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'[EE] Understanding filters and resonator circuits'
2012\06\05@135428 by V G

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Capacitors:

I'm trying to (fully) understand the principle of capacitor reactance and
high pass filters. I'm having trouble forming a mental picture of how a
capacitor in series with a signal blocks DC and has the lowest impedance
for the highest frequency signals on a voltage and charge-on-the-plate
level. I can understand how a capacitor tied to ground acts as a buffer and
smoothes out high frequencies by averaging the voltage. But I don't really
get how the reverse is true for a capacitor in series with a signal. I can
see how it blocks DC after it is fully charged.

1. How does a capacitor in series work to block a DC signal, but pass a
high frequency signal?
2. If you apply a voltage to a capacitor in series (keeping in mind that it
takes time for the voltage from the power source to rise), I understand
that the current before the capacitor is NON-zero. But what about the
current AFTER (on the other side) of the capacitor? Is it zero or non-zero
(considering that a capacitor is just two parallel plates)?


LC circuits:

1. Consider a parallel LC circuit where the bottom is tied to ground and
the top is tied to a mixed-frequency signal source. I understand that the
parallel LC circuit will shunt all frequencies to ground other than those
around its resonant frequency. After the signal source is removed, the
circuit will eventually lose its energy. How do I make an oscillator out of
such an LC circuit that self-starts (that is, I can simulate in LTSPICE
which doesn't take into account external noise coming into an oscillator
circuit, which is required for some oscillator designs)? I've looked at LC
oscillator circuit types like the Hartley oscillator, but I don't see how
this can self start in theory under ideal conditions

2012\06\05@141225 by John Gardner

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Think about transformers - They block DC too..

2012\06\05@175558 by jim

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In answer to your first question regarding how a capacitor blocks DC,
think of this.
A capacitor is basically a pair of METAL plates seperated by an
INSULATOR.
We know that the definition of an insulator is, in it's simplest form,
a NON CONDUCTOR.
So, if we have two conductors, seperated by a NON CONDUCTOR, it would
seem ratheer obvious  that current can't flow through the NON CONDUCTOR.  Therefore, if we
have no current flow,
we say that current flow is blocked bythe capacitor.  More precisely,
the current flow is  blocked by the insulator (dielectric) of the capacitor.

A capacitor stores energy in an electric field between the plates when
a voltage is applied.  Even in DC circuits, with a fully discharged capacitor, the current
that flows the instant  the voltage is applied is high, and is limited only by the series
resistance.  
An AC signal constantly changes polarity, therefore the plolarity of
the stored charge, in the  form of an electric field, changes too.  This allows current to flow at
the ame frequency as the
voltage source.  The higher the frequency, the faster this field
changes, and theefore, the effective
reactance goes down.


 Regards,
  Jim

> ---{Original Message removed}

2012\06\05@180546 by Thomas Sefranek

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On Jun 5, 2012, at 1:54 PM, V G wrote:

> Capacitors:
>
> 1. How does a capacitor in series work to block a DC signal, but pass a
> high frequency signal?

A capacitor is an OPEN circuit at DC.   That means NO DC current flow.
(As you say, "two (insulated) parallel plates.)
A capacitor is an AC resistance dependent on frequency.  So it can pass AC current.

> 2. If you apply a voltage to a capacitor in series (keeping in mind that it
> takes time for the voltage from the power source to rise), I understand
> that the current before the capacitor is NON-zero. But what about the
> current AFTER (on the other side) of the capacitor? Is it zero or non-zero
> (considering that a capacitor is just two parallel plates)?
>
 *
 |  __O    Thomas C. Sefranek  spam_OUTwa1rhpTakeThisOuTspamcomcast.net
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 (*)/ (*)  Bicycle mobile on 145.41 MHz.

ARRL Instructor, Technical Specialist, VE Contact.

http://www.harvardrepeater.org

2012\06\05@181826 by Harold Hallikainen

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>
>> 2. If you apply a voltage to a capacitor in series (keeping in mind that
>> it
>> takes time for the voltage from the power source to rise), I understand
>> that the current before the capacitor is NON-zero. But what about the
>> current AFTER (on the other side) of the capacitor? Is it zero or
>> non-zero
>> (considering that a capacitor is just two parallel plates)?
>>

I = C * dV/dT

where I is the current in amperes, C is the capacity in Farads, and dV/dT
is the rate of change of the voltage across the capacitor in volts per
second.

So, the current goes to zero when the rate of change of the voltage across
the capacitor goes to zero.

Harold


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2012\06\05@190831 by Brent Brown

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Others have posted good replies already, hope they've helped to connect the dots for you.

Once the underlying principals are learned I suspect each of us summarizes and simplifies our own knowledge with a mental picture or general analogy that seems to suit. For me it's this: capacitors and inductors are essentially opposites in function, a capacitor attempts to maintain a constant voltage accross it's terminals and does so by changing the current flowing through itself, an inductor attempts to maintain a constant current flow and does this by changing the voltage across itself.

{Quote hidden}

2012\06\05@192206 by Ruben Jönsson

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This is how I mentally picture a capacitor:

When current goes into one leg it goes out of the other. When it does the voltage over the dielectric also rises more and more until it prohibits any more current flowing (causing the capacitor charge/discharge curve). No electrons are actually moved from one side to the other though (electrostatic repelling forces of charges of equal polarity).
You can compare a capacitor in an electric circuit to a barrel with a flexible water tight membrane in the center (dividing the barrel in two parts) in a water flowing circuit. When water pushes in on one side, the membrane flexes at the same time as water is pushed out on the other side. The membrane flexes as much as the water pressure allows. To high pressure and the membrane breakes causing a short. It is also the membrane that holds the energy in the barrel, allowing it to push back water when the pressure is lowered. Also, the more flexed the membrane is, the more work is needed to flex it even more.

This model also makes it easier to understand why a high frequency can pass easier through the capacitor than a low frequency (where the extreme is DC)..

/Ruben

{Quote hidden}

> theory under ideal conditions

2012\06\05@225402 by RussellMc

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> I'm trying to (fully) understand the principle of capacitor reactance and
> high pass filters. I'm having trouble forming a mental picture of how a
> capacitor in series with a signal blocks DC and has the lowest impedance
> for the highest frequency signals on a voltage and charge-on-the-plate
> level.

Water model - Boxianly imperfect but useful.

Resistors are pipes. Diameter and length vary.

Current is current.

Voltage is pressure or head.

Capacitor is modelled by rubber sheet across a tank in the flow. Fluid
fills the tank on one side of the sheet and pressurises it. Stiffness
of sheet size of tank all have effects. This model handles many
capacitor features.

Inductor is a rubber pipe section. Diameter length and rubber
stiffness vary. Biggest departure is effect of removing voltage. A
water rotor with inertia may be able to provide a better model.


       Russel

2012\06\05@232103 by V G

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Answers from everyone are much appreciated. My understanding has increased
greatly.

I still have to look at it from a charge-on-plate level to complete my
understanding, but I get it now

2012\06\06@184116 by Oli Glaser

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On Tue, Jun 5, 2012 at 6:54 PM, V G <.....x.solarwind.xKILLspamspam@spam@gmail.com> wrote:

>
> 1. Consider a parallel LC circuit where the bottom is tied to ground and
> the top is tied to a mixed-frequency signal source. I understand that the
> parallel LC circuit will shunt all frequencies to ground other than those
> around its resonant frequency. After the signal source is removed, the
> circuit will eventually lose its energy. How do I make an oscillator out of
> such an LC circuit that self-starts (that is, I can simulate in LTSPICE
> which doesn't take into account external noise coming into an oscillator
> circuit, which is required for some oscillator designs)? I've looked at LC
> oscillator circuit types like the Hartley oscillator, but I don't see how
> this can self start in theory under ideal conditions.
>

Simulating oscillators in SPICE can be a little tricky, but if you are
aware of potential problems you can tweak stuff as necessary to simulate
the real world more accurately.
For instance with external noise (or lack of), you can simply add a small
"start up" voltage briefly to get things started, then set it to 0V so it
does not interfere with the circuit (e.g. set a source up for something
like 100 cycles/10mV at required frequency)
However this is rarely necessary (unless things are really slow getting
going such as in some crystal simulations) as selecting "start external DC
supply voltages a 0V" usually does the trick.
IIRC there are a few example oscillator circuits in <install
directory>/examples/educational. Pretty sure there is a Hartley example,
plus Colpitts, Wien Bridge, etc. I'd have a look at these (and their
settings) then try your own out.
Sometimes you need to do stuff like change the maximum timestep, skip
initial DC operating point condition, Integration method etc to make it
work, but I've found LTSpice generally works pretty well (compared to some
other SPICEs I've tried) with such circuits

2012\06\07@174445 by William \Chops\ Westfield
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How your calculus ?  And were you looking for a mathematical treatment, or an intuitive treatment?

> I = C * dV/dT

That's the key definition.

For a "signal", V(t) = sum(X * sin (Y*t), (for various X and Y))
Y is linearly related to the frequency.

dV/dt (sin(Y*t)) = Y*cos(Y*t)

So the magnitude is dependent on the frequency.  Linearly, even...

BillW

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