Wise words. The true load-pulling technique advocated professionally uses a
known model of the output stage using S parameters. Yuk. Not for this soldier.
I'd rather sniff underwear in a geriatric clinic thank you. Whatever one did
in this direction, as an amateur, would doubtless be either wrong or so
inaccurate as to be not even wrong as Dr. Pauli was wont to say.

The amp is single-ended out of an MRF136, so I presume it's Class A. The amp's
designation is H-10 (I bought it surplus). It's rated at around 15W, 0.1 - 30
MHz. The circuit diagram shows no hint of current limiting circuitry.

If one is serious about proper design of a matching network - a network, I
might add, which attaches *directly* (near as dammit) to the Tx output - then
one is all at sea without a proper knowledge of source impedance. See my
comments in the other thread about this.

I tried to use a technique advised by K7ITM to measure source impedance (R +
jX), and it produces a negative value of R by calculation. This is the
technique, based on a load setting of (r + jx) ohms. And once again, there is
no cable, folks. And it wouldn't matter if I did have a few inches of it
either.

1. Determine X.
a) Set r ~= R based on a best guess (which would be R=50 ohms nominally in
many systems).
b) Monitor the voltage, current, or power in the load (r).
c) Adjust (x) to maximise the monitored value.
This setting corresponds to x = -X, and we have determined X.

2. Determine R
a) Leave x = -X set as in step 1,so now the circuit is pure resistive.
b) Monitor the voltage across the resistor.
c) Set r to R(nominal) plus and minus a small percentage, and measure the
monitored voltage at both values of r.
The voltage across r is given by
V/V0 = r / (R + r)
Solving the 2 simultaneous equations to eliminate V0 shows that R is
determined by the equation
R = r1*r2*(V2 - V1) / (V1*r2 - V2*r1)
-------------------------------------

My problems
-----------
1. My measurements of V1,V2 lead to the inescapable conclusion that the above
model fails, because the calculated value of R comes out negative.
Let us assume that we set (r2 r1) and we obtain (V2 V1), which is
predicted from the model, and is also the case for my measurements. Under
these conditions, a negative value of R can only be obtained from the
equation if V2/V1 r2/r1, which in my case is true.
I undertook a full (and rather exhaustive and tedious!) calculation of the
expression for R when the value of (x) was not set correctly to -X, thinking
that perhaps this was the cause of the discrepancy. It turns out in this case
that, if the calculated value of R is negative, it has nothing to do with the
setting of (x), and depends ONLY on the condition V2/V1 r2/r1. Since we know
that the actual value of R cannot be negative, this implies a failure of the
model.
How then can the model fail? Since we are maintaining frequency constant, any
collection of resistances and reactances, however complicated, can be modelled
as (R + jX), so it cannot be that. The only assumption left to question is the
constancy of V0, and this is what the failure must be.
This leaves me with more questions than answers, because the way forward is
now completely unclear.

2. I should also mention another, less serious, problem I had, and that is
with the determination of X. The value of (x) I determine from measurement
would be expected to be constant at a given frequency. It is in my case not
so. The derived value of X appears to depend on
a) the power level setting of my amplifier
b) the value of (r) I set in the circuit when determining X.
Quite probably this second problem relates to the first problem's
identification of the failure of the model, and can probably be subsumed under
that category.

Best,
Andrew



"Richard Clark" wrote in message
...
On Sun, 23 May 2004 16:39:19 GMT, "Lord Snooty" wrote:

Indeed so, because the whole idea of characterising my SWR meter is towards
the goal of measuring the output impedance of my RF amp!
I agree that it isn't 50 ohms - it's R + jX, and I want a bulletproof
procedure to find R and X. Got one?

Cheers,
Andrew
G3UHD

Hi Andrew,

You have asked an inescapable question that will lead to a deluge of
scribbling commemorating the best attempts of Houdini.

There are several many methods to determine exactly what you want to
know. The simplest and certainly the one that contains as much
information necessary is called "load pulling." To even mention this
time and bench proven method will result in hoots from those who would
be the last to offer you a fixed answer; however, we shall proceed.

This requires that you have access to known, but non-standard value
loads capable of sustaining the power you will perform your
measurement at. This is not a trivial requirement. It also requires
that you can in some way defeat your ALC which will attempt to offset
the pull of the non-standard load.

It is simplicity itself that only demands you consider the elements of
a Thevenin model and how to determine the model's source Z (or
likewise, the Norton model's source Z). You will need a means to
measure the voltage across the load, or the current through it. Even
here, proportionality is all that is required as long as the Load is
characterized and thus the tools can be rather spartan.

In the long run, this will mean you have to construct and verify your
own non-standard loads. Take care that through your verification you
confirm their value across all power applications (resistors are very
susceptible to drift with temperature). You should also take care to
insure that all paths and leads are as short as possible. Loading
directly at the terminals will save grief of complex compensation math
(and reduce introducing other errors). However, you can choose to
employ remote loads if you take care to characterize the lines through
which they are attached (this means you should be adept at the Smith
Chart).

There is more to be said, but this enough to offer you a significant
lead to find that, yes, the source exhibits nearly 50 Ohms (the common
Ham transmitter running at rated power will fall between 30 and 70
Ohms) - as specified and designed.

73's
Richard Clark, KB7QHC

On Sun, 23 May 2004 20:42:59 GMT, "Lord Snooty" wrote:


The amp is single-ended out of an MRF136, so I presume it's Class A. The amp's
designation is H-10 (I bought it surplus). It's rated at around 15W, 0.1 - 30
MHz. The circuit diagram shows no hint of current limiting circuitry.

If one is serious about proper design of a matching network - a network, I
might add, which attaches *directly* (near as dammit) to the Tx output - then
one is all at sea without a proper knowledge of source impedance. See my
comments in the other thread about this.

Hi Andrew.

Specification sheets respond to these issues quite well. The MRF136
is a 400MHz device, normally offering 15W max with about 16dB gain at
28Vdc (although rated higher) in a class A configuration (showing
about 60% efficiency).

As would be expected, it covers a lot of turf. In the HF, the output
Z runs easily near 50 Ohms in push-pull circuit configurations;
otherwise it is simpler to describe it in the teens to tens of Ohms
across any number of variables you do not disclose (like frequency,
the actual configuration, additional interface components). 1 - 10
MHz does not bode well towards the best implementation of an UHF
device.

Load pulling is vastly simpler than S-parameters (also specified in
the data sheets for this device, down to 1MHz), why you want to marry
the two is a mystery.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
Load pulling is vastly simpler than S-parameters (also specified in
the data sheets for this device, down to 1MHz), why you want to marry
the two is a mystery.

A laboratory bench setup is simpler than a calculator and a couple
of sheets of paper? Egads Richard, s-parameter analysis was invented
in order to make things "vastly simpler" - and it does!
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil wrote,

Richard Clark wrote:
Load pulling is vastly simpler than S-parameters (also specified in
the data sheets for this device, down to 1MHz), why you want to marry
the two is a mystery.

A laboratory bench setup is simpler than a calculator and a couple
of sheets of paper? Egads Richard, s-parameter analysis was invented
in order to make things "vastly simpler" - and it does!
--
73, Cecil http://www.qsl.net/w5dxp



S-parameters are supposed to work best on small-signal analysis, where
everything is fairly linear. In this case, the MRF 136 is a power fet and
Motorola
says you can use s-parameters as a first approximation. In the data sheet, they
provide a list
of common source scattering parameters from 2.0 Mhz up to 800 Mhz. I suppose
that's useful to someone familiar with s-parameter techniques.
73,
KA6RUH

Tdonaly wrote:
S-parameters are supposed to work best on small-signal analysis, where
everything is fairly linear.

Therefore, should work well on transmission-line analysis, where
everything is linear.
--
73, Cecil http://www.qsl.net/w5dxp



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On 23 May 2004 23:50:02 GMT, (Tdonaly) wrote:

S-parameters are supposed to work best on small-signal analysis, where
everything is fairly linear.

Hi All,

Tom's implication (S-parameters don't fit to power amps) is supported
by a simple comparison of another UHF transistor, 2N3948:
by S-param as measured
input R 9 Ohms 38 Ohms
input X 12nH 21pF
output R 199 Ohms 92 Ohms
output X 4.6pF 5pF

S-params may be useful for comparing devices, but are inadequate for
power application design - unless, of course, you can tolerate 400%
error and inverted reactances.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
S-params may be useful for comparing devices, but are inadequate for
power application design - unless, of course, you can tolerate 400%
error and inverted reactances.

The actual subject of the s-parameter analysis was a linear Z0-match
network point, not an amplifier, so your statements are irrelevant.
--
73, Cecil http://www.qsl.net/w5dxp



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On Sun, 23 May 2004 21:19:23 GMT, Richard Clark
wrote:

[snip

|in a class A configuration (showing
|about 60% efficiency).

Huh?

Wes wrote,

On Sun, 23 May 2004 21:19:23 GMT, Richard Clark
wrote:

[snip

|in a class A configuration (showing
|about 60% efficiency).

Huh?


The data sheet doesn't say it gets 60% efficiency in a class A
configuration. It just says, under one bullet, that "Efficiency = 60%."
Further down the column it says, "Ideally Suited For Class A
Operation."
73,
Tom Donaly, KA6RUH

On 24 May 2004 00:04:50 GMT, (Tdonaly) wrote:

Wes wrote,

On Sun, 23 May 2004 21:19:23 GMT, Richard Clark
wrote:

[snip

|in a class A configuration (showing
|about 60% efficiency).

Huh?


The data sheet doesn't say it gets 60% efficiency in a class A
configuration. It just says, under one bullet, that "Efficiency = 60%."
Further down the column it says, "Ideally Suited For Class A
Operation."
73,
Tom Donaly, KA6RUH


Hi Guys,

I trust both of you appreciate the difference between a vaguely
described system and trying to fit a spec sheet to it. Far easier to
proceed direct from one or the other, but making a fit between the two
is like stepping from one boat to another with a high chance of
standing on neither.

73's
Richard Clark, KB7QHC