math more efficient.

 

[rainbow]

People are not saved by good deeds, they are only saved by faith in JESUS
CHRIST.God is an almighty spirit, who cannot be seen, but knows all things
about us. Jesus died on a cross for the sins of all, 2005 years ago. The
reason why the world has so much trouble is because the lack of faith in
Jesus,(if we don`t believe in Jesus, than we don`t fear Jesus,the son of
GOD. not enough people obey the 10 commandments. maybe because people never
heard of Jesus, or they just refuse to believe in such a huge authority. we
will all be judged after we die.Call on Jesus and ask him to forgive you of
your sins.READ THE BIBLE.unless you are born again, you cannot go to heaven.
the bible says some bad things in life are bound to happen,but who ever
makes bad things happen. HOW BAD it will be for them.


"Ethic" <Ethic@nospam.net> wrote in message
news:42ad9c80$0$1149$5402220f@news.sunrise.ch...

Surgeon General's Warnings :

Religion Causes Hatred Cancer,
Xenophobic Disease and Complicates Life.

Quitting Religion Now Greatly Reduces
Serious Risks to Your Mental Health.

* Ethic *


According to Karl Marx, RELIGION IS USED
BY OPPRESSORS TO MAKE PEOPLE FEEL BETTER ...
This is the origin of his comment that :
" RELIGION IS THE OPIUM OF THE MASSES "

( But, Karl Marx is probably censured in USA )

* Ethic *
integrityethics@NOhotmailSPAM.com

"shred" <shred_monger@hotmail.com> nicely replied :
news:1118500257.667619.124790@g47g2000cwa.googlegroups.com...
if theres one thing i can't stand it's born again christians....if
theres someting else i can't stand it's born again christians in the
wrong place,,...keep it for church

 

[The Phantom]

On Mon, 13 Jun 2005 07:58:46 -0800, floyd@barrow.com (Floyd L.
Davidson) wrote:

*Snip*

Either that or we are back to Don Lancaster's correct statement
that they are meaningless terms anyway. They certainly are if
that is the way they are defined!

Don first said:
---------------------------------------
'"DC" is simply the first (or "offset" term in the Fourier expression
of
any repetitive waveform.

"AC" are all of the remaining components.'
----------------------------------------

Then he said:
----------------------------------------
'"AC" or "DC" are gross and meaningless oversimplifications.'
----------------------------------------

Which are we to believe?

 

[John Fields]

On Mon, 13 Jun 2005 07:37:08 -0800, floyd@barrow.com (Floyd L.
Davidson) wrote:

John Fields <jfields@austininstruments.com> wrote:
On Sun, 12 Jun 2005 16:35:48 -0800, floyd@barrow.com (Floyd L.
Davidson) wrote:
That means that the signal is DC. A varying DC, but DC nonetheless.

If there is such a think as "varying DC", connect a load to
it... through a capacitor. Now, how do you describe the effect
that load has on your "varying DC". The load see's *only* AC,
even according to your definition. That AC came from somewhere,
and it certainly was not generated by the capacitor.

---
It most certainly was!

Do a reality check on what you are saying! Capacitors do *not*
generate AC, and when the rest of your theory depends on the
idea that they do, you've made a mistake.

---
Well, Floyd, Take a look at the schematics below and you may notice
that while the first one (the one without the cap in series with the
load) puts out a sinusoidally varying unipolar signal, (DC) the second
one (the one _with_ the cap in series with the load) puts out a
sinusoidally varying bipolar signal. (AC)

Now, since the only difference between them is the cap and one puts
out a varying DC signal while the other one puts out a true signal,
then the cap _must_ be generating the AC signal. If you have a
problem with 'generating' then perhaps 'converting' would be more to
your liking. I doubt it though, you seem to be in this only for the
argument and I'm sure you'll come up with reason why you're unhappy
with 'convert'.
---

Consider a DC coupled audio amplifier running from a single 12V
supply, with its output set to Vcc/2 and feeding an 8 ohm load with a
4VPP sinusoidal signal. Like this:

+12
|
+--o--+
| AMP |---+-->Vout
+--o--+ |
| [4R]
| |
GND GND

You haven't drawn the schematic of an amplifier. There is no
input. Call it what you like, but it isn't an amplifier.

Add the input, and then we know where the AC originated...

---
LOL, you're grasping at straws!
I already said the amp was DC coupled, so showing an input isn't
necessary. But for you...

+12
|
+--o--+
INPUT>---| AMP |---+-->Vout
+--o--+ |
| [4R]
| |
GND GND

Assume a gain of 1.

Happy now?
---

Now, with phi being equal to the phase angle of the signal and zero
degrees corresponding the voltage halfway between the most positive
and least positive output voltage, the output voltage excursions will
look like this:

phi Vout
-----+------
0° 6V
90° 8V
180° 6V
270° 4V
360° 6V

Clearly AC. (And if you don't treat it as AC, your circuit analysis
will be flawed.)

---
You seem to want to put the cart before the horse in that you're
spouting 'circuit analysis' before you've gotten a grasp of the
basics. But if you want to play that way, OK. It's clearly a
fluctuating DC signal, and if you don't take that _fact_ into
consideration _your_ circuit analysis will be erroneous.
---

Now, connect that magical capacitor between the amp and the load, as
shown below, and watch what happens:

+12
|
+--o--+
| AMP |---[cap]--+-->Vout
+--o--+ |
| [4R]
| |
GND GND

phi Vout
-----+------
0° 0V
90° +2V
180° 0V
270° -2V
360° 0V

Why?

Because you feed an AC signal to the capacitor, and hence you
see an AC signal on the other side.

---
No, look a little more closely and you'll see (well, maybe...) that
the current on the input side never changed direction (remained DC),
while the current into the load alternated between going into the load
and coming out of the load. IOW, it went into the cap as fluctuating
DC and came out AC.
---

What's your point? Capacitors pass AC and block DC. All you've
done is *prove* that there was AC coming out of the AMP (as well
as DC).

---
No, there was _no_ AC coming out of the amp, (no changes in the
direction of charge flow, only changes in the magnitude) just
fluctuating DC which the cap converted into AC.
---

Well,for starters, consider that under quiescent conditions the
left-hand side of the cap will be charged to 6V and the right hand
side will be at zero volts since there is no galvanic path to the
output of the amp through the cap.

Now, imagine that the voltage at the amp's outout starts to go
positive. What will happen is that the amp will start sucking
electrons out of the cap, generating a potential difference across the
cap's plates which causes electrons to flow through the resistor,
making the top of the resistor more positive than the bottom.

And clearly you have an alternating voltage on both sides of the
capacitor, and an AC current passing through it. Not generated
by it, but passing through it.

---
No, there is a fluctuating DC on the input side of the cap which the
cap converts to AC to present to the load.
---

Continuing in time, a point will be reached where the output of the
amp will start forcing electrons _into_ the resistor, at which point
the direction of travel of the electrons will be reversed. This
periodic reversal will cause the polarity of the signal into the
resistor to alternate. This alternating voltage will then give rise
to an _alternating current_ in the resistor.
---

That's because AC is *not* defined by any change in direction,
but only by a rate of movement change.

---
Poppycock. It's precisely the alternations in the direction of charge
flow which cause it to be called "Alternating Current".

It is defined by a differential (which necessarily will have a
sign reversal), not "polarity" reversals.

---
Specious gobbledygook.

A reversal of sign is, by definition, a reversal of polarity.
---

Your way would have it be called AC by assigning some arbitrary rate
of change, irrespective of direction as the delineation point, which
makes no sense at all. That is, what would you specify as the rate of
change which would delineate between between AC and DC? 0.5A/s?
0.001A/s? 0.1V/s?

*Any* rate of change (differential) that you can detect, means
you have detected AC.

---
Only if it's accompanied by a change in the direction of charge flow.
---

..
.. Snipped a lot of irrelevant trash apparently designed to change the
.. subject.
..

(And can the spelling flames. If you haven't got any better
manners than you do logic, you have no place complaining that I
forgot to run the spell check on that article. Your claim that
the referenced statement was not the non-sequitur that I pointed
out it was didn't hold water according to the very definition
*you* supplied!)

---
The point is that you didn't point out a non sequitur. (notice that
there's no apostrophe in there) The definition, which I got from
Webster's College Dictionary and posted for your edification, should
have made that clear.

And, speaking of manners, I suggest that yours need a little trip past
Emily Post.




--
John Fields
Professional Circuit Designer

 

[John Fields]

On Mon, 13 Jun 2005 07:47:51 -0800, floyd@barrow.com (Floyd L.
Davidson) wrote:

John Fields <jfields@austininstruments.com> wrote:
On Sun, 12 Jun 2005 18:28:16 -0800, floyd@barrow.com (Floyd L.
Davidson) wrote:

"operator jay" <none@none.none> wrote:

It is not changing polarity. I would hesitate to call it alternating
current. On the "dc sine wave" issue, I wouldn't even get into that debate.
To me the terms involved are open to too many interpretations. As evidenced
in this thread, I suppose.

Where *do* you get this requirement for changing polarity? We
don't call it "Alternating Polarity", we call it "Alternating
Current". If the current is being altered, it's AC.

---
No, if the direction of charge flow alternates between two states,
then it's Alternating Current.

That fits my definition, but not yours! Are you changing your definition
or is that just a momentary bit of logical thought?

---
Try not to be a stupid fuck. Flames will get you nothing back but more
flames. Is that what you want?
---

The states do *not* have to be plus and minus polarity. Just different
current levels...

---
Go back and read it again. Concentrate especially hard on the part
about the _direction_ of charge flow alternating between two states
and maybe you'll get it.
---

You keep talking about AP, and it isn't the same.

---
Yes, it is. In order for the current in a load to alternate, the
polarity of the generator's output voltage must alternate as well.

Sure. But it doesn't need to change polarity. All it needs to do
is change level.

---
I'm starting to think you're having a real problem with reading
comprehension. I write that for the current in a load to alternate,
the polarity of the generator's output voltage must alternate as well,
and then you agree but state that it doesn't need to change polarity.

Don't you understand that an alternation in polarity means that the
polarity changed???

--
John Fields
Professional Circuit Designer