Infinite capacitance
Derek Abbott
In circuit models as you let capacitance C->oo this can be
thought of as *both* an infinite store of charge and a short circuit.
However, in reality a short circuit does not have infinite capacity.
Therefore there is an imperfect correspondence in *concept* between the
limiting process and reality.
Are there other examples in physics where this happens, as dramatically?
Are there ways of viewing the situation to reduce the philosophical
pain it causes?
Douglas W. Jones,201H MLH,3193350740,3193382879
by dab...@aelmg.adelaide.edu.au (Derek Abbott):
>
> In circuit models as you let capacitance C->oo this can be
> thought of as *both* an infinite store of charge and a short circuit.
> limiting process and reality.
Since infinity never occurs in reality, that's no problem. Only God can
charge or discharge an infinite capacitor.
An infinite capacitor is a voltage source. No amount of current into or
out of the plates will change the voltage across them. If God gives you
an infinite capacitor that is initially discharged, it will behave like
a short circuit. If God gives you an infinite capacitor charged to 5
volts, it will stay at 5 volts no matter how much current you push into
it or pull out of it. Usually, where a schematic calls for an infinite
capacitor, you can get away with substituting a well regulated power
supply.
Doug Jones
jo...@cs.uiowa.edu
Michael Covington
Just remember that it always starts out uncharged, and, because it's
infinite, you can never charge it above 0 volts.
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Greg Wolodkin
> In circuit models as you let capacitance C->oo this can be
> thought of as *both* an infinite store of charge and a short circuit.
Remember -- the infinite capacitance will still block DC..
Z = lim 1/jwC = / infinity or open circuit at DC (w=0) \
C->oo \ zero or short circuit at all other frequencies /
Actually I guess there are different ways to consider the limit, but it
seems reasonable to let C->oo first, then evaluate the limit as w->oo.
Since all finite-valued capacitors block DC, it's only fair that the
infinite one should as well..
> Therefore there is an imperfect correspondence in *concept* between the
> limiting process and reality.
Suppose you have an infinite capacitor C. Define v(0) to be the voltage
across the capacitor initially. If v(0) is nonzero, by (q = Cv) we must
also have an infinite amount of stored charge, just as you mentioned. So
let's consider the (possibly contradictory) notion of short circuit.
Suppose we want to change the voltage across the capacitor:
1 /t
v(t) = - | i(s) ds + v(0)
C /0
Since C is infinite, you can see that it would require an infinite
current or an infinite amount of time in the integral to change v.
Thus the analogy with a short circuit, or zero impedance, since for
all practical purposes v(t) is independent of i(t).
What you are missing is that the capacitor is an open, not a short,
at DC. If it *were* a short circuit at DC, it would not be able to
store any charge, and perhaps there is the confusion. I guess it's
all in how you evaluate the limit.
Best wishes,
Greg
Jeff Verive
>
Easy. The key is to consider the effect of such simplifications in relation
to the other circuit elements. Therefore, at 1 MHz (for example), a 10 uFd
capacitor looks like a short circuit compared to a 1 megohm resistor, but not
when compared to a 1 milliohm resistor.
Blindly following rules of thumb such as those used to simplify analysis can
lead to far more than philosophical pain. It can make learning electronics
incredibly difficult. (SOAPBOX ALERT!!!) I have found that putting maximum
effort into learning and understanding the basic principals of electronics
makes courses like Transmission Line Theory and Electromagnetics tolerable,
even enjoyable. Once you get the principals down, then you can use the
simplifications judiciously (try to explain the operation of an op-amp with
positive feedback by "remembering" that the inverting and non-inverting
inputs are supposed to be at the same potential).
Just my $ .02
Alexander Caldwell Schneider
>Derek Abbot <dab...@aelmg.adelaide.edu.au> writes:
>> In circuit models as you let capacitance C->oo this can be
>> thought of as *both* an infinite store of charge and a short circuit.
>
>Remember -- the infinite capacitance will still block DC..
>
> Z = lim 1/jwC = / infinity or open circuit at DC (w=0) \
> C->oo \ zero or short circuit at all other frequencies /
>
>Actually I guess there are different ways to consider the limit, but it
>seems reasonable to let C->oo first, then evaluate the limit as w->oo.
>Since all finite-valued capacitors block DC, it's only fair that the
>infinite one should as well..
What you are doing here is (arbitrarily) assuming that 0*oo=0. 0*oo cannot
be assumed to be infinite, nor can it be assumed to be zero or finite. I
think the best way to think about this limit is to imagine the behavior
of a capacitor of greater and greater capacitance. Consider this circuit:
||
Vin -----||---------- Vout
|| |
C |
>
> 1 ohm
>
|
|
---
-
Assume Vin is a step voltage at t=0 (Laplace transform 1/s). Vout then has
the Laplace transform 1/s*(1/(1+(1/Cs))) or 1/(s+(1/C)). Let C approach
infinity and you have Vout=1/s=Vin. Thus the capacitor behaves as a short
circuit. This isn't necessarily a rigorous proof, but it gets the point
across.
>Since C is infinite, you can see that it would require an infinite
>current or an infinite amount of time in the integral to change v.
>Thus the analogy with a short circuit, or zero impedance, since for
>all practical purposes v(t) is independent of i(t).
>
>What you are missing is that the capacitor is an open, not a short,
>at DC. If it *were* a short circuit at DC, it would not be able to
>store any charge, and perhaps there is the confusion. I guess it's
>all in how you evaluate the limit.
>
>Best wishes,
>Greg
An ideal open circuit is a *zero* capacitance. It isn't all in how you
evaluate the limit. There is a right answer here.
Alex
P.S. Berkeley, huh?
John Whitmore
>> In circuit models as you let capacitance C->oo this can be
>> thought of as *both* an infinite store of charge and a short circuit.
>Remember -- the infinite capacitance will still block DC..
But, what does that mean? If you only have a few years of
time to spend charging the capacitor and measuring its response,
you won't be seeing the DC response at all, but the response to
frequencies on the order of 10^-8 Hz.
If you want to push the clock back to the beginning
of the universe, you can get to 10^-18 Hz, but you STILL
aren't looking at DC.
>Since C is infinite, you can see that it would require an infinite
>current or an infinite amount of time in the integral to change v.
Yep. Exactly like a short circuit.
>What you are missing is that the capacitor is an open, not a short,
>at DC.
In terms of measurements that can be made at the capacitor
terminals, tell me what that means. Are you going to claim that
this capacitor is capable of being charged to nonzero voltage?
That is equivalent to infinite energy stored (and the universe
doesn't CONTAIN infinite energy). So, we are left with a pair
of terminals that have zero voltage difference, with all the current
forced into terminal 1 coming out of terminal 2.
If this differs from a short circuit, tell me HOW.
John Whitmore
p.s. we test theories with observations; if something we discuss
is NOT observable, it may be impossible to justify using our
theories (because they are untested in such a situation).
Joel Kolstad
>(try to explain the operation of an op-amp with
>positive feedback by "remembering" that the inverting and non-inverting
>inputs are supposed to be at the same potential).
I nailed a genuine PhD'ed professor with this one. After happily analyzing
an op-amp circuit, I asked her why couldn't we just change the inverting
and non-inverting pins and keep the status quo? She had to go home and
think about it before coming back the next day with the answer. :-)
---Joel Kolstad
Greg Wolodkin
> Z = lim 1/jwC = / infinity or open circuit at DC (w=0) \
> C->oo \ zero or short circuit at all other frequencies /
>
>Actually I guess there are different ways to consider the limit, but it
Uh.. let me try that again. The limit is zero for all w, including w=0,
no matter how you consider it.
OTOH, if we restrict C to physically realizable values.. ;-) ;-) ;-)
Cheers,
Greg
John Nagle
>Since infinity never occurs in reality, that's no problem. Only God can
>charge or discharge an infinite capacitor.
You can get real close in SPICE, though. Especially since SPICE
interprets prefixes consistently across component types. Just as
"1M" for a resistor means one megohm, "1M" for a capacitor means
one megafarad. This can lead to very strange circuit behavior and
time constants measured in weeks.
John Nagle
Keith Dow
>
>thought of as *both* an infinite store of charge and a short circuit.
>
>However, in reality a short circuit does not have infinite capacity.
Really?
C=Q/V
No matter how much charge you put in, the voltage
will be zero. You can also take out as much charge
as you want and the voltage is zero. What is the
problem?
--
Keith Dow A silicon valley guy.
Give me a Macintosh or give me death.
"Ditto? Ditto? You provincial putz!" Harvey Corman in Blazing Saddles
Stephen King
>In article <250hgk$7...@agate.berkeley.edu> wo...@cory.EECS.Berkeley.EDU (Greg Wolodkin) writes:
>>Derek Abbot <dab...@aelmg.adelaide.edu.au> writes:
>>> In circuit models as you let capacitance C->oo this can be
>>> thought of as *both* an infinite store of charge and a short circuit.
>
>>Remember -- the infinite capacitance will still block DC..
>
> But, what does that mean? If you only have a few years of
Simply because I want to put in my $.02 worth,
dVc/dt = i/C
If C is infinite, then i/C = 0. Meaning that you would NEVER get a voltage
drop across the thing. If you connected it to a DC source, there would
ALWAYS be a displacement current, and thus source would not be blocked.
--
Se non e` vero, e` ben trovato
...{utzoo|mnetor}!dciem!dretor!king ki...@dretor.dciem.dnd.ca
Greg Thoman
Quick typo check: in all SPICE simulators I've run across
(only a few of the many), 1m and 1M both mean milli-, not mega-.
To get mega- you use 1meg or 1MEG.
-----
Greg Thoman: The opinions expressed herein are mine alone, and I am
solely irresponsible for them.
James Bach
> interprets prefixes consistently across component types. Just as
> "1M" for a resistor means one megohm, "1M" for a capacitor means
> one megafarad. This can lead to very strange circuit behavior and
> time constants measured in weeks.
>
> John Nagle
WRONG!!!
In SPICE (at least in the 3 different "brands" that I've used) an "M" (or "m") means "milli" . . .
you have to type "MEG" to get "mega". . .
I've been bitten so many times by this stupidity that it makes me sick. Just because the
original '70s Berkley SPICE was written in FORTRAN which was too dumb to recognize lower
case letters (i.e. M=Mega m=milli, just like engineers use everyday), why should we
perpetuate the stupidity for a few more decades? With the modern computers we have today
it would be very easy for the software to accept the two "cases" and interpret them the way
humans do. Oh, I forgot the age-old "backwards compatability" argument. Yeh, right. Like
I'm going to wanna dig-out a 5 year old netlist and expect it to run instantly on my newer
SPICE system without ANY modifications (like retyping a few resistor values).
Here's another dumbism that is probably a throwback to the original Berkley SPICE . . .
Try entering an independant current source ("I" type) with "100uA" as the current spec.
Most likely you'll get a 100 Amp current source (I did)! You have to type either "100UA" or
(to be even "safer") "100E-6". Apparently most SPICE engines can't seem to recognize "u"
for micro in the current source definition, even though it has NO PROBLEM with capacitors
and inductors.
--
James C. Bach Ph: (317)-451-0455 The views & opinions expressed
Advanced Project Engr. GM-NET: 8-322-0455 herein are mine alone, and are
Powertrain Strategy Grp Amateur Radio: WY9F NOT endorsed, sponsored, nor
Delco Electronics Corp. Just say NO to UNIX! encouraged by DE or GM.
Wayne Ling
>
An infinite capacitor is best illustrated in a circuit involving
small signals. The infinite capacitor blocks the DC, allowing
both sides of the capacitor to be clamped to certain voltage levels
with voltage dividers. The infinite capacitor allows small AC
signals to pass through without noticeable attenuation. In reality,
the capacitance of the capacitor is made artifically large in relation
to the AC signal.
----------------------------------------------------------
Wayne Ling, P.Eng. wl...@bnr.ca
Bell Northen Research Ltd. Ottawa, Canada.
Programming is my life ? my lifelihood ?