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

[GODISMYSHADOW]

Say I make a simple crystal radio using a 1N34A diode.
If I use it to tune in a radio station 100 miles away,
would it be possible in theory to detect an infinitesimal
increased load at the station when I tune it in with my
crystal set?

 

[vk6kro]

You wouldn't get an increased load because the signal has left the antenna and is radiating away from it..

However, you might get a VERY small signal radiated back to the transmitter from a resonant tuned circuit with an antenna on it.

This is a little like the "chaff" that aircraft eject to confuse radar systems. Each piece of "chaff" is a resonant antenna length and radiates a small part of the radar signal back to the receiver.

It is called a "passive reflector".

To have the remotest chance of hearing it, you would have to turn off the transmitter, of course.

 

[berkeman]

Say I make a simple crystal radio using a 1N34A diode.
If I use it to tune in a radio station 100 miles away,
would it be possible in theory to detect an infinitesimal
increased load at the station when I tune it in with my
crystal set?

No, I don't think so. The energy still radiates away as usual from the transmitting (TX) antenna. You are just intercepting some of the energy that would have kept on going.

You probably can detect a drop in energy at points past your antenna, since you have absorbed some of the energy with your antenna.

 

[GODISMYSHADOW]

The tiny signal from the crystal set radiated back to
the transmitter should generate a minute opposing
EMF in the transmitter's antenna. In theory, would
you not be able to detect this at the station?

 

[berkeman]

The tiny signal from the crystal set radiated back to
the transmitter should generate a minute opposing
EMF in the transmitter's antenna. In theory, would
you not be able to detect this at the station?

That's a different question. An ideal RX antenna and receiver will not put out any signal from the RX antenna. To the extent that some local oscillator energy in a heterodyne receiver makes it out the RX antenna, and to the extent that the RX antenna is not well matched to the receiver's feed line, then yes, you could detect something at the original TX station. But it's due to either a partially reflected signal at the RX antenna, or due to other noise being generated by the receiver. Not because of the energy that is absorbed by the RX antenna and receiver. Makes sense?

 

[sophiecentaur]

The guy with a receiver 'downstream' of you might notice a small difference in his received signal power. But you'd need a large, well matched antenna and be very close to him.

AS far as the transmitter is concerned, your antenna would constitute a small Radar target, of course but radar systems always have to switch off the transmit signal (i.e. pulse) to avoid drowning out the received reflected signals (pulses).
Your clandestine listening is quite undetectable - relax.

 

[GODISMYSHADOW]

That's a different question. An ideal RX antenna and receiver will not put out any signal from the RX antenna. To the extent that some local oscillator energy in a heterodyne receiver makes it out the RX antenna, and to the extent that the RX antenna is not well matched to the receiver's feed line, then yes, you could detect something at the original TX station. But it's due to either a partially reflected signal at the RX antenna, or due to other noise being generated by the receiver. Not because of the energy that is absorbed by the RX antenna and receiver. Makes sense?

I'm taking about a simple crystal set (germanium diode AM radio). When you tune to the
broadcast station, the two circuits will be resonating. When in this state the crystal set
will be drawing power from the transmitter. Nikola Tesla understood this stuff.

 

[berkeman]

I'm taking about a simple crystal set (germanium diode AM radio). When you tune to the
broadcast station, the two circuits will be resonating. When in this state the crystal set
will be drawing power from the transmitter. Nikola Tesla understood this stuff.

They do not resonate together. The RX station circuit is being excited by the incident radiation, which has nothing to do with the TX station after it leaves the TX antenna.

 

[Averagesupernova]

They do not resonate together. The RX station circuit is being excited by the incident radiation, which has nothing to do with the TX station after it leaves the TX antenna.

I'm not so sure. If you get enough antennas I would assume you will load the transmitter differently. Granted, in the real world this doesn't occur for obvious reasons, but my point stands.

 

[berkeman]

I'm not so sure. If you get enough antennas I would assume you will load the transmitter differently. Granted, in the real world this doesn't occur for obvious reasons, but my point stands.

Even at 100 miles away? I was picturing the question in his first post.

 

[sophiecentaur]

I'm taking about a simple crystal set (germanium diode AM radio). When you tune to the
broadcast station, the two circuits will be resonating. When in this state the crystal set
will be drawing power from the transmitter. Nikola Tesla understood this stuff.

The mutual impedance between transmit and receive antenna is negligible. The receiver is not "drawing power" from a distant transmitter - it is merely intercepting the wave which has been launched previously. Let's face it, for a distant receiver, the transmitter could have been switched off and the antenna dismantled long before the receiver actually gets the signal.
Nikola Tesla strikes again, I notice. Will he never lie down?

 

[conway]

That's a different question. An ideal RX antenna and receiver will not put out any signal from the RX antenna.

Actually, to my way of thinking, if the receiving antenna is matched and tuned for maximum power absorption, it must also re-radiate. In fact, in the ideal case, as much power will be re-radiated outwards as will be absorbed at the load. In practise this type of system is not normally seen in AM crystal radio receivers because the parasitic lossed dominate all other circuit parameters.

 

[sophiecentaur]

If no power is absorbed by the load then there is no received signal.

 

[berkeman]

Actually, to my way of thinking, if the receiving antenna is matched and tuned for maximum power absorption, it must also re-radiate. In fact, in the ideal case, as much power will be re-radiated outwards as will be absorbed at the load.

Why do you say that? How much power is reflected off of a correct termination resistor at the end of a transmission line?

 

[conway]

I'm sorry, I didn't mean to contradict you. I didn't notice you were a group moderator when I posted.

 

[Averagesupernova]

Even at 100 miles away? I was picturing the question in his first post.

Yes even at 100 miles. It may take antennas right next to each other in a 360 degree sphere, but I still stand by my original point.
-

Let's face it, for a distant receiver, the transmitter could have been switched off and the antenna dismantled long before the receiver actually gets the signal.

You can say the same thing about sending a pulse down a transmission line and unhooking the pulse generator before the pulse is reflected off of the other end of the line.
-
We all seem to forget that the transmitter relationship to the receiver is quite simply a very lossy pad. An tiny tiny tiny tiny tiny tiny tiny tiny tiny tiny tiny portion of the transmitted signal gets to the receiver, so naturally, one lone receiver is not going to do much to the transmitter. Try this: Run a signal generator at 0 dBm into a directional coupler and sample the reflected port with a spectrum analyzer. Short the output of the coupler and you will have a lot of reflected power as shown on the spectrum analzyer. Properly terminate the directional coupler and you will notice no reflected power on the spectrum analyzer. Well, maybe some due to not being able to perfectly match impedances. Now properly terminate the output but do so through 50 or 60 dB of attenuation. You will again notice very little if any reflected power. Now short the output of the pad. Shorting the output of the pad now doesn't seem to do much to the reflected power as shown on the spectrum analyzer. This is similar to what happens when we transmit from an antenna into space. The space is a natural attenuator. So no matter what we try to do at the receiving end, the isolation between transmitter and receiver minimalizes any effect on the transmitter. But if we put enough receiving antennas out there (similarly, if we hang enough short circuits through enough attenuating pads) the transmitter should see it. It certainly isn't a perfect analogy, nothing is.

 

[berkeman]

I'm sorry, I didn't mean to contradict you. I didn't notice you were a group moderator when I posted.

No, no, don't worry about that. If I'm wrong, then I learn something new. Not an issue.

I just honestly didn't understand your statement. I think it's wrong, but I wanted to ask about it in case my understanding is missing something.

To my understanding, the EM wave going past the RX antenna causes current to flow in the antenna, and through the feed coax to the receiver, which has an input impedance well-matched to the Zo of the feed coax. I'm not seeing where there is reflected energy off of the antenna, but I suppose it's possible.

 

[conway]

No, no, don't worry about that. If I'm wrong, then I learn something new. Not an issue.

Thanks, because sometimes I find I have to watch myself around here.

I just honestly didn't understand your statement. I think it's wrong, but I wanted to ask about it in case my understanding is missing something.

To my understanding, the EM wave going past the RX antenna causes current to flow in the antenna, and through the feed coax to the receiver, which has an input impedance well-matched to the Zo of the feed coax. I'm not seeing where there is reflected energy off of the antenna, but I suppose it's possible.

It's actually a very very cool analysis which I did myself quite a few years ago. Of course to receive power there has to be SOME current flowing in the receiving antenna; and once current is flowing, you have a transmitting antenna, so there must be outgoing power. But I want to show that this is not a trivial or inconsequential aspect of the process, but absolutely essential.

This analysis applies to an ideal antenna without any lossy materials around (e.g. the ground). The antenna doesn't have to be long, it just has to be lossless (a superconductor). You tune it with a coil and choose the resistance that absorbs the most power. That's what we call matching. If you now decide to short out the load, the antenna becomes a pure reflector. The current is exactly double the matched load current.

Think about this and see if it makes sense to you. I can explain it better if necessary. The funny thing is that it doesn't matter how short your antenna is...you can still match it with the right lossless coil. The matching resistance changes as you get shorter in length, but there always is a matching resistance whatever length you work with.

 

[berkeman]

Interesting thoughts. My understanding is that the current in the RX antenna is consistent (via Maxwell's equations, etc.) with the EM field flowing past it from the source, at least if the antenna is well matched to the coax feed line. Yes there is current flowing in the antenna, which means there is a field, but I think it is consistent onloy with the original EM wavy going by, not generating an additional reflection.

But I'm not sure... Maybe Averagesupernova or vk6kro or another E&M/antenna person can chime in...

 

[sophiecentaur]

Think about this and see if it makes sense to you. I can explain it better if necessary. The funny thing is that it doesn't matter how short your antenna is...you can still match it with the right lossless coil. The matching resistance changes as you get shorter in length, but there always is a matching resistance whatever length you work with.

I think that resistance is referred to as the radiation resistance - when treating the antenna as a transmitter, which always seems easier to visualise - to me. It represents the resistive component of the impedance that the feeder sees, or the energy lost out into the aether which could be replaced by putting a sphere of resistive material of 377 Ohms - the 'impedance of free space' around the antenna. The other way round is a bit harder because you have to get a current induced in a receiving antenna in the first place, before it works.
I got round that problem by 'believing in reciprocity'.

 

[conway]

Yes, it's the radiation resistance. I admire your lack of self-deception in the way you admit to adopting your belief in reciprocity more out of convenience than understanding.

It is true that the meaning of radiation resistance seems more obvious as applied to a transmitter than a receiver. Actually, the value of 77 ohms for the half-length dipole can be approximated pretty well by taking a sphere split along the equator and driving it at its resonant frequency with 2*pi amps applied symmetrically to both halves. If the radius of the sphere is 1 meter it's not too hard to ballpark that the voltage at the poles is around 2*pi*70 volts, just by tracing a line of longitude and assuming that the E field is just the H field multiplied by 377. It takes a little physical reasoning to decide that the radiation resistance doesn't change when you collapse the sphere to a line.

I have a totally different way of calculating the radiation resistance of the receiving antenna, but that would be the subject of another post.

 

[Averagesupernova]

Radiation resistance shouldn't have a whole lot to do with what the original poster asked. If we are to truly radiate X number of watts into space, there has to be some kind of resistance seen when looking into the antenna. In our ideal world it doesn't matter what it is as long as it is not zero or infinity. In the real world, well it's a different story.

 

[conway]

If I'm correctly understanding what we were talking about, the question was:

For a receiving antenna tuned and matched to absorb maximum power, is the re-radiated power significant?

 

[Averagesupernova]

I don't think there IS any reradiated power to speak of. If we are to 'believe in reciprocity' as sophie has mentioned, then tell us what happens when transmitting into an antenna. Does half the power get reflected back into the transmitter? Don't think so if it is correctly matched.

 

[GODISMYSHADOW]

The mutual impedance between transmit and receive antenna is negligible. The receiver is not "drawing power" from a distant transmitter - it is merely intercepting the wave which has been launched previously. Let's face it, for a distant receiver, the transmitter could have been switched off and the antenna dismantled long before the receiver actually gets the signal.
Nikola Tesla strikes again, I notice. Will he never lie down?

Someone should establish some rules for "thought experiments."
You're talking about astronomical distances. Here, you take sides
with the receiver and switch to the viewpoint of the transmitter.
Then you jump back to the receiver. Seems dangerous to me,
because information itself can't travel faster than light. I don't like
your thought experiment.

 

[conway]

I don't think there IS any reradiated power to speak of.

That was basically also Berkeman's initial take on the question.

If we are to 'believe in reciprocity' as sophie has mentioned, then tell us what happens when transmitting into an antenna. Does half the power get reflected back into the transmitter? Don't think so if it is correctly matched.

No, half the power doesn't get reflected back to the transmitter, it's just radiated into isotropic space.

I think I have to remind everyone we're talking about a theoretical ideal case. A practical crystal radio doesn't operate anywhere near conditions of ideal matched load. In the ideal crystal radio, you would have an enormous tuning coil that would resonate your (electrically) short antenna up to thousands of volts. With the load short circuited, and otherwise unlimited by parasitic ground losses or copper losses, the ultimate resonance is bounded only by the pure radiation resistance of the receiving antenna. In this state the receiving station is a pure reflector. In order to deliver power to a load, a resistance must be inserted in the circuit. The maximum power is absorbed when the load resistance is matched to the radiation resistance of the antenna. In this condition, the antenna current is exactly half its short-circuit value, and the load power is equal to the re-radiated power. Both of these are exactly one-quarter of the maximum scattered power at short circuit.

Of course this highly resonant circuit would deliver mostly carrier power to a speaker with very little audio.

 

[Averagesupernova]

No, half the power doesn't get reflected back to the transmitter, it's just radiated into isotropic space.

Ok. So ALL of the power is radiated then?

 

[vk6kro]

The original question was about crystal sets.

A crystal set is not a matched load for an antenna. The tuned circuit will hopefully be lightly loaded to get selectivity and the half wave diode means the coil is unloaded by headphones for most of an incoming RF cycle. So quite a bit of the incoming RF will be reflected back along the antenna.

The antenna is not going to be resonant by itself because it will normally be too short.
A quarter wave at 1 MHz is about 75 meters (246 ft). However, being attached to a lightly loaded tuned circuit and this being connected to a passable ground system, there MAY be some re-radiation from such a system.

Accepting that this is Fantasyland {TM} stuff, the signal returning to the transmitter antenna will be phased according to the exact distance traveled to the crystal set antenna and back. If there were many such crystal sets, these signals would be at random phases and so would tend to cancel each other out.

In reality, the signals would be microscopic and easily drowned out by band noise.
Any 246 ft (25 floor) metal framed buildings would return a much stronger signal too.

In general, though, suppose you had a dipole in free space which was uncut.
This is actually a shorted load for a dipole. ie zero ohms.
So, you would expect maximum current to be flowing in such a dipole and hence maximum reradiation. As this load resistance is increased, the current will decrease and you could expect less radiation.

 

[conway]

I don't think I'm following you.

(edit: was reply to averagesupernova, vk5vkro's post got in between)

 

[Averagesupernova]

In general, though, suppose you had a dipole in free space which was uncut.
This is actually a shorted load for a dipole. ie zero ohms.
So, you would expect maximum current to be flowing in such a dipole and hence maximum reradiation. As this load resistance is increased, the current will decrease and you could expect less radiation.

I believe the question right now is whether a properly terminated dipole will reradiate anything at all.

 

[Averagesupernova]

Ok conway, if I am understanding you correctly your claim (in quotes below) is that a perfectly terminated dipole in free space will reradiate the same amount of power that would be absorbed in the load resistor.

In fact, in the ideal case, as much power will be re-radiated outwards as will be absorbed at the load.

So my question is that if this is the case, then would it not be logical that in a transmitting antenna, when we drive the feedpoint should we not have reflected power back into the transmitter by the same logic?

 

[conway]

So my question is that if this is the case, then would it not be logical that in a transmitting antenna, when we drive the feedpoint should we not have reflected power back into the transmitter by the same logic?

Yes, that'swhat happens if you try to deliver the maximum theoretical output of your power amplifier into the transmitting antenna. Of course you don't normally operate an antenna that way. Because you would blow your amplifier.

 

[conway]

My last answer might have been a little glib, although it's technically correct. The point is if you turn a light bulb in your living room, you can analyze the lamp cord as a transmission line and there are all kinds of reflections, mismatches, and standing waves to deal with. It actually all works out in the end but its not an especially helpful way of seeing what's going on.

I don't actually know what the wasted power is in a 10-kW AM radio transmitter, but there's no technical reason why you'd want to match the amplifier impedance to the load.

 

[vk6kro]

I believe the question right now is whether a properly terminated dipole will reradiate anything at all.


I suspect that it will. If the dipole current is allowed to flow at all, then the dipole will still radiate. Like this:

The ultimate case would be two unconnected quarter wave lengths with no current flowing between them. I wouldn't expect this to radiate at all.

Yes, that'swhat happens if you try to deliver the maximum theoretical output of your power amplifier into the transmitting antenna. Of course you don't normally operate an antenna that way. Because you would blow your amplifier.

An amplifier set up to deliver a constant carrier can deliver say 75% of its DC power into a 50 ohm load. If the antenna feedline looks like 50 ohms, then this maximum power will be delivered to the feedline.
If there were no losses, this power would also be delivered to the antenna and all of it would be radiated.
This is the normal way transmitters work without any risk of blowing up the amplifier.

 

[conway]

I don't know where you got your graph from, but the re-radiated power should be quadratic with the antenna current, not linear.