Tam/WB2TT wrote:

"Cecil Moore" wrote:
There is no *net* current in the steady state at the input of a
shorted 1/4WL stub. What do you think the current at the short
is? How did that large amount of current get there without flowing?
--
I think it is more instructive to turn on a DC voltage at t=0, rather than a
sine wave. In either case, the voltage builds up step wise, with smaller and
smaller steps until you don't see them.

DC doesn't tell us anything about a 1/4WL shorted stub which is a
network problem, not a circuit problem. The net current at the mouth
of the stub is close to zero while the voltage is at a maximum. At
the short at the other end of the stub, transmission line theory holds.
The voltage is close to zero while the current is at a maximum. Now
exactly how can maximum current be flowing through that short if
no current is flowing into and out of the stub. Hint: it can't!

The forward current and reflected current cancel at the mouth of
the shorted 1/4WL stub. However, they add in-phase 1/4WL away at
the short, maybe to many amps of RF current at the shorted end.
Since there is no physical impedance at the mouth of a stub, nothing
except superposition of forward and reflected waves happens there
and nothing except a virtual impedance exists there. All of the action
is at the shorted end of the stub where there exists 100% reflection.
--
73, Cecil http://www.qsl.net/w5dxp



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"Cecil Moore" wrote in message
...
The forward current and reflected current cancel at the mouth of
the shorted 1/4WL stub. However, they add in-phase 1/4WL away at
the short, maybe to many amps of RF current at the shorted end.
Since there is no physical impedance at the mouth of a stub, nothing
except superposition of forward and reflected waves happens there
and nothing except a virtual impedance exists there. All of the action
is at the shorted end of the stub where there exists 100% reflection.
--
73, Cecil http://www.qsl.net/w5dxp

I just did the simulation again, and you are right. For a 1V signal feeding
a 50 Ohm line through a 450 Ohm series resistor, there is 20 ma of
current flowing through the short. Don't understand why I didn't see it the
other time. There is 0 current through the 450 Ohm
resistor.

Tam/WB2TT

For Gawd sake, don't tell Cecil he's right. There'll be no stopping him!

Reg Edwards wrote:
For Gawd sake, don't tell Cecil he's right.

You wouldn't expect a Texan to be left, would you, Reg?



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Tam/WB2TT wrote:
"Cecil Moore" wrote:

The forward current and reflected current cancel at the mouth of
the shorted 1/4WL stub. However, they add in-phase 1/4WL away at
the short, maybe to many amps of RF current at the shorted end.
Since there is no physical impedance at the mouth of a stub, nothing
except superposition of forward and reflected waves happens there
and nothing except a virtual impedance exists there. All of the action
is at the shorted end of the stub where there exists 100% reflection.

I just did the simulation again, and you are right. For a 1V signal feeding
a 50 Ohm line through a 450 Ohm series resistor, there is 20 ma of
current flowing through the short. Don't understand why I didn't see it the
other time. There is 0 current through the 450 Ohm resistor.

Install a Bird wattmeter at the 450 ohm resistor and you will read
forward power and reflected power. The *net* current through the 450
ohm resistor is zero. The forward current is not zero. The reflected
current is not zero. Their magnitudes (10 ma?) are equal and their phases
are 180 degrees from each other so they cancel to net zero current.

If we take the forward current at the 450 ohm resistor as a reference of
10 ma at zero degrees, the reflected current will be 10 ma at 180 degrees.
Those two current signals add up (superpose) to zero at the resistor.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil, W5DXP wrote:
"The forward ciurrent and reflected current cancel at the mouth of the
shorted 1/4 WL stub. However, they add in-phase 1/4-WL away at the
short, maybe to many amps of RF current at the shorted end. Since----."

Very clear and concise. I am persuaded that once circulating current is
established, only the current required to supply losses is drawn at the
mouth of the shorted 1/4-WL stub.

Best regards, Richard Harrison, KB5WZI