Dilon Earl wrote:


Where does the loss occur? If you have 3 db of mismatch loss, is it
in the coax, tank circuit?


The loss in "mismatch loss" refers only to the
fact that the power delivered by the generator to
the load is less than it would be if the load
resistance were the same value as the generator
resistance, in other words if the load and
generator were "matched".

The best way to get a handle on this subject is to
draw a diagram of a generator with voltage V=10,
an internal resistance of 50 ohms, and a load
resistor of R ohms. Let R vary from 1 ohm to 100
ohms and calculate the power dissipated in the
generator resistance (50 ohms), the power in the
load resistance (R), and the total power. Plot a
graph of the three quantities. The load power goes
through a maximum when R=50 ohms.

The maximum power dissipated in the generator
resistance is 10^2/50=2 W, which occurs when R=0
ohms. The minimum power dissipated in the
generator resistance is 3.33^2/50=0.22 W which
occurs when R=100 ohms. When R=50 ohms, the load
power is 5^2/50=0.5 W (the maximum value), the
dissipation in the generator resistance is
5^2/50=0.5 W and the total power is 10^2/100=1 W.

Bill W0IYH

On Thu, 17 Jul 2003 09:06:32 -0500, "William E. Sabin"
sabinw@mwci-news wrote:

Dilon Earl wrote:


Where does the loss occur? If you have 3 db of mismatch loss, is it
in the coax, tank circuit?


The loss in "mismatch loss" refers only to the
fact that the power delivered by the generator to
the load is less than it would be if the load
resistance were the same value as the generator
resistance, in other words if the load and
generator were "matched".

The best way to get a handle on this subject is to
draw a diagram of a generator with voltage V=10,
an internal resistance of 50 ohms, and a load
resistor of R ohms. Let R vary from 1 ohm to 100
ohms and calculate the power dissipated in the
generator resistance (50 ohms), the power in the
load resistance (R), and the total power. Plot a
graph of the three quantities. The load power goes
through a maximum when R=50 ohms.

The maximum power dissipated in the generator
resistance is 10^2/50=2 W, which occurs when R=0
ohms. The minimum power dissipated in the
generator resistance is 3.33^2/50=0.22 W which
occurs when R=100 ohms. When R=50 ohms, the load
power is 5^2/50=0.5 W (the maximum value), the
dissipation in the generator resistance is
5^2/50=0.5 W and the total power is 10^2/100=1 W.

Bill W0IYH

Bill;
Thanks, that all makes sense. Can you consider a Transmitter to
have an internal resistance like the generator that changes with the
plate and tune controls?
If I have a 100 watt transmitter and my wattmeter shows 3 watts
reflected. Is 3 watts actually being dissipated in the tank and final
PA?
Sorry to ask such simple questions. I did search through Google
on posts on this subject, just never could find the answer I was
looking for.

On Thu, 17 Jul 2003 17:00:55 GMT, Dilon Earl
wrote:
If I have a 100 watt transmitter and my wattmeter shows 3 watts
reflected. Is 3 watts actually being dissipated in the tank and final
PA?

Hi Dilon,

Does it become 3 watts hotter under the same drive conditions without
the reflected power? You would be surprised how few pundits actually
discuss this in these terms. Of course everyone would be surprised if
anyone attempted to perform this chore.

I like to include this jab at those who rave on about the
impossibility of knowing the internal resistance of a transmitter and
are satisfied to squeak out 100W RF for 250W DC in.

73's
Richard Clark, KB7QHC

On Thu, 17 Jul 2003 17:12:33 GMT, Richard Clark
wrote:

On Thu, 17 Jul 2003 17:00:55 GMT, Dilon Earl
wrote:
If I have a 100 watt transmitter and my wattmeter shows 3 watts
reflected. Is 3 watts actually being dissipated in the tank and final
PA?

Hi Dilon,

Does it become 3 watts hotter under the same drive conditions without
the reflected power? You would be surprised how few pundits actually
discuss this in these terms. Of course everyone would be surprised if
anyone attempted to perform this chore.

I like to include this jab at those who rave on about the
impossibility of knowing the internal resistance of a transmitter and
are satisfied to squeak out 100W RF for 250W DC in.

73's
Richard Clark, KB7QHC

Richard;
I'm not sure if it does get 3 watts hotter. I was always under
the impression that operating a transmitter with a high reflected
power was unhealthy for my PA.

On Thu, 17 Jul 2003 22:56:41 GMT, Dilon Earl
wrote:

Richard;
I'm not sure if it does get 3 watts hotter. I was always under
the impression that operating a transmitter with a high reflected
power was unhealthy for my PA.


Hi Dilon,

Some would suggest not, but then they wouldn't warrant their own
advice.

73's
Richard Clark, KB7QHC

I like to include this jab at those who rave on about the
impossibility of knowing the internal resistance of a transmitter and
are satisfied to squeak out 100W RF for 250W DC in.

73's
Richard Clark, KB7QHC

---------------------------------------------------

Rich, I'm reminded of Laurel and Hardy. Here's another fine mess you've
got yourself into.
---
From your favourite Italian Clown.

Dilon Earl wrote:

Bill;
Thanks, that all makes sense. Can you consider a Transmitter to
have an internal resistance like the generator that changes with the
plate and tune controls?
If I have a 100 watt transmitter and my wattmeter shows 3 watts
reflected. Is 3 watts actually being dissipated in the tank and final
PA?


No.

If the transmitter output is 100 W and the
reflected power is 3 W, then the 100 W is the
difference between 100+3=103 W (forward power) and
3 W (reflected power).

The question "where does the reflected power go?"
never seems to have an acceptable answer. Very
strange.

A good way to look at is as follows: The junction
of the transmitter output jack and the coax to the
antenna is a "node", which is just a "point" or
"location" where the jack and the coax meet. At
this node the voltage is exactly equal to the
voltage output of the amplifier (VPA) and also the
voltage across the input of the coax (VCOAX). The
voltage VCOAX) across the coax is equal to the
phasor sum of a forward voltage wave that travels
toward the antenna and a reverse voltage wave that
is traveling from the antenna backward toward the
transmitter.

Also, at the node, IPA is the current from the PA
and ICOAX is the phasor sum of a current wave that
travels to the antenna and a return current wave
that travels toward the transmitter.

At the node, the IPA current and the ICOAX current
are exactly equal and in the same direction
(toward the antenna). At the node the IPA current
is equal to the ICOAX coax forward current minus
the ICOAX reflected current. In other words there
is an *EQUILIBRIUM* at the node between VPA
voltage and VCOAX voltage, and an *EQUILIBRIUM*
between IPA current and ICOAX (forward and
reflected) current.

This explanation accounts for everything that is
going on at the node. The answer to the question
"where does the reflected power go?" is the
following: "It is a nonsense question that has
caused nothing but misery". The reflected power
does not actually *GO* anywhere. The correct
answer is that forward and reflected coax waves
always combine precisely and exactly with the
voltage and current that is delivered by the PA.
The voltage and current at the junction are
correctly accounted for. The basic principles here
are Kirchhoff's voltage law and Kirchhoff's
current law, as applied to the node. You can study
Kirchhoff's laws in the textbooks.

If we apply these laws and calculate the 100 W
power out of the PA and the 100 W power that is
dumped into the coax, they are exactly equal. They
cannot possibly be unequal. The power delivered is
the real part of the product of VPA and IPA (100
W), which is identical to the real part of the
product of VCOAX and ICOAX (100 W).

Observe carefully the following: We do not need to
know anything about the PA and its circuitry. The
PA is nothing more than an anonymous "black box".
In other words, any 100 W (output) PA will perform
exactly as I have described.

Bill W0IYH

William E. Sabin wrote:
If the transmitter output is 100 W and the reflected power is 3 W, then
the 100 W is the difference between 100+3=103 W (forward power) and 3 W
(reflected power).

If the source is a signal generator equipped with a circulator and
load, the generator is putting out 103 watts, and the circulator
load is dissipating 3 watts, is the generator still only putting
out 100 watts by definition?
--
73, Cecil, W5DXP

On Thu, 17 Jul 2003 14:42:37 -0700, W5DXP
wrote:

William E. Sabin wrote:
If the transmitter output is 100 W and the reflected power is 3 W, then
the 100 W is the difference between 100+3=103 W (forward power) and 3 W
(reflected power).

If the source is a signal generator equipped with a circulator and
load, the generator is putting out 103 watts, and the circulator
load is dissipating 3 watts, is the generator still only putting
out 100 watts by definition?

No, you just said it was putting 103 watts.. :-)

Was that the right answer?

W5DXP wrote:

William E. Sabin wrote:
If the transmitter output is 100 W and the reflected power is 3 W, then
the 100 W is the difference between 100+3=103 W (forward power) and 3 W
(reflected power).

If the source is a signal generator equipped with a circulator and
load, the generator is putting out 103 watts, and the circulator
load is dissipating 3 watts, is the generator still only putting
out 100 watts by definition?

If the sig gen is putting out 100 watts, with 3 watts reflected and 97
watts going to the load, then 3 watts must be going to the circulator.
If 100 watts is going to the load and 3 watts is reflected back to the
circulator, then the sig gen is putting out 103 watts. But if 97 watts
is getting to the load, 3 watts is reflected and there is no circulator
or other load, then how much do you think the sig gen is actually
putting out?

73, ac6xg