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

"Richard Clark" wrote in message
...
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

There are uncharacterised components surrounding my MRF136, and I'd rather
determine their lumped behaviour with the transistor than unsolder them one by
one and try and predict the answer. But that's because I think it's possible
to determine what's going on. One reason I *don't* think this is the following
little bench test this afternoon - with the load jammed directly on the amp
output; no cable, no VSWR meter.
At 8.000 Mhz and a load consisting of 50 ohms carbon 10W 10% in series with a
capacitance trimmer bank, at an output power level of about 2W, a load
capacitor value of 250 +/- 10pF (-j80 ohms @ 8 MHz) was found to produce a
minimum in the total voltage across the load. Also, as capacitance was
increased over the range 100-700pF, the voltage across the load resistor
increased monotonically.

The latter is easy to explain (it means the source reactance is positive, and
smaller than +j28.4 ohms), but the former is beyond my ken.

Best,
Andrew

On Mon, 24 May 2004 05:46:01 GMT, "Lord Snooty" wrote:

At 8.000 Mhz and a load consisting of 50 ohms carbon 10W 10% in series with a
capacitance trimmer bank, at an output power level of about 2W, a load
capacitor value of 250 +/- 10pF (-j80 ohms @ 8 MHz) was found to produce a
minimum in the total voltage across the load. Also, as capacitance was
increased over the range 100-700pF, the voltage across the load resistor
increased monotonically.

The latter is easy to explain (it means the source reactance is positive, and
smaller than +j28.4 ohms), but the former is beyond my ken.

Best,
Andrew

Hi Andrew,

You got me confused too.

Is the "load" the resistor, or the resistor-cap combination when you
measure these voltages?

You describe a voltage minimum across the load for a cap setting of
250pF; but you also maintain that the voltage across the load
increases for the variation in capacitance from 100 to 700pF which
contradicts the first measurement.

Further, the construction of a high power semiconductor does not lend
itself to supporting inductive reactances (the junctions are quite
manifestly capacitive in structure). By specification the device is
characterized as exhibiting 27pF @ 1 MHz (for 28Vdc although there is
not much variation until below 10Vdc). There are a world of other
variables to consider, but none portray inductance within the device.
This alone should provoke you to re-examine your premise.

73's
Richard Clark, KB7QHC

"Richard Clark" wrote in message
...
On Mon, 24 May 2004 05:46:01 GMT, "Lord Snooty" wrote:

At 8.000 Mhz and a load consisting of 50 ohms carbon 10W 10% in series
with a
capacitance trimmer bank, at an output power level of about 2W, a load
capacitor value of 250 +/- 10pF (-j80 ohms @ 8 MHz) was found to produce a
minimum in the total voltage across the load. Also, as capacitance was
increased over the range 100-700pF, the voltage across the load resistor
increased monotonically.

The latter is easy to explain (it means the source reactance is positive,
and
smaller than +j28.4 ohms), but the former is beyond my ken.

Best,
Andrew

Hi Andrew,

You got me confused too.

Is the "load" the resistor, or the resistor-cap combination when you
measure these voltages?

You describe a voltage minimum across the load for a cap setting of
250pF; but you also maintain that the voltage across the load
increases for the variation in capacitance from 100 to 700pF which
contradicts the first measurement.

Further, the construction of a high power semiconductor does not lend
itself to supporting inductive reactances (the junctions are quite
manifestly capacitive in structure). By specification the device is
characterized as exhibiting 27pF @ 1 MHz (for 28Vdc although there is
not much variation until below 10Vdc). There are a world of other
variables to consider, but none portray inductance within the device.
This alone should provoke you to re-examine your premise.

73's
Richard Clark, KB7QHC

To clarify
a) "Load" in my context means "load resistor (r) and load capacitor (reactance
jx) in series"
b) The output transistor feeds to the output port through an inductor. One
would therefore expect X, the source reactance, to be positive.

Andrew

On Wed, 26 May 2004 02:07:50 GMT, "Lord Snooty" wrote:

To clarify
a) "Load" in my context means "load resistor (r) and load capacitor (reactance
jx) in series"
b) The output transistor feeds to the output port through an inductor. One
would therefore expect X, the source reactance, to be positive.

Hi Andrew,

a load capacitor value of 250 +/- 10pF (-j80 ohms @ 8 MHz) was found to produce a
minimum in the total voltage across the load.
What was the voltage?
Also, as capacitance was increased over the range 100-700pF, the voltage across the load resistor
increased monotonically.
What were the voltages?
The latter is easy to explain (it means the source reactance is positive, and
smaller than +j28.4 ohms), but the former is beyond my ken.
as the capacitive reactance falls, you note the voltage climbs, this
hardly requires an inductance to explain this. Simple divider action
serves quite well. You have since revealed the inductor buffered
output, but the data is still pretty skimpy to bless it as the major
contributor to source Z.

What are you using to measure this voltage?

73's
Richard Clark, KB7QHC

"Richard Clark" wrote in message
...
On Wed, 26 May 2004 02:07:50 GMT, "Lord Snooty" wrote:
Hi Andrew,

a load capacitor value of 250 +/- 10pF (-j80 ohms @ 8 MHz) was found to
produce a
minimum in the total voltage across the load.
What was the voltage?

Load voltage lies in the 5 to 20 volt peak range, depending on the power
setting.
This minimum represented about a 15% dip.

Also, as capacitance was increased over the range 100-700pF, the voltage
across the load resistor
increased monotonically.
What were the voltages?

Again, a nominal value between 5 and 20 V pk.

The latter is easy to explain (it means the source reactance is positive,
and
smaller than +j28.4 ohms), but the former is beyond my ken.

as the capacitive reactance falls, you note the voltage climbs, this
hardly requires an inductance to explain this. Simple divider action
serves quite well. You have since revealed the inductor buffered
output, but the data is still pretty skimpy to bless it as the major
contributor to source Z.

Agreed, but if I keep increasing the load C (decreasing the capacitative
reactance ), I will see a peak in the voltage across the load resistor, which
will only happen if a conjugate match is occurring.

What are you using to measure this voltage?

A scope probe set to 10x, which has an unmeasurably high DC resistance and a
capacitance of 22 pF (measured). It's a good idea to use the 10x, not 1x,
setting, since the latter contributes about 80 pF, and will change the AC
response of the circuit.

Best,
Andrew