Can a crystal radio detect an infinitesimal load at a distant radio station?

[conway]

I'm going to support Sophie on this issue. When the receiver is far away from the transmitter, it does not in any way "load down" the transmitter, and basically for the reasons Sophie said. Of course there is a very small reflected wave but that's not what we mean by loading, even in principle. The energy transfer from A to B does not depend on the detection of the reflected wave; the receiver and transmitter would work exactly the same if the reflected wave could somehow be caught and destroyed before it got back to the transmitter.

 

[berkeman]

I'm going to support Sophie on this issue. When the receiver is far away from the transmitter, it does not in any way "load down" the transmitter, and basically for the reasons Sophie said. Of course there is a very small reflected wave but that's not what we mean by loading, even in principle. The energy transfer from A to B does not depend on the detection of the reflected wave; the receiver and transmitter would work exactly the same if the reflected wave could somehow be caught and destroyed before it got back to the transmitter.

And I disagree with your reflected wave concept for a well-terminated RX half-wave dipole antenna, at least for now. We can agree to disagree, until one of us comes up with the math or an experiment to settle that, okay?

 

[sophiecentaur]

I have a feeling that there is some confusion about how an antenna works. It isn't just the metal bits out there in space that are responsible for the power going in or our. An infinitely / very thin wire has more or less the same energy gathering properties as a fat wire. (Width helps to make the bandwidth of the element wider, but that's all) So the antenna must be working due to the fields surrounding it, which can be regarded as being caused by currents flowing in the conducting bit. These fields can be regarded as coupling energy into the feedpoint of the antenna. So why should it be suggested that, just because the antenna is matched, these fields should suddenly have no effect on the passing wave? You can, in principle and in practice, feed a wire anywhere you like - not just its mid point - and have a perfect match.

I was trying to think of an analogous situation and I suggest that a light disc shaped float, floating on the sea would be suitable. If it can float up and down on the waves without taking any energy then, if it were light enough, you would get no significant scattered energy from it. However, if you attempt to take energy from it - using it to pull on a crank, for instance, the float will no longer move directly in step with the water and this will involve waves emanating from it. You could imagine an appropriate value of friction (resistance) applied to the crank that would take a maximum of energy from the system - the waves would still be disturbed; you wouldn't imagine that, suddenly the waves would go past without being affected. So taking energy has involved disturbing the waves. Coupling the float to an appropriate resonating mass / spring system (without loss - representing a resonant length of wire) could produce greater movement of the float and, hence, more scattered waves. But in both cases there will be disturbance. The only time there is none would be when the float is not altering the power flow at all.

 

[conway]

OK, I don't think the math is normally helpful in these kinds of discussion, but here goes:

Take an AM radio station putting out a signal of 377 mV/meter (that's a pretty strong signal, obviously I picked the numbers so that the power density would be ExH=377 microvolts per meter squared). Let the station be at 680 kHz, wavelength approx 440 meters. Let your antenna be an 11 meter dipole. (No ground. Assume there is no ground.)

Now let's go to Wikipedia (http://en.wikipedia.org/wiki/Dipole_antenna) and look up the radiation resistance for a short (L/Lambda=1/40) dipole. Paraphrasing slightly:

R = approx 200 (L/lambda)^2 which gives in this case 0.125 ohms. To match the radiation resistance we use an appropriate coil to cancel the reactance of the antenna and insert a load resistance equal to R. The antenna voltage is ExL = approx 4 volts and the current is therefore 4 volts / 0.25 ohms = 16 amps.

The power to the load is I^2*R(load) = 32 watts and the power re-radiated, substituding R(rad) for R(load), is obviously the same.

 

[sophiecentaur]

The sums have sorted it out, as usual. Thanks conway.

 

[conway]

As usual? It's rare that anything ever really gets sorted out in these kinds of discussions. I wouldn't write this one off yet either...

 

[conway]

I have to say I was wondering if anyone thought my numbers were unrealistic.

 

[sophiecentaur]

Near enough to make the point.

The point seems to be that matching your antenna feed is not the same as replacing an area in space with a 'hole' for energy to flow out of, which is indistinguishable from the rest of the sphere into which the power is radiating. It is more like hanging a load half way along a long, perfectly terminated transmission line. You can get a optimum level of power into the load but that will, inevitably, produce a reflection / mismatch on the line at that point.