 |
Search Sci.Electronics.Basics |
|
|
|
|
|
|
 |
|
|
Sci.Electronics.Basics -> Can a capacitor let DC current through?
There are 54 messages in this thread.
You are currently looking at messages 20 to 40.
|
Author: RadiosrfunDate: 12:55 21-08-07
|
|
"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.
|
|
|
|
Author: Don BoweyDate: 13:12 21-08-07
|
|
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.
|
|
|
|
Author: NobodyDate: 14:10 21-08-07
|
|
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.
|
|
|
|
Author: Anonymous.Date: 14:31 21-08-07
|
|
"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!
|
|
|
|
Author: Don BoweyDate: 14:44 21-08-07
|
|
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.
|
|
|
|
Author: Don BoweyDate: 14:45 21-08-07
|
|
On 8/21/07 11:31 AM, in article fafb1r$vq1$1@aioe.org, "Anonymous."
<me@privacy.net> wrote:
>
> "Don Bowey" <dbowey@comcast.net> wrote in message
> news:C2F06B17.74542%dbowey@comcast.net...
>>
>> Capacitors block DC voltage. AC signals appear to pass through
>> capacitors,
>> but they don't actually do that.
>
> Correct.
>
> It's simple electrostatics.
>
> An electron arriving on one plate of (assumed vacuum dielectric)
> will repel another from the opposite plate.
>
> An electron pulled off one plate will attract an electron onto
> the other.
>
> Pulling and pushing in this way is the essence of AC.
>
> The current doesn't actually flow through, but because the net effect on
> both
> sides is the same, the current appears to flow through!
>
>
Yes!
|
|
|
|
Author: Bob MyersDate: 14:58 21-08-07
|
|
"Anonymous." <me@privacy.net> wrote in message
news:fafb1r$vq1$1@aioe.org...
> Pulling and pushing in this way is the essence of AC.
>
> The current doesn't actually flow through, but because the net effect on
> both
> sides is the same, the current appears to flow through!
No, it really does flow through. The problem here is
that while we typically imagine current to be a
stream of electrons, all nicely charging down a path
(perhaps in single file, and all in step), real "electrical
current" only follows this model in such thing as the
beam of a CRT.
Current is the effective flow of electrical ENERGY;
it involves the motion of charge carriers, and its units
are defined in terms of charges-per-second, but is not
directly equivalent with the motion of a particular charge
carrier or group except under very specific circumstances.
For example, what is the "speed" of electrical currents
through a conductor? For almost all practical concerns,
it makes sense to look at only the propagation speed
of the energy or signal in question - even though the
individual electrons are moving FAR slower than this.
Bob M.
|
|
|
|
Author: Don BoweyDate: 17:42 21-08-07
|
|
On 8/21/07 11:58 AM, in article fafclg$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)
|
|
|
|
Author: Bob MyersDate: 17:50 21-08-07
|
|
"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.
|
|
|
|
Author: John FieldsDate: 19:39 21-08-07
|
|
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
|
|
|
|
Author: John FieldsDate: 19:42 21-08-07
|
|
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
|
|
|
|
Author: Don BoweyDate: 20:02 21-08-07
|
|
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!
|
|
|
|
Author: Rich GriseDate: 21:02 21-08-07
|
|
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
|
|
|
|
Author: Michael A. TerrellDate: 23:36 21-08-07
|
|
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
|
|
|
|
Author: NobodyDate: 23:42 21-08-07
|
|
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".
|
|
|
|
Author: vorangeDate: 02:02 22-08-07
|
|
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.
|
|
|
|
Author: John PopelishDate: 02:22 22-08-07
|
|
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.
|
|
|
|
Author: John LarkinDate: 11:38 22-08-07
|
|
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
|
|
|
|
Author: Don BoweyDate: 12:23 22-08-07
|
|
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...?"
|
|
|
|
Author: Bob MyersDate: 13:19 22-08-07
|
|
"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.
|
|
|
|
|
|
|
|
|
Contact | Electronic Portal
|
|
|
