What is the reason a 2:1 SWR can cause such havoc?

How can I avoid this catastrophic condition?

I feed my dipoles with 450 Ohm ladder line, but the last 20 feet or so is 50
Ohm coax, I guess that makes it work ok. I haven't blown up my finals yet.

Lions and tigers and bears Oh my...

"Steveo" wrote in message
news:nceoaqqpc0a3yzz.280620052102@kirk...
if you have over a 2:1 standing wave you can do damage to your finals
or linear

On Wed, 29 Jun 2005 11:07:17 -0400, "Fred W4JLE"
wrote:

What is the reason a 2:1 SWR can cause such havoc?

How can I avoid this catastrophic condition?

I feed my dipoles with 450 Ohm ladder line, but the last 20 feet or so is 50
Ohm coax, I guess that makes it work ok. I haven't blown up my finals yet.

Lions and tigers and bears Oh my...
*****

Actually can happen if you push the finals to where there is
insufficeint margin to the maximum heat dissapation. Tubes are a bit
more forgiving. Transistor inadequately heatsinked and overdriven,
typical CB usage, often have little of no margin for heat dissapation.

If the transmitter has a refelction coefficient of zero and the load
say .3, then that reflected power from the load is dissapated as heat
in the output circuits and any final transistors or tubes. Now if the
radio has a reflection coefficient other than zero that will lessen
the heat dissapation on the transimiiter. Now you get load and source
reflections convoluting within the transmission line.

You ought to model a 400 Mhz square wave with source and load
refelctions coefficients other than zero. It can get ugly


james

james wrote:
If the transmitter has a refelction coefficient of zero and the load
say .3, then that reflected power from the load is dissapated as heat
in the output circuits and any final transistors or tubes.

Sometimes yes, sometimes no. If the reflected current arrives out
of phase with the forward current, then the final dissipation can
actually be *reduced* by the mismatch.
--
73, Cecil http://www.qsl.net/w5dxp


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On Wed, 29 Jun 2005 16:42:49 -0500, Cecil Moore
wrote:

Sometimes yes, sometimes no. If the reflected current arrives out
of phase with the forward current, then the final dissipation can
actually be *reduced* by the mismatch.
*****

Power is power. Phase is not a problem. Take the mafnitude of the
transmitted power and teh magnitude of the reflected power. The
results are phaseless. The magnitudes add linearly.

QED

james

james wrote:

On Wed, 29 Jun 2005 16:42:49 -0500, Cecil Moore
wrote:


Sometimes yes, sometimes no. If the reflected current arrives out
of phase with the forward current, then the final dissipation can
actually be *reduced* by the mismatch.

*****

Power is power. Phase is not a problem. Take the mafnitude of the
transmitted power and teh magnitude of the reflected power. The
results are phaseless. The magnitudes add linearly.

QED

james


Cecil was talking about current, not power. You can't add
power the way you can voltage and current. If you could, you
could build a very nice perpetual motion machine just by using the
reflections in a transmission line to add power so that the output
was greater than the input.
73,
Tom Donaly, KA6RUH

On Wed, 29 Jun 2005 22:58:51 GMT, "Tom Donaly"
wrote:

james wrote:

On Wed, 29 Jun 2005 16:42:49 -0500, Cecil Moore
wrote:


Sometimes yes, sometimes no. If the reflected current arrives out
of phase with the forward current, then the final dissipation can
actually be *reduced* by the mismatch.

*****

Power is power. Phase is not a problem. Take the mafnitude of the
transmitted power and teh magnitude of the reflected power. The
results are phaseless. The magnitudes add linearly.

QED

james


Cecil was talking about current, not power. You can't add
power the way you can voltage and current. If you could, you
could build a very nice perpetual motion machine just by using the
reflections in a transmission line to add power so that the output
was greater than the input.
73,
Tom Donaly, KA6RUH
******

Tom

The problem is that current is not reflected back from the load, power
is. Thus the you can add magnitudes of power.


james

Tom Donaly wrote:
Cecil was talking about current, not power. You can't add
power the way you can voltage and current.

That's absolutely correct. When one adds powers, one must
include the interference term which takes care of conservation
of energy. The equation is:

Ptot = P1 + P2 + SQRT(P1*P2)cos(theta)

where theta is the phase angle between V1 (associated with
P1) and V2 (associated with P2). The last term is labeled
the "interference term" and is absolutely necessary when
adding powers. If the interference term is positive, the
interference is constructive. If the interference term is
negative, the interference is destructive.

The best reference on interference and the adding of powers
that I have found is Chapter 9 in "Optics", by Hecht.
--
73, Cecil http://www.qsl.net/w5dxp

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The rig has no way of detecting any alleged "reflected power". It can't
tell the difference between a feedline with a lot of "reflected power",
a feedline with no "reflected power", and a plain resistor. It behaves
exactly the same in all cases, provided only that the impedance that
each provides to it is the same.

Anyone not convinced of this should put a couple or more dummy loads in
series or parallel, make up a few lengths of transmission line of
various impedances, and see for himself.

Roy Lewallen, W7EL

james wrote:
On Wed, 29 Jun 2005 16:42:49 -0500, Cecil Moore
wrote:


Sometimes yes, sometimes no. If the reflected current arrives out
of phase with the forward current, then the final dissipation can
actually be *reduced* by the mismatch.

*****

Power is power. Phase is not a problem. Take the mafnitude of the
transmitted power and teh magnitude of the reflected power. The
results are phaseless. The magnitudes add linearly.

QED

james

On Wed, 29 Jun 2005 16:29:26 -0700, Roy Lewallen
wrote:

The rig has no way of detecting any alleged "reflected power". It can't
tell the difference between a feedline with a lot of "reflected power",
a feedline with no "reflected power", and a plain resistor. It behaves
exactly the same in all cases, provided only that the impedance that
each provides to it is the same.

*****

Agreed that a rig cannot detect the difference between forward and
reflected power. If the reflection coeffiecient of the source is zero
then final stage of a transmiter will look purely resistive to any
power reflected by the load. Thereby that refelcted power is
dissapated as heat. Other reflection coefficients at the source will
yield lesser amounts of reflected power from the load as heat.

james

Anyone not convinced of this should put a couple or more dummy loads in
series or parallel, make up a few lengths of transmission line of
various impedances, and see for himself.

Roy Lewallen, W7EL

james wrote:
On Wed, 29 Jun 2005 16:42:49 -0500, Cecil Moore
wrote:


Sometimes yes, sometimes no. If the reflected current arrives out
of phase with the forward current, then the final dissipation can
actually be *reduced* by the mismatch.

*****

Power is power. Phase is not a problem. Take the mafnitude of the
transmitted power and teh magnitude of the reflected power. The
results are phaseless. The magnitudes add linearly.

QED

james

james wrote:
Power is power. Phase is not a problem. Take the mafnitude of the
transmitted power and teh magnitude of the reflected power. The
results are phaseless. The magnitudes add linearly.

Strangely enough, the results are not phaseless. The equation
for reflected power at an impedance discontinuity is:

Pref = P3 + P4 - SQRT(P3*P4)cos(theta)

Where theta is the phase angle between V3 and V4, the
associated reflected interferring voltages.

Reference "Optics", by Hecht, Chapter 9 - Interference

The last term in the equation above is known as the "interference"
term. Unless you take the interference term into account, you
don't have a ghost of a chance of ascertaining where the power
goes.
--
73, Cecil http://www.qsl.net/w5dxp

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