"Henry Kolesnik" wrote in message
.. .
Richard
This is in response to your answer of last night. Before going to bed I
got
out the book REFLECTIONS II by Walt Maxwell W2DU. I'm typing verbatim
from
page 2-2 and 23-1 for those who don't have the book.page 2-2 "Contrary to
what many believe, it is not true that when a transmitter delivers power
into a line with reflections, a returning wave sees an internal generator
resistance as a dissipative load. Nor is the reflected wave converted to
heat and, while at the same time damaging the final amplifier....the
reflected power is entirely conserved...." from page 23-1 "One of the
most
serious misconceptions concerned reflected power reaching the tubes in the
RF amplifier of the transmitter. The prevalent, but erroneous thinking
was
that the reflected power enters the amplifier, causing tube overheating
and
destruction. However, I dispelled this misconception in the above
mentioned
publications, using wave-mechanics treatment, discussed here in greater
detail, by showing that when the pi-network tank is tuned to resonance, a
virtual short circuit to rearward traveling waves is created at the input
of
the network. Consequently, instead of the reflected power reaching the
tubes of the amplifier, it is totally re-reflected toward the load by the
virtual short circuit appearing only to waves at the network input".
I'm guessing it's a virtual short because the pi-network is resonant but
what happens if it is a bit off. Also what happens in a transistor final
with no pi?

73 Hank WD5JFR
Henry,
Here is an example of what you just said. Take a sine wave source, and
connect it to a 1/4 wave section of shorted transmission line through a
series resistor R. The reflected wave will reach this resistor 1/2 cycle
later, and will be in phase with the source. For a lossless transmission
line, there will be *0 Volts across the resistor*. There will be 0 current
through the resistor, and the reflected wave will be re reflected for all
values of R, including R=Z0, because the reflected wave will not "know" what
R is. You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.

Tam/WB2TT

Tam/WB2TT wrote:
You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.

Ever measure the forward and reflected currents halfway into a
shorted 1/4WL stub? How can currents be flowing unimpeded into
and out of an infinite impedance? Hint: it's the net current
that is zero at the input of a 1/4WL stub. The forward current
and reflected current are equal in magnitude and opposite in
phase. Their individual magnitudes can be very large.
--
73, Cecil http://www.qsl.net/w5dxp





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Cecil wrote,

Ever measure the forward and reflected currents halfway into a
shorted 1/4WL stub? How can currents be flowing unimpeded into
and out of an infinite impedance?


In the case of a lossless, 1/4WL stub, (no such thing, except on this
newsgroup) in the steady state you can disconnect the transmitter and
the currents will still be there. Current doesn't flow, Cecil, charge flows.
Current is just the rate of flow of charge, dQ/dt. In a 1/4WL stub, charge,
and the fields associated with it, are in a state of oscillation. Over time,
their average movement in space is zero. It's too bad you have to think of
current
as like the water in a big river that has to flow from one place to another
in order to exist. (That's wrong, too. Current in a river is the rate of flow
of
water past a point, not the water itself.) Sloppy semantics sure screw up
understanding.
73,
Tom Donaly, KA6RUH

Tdonaly wrote:
In the case of a lossless, 1/4WL stub, (no such thing, except on this
newsgroup) in the steady state you can disconnect the transmitter and
the currents will still be there.

Since that configuration doesn't exist in reality, only God
can cause what you are asserting. Why am I not surprised that
you need a supernatural being to prove your arguments?

Current doesn't flow, Cecil, charge flows.

:-) :-) Having to resort to trivial arguments is the sure sign
of a loser. :-) :-)
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

Tam/WB2TT wrote:
You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.

Ever measure the forward and reflected currents halfway into a
shorted 1/4WL stub? How can currents be flowing unimpeded into
and out of an infinite impedance?

The question is a little misleading because the direction of the flow of
current changes every half cycle and is transverse, or orthogonal to the
direction of wave propagation. In a transmission line, the current
flows through Z0, ostensibly, which is essentially the impedance from
one conductor to the other at every point along the transmission line.
Other than that, superposed forward and reflected waves behave just as
you described, naturally.

73, Jim AC6XG

"Jim Kelley" wrote in message
...
Cecil Moore wrote:

Tam/WB2TT wrote:
You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.

Ever measure the forward and reflected currents halfway into a
shorted 1/4WL stub? How can currents be flowing unimpeded into
and out of an infinite impedance?

The question is a little misleading because the direction of the flow of
current changes every half cycle and is transverse, or orthogonal to the
direction of wave propagation. In a transmission line, the current
flows through Z0, ostensibly, which is essentially the impedance from
one conductor to the other at every point along the transmission line.
Other than that, superposed forward and reflected waves behave just as
you described, naturally.

73, Jim AC6XG
There is no current in the steady state. The steady state voltage is
independent of source impedance, which affect only how long it takes to
reach that. I ran a simulation on this, and you can see that as the voltage
builds up, the current decreases

Tam

Tam/WB2TT wrote:

"Jim Kelley" wrote in message
...
Cecil Moore wrote:

Tam/WB2TT wrote:
You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.

Ever measure the forward and reflected currents halfway into a
shorted 1/4WL stub? How can currents be flowing unimpeded into
and out of an infinite impedance?

The question is a little misleading because the direction of the flow of
current changes every half cycle and is transverse, or orthogonal to the
direction of wave propagation. In a transmission line, the current
flows through Z0, ostensibly, which is essentially the impedance from
one conductor to the other at every point along the transmission line.
Other than that, superposed forward and reflected waves behave just as
you described, naturally.

73, Jim AC6XG
There is no current in the steady state. The steady state voltage is
independent of source impedance, which affect only how long it takes to
reach that. I ran a simulation on this, and you can see that as the voltage
builds up, the current decreases

Tam

Hi Tam,

The simulation would be fun to play with. What do you use?

73 de jk

SWCAD from Linear. I also built up a model of an ideal SWR meter.

Tam

Tam/WB2TT wrote:

SWCAD from Linear. I also built up a model of an ideal SWR meter.

Tam

I almost bought LLTC when they came out with that. Should have. Thanks
Tam.

jk

Tam/WB2TT wrote:
There is no current in the steady state.

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



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