Re: Capacitor - Charge- Energy

[email protected]

            i think if there is vaccum between the
plates capacitor can't store energy.I don't think so.  As long as their is an electric field between the 
plates, there is stored energy.  With a dielectric between the 
plates, you can support a larger field, hence store more energy.

Paul

Re: zero bias shotkey diode,myth or real?

Jim Purcell

#1025

Himanshu,

I don't think that there are diodes that operate at 0 volts bias...?But surely what you can do is make ideal diodes using OP-Amps...These Op-Amp diodes give the ideal characterstics which you cann't get from any diode..

True, but those are diodes in name only, what as asked was whether there
was a zero drop shottky and I think the answer is no.

Jim

Re: capacitors and dielectrics

[email protected]

#1026

Now watch this:  We've charged the capacitor to voltage V.  We rewrite the 
definition of capacitance Q/V=C ---> V=Q/C.  This predicts what will happen 
if we reduce the capacitance by sliding the dielectric out from between the 
plates.  Yup: the voltage will go up, because we won't be reducing the 
charge on the plates.  C goes down, V goes up.  You can do the same thing by 
just pulling the plates apart.Voltage is just the potential between plates, kind of like weight.  I 
assume the work we do moving the dielectric goes into changing 
the potential.

The opposite will happen if we increase the capacitance of a capacitor.  If 
we were to charge an air dielectric capacitor to a voltage of 700 volts and 
then slip a piece of glass between the plates such that the glass completely 
fills the space between the plates, we'd find that the voltage has dropped 
to 100 volts.  Thus we say that glass has a relative dielectric constant of 
seven.  The dielectric constant of air and vacuum are equal for most 
practical purposes.As I recall, when a dielectric is placed between the plates, an 
opposing field is setup inside the dielectric.  I guess that opposing 
field is what is reducing the voltage on the plates.

Paul

Re: zero bias shotkey diode,myth or real?

Jim Purcell

#1024

peter,

is there some one who can inform me,as to the avalibility of a zero
volts drop shotkey diode .One that shorted out. :-)   I think if a diode had zero voltage drop
it probably would not be functioning properly. That's an assumption,
I have no proof.

Jim

Re: Capacitor - Charge- Energy

Jim Purcell

#1023

Himanshu,

This is being used in many applications where we need to make high voltages from smaller ones...

Examples please?

Jim

Re: Capacitor - Charge- Energy

Jim Purcell

#1022

Himanshu,

Well.... when you remove the dielectrics then the capacitance decreases...This is as per the formula C=keA/d where k is the dielectric constant...?now since the capacitor is disconnected hence the charge is const.and as V=Q/C...C decrease mean that voltage diff. would increase...

I'm not sure that we can assume that the charge would not
change if the capacitor were dismantled in any way. That's
why, I think, discussing where the charge is stored is purely
academic. Each factor, dielectric constant, plate area and
distance determines the capacitance and thus the amount
of energy stored by influencing how the charging process
progresses. However once the source voltage is removed
I'm not sure any changes made can be assumed to change
energy stored, as to voltage or coulombs. That's just my
'feeling' not anything I know for certain. And, as I said,
probably purely academic.

Jim

Re: The need to know!

Jim Purcell

#1021

Rama,

Call me Rama.
Call me anything but late for dinner.
?
?As you mention, yes - I am referring to space between the plates and dielectric. I thought that was pretty obvious.
YOu seemed to be making a distinction between plate distance and dielectric thickness when you
said they cold differ. If they differ then you must have air as an additional dielectric.
?

Are you talking leakage due to exceeding dielecric strength?
Yeah. Maybe discharge is not the correct term, but what the heck - I am not writing a book :-)

??????????????????? I'm sure that the cap will discharge but worse, it might be
??????????????????? Permanently damaged in any but air capacitors. Air is generally
??????????????????? a self healing dielectric.

If the plates were farther apart than the dielectric thickness you'd have a mixture of dielectrics, the normal dielectric plus some air (or vacuum.) I am not sure of more important or less important - both play a role.

????????????????? Easy to get that from the formula for capacitance as a function of
????????????????? physical design parameters.? In this formula the plate distance is the
????????????????? only term on the bottom and dielectric constant is on top with
????????????????? a constant and plate area. So distance has greater effect on
????????????????? capacitance. And dielectric thickness is generally assumed to be
????????????????? the same as plate distance.
?

Jim
?
?

Re: Capacitor - Charge- Energy

Jim Purcell

#1020

Sunantoro,

Jim, relax please....Actually it was an exhilarating experience, all that techie thinking, it's been
years since I've done that. But I was getting a bit tired last night,

The plates and the dielectric configure a system named capacitor that is
capable of storing energy. So the energy is stored in the Capacitor AS A
SYSTEM, not in the plate nor in the dielectric in isolation. If then there
is a query: in the form of what is the energy stored, the answer is, I
believe, in the form of electrical POTENTIAL-GAP between those two plates.
One plate is very Positive against the other.
How can you accept this as a compromise???I have begun to believe that what we have here is one of those issues that is
very dependent on semantics as well as peoples understanding of terms.
That the capacitor stored energy is a good generalization. It probably doesn't
really matter which component store which, but a better understanding of
what is actually happening and a refinement of understanding terms that
sometimes get bandied about with little understanding can only be a good
thing in the long run.

As for the potential on each capacitor plate, I don't see think you can ever
talk about potential or even polarity at any point in a circuit without a
reference point. So the positive cap. terminal is only positive with respect
to the negative terminal. If the capacitor is in a larger circuit of course
it's terminals may have voltage levels that are referenced to say the
power supply common, a.k.a. ground.

Jim

Re: capacitors and dielectrics

Doug Hale

#1019

Mark,
  I looked at your website - a nobel cause.

   Are you familliar with Marshal Brain    and  ?

   The main reason I am on this list is because electronics hobbiests are an endangered spiecies.

Doug Hale

Mark Kinsler wrote:

Show quoted text

We generally do not normally work in units of electrical charge, so the concept seems a bit strange.  The easiest way to deal with this is to use the relationship I=Q/t, which says that current is the rate at which charge is delivered.  So we rewrite this definition as Q=It, which says that we can calculate the charge we've placed on a capacitor's plates by multiplying the current through the capacitor by the time we've applied that current.  Result: to determine charge, you need a watch and an ammeter.

If we charge a capacitor to a voltage V by pushing a charge Q (=It) onto the plates, we will find that the relationship between the voltage and the charge is Q/V=C, which is the definition of capacitance.  Note that we can thus measure the capacitance of a capacitor with an ammeter, a watch, and a voltmeter.

Now watch this:  We've charged the capacitor to voltage V.  We rewrite the definition of capacitance Q/V=C ---> V=Q/C.  This predicts what will happen if we reduce the capacitance by sliding the dielectric out from between the plates.  Yup: the voltage will go up, because we won't be reducing the charge on the plates.  C goes down, V goes up.  You can do the same thing by just pulling the plates apart.

The opposite will happen if we increase the capacitance of a capacitor.  If we were to charge an air dielectric capacitor to a voltage of 700 volts and then slip a piece of glass between the plates such that the glass completely fills the space between the plates, we'd find that the voltage has dropped to 100 volts.  Thus we say that glass has a relative dielectric constant of seven.  The dielectric constant of air and vacuum are equal for most practical purposes.

M Kinsler
512 E Mulberry St. Lancaster, Ohio USA 740 687 6368

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Re: newbie inquiry

Jim Purcell

#1018

angtengchat,

<<If in doubt use the gloves. :-).

Which brand of gloves would be good ! What must i do to make sure that the
gloves is Good / insulated. I am always afraid of electricity.Forgive me if this seems like a bogus message. A healthy respect for
electricity is
good for an electronics technician. Fear is not, choose another field if it
scares
you, you will do more harm because of your fear.

Jim

gloves

Mark Kinsler

#1017

Do NOT try to use gloves when working with high voltage equipment.  While it is true that electric linemen can work with high voltages (typically up to 2kV or so) with gloves, these gloves are changed and tested regularly.  The gloves are thick rubber and they are expensive.  The lineman checks them himself every time he puts them on, and then covers them with heavy leather gloves to protect the rubber.

The power company will periodically collect the gloves, stuff them into a 55-gallon drum, and ship them off to a facility which will test them: the gloves are filled with salt water and lowered into a vat containing more salt water.  An electrode is placed inside the glove and a high voltage is applied across this electrode and the salt water vat.  If the glove withstands 10kV for a while (I forget how long) it passes the test and a rubber certification seal is affixed to the glove.

Many gloves do not pass the test, and the site of the breakdown is typically impossible to see without very close examination.

All of this is to emphasize that you must work with insulated tools, not gloves, around energized high-voltage apparatus.  In general, this is how it's done in industry and in the electric power business.  Do a search of 'hot stick' on Google to see how it's done.

For the most part, high-voltage laboratory people simply discharge the equipment and apply grounds to assure that it stays that way while working on it.  Then the ground sticks are withdrawn, everyone retires to a protective cage or other safe place, and the equipment is re-energized.

Nobody fools with gloves.

M Kinsler

512 E Mulberry St. Lancaster, Ohio USA 740 687 6368

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Re: newbie inquiry

Jim Purcell

#1016

Keith,

IIRC, the dangerous levels are such as would deliver energy of 10 Joules
along a path through the Heart, i.e. across the Thorax or via Head to
Foot.   A large capacitance could deliver this!It all depends on the capacitor voltage. Capacitors don't suspend Ohms Law.
If the capacitor charge is at lower voltage then it won't be dangers to a
person
who grabbed the leads. That Joules figure is for the amount of energy in the
capacitor, not how much will reach the unsuspecting should who makes contact
with the capacitor.

Jim

Demonizing Hackers

Charles

#1015

I don't know how the group got diverted on to the topic of hacking, 
other than my invitation to come and check out my yahoo group, 
hackerz_n_hickory... but anyways, to address your fears...
Hackers, for the most part are really good at one thing, are 
reclusive, masturbate more than most people, and are anti social, so 
the standard hacker is basicly a technologically capable psychopath.. 
Pychopaths are a good thing, and the strongest mentally...
Hackers are not evil, nor do they wish to destroy everything, sure, 
they tear up stuff, to find out how it works, they probe things, ask 
questions, and pull out there soldering iron in an attempt to make 
things better.. not to destroy them... If it weren't for hackers, 
99.999 percent of you out there wouldn't be working on your windows 
based PC's... Personal computers were actually developed because of 
hackers!! No one but me remembers the Alto??? Read a book called 
dealers in lightening, if you want to find out where PC's really came 
from, excentric.. intelligent, hackers.. who wanted to make things 
better, and faster... and I believe they have... the 
malicious "hackers" you refer to are called Crackers in the hacking 
communiity... they crack into sites, steal credit card numbers, etc.. 
but the rest of us, like me, are out here wondering how to wire 32 2 
gigahertz pentiums together, and install 30 gigs of memory in our new 
P4's.... (Intel already is developing a 20 gigahertz chip....) Were 
the people that modify our computer cases.. put neon lights in the 
case... have them running faster than anything you can get mass 
market... we use scsi drives because we know they are faster, more 
reliable, and better made than the mass market crap you can buy off 
the shelf.. we rip apart the bios code of our systems, just to free 
up an extra bit or two of memory.. we are the true techno freaks of 
the cyberage.. the dreamers.. the revolutionaries.. the people who 
take a old ford pinto and drop a 455 cubic inch engine in the 
thing... that is the equivalent of what we do... or at least, what I 
do.. so, yeah.. we rip things apart, we look into the bios code, we 
find ways to make our computers run better, to make our cd players 
and recorders play new types of media (I remember when you could 
order a home made MP3 player from a few hardware hackers online about 
a year before they made it to market in the us....I still have my 233 
hooked up in the back of my car for just such a purpose...40 gig hard 
drive, filled with music!) What I'm saying is, hackers, hardware, 
software, and otherwise, eventually end up developing new devices 
that make your lives easier, faster, or cooler... so don't demonize 
them..Just sorta like saying all Christians are like Pat Robertson, 
or that all Muslims are like Osama Bin Laden..Generalizations suck... 
and for the most part.. hackers are the reason we have these cool ass 
lil P4's and macs to play with, talk trash to women on the internet, 
etc.... So, thank goodness for them... and if someone does something 
illegal, treat them like the crooks they are, I'm down with that.. 
what I'm not ok with, is everyone who likes electronics/ computers/ 
etc, downgrading all hackers into the realm of criminals... 

Just my humble opinion... and hey, opinions are like assholes, 
everyone has them, and they all stink :P

Charles

Parazite@...
hazenoff@...
hazenoff@...
hazenoff@...

Re: Capacitor - Charge- Energy

werrr fggfg

#1014

hi jim,
      i think because of the deformation of the
dielectric  molecular configuration  capacitor stores
charge.
even if there is no dielectric there is air as
dielectric.  the charge storage capacity of the
capacitor depends on the dielectric material.the
better
it deforms better will be the charge storage
capability.
            i think if there is vaccum between the
plates capacitor can't store energy.
                                        regards
                                          rudheesh



 --- Sunantoro <SUNANTORO@...> wrote: 
<HR>
<html><body>
<tt>
Jim, relax please....<BR>
The plates and the dielectric configure a system named
capacitor that is<BR>
capable of storing energy. So the energy is stored in
the Capacitor AS A<BR>
SYSTEM, not in the plate nor in the dielectric in
isolation. If then there<BR>
is a query: in the form of what is the energy stored,
the answer is, I<BR>
believe, in the form of electrical POTENTIAL-GAP
between those two plates.<BR>
One plate is very Positive against the other.<BR>
How can you accept this as a compromise???<BR>
<BR>
SUNAN<BR>
<BR>
<BR>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; -----Original
Message-----<BR>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
From:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Jim Purcell
[SMTP:jpurcell@...]<BR>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
Sent:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Tuesday, October
30, 2001 11:25 AM<BR>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
To:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
Electronics_101@...<BR>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
Subject:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Re:
[Electronics_101] Capacitor - Charge- Energy<BR>
<BR>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Sunantoro,<BR>
<BR>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; > When we charge a
capacitor, we actually pull-out the<BR>
electrons from one<BR>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; > plate and at the
same time provide more electrons on the<BR>
other plate.<BR>
<BR>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; All this thinking is
making my head hurt (figure of speech only). I<BR>
haven't t&#92; thought this much about 'tonix in ages. I
retired last May but it<BR>
was even before that when I last brain stormed about
it.<BR>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; OK, here's an
experiment. Take a conductor and charge it. I can
do<BR>
charge some dielectric materials, but not conductors.
The if the charge is<BR>
really on the plates it would be there when the
dielectric is removed. Now<BR>
if you mean that you can measure the potential of the
charge at the plates,<BR>
I can buy that.<BR>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Jim<BR>
</tt>

<br>

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capacitors and dielectrics

Mark Kinsler

#1013

We generally do not normally work in units of electrical charge, so the concept seems a bit strange.  The easiest way to deal with this is to use the relationship I=Q/t, which says that current is the rate at which charge is delivered.  So we rewrite this definition as Q=It, which says that we can calculate the charge we've placed on a capacitor's plates by multiplying the current through the capacitor by the time we've applied that current.  Result: to determine charge, you need a watch and an ammeter.

If we charge a capacitor to a voltage V by pushing a charge Q (=It) onto the plates, we will find that the relationship between the voltage and the charge is Q/V=C, which is the definition of capacitance.  Note that we can thus measure the capacitance of a capacitor with an ammeter, a watch, and a voltmeter.

Now watch this:  We've charged the capacitor to voltage V.  We rewrite the definition of capacitance Q/V=C ---> V=Q/C.  This predicts what will happen if we reduce the capacitance by sliding the dielectric out from between the plates.  Yup: the voltage will go up, because we won't be reducing the charge on the plates.  C goes down, V goes up.  You can do the same thing by just pulling the plates apart.

The opposite will happen if we increase the capacitance of a capacitor.  If we were to charge an air dielectric capacitor to a voltage of 700 volts and then slip a piece of glass between the plates such that the glass completely fills the space between the plates, we'd find that the voltage has dropped to 100 volts.  Thus we say that glass has a relative dielectric constant of seven.  The dielectric constant of air and vacuum are equal for most practical purposes.

M Kinsler
512 E Mulberry St. Lancaster, Ohio USA 740 687 6368

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Re: zero bias shotkey diode,myth or real?

Doug Hale

#1011

Shotkey diode forward bias drops are usually around  .2 volts
germainim - .3
silicon - .6 to .7
selenium 2 to 5

DOug Hale

peter wrote:

Show quoted text

is there some one who can inform me,as to the avalibility of a zero volts drop shotkey diode .

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Re: Capator tester

Doug Hale

#1012

I have never design and/or built a capacitor tester but the concepts are strait forward.

There are a few parameters to test - some basic ones and some not so basic ones.

      Leakage: test this with a mega ohm meter
           Capacitance:
             This is indirectly tested in a few ways:
               AC steady State:
                    Xc = 1/(2 Pi F C)   =>   C = 1/(2 Pi F Xc)

                 Where:
                       Xc = Capacative Reactance (AC resistance) in ohms
                       C =  Capacitance in farads
                       F = Frequency in hertz
                       Pi = Pi .. 3.14159....

                This is basicly an AC ohm meter with the equation computed into the meter scale.

             Transient Response:
                     I = C dV/dt     =>    C =  I dt/dV
                   To test using this methode you would do one of two things:
                                1) Provide a step current and measure the slope of the voltage change
                                2) provide a sloped voltage across the cap and test the contant current

                    this is a complex process - could be done fairly easily with a small micro-controller to controll the process

  There are other things ways and other things to test for - like:
          dielectric break down voltage - usualy destructive - basicaly a hipot tester
          parasitic inductance and resistance - variations in the above tests
          themal sensitrivity tests - test above under different thermal conditions


Or just search the web for an already designed curcuit!!!


Doug Hale


angtengchat wrote:

Show quoted text

<<Could anybody point me in the right direction to design a capacitor tester.

I have not done this. However a good lead would be to use the principle of RC constant.

I normally use a simple sanwa multi-meter to check if a capacitor is still "s".

 
    ----- Original Message -----

    From:Jim Purcell <mailto:jpurcell@...>

    To: Electronics_101@...
    <mailto:Electronics_101@...>

    Sent: Tuesday, October 30, 2001 12:59 AM

    Subject: Re: [Electronics_101] Capator tester


    Michael,

    Could anybody point me in the right direction to design a
    capacitor tester.

    I assume you mean an instrument that will measure the capacitance
    as well as
    test whether the capacitor is good. Many years ago I built an
    Eico kit capacitance
    bridge. It use the principle of a wheatstone brige to measure
    capacitnace. Of
    course it was the AC version of the bridge and the user manually
    ajusted the
    bridge for a balance and read the capacitance according to where
    the pointer
    pointed. That kind of instrument used the principle of capacitive
    reactance.
    I think one side of the bridge consisted of one known and the unknown
    capacitance and the other a fixed and a variable resistance. I
    confess
    that Ihave not read any description of how the modern solid state
    instruments work but I guess they are just automated implementations
    of the old bridge circuits. In any they most likely depend on passing
    AC through the capacitor and determining capacitance accoring to
    a voltage drop across a component.
    Jim

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Re: Cable TV signal amplification

Doug Hale

#1009

Radio Shack , and others sell a TV distribution amplifier. It amplifies and splits.

Doug

av1a@... wrote:

Show quoted text

Hi,
when i split the cable in two by A/B connector , one for my tv & other for
PC tv tuner card , i started to get noise.
Is there any simple way to amplify these signals.

Imran

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chat

carlos diaz

#1010

IF YOU CAN DONT SEND ME ANY MORE INFO THANKU SONIC626UK@...

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from electronics and computer engineers association

shaifali

#1008

开云体育

hi guys and gals, we are the leading software company in Uttar Pradesh(India)for the wireless LAN solutions. we are the group of 16 engineers those are working on wireless field and we also start a association of electronics and computer engineers.and we want that all electronics and computer engineers register themselve to this association to get a proper plate form in India.if anybody interested to join this association than mail me with your address so that we can send you the detail of the association. in this association we have three types of membership 1-student membership 2-teacher membership and 3-professional membership we provide the quarterly journals and latest information about industry trends and lots of more. with regards shaifali

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