My daughter got one of those “build-your-own” crystal radios as a present the other day. Of course, a bunch of questions regarding radios came up that I’m not equipped to answer.
In AM transmission/reception, how does the receiver extract the original signal from the carrier wave?
and, a related question,
How does the diode help extract the signal simply by eliminating one half of the received, amplitude-modulated wave (say the negative half)?
I’ve checked a few sites on the Net and they’ve been helpful. Still, I’m foggy about the answers to the questions above.
This is my understanding, as I remember from high school physics class. The pictures are pretty shoddy - I drew them myself.
Figure 1, the original signal. This is a sound wave as captured from the microphone. The sound wave is modulated with the carrier to produce figure 2, the modulated signal, and this is broadcast over the air.
The receiver uses a tuning circuit to extract the modulated signal from the air, then runs it through a diode to produce figure 3, the rectified signal.
Now, how do we change the rectified signal into the original sound wave? We don’t have to. Since the speaker isn’t capable of vibrating at the carrier frequency (around 1 MHz), it will follow the peaks of the wave, shown by the dark line at the top of the image. Notice that the dark line is approximately the same as the original signal.
Radio speakers are essentially diaphrams controlled by electromagnets. The speaker electromagnets are operated by DC electrical currents. Thus, for example, if the DC current, running through the speaker electromagnet wiring, changes strength 500 times per second, the electromagnet strengthens and weakens 500 times per second, and pulls the speaker diaphram back-and-forth at the same rate. Thus you hear a 500 hertz tone.
Let’s say the AM station is transmitting on 850 kilohertz, and sending out a steady 500 hertz tone. This means the station’s signal is strengthening and weakening 500 times a second. Thus, in your receiver’s antenna, you have an alternating current with a frequency of 850 kilohertz, which is also strengthening-and-weakening 500 times per second.
In theory, you could run this received current directly through the speaker and hear a 500 hertz tone, since the overall signal strength is fluxuating at that rate. The problem with the above is that the high frequency 850 kilohertz AC current, when it runs through the speaker electromagnet wiring, causes self-inductance, which immobilizes everything.
The solution is to convert the received current to something approaching DC, in order to gets things to operate properly. Thus the diode blocks one-half of the AC current, which results in a kind of pulsating DC current. This current still strengthens 500 times-per-second, so the electromagnet now moves the speaker diaphram 500 times-per-second, thus making the signal audible.
Main source: “Letters of a Radio-Engineer to his Son”, 1922, John Mills
You already seem to understand that the diode eliminates (or blocks) either the positive or negative excursion of the AM envelope, so I won’t give you the answer that you seem already to know.
Your questions seems to me to be how does the diode do this? You really need to dive deep into the world of semiconductor physics to understand how, because the action takes place at the sub-atomic level. But basically diodes allow current to flow in only one direction. It does this by being sensitive to a certain polarity of voltage, so depending on its placement and orientation the diode either blocks or allows to pass one half of the AM envelope. (Nit pick: This process is not called rectification, it is called either detection or demodulation. Rectification is the diode doing this in a power supply.)
After detection, we are left with the signal seen in Mr2001’s fine rendition figure 3. The dark line on top represents the original audio wave and the thin lines inside are what’s left over from the I.F. carrier. In AM radios, this carrier is always at a frequency of 455KHz and does need to be bypassed. This is easily accomplished with a small capacitor, one side of which is usually connected to the output side of the detector diode & the other side goes to ground. We call this the RF Bypass Capacitor. If RF is allowed to get into the audio stages, it may cause unwanted oscillations or become amplified & reflect back into the RF stages.
Demodulation is the main role of a radio receiver, all of the other stages except for the front end are just amplification. You can hear strong local broadcasts with a simple detector made from two wires twisted together & allowed to oxidize thereby creating a rectifying junction similar to what’s inside a diode. The result can be heard through high impedance earphone powered by a single 1.5v cell.
CarlGauss said “How does the diode help extract the signal simply by eliminating one half of the received, amplitude-modulated wave (say the negative half)?”
The positive half of the wave form is the same shape as the the negative half. When the unrectified waveform is applied to a headphone the two halves cancel each other. The rectified waveform is a duplicate of the original modulating waveform and is an audio frequency. The carrier component of the waveform is at such a high frequency that it is not seen by the high impedence (high impedence at the carrier frequency) headphones.
It’s a simple thing to explain in person with a pencil in hand but not so easy without pictures.
Mr2001’s figures are really pretty good. But regarding Fig 2, a line tracing the variations on the top half should be a mirror image of one tracing the bottom half. That would make it a little easier to understand, because you could see how the cancelation would occur.
The key things that I learnt, things that had been holding me back were:
Since the speaker isn’t capable of vibrating in the high kilohertz or megahertz range, we don’t have to worry about subtracting the original sine wave.
The importance of the diode in eliminating half the wave is that this eliminates self-inductance which would have ‘frozen’ the speaker.
