On a recent evening at the W1YU Club at Yale, we had a program on QRP operation. That got me thinking. What equipment do I have for QRP work?
Well, I have an Elecraft XG-2. It puts out a fairly well calibrated 50 μVolts into 50 ohms. That is about 50 picoWatts on 20, 40, or 80 meters.
So here is my QRP transmitter. All it needs is a CR-2032 battery. I find it is tricky to send CW with the power switch, so I attached an old straight key. The coax goes to a 3-element SteppIR at 40 ft.
What can we do with 50 pW on 20 meters? First, try listening for it on my main rig, the Orion using a 40 M dipole. There it is, about S2 on 14.060 MHz. It is stronger if I aim the beam south, toward the dipole.
I tried a quick "CQ" just in case... But the band was dead. No response, no surprise!
Next I tried using my Icom R8500 set up temporarily in the car, with a 15 foot wire strung out over the trunk. The '8500 is not a great CW rig - it doesn't even have the narrow filter, but otherwise it fills the bill.
First test: yes I can hear the signal in my driveway, about 60 ft from the beam. I drove away to a point about 300 M away and gave a listen. (Auto noise was hopeless - I had to shut off the Acura completely before I could get near the noise floor.) Nothing at 300 M. Drove closer and closer, but still nothing positive. Finally, the signal was there at S1 about 2 houses away from mine -- maybe 50 M. It would have been usable for a QRS CW QSO.
(If we had a more appropriate receiver filter system and receive antenna, we might have gotten 10 dB more sensitivity.)
OK - if 50 meters is the range for 50 pW, what can we calculate? How about if we had 50 microwatts instead? That's a million times more power, and the square root of a million is 1,000. (We assume the inverse square law works, although really we're in the near field at only 50 M separation.) So what range would we expect for 50 μW? A thousand times more -- 50 kM or about 30 miles, at least for free space line of sight.
One day, I may throw together a 50 μW "QRO" rig to check this prediction.
Another calculation: 50 M is 28.6 milli-miles, so we have 571 million miles per Watt. It wasn't a two-way QSO, but would this be a record?
[Miles per Watt, as others have noted, is a nonsense ratio. For a constant signal to noise ratio, "miles" will vary with the square root of power, not linearly in power. In other words, the record should go to the QSO with the highest "miles per square root of Watts".]
[5/22/08: Photo restored.]