"Don Bowey" <dbowey@comcast.net> wrote in message
news:C2F0AA5A.745BB%dbowey@comcast.net...

>> No, it really does flow through.
>
> Oh Hum.
>
> Nope.

I have a motor that runs on 60 Hz AC. I
put a big honkin' capacitor in series with the
line. The motor keeps running. I put an ammeter
also in series, first on one side of the cap and then
on the other, and get the same reading both
times. So what isn't "flowing through," in your
opinion?

Or are you thinking that the very electrons that
are passing through my desk lamp at this very
moment actually made a visit to the nearest
power generating station in recent memory?

Bob M.

On Tue, 21 Aug 2007 19:10:16 +0100, Nobody <nobody@nowhere.com>
wrote:

>On Tue, 21 Aug 2007 10:12:39 -0700, Don Bowey wrote:
>
>> My thought about this is... No, capacitors do not let DC current through,
>> nor AC currents either.
>>
>> Capacitors block DC voltage. AC signals appear to pass through capacitors,
>> but they don't actually do that.
>>
>> Flame away.
>
>The *signal* definitely passes through the capacitor. If it "appears" to
>pass, then it does; that's the nature of a "signal".
>
>Whether or not the current "passes" is largely a question of semantics.
>Certainly, the electrons themselves don't pass, but at any moment
>in time, current flows into one lead and the same amount of current
>simultaneously flows out of the other.

---
No, _charge_ flows into one lead and out of the other.


--
JF

On Tue, 21 Aug 2007 10:12:39 -0700, Don Bowey <dbowey@comcast.net>
wrote:

>
>My thought about this is... No, capacitors do not let DC current through,
>nor AC currents either.
>
>Capacitors block DC voltage. AC signals appear to pass through capacitors,
>but they don't actually do that.
>
>Flame away.

---
The flame of truth burns eternal.


--
JF

On 8/21/07 4:42 PM, in article d0umc31sm35vt90a192ppkio88locrk5c6@4ax.com,
"John Fields" <jfields@austininstruments.com> wrote:

> On Tue, 21 Aug 2007 10:12:39 -0700, Don Bowey <dbowey@comcast.net>
> wrote:
>
>>
>> My thought about this is... No, capacitors do not let DC current through,
>> nor AC currents either.
>>
>> Capacitors block DC voltage. AC signals appear to pass through capacitors,
>> but they don't actually do that.
>>
>> Flame away.
>
> ---
> The flame of truth burns eternal.
>

And I am its bearer. Shazam!

On Sun, 19 Aug 2007 19:54:16 -0700, vorange wrote:

> 1st question : Till now, I believed that capacitors only let AC
> signals through while blocking DC. But then, I saw a schematic whee
> they put a capacitor on the output line of an opamp. The signal into
> the opamp was a square wave signal (which I imagine is DC and not
> AC).

Well, you imagine wrong. It's "pulsating DC", yes, but it's composed
of many AC components riding on a DC reference. The way it gets through
the capacitor is that at the rising and falling edge of the waveform,
the capacitor suddenly finds itself with a voltage difference across
it, and charge wants to flow so as to minimize that potential.

So, essentially, for this application, the square wave can be considered
AC.

In fact, with the right instruments, when you first apply DC to one
terminal of a cap, that change in voltage is coupled to the other
plate by the capacitance, and the capacitor charges according to
T=RC (google "time constant".)

Hope This Helps!
Rich

John Fields wrote:
>
> On Tue, 21 Aug 2007 10:12:39 -0700, Don Bowey <dbowey@comcast.net>
> wrote:
>
> >
> >My thought about this is... No, capacitors do not let DC current through,
> >nor AC currents either.
> >
> >Capacitors block DC voltage. AC signals appear to pass through capacitors,
> >but they don't actually do that.
> >
> >Flame away.
>
> ---
> The flame of truth burns eternal.


Unfortunately, so do the sparks of stupidity.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

On Tue, 21 Aug 2007 19:31:19 +0100, Anonymous. wrote:

>> Capacitors block DC voltage. AC signals appear to pass through
>> capacitors,
>> but they don't actually do that.
>
> Correct.
>
> It's simple electrostatics.
>
> An electron arriving on one plate of (assumed vacuum dielectric)
> will repel another from the opposite plate.
>
> An electron pulled off one plate will attract an electron onto
> the other.
>
> Pulling and pushing in this way is the essence of AC.
>
> The current doesn't actually flow through, but because the net effect on both
> sides is the same, the current appears to flow through!

You can say the same thing about a length of wire. The electrons which
flow into one end aren't the same ones which flow out the other end,
unless the current is DC and you are prepared to wait a long time.

When it comes to AC, the question of whether a current flows *through* a
component is the same whether that component is a capacitor or a resistor.

An alternating current either flows "through" both, or it flows through
neither, or you are arbitrarily changing the definition of "through".

On Aug 22, 2:02 am, Rich Grise <r...@example.net> wrote:
> Well, you imagine wrong. It's "pulsating DC", yes, but it's composed
> of many AC components riding on a DC reference.

Thank you all for your response.

But a few more (perhaps dumb) questions :

1) What is the output from a microcontroller's pwm that is generating
square waves considered (+5v -> 0v -> +5v -> 0v...etc)? Is it
considered an AC or DC signal or a combination of both? After reading
all the replies, I'm left with the impression that "pulsed DC" is
considered as AC by some folks here even though it does not reverse
direction (go negative) when it goes low (0v).

2) Should not the output from a capacitor be HIGH if it encouters a
steady DC current? I know you've said that capacitors 'block' steady
DC currents but why? Mentally, I imagine when the dc current first
hits the capacitor's plate like a tsunami, it charges up the plate and
pushes electrons on the opposite plate away. I imagine those (pushed
away) electrons then go "racing away" from the capacitor perhaps
towards a load which should be driven HIGH by those electrons so long
as the capacitor's opposite plate is charged (which is what a DC
current should keep doing). Why then is this not the case ? Somehow
the mental model just does not fit. I can imagine how AC passes
after reading the descriptions you guys have provided but why DC does
not generate a contiuous high.

If only I could watch cartoons of what the electrons were doing, it
would all be clear to me.

vorange wrote:

> 1) What is the output from a microcontroller's pwm that is generating
> square waves considered (+5v -> 0v -> +5v -> 0v...etc)? Is it
> considered an AC or DC signal or a combination of both? After reading
> all the replies, I'm left with the impression that "pulsed DC" is
> considered as AC by some folks here even though it does not reverse
> direction (go negative) when it goes low (0v).

DC has two common meanings. One is unidirectional current
or voltage, regardless of how it varies with time, and the
other is steady current or voltage. I think most circuit
designers (who are familiar with Fourier and LaPlace
analysis) tend to think in the frequency domain, at least
part of the time, and are more likely to think of DC as a
steady current or voltage, essentially a zero frequency signal.

By that frequency domain way of thinking, that pulsing PWM
unidirectional voltage has some DC component (the average
voltage, and a whole series of AC components with various
magnitudes and phases, relative to the pulse timing, that
are all harmonics of the pulse frequency. Filters with
various frequency responses will pass varying amounts of all
those components.

> 2) Should not the output from a capacitor be HIGH if it encouters a
> steady DC current?

Very. Infinite, if you wait an infinitely long time. The
only way to get current through a capacitor is to have the
voltage across it change at some rate. The formula that
relates current to rate of change of voltage is :

I=C*(dv/dt) with I in amperes, C in farads, and dv/dt in
volts per second.

So, the only way to get 1 ampere of DC to pass through an
ideal 1 uF capacitor is to have the voltage across the
capacitor climb at 1 million volts per second, and keep
climbing at that rate for as long as the current must occur.
Since few of us have voltage sources, that can climb
toward infinity, sitting around, we generally think of
capacitors as devices that cannot pass DC.

> I know you've said that capacitors 'block' steady
> DC currents but why? Mentally, I imagine when the dc current first
> hits the capacitor's plate like a tsunami, it charges up the plate and
> pushes electrons on the opposite plate away.

That's good. But that sudden onset of voltage is not DC in
the frequency domain sense, but DC plus an infinity of AC
frequencies all added together. The sudden high rate of
change of voltage across the capacitor accounts for the
high, momentary current.

> I imagine those (pushed
> away) electrons then go "racing away" from the capacitor perhaps
> towards a load which should be driven HIGH by those electrons so long
> as the capacitor's opposite plate is charged (which is what a DC
> current should keep doing). Why then is this not the case ?

Sounds pretty good to me. The output DC will be maintained,
if there is no load current. So static DC voltage is
possible through a capacitor. This is the same as the DC on
rubbed balloons. But you can't get a steady current from
this effect, only steady, insulated voltage.

> Somehow
> the mental model just does not fit. I can imagine how AC passes
> after reading the descriptions you guys have provided but why DC does
> not generate a contiuous high.

AC supplies repeated (and alternating) rate of change of
voltage, so alternating current can be passed.

> If only I could watch cartoons of what the electrons were doing, it
> would all be clear to me.

It is a shame we can't put on special glasses and watch the
little buggers. All this would be a lot less abstract.

On Tue, 21 Aug 2007 23:02:53 -0700, vorange <orangepic@yahoo.com>
wrote:

>On Aug 22, 2:02 am, Rich Grise <r...@example.net> wrote:
>> Well, you imagine wrong. It's "pulsating DC", yes, but it's composed
>> of many AC components riding on a DC reference.
>
>Thank you all for your response.
>
>But a few more (perhaps dumb) questions :
>
>1) What is the output from a microcontroller's pwm that is generating
>square waves considered (+5v -> 0v -> +5v -> 0v...etc)? Is it
>considered an AC or DC signal or a combination of both? After reading
>all the replies, I'm left with the impression that "pulsed DC" is
>considered as AC by some folks here even though it does not reverse
>direction (go negative) when it goes low (0v).

"AC" and "DC" are extremely vague terms, and a capacitor doesn't care
about terminology; most capacitors are made in foreign countries and
don't even understand English. Whether a signal is one or the other
depends on the time frame over which it's observed. A 1-cycle-per-year
sine wave sure looks like DC if you observe it for an hour.

For practical purposes, in this case, look at the signal as having a
longterm average value and call that the DC component. If you subtract
that from the waveform, what's left is the AC component. So a general
signal, like your square wave, has both AC and DC components. A
capacitive coupling circuit, if the capacitor is the right value, will
block the longterm average ("DC") and pass through the short-term
wiggles (the "AC" component).

If the square wave is slow enough, the capacitor will get confused and
think that the high and low parts of the wave are actually DC, so will
lose interest in passing them, so the coupled waveform will start to
droop. Capacitors aren't terrible bright, and have short attention
spans.

John

On 8/21/07 8:42 PM, in article pan.2007.08.22.03.42.26.0@nowhere.com,
"Nobody" <nobody@nowhere.com> wrote:

> On Tue, 21 Aug 2007 19:31:19 +0100, Anonymous. wrote:
>
>>> Capacitors block DC voltage. AC signals appear to pass through
>>> capacitors,
>>> but they don't actually do that.
>>
>> Correct.
>>
>> It's simple electrostatics.
>>
>> An electron arriving on one plate of (assumed vacuum dielectric)
>> will repel another from the opposite plate.
>>
>> An electron pulled off one plate will attract an electron onto
>> the other.
>>
>> Pulling and pushing in this way is the essence of AC.
>>
>> The current doesn't actually flow through, but because the net effect on both
>> sides is the same, the current appears to flow through!
>
> You can say the same thing about a length of wire. The electrons which
> flow into one end aren't the same ones which flow out the other end,
> unless the current is DC and you are prepared to wait a long time.
>
> When it comes to AC, the question of whether a current flows *through* a
> component is the same whether that component is a capacitor or a resistor.
>
> An alternating current either flows "through" both, or it flows through
> neither, or you are arbitrarily changing the definition of "through".
>

A resistor has no dielectric barrier. A capacitor does.

Why do you believe "An alternating current either flows "through" both, or
it flows through neither...?"

"Don Bowey" <dbowey@comcast.net> wrote in message
news:C2F1B127.747C2%dbowey@comcast.net...
> A resistor has no dielectric barrier. A capacitor does.
>
> Why do you believe "An alternating current either flows "through" both, or
> it flows through neither...?"

In a sense, though, there's no real difference between
the two from the "dielectric barrier" angle. To examine
this further, let's for the moment ignore the "resistor"
question and just look at the difference between AC
conduction through a capacitor vs. a plain conductor.

Electrical energy passes through any "conductive" path by
virtue of the interaction of the fields of charged particles.
In a conductor, this occurs at the atomic/subatomic
level; in a capacitor, the interaction-through-fields
clearly happens on a physically gross level, through the
dielectric, and individual charge carriers cannot pass
through the dieletric. But that really doesn't matter -
to pass electrical energy or an electrical signal, it is
the motion of carriers "downstream," induced by a
similar motion "upstream," that matters - not that a
particular carrier physically passes through the entire
length of the conductive path.

In the case of AC, what goes on within a conductor
in terms of the charge carrier motion is very interesting.
Imagine a perfect conductor as being a frictionless pipe
filled with ping-pong balls which just fit inside the pipe.
If you push in a ball at one end, a ball pops out the other
end (with the time between these two events governed
by the physical attributes - elasticity and such - of the
balls). There has been a transfer of energy, even though
the ball you put in at the one end really didn't get very
far and is certainly not the same ball that popped out
at the far end. If we model AC this way, then you take
one ball at the near end and alternately push it in and
somehow suck it out AT THAT END. And at the far
end, we have a ball which is behaving in exactly the same
way, alternately popping out and being sucked back in.
We can again transfer energy (or information) through this
process, even though the ball we're pushing/pulling on
at the near end NEVER makes it beyond that point!

Going back to actual electricity, let's further note that if we
have a capacitor of sufficient size, there is no way at all to
distinguish the capacitor-in-series case from the "straight
conductor" case, if all we have to look at is the situation
"downstream" of the capacitor. The only way the two
cases could be distinguished in any event is through the
capacitor's effect on the phase relationship between
current and voltage, which, for a sufficiently large
capacitance, becomes negligible. (The only other means
you could use to distinguish these cases at all, given
access to any information you want, would be to somehow
"tag" individual electrons at the "upstream" side, and then
wait a sufficiently long time on the "downstream" side to
see if those particular carriers are coming through. But
you'd have to wait a very long time to be certain...)

Another way to say this is that an infinite capacitance is
indistinguishable from a "short circuit" (a "perfect"
conductive path), again unless you have the ability to
tag individual charge carriers. This makes sense because
a truly infinite capacitance would always have the ability
to make the corresponding change in charge on the
"downstream" plate as ANY amount of charge enters
or leaves the "upstream" plate. You can envision an
"infinite" capacitance as either possessing plates of
infinite area (if you can ignore concerns re the propagation
times across the plate itself) or (possibly better) as
having an infinitely thin dielectric (which equates to
saying we have a zero-thickness "magic barrier"
inserted between two conductors, such that individual
carriers cannot pass through but still have an effect
on the carriers on the other side, as if the barrier were
not there).

In the real world, of course, we can't have infinite capacitances
(or at the very least, you can't easily go down to Radio
Shack and buy one...), so we have to rely on the
frequency of the AC, relative to the capacitance, to
make the effects of the capacitor effectively "drop out"
of the circuit. In more familiar terms, we would say that
for a sufficiently low capacitive reactance has no
significant effect when inserted in series into an AC
circuit; there is no way to discern its presence by looking
at the conditions "downstream" of the capacitor. In any
practical sense of the words, then, we would have to say
that yes, a capacitor DOES "pass AC."

Bringing resistance into the picture, as opposed to a perfect
conductor, is only relevant if we want to compare the effects
of resistance to a comparable capacitive reactance. And
in this case, we would fall back to the fundamental difference
between these two forms of impedance: resistances dissipate
energy, while reactances merely store it and return it to the
circuit later in time (which results in the voltage/current phase
effects in a reactive circuit). But there's still nothing going on
here that would cause us to say that the resistor is actually
"passing AC," while the capacitor (reactance) is not.

Bob M.

"Anonymous." <me@privacy.net> wrote in message news:fahsc5$21i$1@aioe.org...

> Which raises ethical questions about the behaviour of the
> power companies! 50 times per second (60 in Yankland)
> they give you some electrons with the right hand and then
> take them back with the left hand.

Precisely! Or, as a sign in the men's room at a local
drinking establishment succinctly puts it:

"We don't sell beer, we only rent it."

Bob M.

On Wed, 22 Aug 2007 18:39:14 +0100, Anonymous. wrote:
> "Bob Myers" <nospamplease@address.invalid> wrote in message
> news:fahr76$r0e$1@usenet01.boi.hp.com...
>> .....If we model AC this way, then you take
>> one ball at the near end and alternately push it in and
>> somehow suck it out AT THAT END. And at the far
>> end, we have a ball which is behaving in exactly the same
>> way, alternately popping out and being sucked back in.....
>
> Which raises ethical questions about the behaviour of the
> power companies! 50 times per second (60 in Yankland)
> they give you some electrons with the right hand and then
> take them back with the left hand.
>
> As you ain't allowed to keep them, and they are only on loan
> for such a brief period, then it seems morally wrong for them
> to charge you for the privilege!

But on their trip to your house and back, they do work, like
lighting light bulbs, cooking, heating water, etc. - so what
you're paying for is the power you dissipate, or, over the
course of a month, the energy that the power company sends
to you, you use for awhile, and throw away.

Cheers!
Rich

On Wed, 22 Aug 2007 12:42:45 -0600, Bob Myers wrote:
> "Anonymous." <me@privacy.net> wrote in message news:fahsc5$21i$1@aioe.org...
>
>> Which raises ethical questions about the behaviour of the
>> power companies! 50 times per second (60 in Yankland)
>> they give you some electrons with the right hand and then
>> take them back with the left hand.
>
> Precisely! Or, as a sign in the men's room at a local
> drinking establishment succinctly puts it:
>
> "We don't sell beer, we only rent it."

Ick! Who wants to be the second cusomer to rent a particular pint?

Ewww!

Thanks,
Rich

On Tue, 21 Aug 2007 23:36:31 -0400, Michael A. Terrell wrote:
> John Fields wrote:
>> On Tue, 21 Aug 2007 10:12:39 -0700, Don Bowey <dbowey@comcast.net>
>> >
>> >My thought about this is... No, capacitors do not let DC current through,
>> >nor AC currents either.
>> >
>> >Capacitors block DC voltage. AC signals appear to pass through capacitors,
>> >but they don't actually do that.
>> >
>> >Flame away.
>>
>> The flame of truth burns eternal.
>
> Unfortunately, so do the sparks of stupidity.

We've noticed. Thanks for the reminder, Michael.

Cheers!
Rich

On Tue, 21 Aug 2007 23:02:53 -0700, vorange wrote:
....
> If only I could watch cartoons of what the electrons were doing, it
> would all be clear to me.

OK.

http://www.talkingelectronics.com/html/Page02.html
(scroll down a bit)

Cheers!
Rich

John Larkin wrote:

SNIP
>
> "AC" and "DC" are extremely vague terms, and a capacitor doesn't care
> about terminology; most capacitors are made in foreign countries and
> don't even understand English. Whether a signal is one or the other
> depends on the time frame over which it's observed. A 1-cycle-per-year
> sine wave sure looks like DC if you observe it for an hour.

SNIP

Part of the problem is that AC and DC
are, as John said, extremely vague
terms. They have no fundamental basis in
theory other than as a shorthand in
those cases where the context makes
their meaning clear.

AC and DC are descriptive terms that can
cause confusion whenever applied to
anything other than a pure, time-varying
sinusoidal voltage or a time-invariant
voltage, respectively. That these are
quasi-bogus terms becomes clear when we
recall that they have no units! They
don't appear in formulas. We can't
measure the AC-ness or DC-ness of a
circuit or device. Would electronics
theory even miss AC and DC?

I wonder whether there are other
descriptive terms used in electronics
the way AC and DC are used. Offhand, I
can't think of any at all!

As someone else pointed out, the
confusion disappears when we consider
the basic definitions of capacitors and
inductors (which do not include AC or
DC) rather than "capacitors pass AC but
not DC".


Chuck

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Hi,

could you take a look at the first animation on this page (below). Is
that what happens when (steady state) DC encounters a capacitor for
the first time. Does the capacitor pass a small "blip" of a charge
INITIALLY on startup and then block all further DC current.

http://www.talkingelectronics.com/html/Page03.html

If so then my mental model is right. I liken it to blowing air out of
my lungs (in one breath) onto a paper fan. The fan turns at first (a
blip) however I eventually run out of air in my lungs to blow out
(just like the opposite plate of the cap runs out of free electrons to
push away). And that's why the load (the paper fan) stops turning.

But there is that initial blip is there not? If so one should be
careful when using caps.... what if the input pin of a microcontroller
registers that 'blip' as a HIGH input signal and does something in
response to it!? Would it be wise to place a resistor ahead of the
cap if its going into some low impedance input pin of some digital/
logic chip?

Thanks to all for your help.




> On Aug 22, 7:22 am, John Popelish <jpopel...@rica.net> wrote:

vorange wrote:
> Hi,
>
> could you take a look at the first animation on this page (below). Is
> that what happens when (steady state) DC encounters a capacitor for
> the first time. Does the capacitor pass a small "blip" of a charge
> INITIALLY on startup and then block all further DC current.

Yep. Once the voltage stops changing, the current fades
toward zero.

> http://www.talkingelectronics.com/html/Page03.html
>
> If so then my mental model is right. I liken it to blowing air out of
> my lungs (in one breath) onto a paper fan. The fan turns at first (a
> blip) however I eventually run out of air in my lungs to blow out
> (just like the opposite plate of the cap runs out of free electrons to
> push away). And that's why the load (the paper fan) stops turning.
>
> But there is that initial blip is there not?

Yes, there is. And if you look further down, it shows that
you can extract the reverse of that blip by connecting a
load across the capacitor.

> If so one should be
> careful when using caps.... what if the input pin of a microcontroller
> registers that 'blip' as a HIGH input signal and does something in
> response to it!? Would it be wise to place a resistor ahead of the
> cap if its going into some low impedance input pin of some digital/
> logic chip?

The main capacitor you have to worry about is your body. If
you walk across a rug, you may mechanically charge up your
body capacitance and when you touch the processor, you can
dump that charge into a pin. It can be a large enough pulse
to damage the device.

John Popelish wrote:

SNIP
>
> The main capacitor you have to worry about is your body. If you walk
> across a rug, you may mechanically charge up your body capacitance and
> when you touch the processor, you can dump that charge into a pin. It
> can be a large enough pulse to damage the device.

Well, the body is a capacitor in the
sense that distributed somewhere in the
universe is the "other plate" containing
the opposite charge. The resulting
electric field created by that capacitor
is usually not strong enough to overcome
the repulsive effects of the charges on
the body when the body comes into
contact with another object.

The problem with describing the body as
a capacitor is that one might then
imagine that touching one plate of a
"conventional capacitor" will discharge it.

Pedagogically, it is probably better to
describe the body as a charged object,
which is not the way we normally
describe a capacitor.

Chuck

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On Wed, 22 Aug 2007 18:58:59 -0700, vorange wrote:

> Hi,
>
> could you take a look at the first animation on this page (below). Is
> that what happens when (steady state) DC encounters a capacitor for
> the first time. Does the capacitor pass a small "blip" of a charge
> INITIALLY on startup and then block all further DC current.
>
> http://www.talkingelectronics.com/html/Page03.html
>
> If so then my mental model is right. I liken it to blowing air out of
> my lungs (in one breath) onto a paper fan.

Think of it more like a chamber with a rubber diaphragm stretched
across the middle:

----------------------
| / |
| \ |
| / |
------- \ -------
------- / -------
| \ |
| / |
| \ |
----------------------

When you blow in the left, at first the diaphragm moves to the right,
and displaces the air on its right. Eventually, the diaphragm "bottoms
out", and the flow stops (or, you're blowing as hard as you can, even
though the diaphragm is only part-way, but the back-pressure stops you).
When you release the pressure, it blows the air back in your face. That's
like the cap discharging, except that for the charge to flow like the air
does here, you need wires and a complete circuit.

That's the flaw I see in the fluid models of electricity - the fluid
needs to be in a pipe. If you break the pipe, all the fluid spills out.
If you break a wire, the flow stops. So they're kind of opposites in
that respect, but otherwise it seems to work well.

Hope This Helps!
Rich

On Aug 19, 10:54 pm, vorange <orange...@yahoo.com> wrote:
> 1st question : Till now, I believed that capacitors only let AC
> signals through while blocking DC. But then, I saw a schematic whee
> they put a capacitor on the output line of an opamp. The signal into
> the opamp was a square wave signal (which I imagine is DC and not
> AC). How then does the output of the opamp (presumably DC as well)
> pass through the capacitor? This has confused the hell out of me.
> Is there something I'm missing here?
>
> 2nd question : Is it fair to say that if a signal goes from say +5 to
> -5 volts and then back to +5...etc that is is an AC signal because its
> reversing its direction. But if it goes from +10 to 0 volts and then
> back to +10 that it is a DC signal because its not reversing
> direction?
>
> I'm confused

It's actually a little more complicated. it helps if you understand a
little calculus. The capacitor passes the derivative of the signal,
i.e. the rate of change. So, with DC, there is no change, there is no
current throughput. With your square wave, the signal changes where it
rises and falls, so at those points you will get spikes through, on
the flat parts, nothing, i.e. a spike waveform plus and minus around a
baseline of zero. The reason "capacitors pass AC" is that AC as people
generally think of it is made of sine waves, and the derivative of a
sine wave turns out to be the same sine wave, just shifted a bit in
time.