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)