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

John Fields wrote:
On Sun, 12 Jun 2005 16:35:48 -0800, (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