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.
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)