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Cpacitor discharge

M

mkaras

Jan 1, 1970
0
You missed the position I was taking in the thread in answer to the
Op's question.
Just read it carefully again.

Besides, without all those fancy 7 components and exoticly Pulsed
Power supplies included in the circuit given above I could get you
the linear voltage output your circuit gives
with only 2 components plus a capacitor.

Careful though on your statements. Before telling anyone that a
linear output is not an exponential watch out though on one point. I
have sometimes used a linear output as an approximation to an
exponential and it works excellently. Check on the Math of
exponentials to see why.



OK Andrew:
The full truth . . . you are an idiot. And sadly, without the ears to
hear
it or the humility to acknowledge it, you will charge forth continuing
to
prove it.
- mkaras
 
B

BFoelsch

Jan 1, 1970
0
Jan Panteltje said:
The only way way you can get a current out of a capacitor is by using
a resistor.

No way, 'current' equals electrons.
There are many ways other then 'resistors'.
Look up vacuum tube, transistor, semiconductor, and in this context
especially 'current source'.

To make it a bit more clear, in a _resistor_ the current decreases linear
with the increase in resistance.

In something that allows control of the current, say a transistor, or tube,
the current does _not_ rise proportionally to the applied voltage.

In a (junction) transistor the Ic becomes greatly independent above some minimum
voltage, and then only depends on Ib.
Same for a 'penthode' tube.
Many integrated circuits use current sources and current mirrors, and as such
Isupply may depend little on Usupply.

This is one for [sci].electronics.basics.

My favorite is from Physics 101.

Put a 365 pF variable capacitor (set to maximum C) in parallel with a 365 pF
fixed capacitor.
Charge the combination to 100 VDC.
Adjust the variable capacitor to 10 pF.

1. What is the resultant voltage across the combination?
2. What is the stored energy before and after?
3. How much work did it take to move the capacitor plates?

Charge is moved, no resistance involved.
 
E

ehsjr

Jan 1, 1970
0
Tom said:
Andrew Edge wrote:

Lessee...the OP stated that the current is constant, not i=v/R.
For your information a Capacitor is an open circuit to a constant
current. So a capacitor generating constant current is impossible.

So you are saying this circuit won't work,
within any time limits, whatsoever:

/
+12 ----o o-----+--->|---+
| |
[R] |
| |
+--->|---+ I -> ~ 10 mA
| | -----
[1F] +---In|LM317|Out---+
| ----- |
| Adj [120R]
| | |
| +---------+
| |
| [R]
| |
Gnd -----------+---------------------------+

Check on Kirchoff's current law on the capacitor leads.
I hoped someone would have told him that.

The op is discharging a cap through an active load
which he says draws constant current. Such a circuit
is shown above.

After the switch has been closed for sufficient time
to fully charge the capacitor, it is opened. Will
the discharge be constant current, or does the circuit
suddenly lose the ability to regulate?

Ed

<snip>



You don't really think your logic will matter to him, do you, Ed?


I suppose not.
Note that he wrote, "For your information a Capacitor is an open
circuit to a constant current." That would mean that in a loop
comprising a constant current source, a capacitor, and a resistor,
there would be no current through the resistor...

Sigh.

Yup. I still have't figured out how to charge a
capacitor from a constant current source using
his gem of an idea.

Maybe it's a flux capacitor? :)

Ed
 
M

Michael A. Terrell

Jan 1, 1970
0
ehsjr said:
Maybe it's a flux capacitor? :)


Do you call it "All fluxed up" when it fails? ;-)


--
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
 
A

Andrew Edge

Jan 1, 1970
0
Jan Panteltje said:
The only way way you can get a current out of a capacitor is by using
a resistor.

No way, 'current' equals electrons.
There are many ways other then 'resistors'.
Look up vacuum tube, transistor, semiconductor, and in this context
especially 'current source'.

To make it a bit more clear, in a _resistor_ the current decreases linear
with the increase in resistance.

In something that allows control of the current, say a transistor, or tube,
the current does _not_ rise proportionally to the applied voltage.

In a (junction) transistor the Ic becomes greatly independent above some minimum
voltage, and then only depends on Ib.
Same for a 'penthode' tube.
Many integrated circuits use current sources and current mirrors, and as such
Isupply may depend little on Usupply.

This is one for [sci].electronics.basics.

My favorite is from Physics 101.

Put a 365 pF variable capacitor (set to maximum C) in parallel with a 365 pF
fixed capacitor.
Charge the combination to 100 VDC.
Adjust the variable capacitor to 10 pF.

1. What is the resultant voltage across the combination?
2. What is the stored energy before and after?
3. How much work did it take to move the capacitor plates?

Charge is moved, no resistance involved.

You forgot. Capacitors have internal resistance. Wires connecting
resistance. Quite a lot of resistance is involved.

Andy
 
A

Andrew Edge

Jan 1, 1970
0
Andrew Edge wrote:

Lessee...the OP stated that the current is constant, not i=v/R.
For your information a Capacitor is an open circuit to a constant
current. So a capacitor generating constant current is impossible.

So you are saying this circuit won't work,
within any time limits, whatsoever:

/
+12 ----o o-----+--->|---+
| |
[R] |
| |
+--->|---+ I -> ~ 10 mA
| | -----
[1F] +---In|LM317|Out---+
| ----- |
| Adj [120R]
| | |
| +---------+
| |
| [R]
| |
Gnd -----------+---------------------------+
Check on Kirchoff's current law on the capacitor leads.
I hoped someone would have told him that.

The op is discharging a cap through an active load
which he says draws constant current. Such a circuit
is shown above.

After the switch has been closed for sufficient time
to fully charge the capacitor, it is opened. Will
the discharge be constant current, or does the circuit
suddenly lose the ability to regulate?

Ed

<snip>


You don't really think your logic will matter to him, do you, Ed?

Note that he wrote, "For your information a Capacitor is an open
circuit to a constant current." That would mean that in a loop
comprising a constant current source, a capacitor, and a resistor,
there would be no current through the resistor...

Sigh.


YES ...
Z(Capacitor) = 1/jwC

w=0 for a direct current source =>
Z = Infinity =>No current flows through circuit.
Can it be any easier.

Got a better argument .
Lets listen to it.
 
A

Andrew Edge

Jan 1, 1970
0
Andrew Edge wrote:

Hi Tom.



For your information a Capacitor is an open circuit to a constant
current. So a capacitor generating constant current is impossible.

So you are saying this circuit won't work,
within any time limits, whatsoever:

/
+12 ----o o-----+--->|---+
| |
[R] |
| |
+--->|---+ I -> ~ 10 mA
| | -----
[1F] +---In|LM317|Out---+
| ----- |
| Adj [120R]
| | |
| +---------+
| |
| [R]
| |
Gnd -----------+---------------------------+
Check on Kirchoff's current law on the capacitor leads.
I hoped someone would have told him that.

The op is discharging a cap through an active load
which he says draws constant current. Such a circuit
is shown above.

After the switch has been closed for sufficient time
to fully charge the capacitor, it is opened. Will
the discharge be constant current, or does the circuit
suddenly lose the ability to regulate?

Ed



<snip>
Hi Ed.
The Op said he wass discharging through a Microcontroller lead, Not
through a Voltage regulator chip.

Andy
 
B

BFoelsch

Jan 1, 1970
0
Jan Panteltje said:
On a sunny day (Tue, 13 Feb 2007 18:49:41 +0100) it happened Andrew Edge
<[email protected]>:

The only way way you can get a current out of a capacitor is by using
a resistor.

No way, 'current' equals electrons.
There are many ways other then 'resistors'.
Look up vacuum tube, transistor, semiconductor, and in this context
especially 'current source'.

To make it a bit more clear, in a _resistor_ the current decreases linear
with the increase in resistance.

In something that allows control of the current, say a transistor, or tube,
the current does _not_ rise proportionally to the applied voltage.

In a (junction) transistor the Ic becomes greatly independent above some minimum
voltage, and then only depends on Ib.
Same for a 'penthode' tube.
Many integrated circuits use current sources and current mirrors, and as such
Isupply may depend little on Usupply.

This is one for [sci].electronics.basics.

My favorite is from Physics 101.

Put a 365 pF variable capacitor (set to maximum C) in parallel with a 365 pF
fixed capacitor.
Charge the combination to 100 VDC.
Adjust the variable capacitor to 10 pF.

1. What is the resultant voltage across the combination?
2. What is the stored energy before and after?
3. How much work did it take to move the capacitor plates?

Charge is moved, no resistance involved.

You forgot. Capacitors have internal resistance. Wires connecting
resistance. Quite a lot of resistance is involved.

Andy

I didn't forget anything, in contrast to you, who never knew anything.

PLONK
 
A

Andrew Edge

Jan 1, 1970
0
On a sunny day (Tue, 13 Feb 2007 18:49:41 +0100) it happened Andrew Edge
<[email protected]>:

The only way way you can get a current out of a capacitor is by using
a resistor.

No way, 'current' equals electrons.
There are many ways other then 'resistors'.
Look up vacuum tube, transistor, semiconductor, and in this context
especially 'current source'.

To make it a bit more clear, in a _resistor_ the current decreases linear
with the increase in resistance.

In something that allows control of the current, say a transistor, or
tube,
the current does _not_ rise proportionally to the applied voltage.

In a (junction) transistor the Ic becomes greatly independent above some
minimum
voltage, and then only depends on Ib.
Same for a 'penthode' tube.
Many integrated circuits use current sources and current mirrors, and as
such
Isupply may depend little on Usupply.

This is one for [sci].electronics.basics.

My favorite is from Physics 101.

Put a 365 pF variable capacitor (set to maximum C) in parallel with a 365 pF
fixed capacitor.
Charge the combination to 100 VDC.
Adjust the variable capacitor to 10 pF.

1. What is the resultant voltage across the combination?
2. What is the stored energy before and after?
3. How much work did it take to move the capacitor plates?

Charge is moved, no resistance involved.

You forgot. Capacitors have internal resistance. Wires connecting
resistance. Quite a lot of resistance is involved.

Andy

I didn't forget anything, in contrast to you, who never knew anything.

PLONK

Oh sounds like you let go of something.
Thank God. Explains a lot of things. Now get to the flushing part,
which sounds just like your name.

Andy
 
K

Klaus M.

Jan 1, 1970
0
OK Andrew:
The full truth . . . you are an idiot. And sadly, without the ears to
hear
it or the humility to acknowledge it, you will charge forth continuing
to
prove it.

Makras
you seem to be the bigger idiot. You post a link that is totally
unrelated to what he is saying. He answers and kindly gives you an
explanantion.
And you call him an idiot.
Just piss off Makras troll.


Klaus
 
J

Jamie

Jan 1, 1970
0
Andrew said:
On a sunny day (Tue, 13 Feb 2007 18:49:41 +0100) it happened Andrew Edge
<[email protected]>:


The only way way you can get a current out of a capacitor is by using
a resistor.

No way, 'current' equals electrons.
There are many ways other then 'resistors'.
Look up vacuum tube, transistor, semiconductor, and in this context
especially 'current source'.

To make it a bit more clear, in a _resistor_ the current decreases linear
with the increase in resistance.

In something that allows control of the current, say a transistor, or
tube,

the current does _not_ rise proportionally to the applied voltage.

In a (junction) transistor the Ic becomes greatly independent above some
minimum

voltage, and then only depends on Ib.
Same for a 'penthode' tube.
Many integrated circuits use current sources and current mirrors, and as
such

Isupply may depend little on Usupply.

This is one for [sci].electronics.basics.

My favorite is from Physics 101.

Put a 365 pF variable capacitor (set to maximum C) in parallel with a 365 pF
fixed capacitor.
Charge the combination to 100 VDC.
Adjust the variable capacitor to 10 pF.

1. What is the resultant voltage across the combination?
2. What is the stored energy before and after?
3. How much work did it take to move the capacitor plates?

Charge is moved, no resistance involved.


You forgot. Capacitors have internal resistance. Wires connecting
resistance. Quite a lot of resistance is involved.

Andy
I think he was referring to an ideal capacitor.
that usually works in the ideal world how ever,
and good designer has experience and knows where
the ESR and leakage problems lye and there for, learn
over time how to calculate that.
I find that while doing courses, ideal factors are normally
what they are looking for. Q then comes in later on.
 
E

ehsjr

Jan 1, 1970
0
Andrew said:
Andrew Edge wrote:

Hi Tom.



Lessee...the OP stated that the current is constant, not i=v/R.

For your information a Capacitor is an open circuit to a constant
current. So a capacitor generating constant current is impossible.

So you are saying this circuit won't work,
within any time limits, whatsoever:

/
+12 ----o o-----+--->|---+
| |
[R] |
| |
+--->|---+ I -> ~ 10 mA
| | -----
[1F] +---In|LM317|Out---+
| ----- |
| Adj [120R]
| | |
| +---------+
| |
| [R]
| |
Gnd -----------+---------------------------+

Check on Kirchoff's current law on the capacitor leads.
I hoped someone would have told him that.

The op is discharging a cap through an active load
which he says draws constant current. Such a circuit
is shown above.

After the switch has been closed for sufficient time
to fully charge the capacitor, it is opened. Will
the discharge be constant current, or does the circuit
suddenly lose the ability to regulate?

Ed



<snip>

Hi Ed.
The Op said he wass discharging through a Microcontroller lead, Not
through a Voltage regulator chip.

Andy

Hi Andy,

It doesn't matter. The concept you offered
was that "a capacitor is an open circuit to
a constant current. So a capacitor generating
constant current is impossible."

Is that really what you mean to say?

The diagram I drew shows a capacitor feeding
a constant current regulator. Per your theory,
this could not work - but it does.

Ed
 
T

Tom Bruhns

Jan 1, 1970
0
Andrew said:
Andrew Edge wrote:
<snip>
Hi Tom.
Lessee...the OP stated that the current is constant, not i=v/R.
For your information a Capacitor is an open circuit to a constant
current. So a capacitor generating constant current is impossible.
So you are saying this circuit won't work,
within any time limits, whatsoever:
/
+12 ----o o-----+--->|---+
| |
[R] |
| |
+--->|---+ I -> ~ 10 mA
| | -----
[1F] +---In|LM317|Out---+
| ----- |
| Adj [120R]
| | |
| +---------+
| |
| [R]
| |
Gnd -----------+---------------------------+
Check on Kirchoff's current law on the capacitor leads.
I hoped someone would have told him that.
The op is discharging a cap through an active load
which he says draws constant current. Such a circuit
is shown above.
After the switch has been closed for sufficient time
to fully charge the capacitor, it is opened. Will
the discharge be constant current, or does the circuit
suddenly lose the ability to regulate?
Ed
<snip>
Hi Ed.
The Op said he wass discharging through a Microcontroller lead, Not
through a Voltage regulator chip.

Hi Andy,

It doesn't matter. The concept you offered
was that "a capacitor is an open circuit to
a constant current. So a capacitor generating
constant current is impossible."

Is that really what you mean to say?

The diagram I drew shows a capacitor feeding
a constant current regulator. Per your theory,
this could not work - but it does.

Ed

Looks to me, Ed, like Andy doesn't quite get the concept of the
difference between Mr. LaPlace and Mr. Fourier.

Cheers,
Tom
 
M

Michael A. Terrell

Jan 1, 1970
0
ehsjr said:
I'm glad to see the correct technical term! :)


Always happy to be able to help! ;-)


--
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
 
A

Andrew Edge

Jan 1, 1970
0
Andrew said:
Andrew Edge wrote:

<snip>


Hi Tom.



Lessee...the OP stated that the current is constant, not i=v/R.

For your information a Capacitor is an open circuit to a constant
current. So a capacitor generating constant current is impossible.

So you are saying this circuit won't work,
within any time limits, whatsoever:

/
+12 ----o o-----+--->|---+
| |
[R] |
| |
+--->|---+ I -> ~ 10 mA
| | -----
[1F] +---In|LM317|Out---+
| ----- |
| Adj [120R]
| | |
| +---------+
| |
| [R]
| |
Gnd -----------+---------------------------+


Check on Kirchoff's current law on the capacitor leads.
I hoped someone would have told him that.

The op is discharging a cap through an active load
which he says draws constant current. Such a circuit
is shown above.

After the switch has been closed for sufficient time
to fully charge the capacitor, it is opened. Will
the discharge be constant current, or does the circuit
suddenly lose the ability to regulate?

Ed



<snip>

Hi Ed.
The Op said he wass discharging through a Microcontroller lead, Not
through a Voltage regulator chip.

Andy

Hi Andy,

It doesn't matter. The concept you offered
was that "a capacitor is an open circuit to
a constant current. So a capacitor generating
constant current is impossible."

Is that really what you mean to say?

At least not in the conditions given by the Op is that possible.
The diagram I drew shows a capacitor feeding
a constant current regulator.
The capacitor is not feeding a constant regulator circuit. Its use
there is totally different . You should know better as you suggested
this.

The LM317 works even as a current regulator even without the capacitor
that proves its presence is not critical. I was under the impression
(and possibly you too) the Op was asking about Capacitors generating
constant currents.

Per your theory,
this could not work - but it does.
I suggested a circuit earlier on where it could have been possible.

Andy
 
A

Andrew Edge

Jan 1, 1970
0
Oh, you mean he lives in the US? :)

-:)

Interesting though (that from the Ip addresses) of mkaras and ehjsr
who share the same thoughts on this subject both come from Reston in
Virginia. What does that suggest?

Andy
 
E

Eeyore

Jan 1, 1970
0
Andrew said:
I was under the impression (and possibly you too) the Op was asking about
Capacitors generating constant currents.

Why don't you read what he actually wrote, you utter pea-brain ?

Graham
 
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