Radio913 wrote:
How does a capacitor reflect more power than you feed it?

With an inductor in the circuit, the voltage on a capacitor can
be greater than the source voltage. Consider the following series
resonant circuit. What are the voltages on the cap and coil at
resonance?

100W source========1wl feedline===cap===50 ohm===coil
--
73, Cecil http://www.qsl.net/w5dxp



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On Mon, 22 Sep 2003 22:49:22 -0500, Cecil Moore
wrote:
So what?

Hi Cecil,

Exactly, you don't have a clue by how much do you?

73's
Richard Clark, KB7QHC

On Tue, 23 Sep 2003 07:24:12 +0100, "Ian White, G3SEK"
wrote:

Richard Clark wrote:
Hi Ian,

A Directional Coupler consists of two transmission lines.

You say that after cutting out all the examples that I gave of
directional couplers that don't.

There is NO material difference offered in your original to merit its
inclusion in the first place.

Transmission Lines are the media through which B/H waves migrate
inexorably fixed together. The premise (which you alone bring as a
clouded presumption) that the Bruene bridge somehow works with
independence from this is simply a convenience in discussing its
operation, a convention of discussion at best and not a reality.

The waves migrate along the *main* transmission line - and obviously
every directional coupler has to contain one of those. But many do not
contain any secondary transmission line, in any realistic physical sense
of that term.

Ian, you exhibit a whole lack of experience into the matter. There
are more such examples of Directional Couplers that fit the most
precise definition of transmission line than not. Those that do not
(your Bruene bridges) are simply lumped equivalents that still
maintain classic formalisms that observe all the strictures of wave
mechanics. That they can be described by simpler metaphors does not
diminish either their utility, nor their intellectual purity.

Your rejections of their application are preposterous examples of
pedagogical minutia.

73's
Richard Clark, KB7QHC

Cecil Moore wrote in message ...
Radio913 wrote:
How does a capacitor reflect more power than you feed it?

With an inductor in the circuit, the voltage on a capacitor can
be greater than the source voltage. Consider the following series
resonant circuit. What are the voltages on the cap and coil at
resonance?

100W source========1wl feedline===cap===50 ohm===coil


Is that a 50 ohm resistor? And where is the ground?

Could you re-draw this, Cecil?


I was hoping Keith would measure the end of the inductor with the cap
removed, so we could get an idea of the incident voltage wave.

Slick

Richard Clark wrote of:
preposterous examples of pedagogical minutia.

Eschew sesquipedalianism!

--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

Newton may not have been EXACTLY right but he had a damned near-enough
practical approximation.

Much better than a rough-and-ready Bird.

Radio913 wrote:

You never measured the incident voltage. And you refused to measure the
end of the inductor, with the capacitor removed (even with the 15 pF, it should
tell us something about Vi).

I have done my lab work and produced results consistent with
classic rho.

You have not accepted my results, possibly because they are
inconsistent with your world view.

It is now your turn to hit the lab. You will, barring error,
obtain results consistent with mine. You will be able to
measure any parameter you wish, even do other experiments,
and you will find the results are always consistent with
classic, not revised rho.

The benefit of going to the lab is all yours. You will
learn how it works. Alternatively, perhaps, you will demonstrate
that classic rho is all wrong and revised rho rules. In this
case, if YOU have done the lab work, YOU will get (and deserve)
all the glory of a major revision to transmission line theory.

If I went back to the lab you are unlikely to accept any new
results from me any more than you have accepted those to date.
Sometimes seeing is believing.

.......On the other hand, perhaps you can convince me.

Predict what the measurement will be and what it will mean.
Tell me how you did the prediction. And allow some error
bounds. Say we assume the probe is between 15 and 30 pf.

Then we'll see.

This may be true, but are you saying that a capacitor can reflect an
RMS voltage wave that is greater than the one that charges it?

Yes indeed. Resonant circuits achieve this with ease.

...Keith

Absolutely incorrect! If capacitance is defined as Coulombs/Volt, then
how are you getting more coulombs than you put in? Remember, i said Root Mean
Square voltage.

How does a capacitor reflect more power than you feed it?

It's almost time for me to cut out of this discussion, if you still don't
understand me.

I can only suggest that you go to the lab. Given your statements
above, there is a great opportunity for hands on learning here.

....Keith

Dr. Slick wrote:

Cecil Moore wrote in message ...

Radio913 wrote:

How does a capacitor reflect more power than you feed it?

With an inductor in the circuit, the voltage on a capacitor can
be greater than the source voltage. Consider the following series
resonant circuit. What are the voltages on the cap and coil at
resonance?

100W source========1wl feedline===cap===50 ohm===coil

Is that a 50 ohm resistor? And where is the ground?

Yes, a 50 ohm resistor. No ground, it is a balanced system. Note
the two parallel wires.

Could you re-draw this, Cecil?

Sigh ...

+------1WL feedline--(-j500)--+
| |
100W source (50+j0)
| |
+------1WL feedline--(+j500)--+

--
73, Cecil http://www.qsl.net/w5dxp



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Reg Edwards wrote:
Newton may not have been EXACTLY right but he had a damned near-enough
practical approximation.

What was his approximation for the orbit of Mercury? :-)
--
73, Cecil http://www.qsl.net/w5dxp



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wrote:

power = volts * amps
= (Vfwd + Vref) * (Ifwd + Iref)
= Vf*If + Vf*Ir + Vr*If + Vr*Ir

Seems you've lost a couple of terms in there.

This is why, in general (using my definition of general), superposition
does not hold for power. Those extra terms get in the way.

I suspect that Vf*Ir and Vr*If have absolutely no physical existence.
Light waves traveling in opposite directions have no effect on each
other. Why should RF EM waves traveling in opposite directions have
any effect on each other?
--
73, Cecil http://www.qsl.net/w5dxp



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