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

"vorange"
>
> 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).


** That is quite wrong.


> 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?


** A clue.


> 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.


** Yes - that is pure AC.

> 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?


** That is a signal with bath AC and DC components .

The DC component is 5 volts - ie the average value, as read on a DC meter.

The AC component is +/-5 volts peak, with the DC one removed.

Which is just what a series capacitor will do.




......... Phil

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). 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
>
Don't be, dig into some 'very' serious books and read the definition of
capacitor.
You will find that the transfer function relates to dv/dt, meaning the
voltage change during time slice.
A digital wave has rising and falling edges, those are passed by the
capacitor, the DC component which does not change in time is blocked.

HTH

Stanislaw.

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.

That's correct for a perfect capacitor. Real capacitors will "leak" some
DC, but it's normally negligible (if it isn't, you've probably chosen the
wrong sort of capacitor).

> 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).

Any kind of "wave" is AC, although it may have a DC component.

> 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?

The AC passes through, the DC doesn't.

> 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?

No. The first one is a 5V-peak (10V peak-to-peak) AC signal with no DC
component (assuming that it's symmetrical, e.g. sine wave or 50% square
wave), the second is a 5V-peak AC signal with a 5V DC component.

In both cases, what will come "out" of the capacitor is a 5V-peak (10V
peak-to-peak) AC signal with no DC component.

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).

Big mistake.

Graham

Stanislaw Flatto wrote:

> You will find that the transfer function relates to dv/dt, meaning the
> voltage change during time slice.

And you think a rank beginner will even remotely understand those terms do you ?

Graham

"Eeyore" <rabbitsfriendsandrelations@hotmail.com> wrote in message
news:46C9BD07.E9C4EFBB@hotmail.com...
>
>
> 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).
>
> Big mistake.

Whether it was a mistake or not depends on the intent
of that particular design, the size of the cap, the frequency
of the square wave, etc., etc., etc....

Bob M.

On Mon, 20 Aug 2007 17:12:11 +0100, Eeyore
<rabbitsfriendsandrelations@hotmail.com> wrote:

>
>
>Stanislaw Flatto wrote:
>
>> You will find that the transfer function relates to dv/dt, meaning the
>> voltage change during time slice.
>
>And you think a rank beginner will even remotely understand those terms do you ?

---
If you do I don't see why he should have any difficulty with them.


--
JF

On Mon, 20 Aug 2007 08:49:31 +0100, "Anonymous." <me@privacy.net>
wrote:

>
>"vorange" <orangepic@yahoo.com> wrote in message
>news:1187578456.973349.156600@r34g2000hsd.googlegroups.com...

>> 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?
>
>You are describing an AC signal which is a square wave of +/- 5 volts
>imposed upon a DC signal of +5 volts.
>
>When you look at the output of the op amp, (assuming a gain of unity
>and a good high frequency response) you will see only an AC signal
>of +/- 5 volts.
>
>1. The DC signal has not passed through the capacitor, but the AC signal
>has.
>
>2. The output of the capacitor is no longer referenced to 0V.

---
Yes, it is, but its output swings from 5V more positive than 0V to
5V more negative than 0V.


--
JF

Eeyore wrote:
>
> Stanislaw Flatto wrote:
>
>> You will find that the transfer function relates to dv/dt, meaning the
>> voltage change during time slice.
>
> And you think a rank beginner will even remotely understand those terms do you ?
>
> Graham
>
From my failing memory, many times such chance remark stays with the
reader in a corner of brain to work for years as a whip to learn.
HTH.
_Understanding_ comes much later.

Stanislaw.

Eeyore wrote:
>
> 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).
>
> Big mistake.


Only if he actually listens to you, Donkey.


--
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

"vorange" <orangepic@yahoo.com> wrote in message
news:1187578456.973349.156600@r34g2000hsd.googlegroups.com...
> 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?
>

There are a few ways to think about this. One way is to realize that a
square wave(or any wave) is composed of a series(which means a sum) of
simple sine waves. A capacitor passes these sine waves. It does attenuate
each frequency but as long as certain conditions are met the ouput will look
like the input.

A second way is to realize that the capacitor doesn't "block" except in
steady state conditions(and this is where your confusion is comming from).
The capacitor will charge and discharge on each cycle of the square wave and
as long as this charging is fast enough the output will still look like a
square wave.


> 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 would say this would technically be true. AC means alternating current. If
you have a square wave that goes from -5V to +5V then it must in some way
alternate the current when it changes from - to + or + to -. (even if its
just a fA)

If you take the square wave from 0 to +10 then the current never alternates
but just changes in magnitude(but not sign). -5 to 5 never changes
magnitude(for an ideal signal) but only sign.

Jon

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). 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
>
well, In the circuit you were looking at, it still is blocking DC how
ever, the change in levels on that DC will reflect on the out side of
the capacitor.
To break it down in simple terms.

Picture the capacitor as a battery. when a battery is discharged and
you connect a charger with an AMP meter on it. You'll see ampere's being
displayed on the meter until the battery fully absorbs all that its
going to take. At that point, the amp meter will show its lowest
reading, indicating the battery is charged.
When the source and absorbing device(Capacitor/battery) are at equal
voltages. No current flows.

Now lets translate that to an OP-AMP output.

The Op-Amp goes high to lets say 10 volts and you have a capacitor
connected to the output in series to a voltage meter for example.

While the capacitor is charging, current is being generated which will
allow the voltage meter to register a reading. When the capacitor
reaches it's full charge equal to what the op-amp voltage is, the
current will reduce to virtually zero. This will indicate on the voltage
meter no current which means no voltage will display. at this point, the
OP-AMP output can remain at the DC 10 volts and you'll see no voltage
at all on the meter because the 10 volts of the op-amp is equal to the
10 volts stored in the capacitor.
Just put in your mind 2 batteries connected in series back to back.
when you take a volt reading from the ends of these 2 in series. You'll
get 0 volts because the polarity of each are canceling each other.
this is what happens to a capacitor when it becomes fully charged to
the source in series..

Now, picture the op-amp going to ground or low after the capacitor has
fully charged on the + cycle. what we get now is something you may not
expect.
Think of taking a fully charged battery and reversing the polarity
connections.
In this case, the capacitor lead connected to the Op-Amp output was
fully charged to + volts, and now since the op-amp has shifted to
low/common, it has in effect, reversed the connections of that capacitor
so that the + side charge is now grounded.
since the output side of the capacitor is above the ground potential,
you will see the charge in the capacitor now discharging in reverse
which will give you a real (-) voltage from a system that only had 0 or
10 volts+
This is how below 0 volt AC signals are formed from a DC pulse for
example.
This voltage will be there until the capacitor has fully discharged
it's energy through the output load, in which case, would be your volt
meter.
And thus, starts the cycle of a fully discharged capacitor on the +
side again!



Hope you got something out of that.


--
"I'm never wrong, once i thought i was, but was mistaken"
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5

On Mon, 20 Aug 2007 19:03:26 -0500, "Jon Slaughter"
<Jon_Slaughter@Hotmail.com> wrote:

>
>"vorange" <orangepic@yahoo.com> wrote in message
>news:1187578456.973349.156600@r34g2000hsd.googlegroups.com...
>> 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?
>>
>
>There are a few ways to think about this. One way is to realize that a
>square wave(or any wave) is composed of a series(which means a sum) of
>simple sine waves. A capacitor passes these sine waves. It does attenuate
>each frequency but as long as certain conditions are met the ouput will look
>like the input.
>
>A second way is to realize that the capacitor doesn't "block" except in
>steady state conditions(and this is where your confusion is comming from).
>The capacitor will charge and discharge on each cycle of the square wave and
>as long as this charging is fast enough the output will still look like a
>square wave.
>
>
>> 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 would say this would technically be true. AC means alternating current. If
>you have a square wave that goes from -5V to +5V then it must in some way
>alternate the current when it changes from - to + or + to -. (even if its
>just a fA)
>
>If you take the square wave from 0 to +10 then the current never alternates
>but just changes in magnitude(but not sign).

---
Yup.

What we old fuckers used to call "pulsating DC"



--
JF

On Mon, 20 Aug 2007 20:27:54 -0400, Jamie
<jamie_ka1lpa_not_valid_after_ka1lpa_@charter.net> wrote:

>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). 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
>>
>well, In the circuit you were looking at, it still is blocking DC how
>ever, the change in levels on that DC will reflect on the out side of
>the capacitor.
> To break it down in simple terms.
>
> Picture the capacitor as a battery. when a battery is discharged and
>you connect a charger with an AMP meter on it. You'll see ampere's being
>displayed on the meter until the battery fully absorbs all that its
>going to take. At that point, the amp meter will show its lowest
>reading, indicating the battery is charged.
> When the source and absorbing device(Capacitor/battery) are at equal
>voltages. No current flows.
>
> Now lets translate that to an OP-AMP output.
>
> The Op-Amp goes high to lets say 10 volts and you have a capacitor
>connected to the output in series to a voltage meter for example.
>
> While the capacitor is charging, current is being generated which will
>allow the voltage meter to register a reading. When the capacitor
>reaches it's full charge equal to what the op-amp voltage is, the
>current will reduce to virtually zero. This will indicate on the voltage
>meter no current which means no voltage will display. at this point, the
>OP-AMP output can remain at the DC 10 volts and you'll see no voltage
>at all on the meter because the 10 volts of the op-amp is equal to the
>10 volts stored in the capacitor.
> Just put in your mind 2 batteries connected in series back to back.
> when you take a volt reading from the ends of these 2 in series. You'll
>get 0 volts because the polarity of each are canceling each other.
> this is what happens to a capacitor when it becomes fully charged to
>the source in series..
>
> Now, picture the op-amp going to ground or low after the capacitor has
>fully charged on the + cycle. what we get now is something you may not
>expect.
> Think of taking a fully charged battery and reversing the polarity
>connections.
> In this case, the capacitor lead connected to the Op-Amp output was
>fully charged to + volts, and now since the op-amp has shifted to
>low/common, it has in effect, reversed the connections of that capacitor
>so that the + side charge is now grounded.
> since the output side of the capacitor is above the ground potential,
>you will see the charge in the capacitor now discharging in reverse
>which will give you a real (-) voltage from a system that only had 0 or
>10 volts+
> This is how below 0 volt AC signals are formed from a DC pulse for
>example.
> This voltage will be there until the capacitor has fully discharged
>it's energy through the output load, in which case, would be your volt
>meter.
> And thus, starts the cycle of a fully discharged capacitor on the +
>side again!
>
>
>
> Hope you got something out of that.

---
I suggest that since there's no galvanic connection between the
windings of a capacitor, its secondary can be biased at any
convenient voltage and its output will swing about that point.


--
JF

John Fields wrote:

> On Mon, 20 Aug 2007 20:27:54 -0400, Jamie
> <jamie_ka1lpa_not_valid_after_ka1lpa_@charter.net> wrote:
>
>
>>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). 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
>>>
>>
>>well, In the circuit you were looking at, it still is blocking DC how
>>ever, the change in levels on that DC will reflect on the out side of
>>the capacitor.
>> To break it down in simple terms.
>>
>> Picture the capacitor as a battery. when a battery is discharged and
>>you connect a charger with an AMP meter on it. You'll see ampere's being
>>displayed on the meter until the battery fully absorbs all that its
>>going to take. At that point, the amp meter will show its lowest
>>reading, indicating the battery is charged.
>> When the source and absorbing device(Capacitor/battery) are at equal
>>voltages. No current flows.
>>
>> Now lets translate that to an OP-AMP output.
>>
>> The Op-Amp goes high to lets say 10 volts and you have a capacitor
>>connected to the output in series to a voltage meter for example.
>>
>> While the capacitor is charging, current is being generated which will
>>allow the voltage meter to register a reading. When the capacitor
>>reaches it's full charge equal to what the op-amp voltage is, the
>>current will reduce to virtually zero. This will indicate on the voltage
>>meter no current which means no voltage will display. at this point, the
>>OP-AMP output can remain at the DC 10 volts and you'll see no voltage
>>at all on the meter because the 10 volts of the op-amp is equal to the
>>10 volts stored in the capacitor.
>> Just put in your mind 2 batteries connected in series back to back.
>> when you take a volt reading from the ends of these 2 in series. You'll
>>get 0 volts because the polarity of each are canceling each other.
>> this is what happens to a capacitor when it becomes fully charged to
>>the source in series..
>>
>> Now, picture the op-amp going to ground or low after the capacitor has
>>fully charged on the + cycle. what we get now is something you may not
>>expect.
>> Think of taking a fully charged battery and reversing the polarity
>>connections.
>> In this case, the capacitor lead connected to the Op-Amp output was
>>fully charged to + volts, and now since the op-amp has shifted to
>>low/common, it has in effect, reversed the connections of that capacitor
>>so that the + side charge is now grounded.
>> since the output side of the capacitor is above the ground potential,
>>you will see the charge in the capacitor now discharging in reverse
>>which will give you a real (-) voltage from a system that only had 0 or
>>10 volts+
>> This is how below 0 volt AC signals are formed from a DC pulse for
>>example.
>> This voltage will be there until the capacitor has fully discharged
>>it's energy through the output load, in which case, would be your volt
>>meter.
>> And thus, starts the cycle of a fully discharged capacitor on the +
>>side again!
>>
>>
>>
>> Hope you got something out of that.
>
>
> ---
> I suggest that since there's no galvanic connection between the
> windings of a capacitor, its secondary can be biased at any
> convenient voltage and its output will swing about that point.
>
>
Hmm, You in a cocky mood these days John



--
"I'm never wrong, once i thought i was, but was mistaken"
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5

John Fields wrote:

> On Mon, 20 Aug 2007 19:03:26 -0500, "Jon Slaughter"
> <Jon_Slaughter@Hotmail.com> wrote:
>
>
>>"vorange" <orangepic@yahoo.com> wrote in message
>>news:1187578456.973349.156600@r34g2000hsd.googlegroups.com...
>>
>>>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?
>>>
>>
>>There are a few ways to think about this. One way is to realize that a
>>square wave(or any wave) is composed of a series(which means a sum) of
>>simple sine waves. A capacitor passes these sine waves. It does attenuate
>>each frequency but as long as certain conditions are met the ouput will look
>>like the input.
>>
>>A second way is to realize that the capacitor doesn't "block" except in
>>steady state conditions(and this is where your confusion is comming from).
>>The capacitor will charge and discharge on each cycle of the square wave and
>>as long as this charging is fast enough the output will still look like a
>>square wave.
>>
>>
>>
>>>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 would say this would technically be true. AC means alternating current. If
>>you have a square wave that goes from -5V to +5V then it must in some way
>>alternate the current when it changes from - to + or + to -. (even if its
>>just a fA)
>>
>>If you take the square wave from 0 to +10 then the current never alternates
>>but just changes in magnitude(but not sign).
>
>
> ---
> Yup.
>
> What we old fuckers used to call "pulsating DC"
>
>
>
Old, I'd like to consider my self a young guy!

Gray hair are loved by many young ladies!

I don't know if it has anything to do with lack of
threat or what ever!

--
"I'm never wrong, once i thought i was, but was mistaken"
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5

Jamie wrote:
>
> Old, I'd like to consider my self a young guy!
>
> Gray hair are loved by many young ladies!
>
> I don't know if it has anything to do with lack of
> threat or what ever!


What does gray hair have to do with age? Mine started to turn while
I was still in high school, and I was completely gray by the time I was
20.


--
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

"Michael A. Terrell" <mike.terrell@earthlink.net> wrote in message
news:46CB111C.569C1A1F@earthlink.net...
> Jamie wrote:
>>
>> Old, I'd like to consider my self a young guy!
>>
>> Gray hair are loved by many young ladies!
>>
>> I don't know if it has anything to do with lack of
>> threat or what ever!
>
>
> What does gray hair have to do with age? Mine started to turn while
> I was still in high school, and I was completely gray by the time I was
> 20.
>
>
> --
> 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

My brother had half gray hair - half his normal color by the time he was
like 30.

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.

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.

On 8/21/07 11:10 AM, in article pan.2007.08.21.18.10.15.843000@nowhere.com,
"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.
>

But it's not a matter of semantics - it is a matter of what is actually
happening with AC signals. Practically speaking, it doesn't usually make
much difference, however.

On 8/21/07 11:31 AM, in article fafb1r$vq1$1@aioe.org, "Anonymous."
<me@privacy.net> wrote:

>
> "Don Bowey" <dbowey@comcast.net> wrote in message
> news:C2F06B17.74542%dbowey@comcast.net...
>>
>> 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!
>
>

Yes!

"Anonymous." <me@privacy.net> wrote in message news:fafb1r$vq1$1@aioe.org...


> 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!

No, it really does flow through. The problem here is
that while we typically imagine current to be a
stream of electrons, all nicely charging down a path
(perhaps in single file, and all in step), real "electrical
current" only follows this model in such thing as the
beam of a CRT.

Current is the effective flow of electrical ENERGY;
it involves the motion of charge carriers, and its units
are defined in terms of charges-per-second, but is not
directly equivalent with the motion of a particular charge
carrier or group except under very specific circumstances.
For example, what is the "speed" of electrical currents
through a conductor? For almost all practical concerns,
it makes sense to look at only the propagation speed
of the energy or signal in question - even though the
individual electrons are moving FAR slower than this.

Bob M.

On 8/21/07 11:58 AM, in article fafclg$jc6$1@usenet01.boi.hp.com, "Bob
Myers" <nospamplease@address.invalid> wrote:

>
> "Anonymous." <me@privacy.net> wrote in message news:fafb1r$vq1$1@aioe.org...
>
>
>> 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!
>
> No, it really does flow through.

Oh Hum.

Nope.

(snip)