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Re: zero bias shotkey diode,myth or real?
Himanshu Sharma
开云体育Hi ,
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I don't think that there are diodes that operate at
0 volts bias...
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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..
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Regards :-),
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--himanshu sharma
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Re: Light Activated Alarm
Himanshu Sharma
开云体育Hey ,
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What so funny about this...?? if there is then
please do let me know of it...:-)
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Regards :-),
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--himanshu sharma
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Re: Capacitor - Charge- Energy
Himanshu Sharma
开云体育Hey ,
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This is being used in many applications where we
need to make high voltages from smaller ones...
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Regards :-),
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--himanshu sharma
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Re: Capacitor - Charge- Energy
Sunantoro
Himanshu Sharma,
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Despite you explain scientifically, I still don't feel quite comfortable with the 'voltage increase'. Perhaps a parameter of the formula is being left behind.......I don't know. Anyway..thanks SUNAN -----Original Message-----
From: Himanshu Sharma [SMTP:hs_ramdev@...] Sent: Tuesday, October 30, 2001 2:16 PM To: Electronics_101@... Subject: Re: [Electronics_101] Capacitor - Charge- Energy hey , 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... Regards :-), --himanshu sharma ----- Original Message ----- From: Sunantoro <mailto:SUNANTORO@...> To: 'Electronics_101@...' <mailto:'Electronics_101@...'> Sent: Tuesday, October 30, 2001 12:25 PM Subject: RE: [Electronics_101] Capacitor - Charge- Energy Himanshu Sharma, Good point! Once a capacitor is charged, the connections are removed, then you can dismantle those plates individually without worry of reducing the "charge" already stored in each of the plates. This is also the proof of the statement that the energy being stored in the plates, not in the dielectric. Can you please explain why is the voltage increase? Increase against what? SUNAN |
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Re: Capacitor - Charge- Energy
Himanshu Sharma
开云体育hey ,
?
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...
?
?Regards :-),
?
--himanshu sharma
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Re: Capacitor - Charge- Energy
Sunantoro
Himanshu Sharma,
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Show quoted text
Good point! Once a capacitor is charged, the connections are removed, then you can dismantle those plates individually without worry of reducing the "charge" already stored in each of the plates. This is also the proof of the statement that the energy being stored in the plates, not in the dielectric. Can you please explain why is the voltage increase? Increase against what? SUNAN -----Original Message-----
From: Himanshu Sharma [SMTP:hs_ramdev@...] Sent: Tuesday, October 30, 2001 1:27 PM To: Electronics_101@... Subject: Re: [Electronics_101] Capacitor - Charge- Energy hey , here we go....thanks for the proof.... When you remove the dielectric slab then as per the formula C=keA/d the capacitance would decrease and hence the voltage would increase... Since the charge is constant...Q=CV Try that...!! So in short the charge stays on the plates...(I did this 2 years back..) Regards :-), --himanshu sharma . |
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Re: The need to know!
G Ramasubramani
开云体育?
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Re: Capacitor - Charge- Energy
Sunantoro
Jim, relax please....
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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??? SUNAN -----Original Message-----
From: Jim Purcell [SMTP:jpurcell@...] Sent: Tuesday, October 30, 2001 11:25 AM To: Electronics_101@... Subject: Re: [Electronics_101] Capacitor - Charge- Energy Sunantoro, > When we charge a capacitor, we actually pull-out the electrons from one > plate and at the same time provide more electrons on the other plate. All this thinking is making my head hurt (figure of speech only). I haven't t\ thought this much about 'tonix in ages. I retired last May but it was even before that when I last brain stormed about it. OK, here's an experiment. Take a conductor and charge it. I can do charge some dielectric materials, but not conductors. The if the charge is really on the plates it would be there when the dielectric is removed. Now if you mean that you can measure the potential of the charge at the plates, I can buy that. Jim |
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Re: Capacitor - Charge- Energy
Himanshu Sharma
开云体育hey ,
?
here we go....thanks for the proof....
?
When you remove the dielectric slab then as per the
formula
?
C=keA/d the capacitance would decrease and hence
the voltage would increase...
Since the charge is constant...Q=CV
?
Try that...!!
?
So in short the charge stays on the plates...(I did
this 2 years back..)
Regards :-),
?
--himanshu sharma
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Re: Automation
How do I get to translate the site?
-Saad --- av1a@... wrote: This site has some great projects like Stepper __________________________________________________ Do You Yahoo!? Make a great connection at Yahoo! Personals. |
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Re: PCI Bus configuration
Jonathan Luthje
Saad,
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This is not quite as easy as it may sound - the PCI slot is quite a complex little piece of hardware - if you hard thinking that it's just a "different sized ISA slot", then think again! The PCI card needs a PCI-PCI bridge interface to work properly in the slot - they aren't as simple as the old "plug it in and switch it on" logic based ISA cards. However, the information you need can be obtained from Regards, Jonathan ----- Original Message -----
From: <saad_75@...> To: <Electronics_101@...> Sent: Tuesday, October 30, 2001 3:53 PM Subject: [Electronics_101] PCI Bus configuration Could anyone please provide me the bus pinouts of the PCI. I plan to |
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Re: Capacitor - Charge- Energy
Jim Purcell
Sunantoro,
When we charge a capacitor, we actually pull-out the electrons from oneAll this thinking is making my head hurt (figure of speech only). I haven't t\ thought this much about 'tonix in ages. I retired last May but it was even before that when I last brain stormed about it. OK, here's an experiment. Take a conductor and charge it. I can do charge some dielectric materials, but not conductors. The if the charge is really on the plates it would be there when the dielectric is removed. Now if you mean that you can measure the potential of the charge at the plates, I can buy that. Jim |
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Re: Capacitor - Charge- Energy
Sunantoro
Jim,
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When we charge a capacitor, we actually pull-out the electrons from one plate and at the same time provide more electrons on the other plate. And then we disconnect the capacitor. The capacitor is now charged. If then we connect the two leads, using a resistor, the electrons flow from the excessed electrons plate to the other. Then the capacitor is being discharged. Very easy to explain and to analyse. In most electrical circuitry, this conventional approach is still good to be used. Am I too "FLAT"? Thanks, SUNAN -----Original Message-----
From: Jim Purcell [SMTP:jpurcell@...] Sent: Tuesday, October 30, 2001 10:53 AM To: Electronics_101@... Subject: Re: [Electronics_101] Capacitor - Charge- Energy Sunantoro, > When people discuss about capacitor, they use "Charge" and "Energy" > interchangeably. This is rather confusing or ambiguous (to me). > Can we simply change it with "Electrons" which flow and accumulate in one > plate when the opposite plate becomes lack of it (electrons)? Don't think any electrons accumulate on the plates. Where would they stay. I'm thinking now that charge is the wrong term to use for the resulting stored energy. We often say that a capacitor stores a charge, and it certainly has to be charged, and the text books talk coulombs something fierce when they get to capacitors. I still can't see the energy in a cap as stored electrons, that's particles. And fields are not supposed to be particles, or am I wrong there too. > > By using this understanding, I believe there is no need to elaborate further > on "Charge" or "Energy", Actually, I have a problem with the term energy to describe what is stored too since energy includes time, i.e. watt seconds, joules. But the stored energy is at rest. OR are we talking about the amount of joules it took to charge... oops, to store the energy. Jim |
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trim your posts
Mark Kinsler
Yo:
These messages would be a great deal easier to read if posters would eliminate all but the most pertinent quoted material. It's getting difficult to search through all of the past correspondence to find out what's new, so generally I find I have to skip messages that contain lots of old stuff in them. Thanks. Mark Kinsler 512 E Mulberry St. Lancaster, Ohio USA 740 687 6368 _________________________________________________________________ Get your FREE download of MSN Explorer at |
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Re: The need to know!
Jim Purcell
G,
??????????? As far as I can see, unless there is a space between the plates and the dielectric, theYes, the parameters of the field are determined by the size of the plates, their distnce apart, i.e. the thickness of the dielectric and the dielectric constant. ?????????? distance between the plates and the dielectric thickness are the same. I don't see how ?????????? you can relate dielectric constant with the plate distance. They are apples and oranges. ?????????? One has units, the other is only a number. ??????????? Yes dielectric strength, not dielectric constant. And until the dielectric breaks downWell. You have 2 plates. Positive charge on one, negative on another. How high can this charge build up before? the dielectric breaks down? That is determined by the dielectric. ??????????? it has little to do with the energy storage, etc. ??????????? Oops, you repeated yourself. I assume that one of those above was supposedIn case the amount exceeds this level, then the dielectric breaks down ??????????? to be dielectric constant. ??????????? The two parameters have little to do with each other. ???????????? It's a lot more serious than discharge. If breakdown is reached the cap. is probably destroyed.and the capacitor gets discharged real fast. ?????????? I can't remember either to be truthful. My point was that if the plate distance was different from the dielectric thickness then there must be something between the plates and the dielectric, so you have a mixture of dielectrics, i.e. the original and air or whatever.I think that the distance between the plates is more important than dielectric thickness. Jim |
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Re: vacuum and charge
Doug Hale
How about inductors:
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The engineering equation for inductors is simular th that of capacitors. L = inductance in henri V = voltage in volts I = current in amperes t = time in seconds V = L dI/dt which means that the current through a one henrie inductor will change at the rate of one ampere per second with a one volt drop across it. Doug Hale Doug Hale wrote: Lets look at some of the mathamatical relationships: |
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