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.

Paul, please look over my various posts on this issue and tell me where I
failed to point out that you can get more power from a bare RF output stage
by loading it differently.

I'm gonna go fertilize my garden!

"Paul Burridge" wrote in message
...
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

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.

Paul, please look over my various posts on this issue and tell me where I
failed to point out that you can get more power from a bare RF output stage
by loading it differently.

I'm gonna go fertilize my garden!

"Paul Burridge" wrote in message
...
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

"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

On Sat, 3 Jan 2004 16:50:52 -0800, "Tim Wescott"
wrote:

Normally you'd expect the final amplifier to be fairly "stiff" - the output
power should be inversely proportional to the resistor you put on it.

??? Shirley, some mistake.
--

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

Tim Wescott wrote:

snip

What this means is that in general your transmitter, like your wall socket,
can deliver more than it's rated power if you put the right load on it.
Normally you'd expect the final amplifier to be fairly "stiff" - the output
power should be inversely proportional to the resistor you put on it. Your
circuit doesn't act exactly like that, even if it doesn't follow the same
curve for a generator with a linear impedance. I suspect that with your
circuit the 7-pole filter is evening things out quite a bit. There's also a
good chance that the "strange results" you see with a load below 10 ohms are
the final amplifier going unstable and oscillating, or your output
transistor heating up and changing characteristics on you.

Yes there were bad things going on with very low value loads, the output was
collapsing and re-establishing. The srtange results may also be caused by the
loads with were high wattage resistors - no idea what inductance they might
have had.

So what it boils down to is that it is very important that your output stage
_sees_ the right impedance, but you shouldn't expect the output stage to
_deliver_ the same impedance that it needs to see. As long as you're
getting power out and your output transistor isn't letting all the smoke out
then you're fine.

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?

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.

--Gary

As I understand it the output impedance approximates to Vcc*Vcc/Po, where Po
is output power and Vcc is the supply voltage
It looks like your measurements point to 50 ohms as about right. How were
you measuring the voltage? - this might affect the measurement

Richard

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)