A nanowatt is one billionth of a watt; yes, one billionth! On that scale a QRP transmitter puts out five billion nanowatts and a standard ham rig puts out 100 billion nanowatts. But is a nanowatt enough to work DX? Yes. But, of course, manufacturers don’t make radios that only put out a nanowatt and besides, it isn’t really that straightforward – let me explain.
Suppose that you want to work an antipodean station – that means a station halfway around the world. The circumference of the Earth is approximately 40,000 kilometers, so your target station is, say, 20,000 kilometers away. How much power should you use? Would QRP (5 watts) be enough? Would your signal make the trip if you used 100 watts? Maybe you should fire up an amplifier and blast out the legal maximum power?
The Bare Minimum
The answer depends on a number of factors so let’s assume you are not fighting a pileup and propagations conditions are good. What is the lowest power the antipodean receiving station could detect? To answer that question I looked up the sensitivity of the receiver in one of my radios. I chose my Yaesu FT-891. Yaesu specifies a receiver sensitivity of 0.16 microvolts. Microvolts? Okay, we can work with that. The input impedance of the receiver is 50 ohms so my FT-891 will detect a signal as low as V^2/R = 0.0005 microwatts – that’s one half of a nanowatt.
Okay, what does this all mean? Well, of course, nobody is really going to be using a transmitter with an output power of one nanowatt. Why don’t we assume a ham in North America has a transmitter that can put out 5 watts, which is the standard definition for a QRP CW signal. He wants to make a CW contact with a station 20,000 kilometers away. Is he going to be successful?
Let’s Get Physical
Now we have to introduce some high school level physics – the Inverse Square Law. The Inverse Square Law tells us that the power of the signal at the receiving station will be inversely proportional to the distance from the transmitting station. Is it game over for QRP? Let’s do the math and find out.
Let’s call the signal strength at the transmitting station P1, and the received signal strength P2. The signal path distance is L. Evaluating P2 = P1/L^2 gives a received signal strength of 1.25 nanowatts – two and half times (but less than 1 S-unit) the receiver sensitivity! Yay, bring out the beer and party balloons! No wait; there’s a bit more to it yet.
Hug the Planet
The signal doesn’t hug the Earth’s surface does it? It goes up to the ionosphere, bounces back down again and repeats until it gets to the distant station. Is that significant? If we assume the F2 layer is 300 kilometers above the Earth’s surface (it varies) and the propagation angle of the transmitting antenna is 10 degrees, the signal will travel six “hops” to reach the destination. Each hop involves a 3500km round trip so the total distance travelled is actually 21,000 kilometers. No biggie eh? Party balloons now?
What about D-layer absorption? Depends on time-of-day and all manner of unknowns, but could be a problem. If propagation conditions are less than optimal we may lose some, or all of the signal on each hop. If there is local QRM the received signal may be swamped; remember it is less than one S-unit above the absolute minimum at the receiving station.
So, in conclusion, yes, one nanowatt – at the receiving end – is theoretically enough to make the contact. But don’t bank on it. That’s what makes ham radio such a fascinating hobby.
This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.