Richard Clark wrote:

Cecil Moore wrote:
At 3 GHz, one inch of wire is close to 1/4WL

At 3 GHz nobody uses wire.

The "why" of that statement proves my point.



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On Tue, 15 Jun 2004 21:10:54 -0500, Cecil Moore
wrote:
At 3 GHz, one inch of wire is close to 1/4WL
At 3 GHz nobody uses wire.
The "why" of that statement proves my point.
It is Tom's match point.

Richard Clark wrote:
Cecil Moore wrote:

At 3 GHz, one inch of wire is close to 1/4WL

At 3 GHz nobody uses wire.

The "why" of that statement proves my point.

It is Tom's match point.

Doubt it. EM waves travel the same speed at 1 MHz or 3 GHZ.
If you get a reflection at 3 GHz, you certainly get a
reflection at 1 MHz. All the steady-state shortcuts in
the world won't change that fact of physics. What's sad
is you don't even realize it.
--
73, Cecil http://www.qsl.net/w5dxp



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On Tue, 15 Jun 2004 23:05:23 -0500, Cecil Moore
wrote:
don't even realize it
;-)

Henry, WD5JFR wrote:
"Shorts can`t dissipate power, so how does a stub work?"

Henry also wrote:
"I know that a shorted 1/4 wave stub exhibits a very high impedance. But
for the 2nd harmonic it`s a 1/2 wave stub and exhibits a very low
impedance or a short."

Henry is correct.

Connect a resistance directly to a transmitter. How much energy is
absorbed by the resistance? Ohm`s law is the first approximation.
Current is directly proportional to the applied voltage if the
transmitter`s internal reistance is negligible.

The shorted 1/4-wave stub exhibits an open circuit at its mouth and
accepts only enough current to supply its losss which are none in the
perfect stub. So, it takes no power from the transmitter afer its
circulating current is etablished.
At 2X the 1/4-wave freqency, the stub is 1/2 wavelengh and does not
transform the short at one end to an open circuit at its other end.
Instead, the 1/2-wave directly presents the short circuit at its far
end.

How much curent can the transmitter supply to a short circuit? It
depends on the internal impedance of the transmitter.

"Transmission Lines, Antennas, and Wave Guides" by King, Mimno, and Wing
explains how "Even Harmonics" are suppressed by the 1/4-wave
short-circuited stub on page 29. The gist is that the stub is imperfect
and its resistance saps harmonic energy which is allowed into the stub
by its low impedance.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
The shorted 1/4-wave stub exhibits an open circuit at its mouth and
accepts only enough current to supply its losss which are none in the
perfect stub.

Richard, I think you would be surprised if you measured the RF current
through the short at the shorted end. It will be the in-phase sum of
the forward current and reflected current and is quite high. I once
melted the insulation at the end of a shorted 1/4WL piece of RG8X.
The heat came from high I^2*R losses at the short.

The SWR inside a perfect stub is infinite.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil, W5DXP wrote:
"Richard, I think you would be surprised if ypu measured RF current
through the short at the shorted end."

I expect high circulating current but without loss not much current is
required of the power source.

In his 1955 edition on page 106 Terman says:
"Thus, if the line is short-circuited at the load, then at frequencies
in the vicinity of a frequency for which the length is an odd number of
quarter wavelengths long, the impedance will be high and will vary with
frequency in the vicinity of resonance (i.e., frequency corresponding to
quarter wavelength) in exactly the same manner as does the impedance of
an ordinary parallel resonant circuit. It is therefore possible to
describe resonance on a transmission line in terms of impedance at
resonance and the equivalent Q of the resonance curve.

On page 107, Terman gives a 200 MHz example. 2-inch air-dielectric coax
is used for 1/4-wave short-circiuited stubs about 15 inches long. The
resonant impedance is more than 250,000 ohms with a Q of 3000.

How much current flows into an impedance of more than 1/4-million ohms?

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
How much current flows into an impedance of more than 1/4-million ohms?

As you know, transmission lines transform impedances. That stub likely
has an SWR around 5000:1. Assuming a driving voltage of 250V, the
current at the 1/4-million ohm point is about 0.001 amps. However,
1/4WL away, at the short in the 1/4WL stub, the current will be
~5 amps with a voltage of ~0.05V. That's an impedance of ~0.01 ohm. The
shorted stub has transformed the impedance from 1/4-million ohms at
the mouth to ~0.01 ohm at the short. That's what transmission lines do.

The question is not, "How much current flows into an impedance of more
than 1/4-million ohms?" The question is: How much current flows 1/4WL
away from that point at the shorted end of the stub? The answer is the
sum of the forward current and reflected current. The voltage at the
shorted end of a 1/4WL stub is the difference between the forward voltage
and the reflected voltage.

The voltage at the mouth of the stub is the sum of the forward voltage
and reflected voltage. The current at the mouth of the stub is the
difference between the forward current and reflected current. In the
example above, that difference is ~0.001 amp. The forward current flowing
inside the stub is ~2.505 amps and the reflected current flowing inside
the stub is ~2.495 amps.

If you don't believe the above, simply measure the RF current at the
shorted end of the stub. Someone modeled it the other day and even
the modeling program indicated that the current was sky high at
the shorted end of a 1/4WL stub.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil, W5DXP wrote:
"Someone modeled it the other day and even the modeling program
indicated that the current was sky high at the shorted end of a 1/4 WL
stub."

Agreed that the resistance of the short is small and allows a large
current even when the volts are low. The small load resistance, ZL is
transformed into another resistance, ZS that is inversely proportional
to ZL. Terman expresses this in equation (4-31) as:

ZS = Zo squared / ZL

So, the smaller ZL, the larger ZS. As ZL goes to zero, ZS goes to
infinity.

A high impedance means: accepts little current for a given voltage. The
open end of a good 1/4-wave short-circuited stub is defined as a high
impedance. So much so that it is also called a "metallic insulator".

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
A high impedance means: accepts little current for a given voltage. The
open end of a good 1/4-wave short-circuited stub is defined as a high
impedance. So much so that it is also called a "metallic insulator".

If it was indeed a high physical impedance, like a 250K resistor, it could
be removed and no much would happen at its resonant frequency. Unfortunately,
it is not a physical impedance and is merely a V/I ratio, a virtual impedance.
There is nothing at the mouth of the stub capable of causing reflections.
All the reflections occur at the shorted end of the stub.

The forward and reflected voltages add in phase and the forward and
reflected currents add 180 degrees out of phase at the mouth of the stub.
The ratio of forward voltage to forward current is the Z0 of the stub as
is the ratio of reflected voltage to reflected current.
--
73, Cecil http://www.qsl.net/w5dxp



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