Converting SSTRAN3000 to drive 50 ohm antenna

In an earlier post I stated that the output impedance of the SSTRAN3000 prior to the PI matching network (across C23) was approximately 820 ohms.

I did more experimentation and found this value to be incorrect.

The actual impedance value is closer to 2,600 ohms. This is due to the limitations of the current source circuit Q1/Q2/Q3 which attempts to maintain a 100 mW power input to the amplifier stage (Q4,Q5).

The constant current source can maintain this power level only at impedances larger than 2,600 ohms. If the impedance drops below this value, the drive current drops considerably.

Therefore I've calculated new values for C23, C5, and L7 in order to convert the output impedance to 50 ohms.

For operation around 1400 KHz

Change these parts:

Change C23 from 820pF to 560pF.

Replace C5 with 3900pF fixed capacitor, (polycarbonate 3900J).

Remove L4, L5, L6, L7 and replace with a single toroid coil, 22 uH, 40 turns wound on a T106-2 iron core.

Set all S5 switch position to OFF.

I am able to get about 50 mW output at J4/P4 (actual measured power output) when driving a 50 dummy load. This is about 50% efficiency; not great but not bad considering what I threw together. The Q of the new Pi network is not yet optimized so I suspect that there is significant loss. I should be able to get the efficiency close to 80%.

Even at 50 mW output, this gives one a good starting point for driving a proper base-loaded or center-loaded antenna. Given the poor-efficiency of the stock antenna configuration, I believe this 50 ohm configuation is still a significant overall boost.

The weather here has not been cooperating with me so I've not got the outdoor center-loaded antenna installed.

Stay tuned!

Mike

... Therefore I've calculated new values for C23, C5, and L7 in order to convert the output impedance to 50 ohms.  ...

Mike - Which condition are you referring to (other things equal)?

• The transmitter is designed/modified so that the source impedance looking back into the antenna output port of the transmitter is 50 ohms, or

• The transmitter is designed/modified to provide ~ maximum r-f output power into a 50 ohm, non-reactive load connected to its antenna output port.

Rich,

I'm designing for the transmitter to provide ~ maximum r-f output power into a 50 ohm, non-reactive load connected to its antenna output port.

Mike

The SSTRAN 3000 can be modified to drive a 50 ohm antenna (non-reactive impedance) by making the following component changes:

Freq (KHz) 1440
R18 (ohms) 2700
L8 (uH) 1000
C23 (pF) 220
L4 (uH) 0
L5 (uH) 0
L6 (uH) 22
L7 (uH) 22
C5 removed

Freq (KHz) 1640
R18 (ohms) 2700
L8 (uH) 1000
C23 (pF) 180
L4 (uH) 10
L5 (uH) 10
L6 (uH) 22
L7 (uH) 0
C5 removed

Modifications are based on calculated transmitter efficiency of 85% at 100 milliwatts input drive with a 3825 ohm load at collector of Q5.

These modifications allow the transmitter to achieve near 100 milliwatts peak output into 50 ohms with 100% modulation driven by 1 KHz 300 mV audio signal.

Measurements made and conformed using the following test equipment:

Siglent SDS 1052DL digital oscilloscope and frequency counter.
MFJ-260C 50 ohm dummy load.
Fluke model 75 DVM
Lodestar SG-4160B signal generator
DX-370 receiver
Keuwlsoft audio function generator
Palstar ZM-30 digital antenna bridge
Sony Discman D-11 (playing Beach Boys Greatest Hits)

You used a calculated transmitter efficiency  in your report but  could you measure the voltage across the 50 ohm dummy load and calculate the actual output power from this?  This would remove speculation about the efficiency.

Neil

Thanks for providing the additional data.

Should the right hand column header be RMS power? (RMS power is derived from RMS voltage but using average voltage gives zero for average power.Using RMS voltage would give the true power)

The relationship VRMS = VPEAK/sqrt 2 only applies to a sine wave.  Using this to calculate the RMS from VPEAK of the modulated envelope would be in error.  At 100 percent modulation with a single sine wave and undistorted the RMS power would be 1.5 times the unmodulated carrier power.  From data in the first table this should be 1.5 x 22 mW = 33 mW.

Knowing the power delivered to a stated load for a transmitter is useful in assessing the efficiency and for comparing transmitters.

Neil

Neil,

I included the "average" column simply as a qualitative reference. The true measurement of power is as an instantaneous value at a single point in time. It's the PEP value that really matters. The PEP value tells you the maximum power output capability of the transmitter.

The "average" power column is based on the time-averaged RMS voltage.

The "average" power value is simply saying that if one measures power repeatedly over a period of several sinusoidal cycles and then takes the average value of all those measurements, one would get that "average power" value.

And, yes, the waveform (envelope) of a 1640 KHz carrier modulated by a 1000 Hz audio sinewave is also a sinewave (as seen on an oscilloscope), hence the calculation of Vp/sqrt(2).

Mike

Mike,

Thanks for the response to my post which I appreciate but still take exception to a few points.

To get the instantenous peak power would be done by measuring the peak voltage, squaring this voltage, and dividing by the resistance. Perhaps the problem in with the column label?

RMS is already a time averaged parameter so it is not necessary to refer to a "time averaged RMS voltage", it is simply RMS voltage.  Again, probably a label problem.

The envelope waveform appears as a sine wave on a scope but the actual waveform is the product of the 1 + modulating signal and the carrier.  The sine wave seen on the envelope is actually the point by point plot of the peaks of the carrier sine wave.  Thus the apparent envelope sine wave is not actually a sine wave because the instaneous voltage is changing at the carrier frequency and this is why the sqrt of 2 factor cannot be applied.  If the RMS voltage of the modulated waveform is desired it would need to be calculated by applying the definition equation of RMS  (integral of the voltage squared over a time interval divided by the time interval)  to the product of the carrier and the modulating tone. A shortcut can be used if the modulating tone is an undistorted sine by knowing that at 100% the total power is 1.5 times the unmodulated carrier power.

But as you point out the PEP is probably more useful and would be the peak voltage squared divided by the resistance which I think we both are saying.    I think there would be less confusion (for me) if the right side column label on the modulated table entry is "Peak Envelope Power" and the present Average Power column be deleted.

Stating the power of the unmodulated carrier calculated from the peak voltage / sqrt 2 is correct (since the carrier is a sine wave) and is a meaningful number which can be used to determine the transmitter efficiency.

Neil

.

Neil,

Yes, proper labeling as you suggested does make it clearer.

Thanks!

I hope to soon have the transmitter on-the-air with center-loaded antenna and see what this thing can really do!

Mike

Mike,

Keep us posted on the progress of your project.  It will be interesting to read about the results.

Also, I dug up a post I made a while back on how to measure antenna current and the power delivered and the impedance.  Link is http://www.part15.us/forum/part15-forums/general-discussion/rf-current-transformer-antenna-tuning

Maybe you could add this to your tool kit.

Neil