When you transmit signals at radio frequency why are the antennas open ended (open circuit)? I can understand that changing the electric field causes changes in the magnetic field and hence the wave will propagate. But what I don't understand is how you can have an electric field if the circuit is open ended in the first place because the impedance should be infinite?
Electronic – Radio wave transmission and receiving antenna
radioRF
Related Solutions
As Leon Heller said, this is not RF. However, it sure is an interesting experiment.
You have noticed that the magnetic field of the primary coil isn't strong enough to transfer energy over such a distance. Amplifying is a good idea indeed, but the question is: how much do you need to amplify?
The transistor you're using in your circuit needs a specific voltage in order to start conducting. The secondary coil probably won't give such voltage. What you can do, is use the transistor as an amplifier:
As you can see, a pull-up (R1) and a pull-down (R2) are used to give the NPN transistor the minimum voltage it needs. With this circuit, even a tiny fluctuation in Vin will affect the current through collector and emitter. Vout is Vin, but amplified (and inverted, but that's not a problem here). You can use Vout to feed a transistor as a switch, as your circuit shows.
However, this is theory. How much you have to amplify heavily depends on the distance between the coils, and you might need to amplify so much, that it isn't worth trying.
Do you have an oscilloscope? I would recommend you making a graph of the amplitude of the voltage on the secondary coil as a function of the distance between the coils. I'm guessing here, but I think this will be an exponential function. When the voltage is nice AC, you might be able to do this with a multimeter as well. Now you have some data and you can calculate the amplification you need at a specific distance. The needed amplification will dramatically increase when increasing the distance, is my guess. That makes this setup not very useful on further distances, and that's why we use RF.
To get you started in RF, I can recommend you the book Crystal Sets to Sideband by Frank W. Harris, K0IYE. Skip or scan chapter one about the history of radio. Chapter 2 is basic knowledge which I think you already have, so also scan it. Chapter 3 is some blahblah about a workspace, which I found demotivating because Harris expects you to have a lot. In chapter 4, the fun starts, with a crystal set.
Tuning a radio to a particular station using a tuning capacitor or any other method selects the carrier frequency.
The frequency-selective circuits in a radio are not "perfect" - they allow a band of frequencies to pass, not just a single discrete frequency, so the carrier and the sidebands generated by the modulating signal (audio) can apss through the RF and IF stages of the receiver.
Best Answer
It's a bit unclear what exactly you are asking. I suppose you have seen a antenna like a dipole and it looks at first glance like just two wires that aren't connected to anything, therefore the question is how can current flow and power be drawn? If so, it would help if you clarified that. Some antennas, like a loop or folded dipole, are exactly the opposite in that they appear to be dead shorts at first glance.
In any case (if my interpretation of your question is correct), what you are missing is that antenna is no longer a open or short at its intended frequency. Antennas are not lumped systems, which means that different parts will be at different phases of the signal at the same time. In fact, antennas exploit this to help produce the large voltages and current it takes to radiate significant power.
Often resonance is involved. Fill a bathtub partway with water. Now put your hand in the middle and move it back and forth only a short distance in the end to end direction. Once you find the right frequency, you will see that you can get a lot of water to slosh back and forth despite only a relatively small motion of your hand. Note that at the peaks, the water at one end of the tub is high and the other low, and nothing is flowing. In between the water is roughly level but is flowing strongly in the middle. Also note that it takes very little force from your hand to cause this and keep it going, but you have to be moving your hand at just the right frequency. A little faster or slower and it doesn't work anymore.
That was resonance, and is exactly what is happening in a dipole. In the case of a dipole, that water level becomes voltage and the flow rate of the water becomes current. The feedpoint in the middle of the dipole is where a little current of just the right frequency is fed in, which causes resonant sloshing of charges back and forth in the antenna. The voltages created at the ends of the antenna can therefore be a lot higher than anything you put in.
This sloshing of current back and forth makes high voltages at the ends. Together with the high current in the middle, the antenna disturbs the local E and B fields in such a way that power is lost from the antenna into those fields. That power eventually organizes itself into a self-propagating wave we call radio.
Since real power is lost, the antenna must appear to have a resistive component from the driving circuit point of view. A ideal antenna used at exactly the right frequency will appear to be purely resistive, meaning all the power dumped into it gets transmitted. This happens very close to the best resonant sloshing frequency for most antennas. That also explains why antennas often only work well for a narrow frequency range.
So getting back to your question, the problem is you are analyzing the antenna at DC, which is completely irrelevant. You have to analyze antennas at the frequencies they are intended to radiate at. At those frequencies, a lot of other stuff happens so that they don't look like opens or shorts as they do at DC.