OK, my point is (and you can test this yourself), that (1) the concept of Rs
is invalid for a class C stage, because it's not a linear device. You can
fool yourself into thinking it is, but it's not. (2) the typical RF output
stage, sans output filter, which consists of a transistor fed by a choke or
transformer and which pulls the collector voltage nearly down to the emitter
acts like an RF voltage source with a fairly low series resistance. I've
_done_ this. I've _measured_ this. I've _burnt_ transistors.

You _can_ get much more power from your output stage than it's "normal"
output power, for that brief (and often shining) moment before the output
transistor lets the smoke out and reverts to the sand and basic
petrochemicals from which it was originally made. This is how you know that
you aren't running it at the Rs = Rl point -- rather, you're running it at
the point where the junction temperature of the transistor is kept to a
level that will let it stay a transistor for a satisfying amount of time.

This is why in my original reply I compared the output amp to a wall socket.
The equivalent Rs of your average wall socket is probably on the order 1/2
to 4 ohms (on the low end for 120V, high for 230). Since I live in the US
and have fairly good feed, I'll use the 1-ohm 120V case, and I'll match Rs
with Rl. Lessee. 120V, 1-ohm, that's 14400 watts -- good! But my circuit
is only is only good for 15 amps and I'm asking for 120?!? Dang! What
happened to the lights?

So you see, every time you plug anything into the wall, you're failing to
match Rs with Rl, and that's a _very_ good thing. It's for exactly the same
reason that you don't match Rs with Rl in your output stage.

"Paul Burridge" wrote in message
...
On Sun, 04 Jan 2004 15:44:04 +0000, Leon Heller
wrote:
-- snip --
My - admittedly little - understanding of the situation is that the
power transferred into a resistor (regardless of power level) will be
maximised when Rs=Rl,. whereupon half the power is dissipated in the
final device and half in the load. Tim's posts seems to refute this
basic truism. Clearly the statement that the output power should be
inversely proportional to the load resistance *must* be false.
--

"I expect history will be kind to me, since I intend to write it."
-
Winston Churchill

"Gary Morton" wrote in message
...

-- snip --

So perhaps the fact that the transmitter is connected to a 50 ohm load and
the
output transformer is 2:1 turns ratio means that the transistor is
"seeing"
200 ohms?

Exactly.

Nevertheless I am still interested in how to verify that any circuit which
I
design (copy!) does have a 50 ohm output impedance. There must be some way
to
verify this figure.

Please see my reply to Paul Burrage in the next branch over -- hopefully I'm
clarifying that for a power output stage you _don't_ want to see matched
impedances.

For a stage (such as an intermediate, class A stage) where you _do_ want to
verify the output impedance you can either find the resistance _and
reactance_ that maximizes the power output, or you can build an RF impedance
bridge (I use a noise bridge) and measure the stage output impedance
directly. Just don't let it break into oscillation on the output of that
poor impedance bridge!

--Gary

"Gary Morton" wrote in message
...

-- snip --

So perhaps the fact that the transmitter is connected to a 50 ohm load and
the
output transformer is 2:1 turns ratio means that the transistor is
"seeing"
200 ohms?

Exactly.

Nevertheless I am still interested in how to verify that any circuit which
I
design (copy!) does have a 50 ohm output impedance. There must be some way
to
verify this figure.

Please see my reply to Paul Burrage in the next branch over -- hopefully I'm
clarifying that for a power output stage you _don't_ want to see matched
impedances.

For a stage (such as an intermediate, class A stage) where you _do_ want to
verify the output impedance you can either find the resistance _and
reactance_ that maximizes the power output, or you can build an RF impedance
bridge (I use a noise bridge) and measure the stage output impedance
directly. Just don't let it break into oscillation on the output of that
poor impedance bridge!

--Gary

Here's an experament: Build a power output stage _only_. Don't put a
filter on it. You should have a transistor with the emitter grounded, a
choke to +12V, and some minimal network on the base to keep it biased. You
should be able to pick this out of your Tuna Tin II design. Capacitor
couple an appropriate resistor to the collector (200 ohms if you use your
Tuna Tin II).

Now feed it with enough RF so it's really into class C, and measure the RF
voltage on the resistor. It should be around 8V (supply * 0.35) give or
take a bit.

Now change the output resistor by a factor of two and measure the RF voltage
again. It should be about the same as before assuming that you've sized
your choke correctly (4x the design resistance). This shows that your
output stage is stiff (i.e. has a low equivalent Rs).

What this all means is that your average well-designed output stage does
_not_ have a matched Rs! Why? Because when Rs = Rl the power in Rl is
maximized _for that Rs_, but the power burnt into heat inside of Rs is also
maximized! Since you want the power in Rl to be much higher than the power
used to heat up your shack (unless it's really cold inside) this is a good
and desireable thing.

"Gary Morton" wrote in message
...
Over the Xmas break I constructed a simple 2 transistor QRP transmitter
using
the Tuna Tin II design.

Using a power meter and a 50 ohm dummy load it appears to show around
200mW, a
little less than the article suggested. Then again I had substituted the
final
stage 2N2222 transistor for a beefier (size wise) BFY51. I'm not too
sure
what exact difference this will make.

In the article it says that the 200 ohm output impedance of the final
stage is
transformed down to around 50 ohm (using a 2:1 turns ratio untuned
transformer wound on a toroid).

I have added a 7th order low pass filter to reduce the harmonics.

I am interested in verifying the output impedance.

If I understand the theory, when the output is terminated with a resistor
which matches the output impedance then the power transferred to the load
will
be maximised.

I set about loading the output will a series of resistors and measuring
the
peak to peak voltage across them in order to calculate the r.m.s. voltage
and
hence the power dissapation.

Resistors of value less than 10 ohm gave strange results.

load r.m.s. power (V*V/R)
voltage

10 1.272 0.162
15 1.767 0.208
27 3.005 0.334
33 3.253 0.321
39 3.676 0.346
47 4.066 0.352
56 4.384 0.343
68 4.666 0.320
100 5.303 0.281

The numbers work reasonably well and point towards 50 ohms-ish.

The power value looks too high - so I might have made a mis-calculation
somewhere!

Can anyone pass comments on the above?

I'm interested in where the 200 ohm figure comes from.

Any suggestions as to other methods of measuring the output impedance?

regards...

--Gary (M1GRY)

Here's an experament: Build a power output stage _only_. Don't put a
filter on it. You should have a transistor with the emitter grounded, a
choke to +12V, and some minimal network on the base to keep it biased. You
should be able to pick this out of your Tuna Tin II design. Capacitor
couple an appropriate resistor to the collector (200 ohms if you use your
Tuna Tin II).

Now feed it with enough RF so it's really into class C, and measure the RF
voltage on the resistor. It should be around 8V (supply * 0.35) give or
take a bit.

Now change the output resistor by a factor of two and measure the RF voltage
again. It should be about the same as before assuming that you've sized
your choke correctly (4x the design resistance). This shows that your
output stage is stiff (i.e. has a low equivalent Rs).

What this all means is that your average well-designed output stage does
_not_ have a matched Rs! Why? Because when Rs = Rl the power in Rl is
maximized _for that Rs_, but the power burnt into heat inside of Rs is also
maximized! Since you want the power in Rl to be much higher than the power
used to heat up your shack (unless it's really cold inside) this is a good
and desireable thing.

"Gary Morton" wrote in message
...
Over the Xmas break I constructed a simple 2 transistor QRP transmitter
using
the Tuna Tin II design.

Using a power meter and a 50 ohm dummy load it appears to show around
200mW, a
little less than the article suggested. Then again I had substituted the
final
stage 2N2222 transistor for a beefier (size wise) BFY51. I'm not too
sure
what exact difference this will make.

In the article it says that the 200 ohm output impedance of the final
stage is
transformed down to around 50 ohm (using a 2:1 turns ratio untuned
transformer wound on a toroid).

I have added a 7th order low pass filter to reduce the harmonics.

I am interested in verifying the output impedance.

If I understand the theory, when the output is terminated with a resistor
which matches the output impedance then the power transferred to the load
will
be maximised.

I set about loading the output will a series of resistors and measuring
the
peak to peak voltage across them in order to calculate the r.m.s. voltage
and
hence the power dissapation.

Resistors of value less than 10 ohm gave strange results.

load r.m.s. power (V*V/R)
voltage

10 1.272 0.162
15 1.767 0.208
27 3.005 0.334
33 3.253 0.321
39 3.676 0.346
47 4.066 0.352
56 4.384 0.343
68 4.666 0.320
100 5.303 0.281

The numbers work reasonably well and point towards 50 ohms-ish.

The power value looks too high - so I might have made a mis-calculation
somewhere!

Can anyone pass comments on the above?

I'm interested in where the 200 ohm figure comes from.

Any suggestions as to other methods of measuring the output impedance?

regards...

--Gary (M1GRY)

On Sun, 4 Jan 2004 16:06:53 -0800, "Tim Wescott"
wrote:

For a stage (such as an intermediate, class A stage) where you _do_ want to
verify the output impedance you can either find the resistance _and
reactance_ that maximizes the power output, or you can build an RF impedance
bridge (I use a noise bridge) and measure the stage output impedance
directly. Just don't let it break into oscillation on the output of that
poor impedance bridge!

Bull****. In any class of operation, maximum power will be transferred
when the impedances are matched. Period.
--

"I expect history will be kind to me, since I intend to write it."
- Winston Churchill

On Sun, 4 Jan 2004 16:06:53 -0800, "Tim Wescott"
wrote:

For a stage (such as an intermediate, class A stage) where you _do_ want to
verify the output impedance you can either find the resistance _and
reactance_ that maximizes the power output, or you can build an RF impedance
bridge (I use a noise bridge) and measure the stage output impedance
directly. Just don't let it break into oscillation on the output of that
poor impedance bridge!

Bull****. In any class of operation, maximum power will be transferred
when the impedances are matched. Period.
--

"I expect history will be kind to me, since I intend to write it."
- Winston Churchill

Bull****. In any class of operation, maximum power will be transferred
when the impedances are matched. Period.

Your arrogant ignorance overwhelms me!!

Listen to those who know more than you.
Read a few engineering texts.

I am not willing to explain this again to a fool.

John - K6QQ

Bull****. In any class of operation, maximum power will be transferred
when the impedances are matched. Period.

Your arrogant ignorance overwhelms me!!

Listen to those who know more than you.
Read a few engineering texts.

I am not willing to explain this again to a fool.

John - K6QQ

Wow. Paul...have you ever thought about trying caffeine free?


"Paul Burridge" wrote in message
...

Bull****. In any class of operation, maximum power will be transferred
when the impedances are matched. Period.