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)

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)

On Sat, 3 Jan 2004 21:18:33 +0800, "Richard Hosking"
wrote:

As I understand it the output impedance approximates to Vcc*Vcc/Po, where Po
is output power and Vcc is the supply voltage

I'd be interested to see the derivation of this.

It looks like your measurements point to 50 ohms as about right. How were
you measuring the voltage? - this might affect the measurement

I did the same thing (using different resistor values and then
calculating by V^2/R like the OP has done. It worked for me. The Zout
of my particular design was 140 ohms, though - just right for a folded
dipole. I used 2n3904 transistors; they appear to give a better gain
than 2n2222s.
--

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

Richard Hosking wrote:

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


Vcc * Vcc / Po = 13.8 * 13.8 / .352 = 540 ohms (for a 47 ohm load).

This value is certainly in the right ballpark.

I guess that I should also measure the current taken by the final stage and
see if this ties in too.

I connected various 1/4w resistors across the output using a couple of
adaptors, but being aware to keep wire lengths as short as possible. I then
simply used a scope probe and scope to make the measurement. Scope probe was a
standard x10 10Mohm impedance and presumably small C.

Interestingly I was surprised at the magnitude of the voltages for such a tiny
power.

regards..

--Gary

On Sat, 3 Jan 2004 21:18:33 +0800, "Richard Hosking"
wrote:

As I understand it the output impedance approximates to Vcc*Vcc/Po, where Po
is output power and Vcc is the supply voltage

I'd be interested to see the derivation of this.

It looks like your measurements point to 50 ohms as about right. How were
you measuring the voltage? - this might affect the measurement

I did the same thing (using different resistor values and then
calculating by V^2/R like the OP has done. It worked for me. The Zout
of my particular design was 140 ohms, though - just right for a folded
dipole. I used 2n3904 transistors; they appear to give a better gain
than 2n2222s.
--

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

Richard Hosking wrote:

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


Vcc * Vcc / Po = 13.8 * 13.8 / .352 = 540 ohms (for a 47 ohm load).

This value is certainly in the right ballpark.

I guess that I should also measure the current taken by the final stage and
see if this ties in too.

I connected various 1/4w resistors across the output using a couple of
adaptors, but being aware to keep wire lengths as short as possible. I then
simply used a scope probe and scope to make the measurement. Scope probe was a
standard x10 10Mohm impedance and presumably small C.

Interestingly I was surprised at the magnitude of the voltages for such a tiny
power.

regards..

--Gary

If you want to obtain the maximum power transfer between a generator and a
load, and if all the pieces behave like ideal linear components, then you
need to match the impedance of the two. But this is not what you are doing
here. First, you want to obtain the maximum power out without burning up
you output device, and second, a class-C output stage isn't a linear
circuit!

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.

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.

"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 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

Paul Burridge wrote:
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.

Both AF and HF power amplifiers generally have a very low output impedance.

Motorola's AN721 app note on impedance matching networks for RF
transistors explains this.

This doesn't apply to QRP output stages, though.

Leon
--
Leon Heller, G1HSM
Email:
My low-cost Philips LPC210x ARM development system:
http://www.geocities.com/leon_heller/lpc2104.html

On Sun, 04 Jan 2004 15:44:04 +0000, Leon Heller
wrote:



Paul Burridge wrote:
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.

Both AF and HF power amplifiers generally have a very low output impedance.

Motorola's AN721 app note on impedance matching networks for RF
transistors explains this.

This doesn't apply to QRP output stages, though.

Hi, Leon,

Do you know Brian (G4BCO) by any chance?
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