Consider the simple AM detector below:

enter image description here

Inductor L1 and capacitor C1 builds the "tank circuit", which purpose is to filter all frequencies, except the selecteded one. However, I'm having trouble to fully understand this circuit.

In my understanding, the circuit above would not work, because the antenna is shorted with the diode D1. The diode is receiving the same voltage as the antenna, so the tank circuit is, theoretically, not interfering in the output.

If I was trying to build a tank circuit to select a specific frequency, I would do something like this:


simulate this circuit – Schematic created using CircuitLab

With this configuration, the resistor acts like a "voltage divider", and the frequency plot would be similiar to this:

enter image description here

(To select frequencies near 600kHz)

Of course, if I try to simulate the same circuit, but without the resitor, Vout becomes the same as Vin, since there would be a short between them. But this is exactly the case in the AM detector! How it works, then?

Any help is appreciated, Thanks!

Obs.: The AM detector I showed is used in almost every AM received schematic I found.

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    \$\begingroup\$ Only if the tank circuit builds up to 0.3v (germanium) or 0.5v (silicon) will the diode begin to affect the tank stored energy. In past decades, using magnetic coil (1,000 ohms of wire) earphones, the huge earphone inductance was a very light load upon the RF tank even when diode was "on". \$\endgroup\$ – analogsystemsrf Apr 8 '17 at 13:19

Perhaps you are misunderstanding the word 'detector', which is what rectifiers were called in the 'olden days'.

In your first circuit, the tuned circuit filters the incoming signal. The antenna has a finite impedance, it's not a voltage source, so allows the tuned circuit to suppress the response at higher and lower frequencies.

Amplitude modulation is a way of transmitting a low frequency signal which is coded into the amplitude of an RF signal. Mid level RF signal means zero signal. Lower level RF means negative signal. Larger amplitude RF signal means positive signal. However, all this time, the RF signal is swinging positive and negative, with zero average.

D1 is shown without a load, but in reality it will be followed by a resistive load to ground. When the incoming RF signal is larger, the peaks of the RF signal will be rectified by D1 and produce a large current in its load. A small RF signal produces little current into the load. The changing load current represents the signal modulated onto the carrier.

In your second circuit, you have the tuned circuit behaviour, but no rectifier, no detector. For a big signal, the output swings a lot above and below ground, for zero average output. For a small signal, the output swings a little above and below ground, for zero average output. The average output then is always zero, regardless of the modulation. You have 'filtered' the carrier, but you have not 'detected' it.

  • \$\begingroup\$ Thanks for the answer. Yeah, all you said make sense. Actually, the circuit I designed was only to show what I was not understanding about the resonance part of the circuit (inductor + capacitor). I was not taking the diode and the demodulation into consideration. Based on what you've said, the filtering works because the antenna has a finite impedance and does not behave like a voltage source. So, what would be the correct way to model the antenna in the circuit so I could simulate it and study the output? \$\endgroup\$ – felipeek Apr 8 '17 at 3:22
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    \$\begingroup\$ @felipeek if you're receiving a 1 MHz signal with a monopole of a few metres, as in an AM receiver, then the antenna source impedance is very high. Model it as a voltage source in series with 20 pF to start with. Voltage is the E-field times effective height, half your antenna height. See here for impedance equations. \$\endgroup\$ – tomnexus Apr 8 '17 at 8:16
  • \$\begingroup\$ Interesting, I was wondering why instead of having a diode-capacitor-inductor or diode-lpf configuration, only a lpf/lc filter could be present which could simply filter the RFs and leave only the audio signal. The modulated signal is periodic, while the audio signal is not. That is why one side of the whole modulated signal must remain, before it is filtered of RF remains, right? \$\endgroup\$ – Daniel Tork Oct 21 '18 at 16:40
  • \$\begingroup\$ Am I right about my conclusion? \$\endgroup\$ – Daniel Tork Oct 23 '18 at 21:48
  • \$\begingroup\$ @DanielTork maybe, I'm not sure I understand exactly what your conclusion is. The AM'd RF signal is zero mean, regardless of its amplitude, which is why merely filtering it yields a zero mean result. Using a diode to eliminate one side of the signal results in one whose mean is a function of the amplitude and so the modulation. A lowpass filter on this removes the remaining RF, leaving you the mean which is the modulation. \$\endgroup\$ – Neil_UK Oct 24 '18 at 5:42

The tuned circuit is only loosely coupled to the transmitter by the antenna, this has a similar effect to the resistor.

You can model it by replacing the resistor (R1) in your simulation with a capacitor of value say 10pF.

The diode will load down the tuned circuit and reduce its Q, often the inductor is tapped to provide a lower impedance drive to the diode that will reduce the loading effect.

The antenna may also couple too tightly to the tuned circuit as well and so may be connected by a small capacitor or to a tap part way down the inductor.

  • \$\begingroup\$ Thanks for the answer. Really, the simulation using a 10pF capacitor generates very good results. But could you explain why the capacitor? In other words, why would I model the antenna with a small capacitor connected to it? \$\endgroup\$ – felipeek Apr 8 '17 at 3:13
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    \$\begingroup\$ It is not modeling the antenna as having a capacitor, it is treating the the air between the transmit antenna and the receive antenna as a capacitor, with the antennas as the "plates". That model is not correct in many ways, but it is sufficient to answer your question. The capacitor drives the LC circuit with a high impedance similar to what you get from an antenna. \$\endgroup\$ – Evan Apr 8 '17 at 3:32

The diode is receiving the same voltage as the antenna, so the tank circuit is, theoretically, not interfering in the output

An antenna always presents an impedance to the circuit it connects to therefore the tank circuit does become a selective filter. A quarter wave monopole presents an impedance of about 37 ohms when it receives a carrier that has a wavelength precisely 4 times the length of the antenna. See this: -

enter image description here

When height is referred to in the above graph it means the length dimension of a vertical monopole and not the height that the antenna is placed above some altitude reference point. Height in this graph means length.

At 1 MHz (wavelength = 300 m), a 75 m monopole presents about 37 ohms. However, back in the days of crystal sets, nobody ever used an antenna that long so, practicality meant that the antenna was "short". Look at the impedance presented by an antenna that is one fifth of the length of a quarter wave monopole (15 m in this example). It has a theoretical impedance that is capacitive and about 1000 ohm reactive. At 1 MHz that's about 160 pF. The blue line in the graph indicates that at 0.05 wavelength the resistive part of the impedance is just an ohm or so (there are formula to get exact values of course).

The bottom line is that an antenna converts the impedance of free space (377 ohms) to an electrical impedance that can vary significantly and be complex in nature. This forms the impedance by which your tuned circuit is allowed to resonate and block those frequencies not wanted and allow those that are wanted.

  • \$\begingroup\$ Wait a second. The diagram doesn't seem to say anything about impedance versus length, but rather, also as titled, impedance versus height. That would mean that the length is fixed (hopefully in terms of wavelengths). If you are going to argue that height and length are interchangable (or that in this case, "height" actually means "monopole length"), then please make that point explicitly. \$\endgroup\$ – a CVn Apr 8 '17 at 12:02
  • \$\begingroup\$ @MichaelKjörling given that most monopoles are vertical, height becomes length but I take your point that height could be mistaken for the height of a fixed length monopole above some arbitrary altitude. \$\endgroup\$ – Andy aka Apr 8 '17 at 12:08
  • \$\begingroup\$ @Andyaka Amazing answer. Thank you very much. Your answer brought me another question: If I was trying to select a frequency that would make my particular antenna have a low impedance (for example, the situation you mentioned: 1MHz frequency and 75m monopole, resuting at 37Ohm impedance) - this would be bad, since the filtering circuit would not suppress a large band of frequencies, right? So, would it be a good solution to manually add a capacitive impedance to the antenna to detect only the desired frequency? \$\endgroup\$ – felipeek Apr 8 '17 at 16:28
  • \$\begingroup\$ Let's use the FM band as an example. 100 MHz is 3 metres and the monopole on your transistor radio is about one quarter wavelength hence it will present an impedance of about 37 ohms with very little added reactance. What do transistor radios do to interface the antenna? Answer there is a parallel tuned circuit generally but, the antenna usually feeds into the inductor at about 10% of its winding above localised earth (0 volts). This roughly matches the antenna and circuit impedance but, importantly you get a nice voltage amplification to the top of the tuned circuit and give good selectivity. \$\endgroup\$ – Andy aka Apr 8 '17 at 16:43
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    \$\begingroup\$ Therefore, adding reactance in series with the antenna in order to form a tuned circuit is just another method. In fact, on really early crystal sets with a short capacitive antenna, just a variable inductor was used to tune in the desired station. There is no magic about antenna tuning; the real magic lies in the laws of the universe that permit almost disparate E and H fields local to a transmit antenna becoming a combined EM wave whose power diminishes with distance as a square law (compared to disparate E and H fields whose power diminishes as the power of 6). \$\endgroup\$ – Andy aka Apr 8 '17 at 16:54

In my view, a very simplistic way to look at your circuit would be to interpret it as being two separate circuits. The first part would be the Antenna:Coil:Capacitor, which would represent your Resistor:Coil:Capacitor from your second circuit.
The Antenna in your initial circuit is just a converter of electromagnetic radiation into a source of alternating voltage, loosely coupled to the RF transmitter far away. this is why we can consider that antenna like some kind of resistor between the RF source (transmitter) and the RF receptor (antenna). Since the two are quite far apart, the coupling is low hence, resistor.
Now the second part of the circuit is the diode (can also be called the detector). That diode will simply use the top half of the RF AC signal and turn it into an RF AC rectified signal, composed of only the top part of the signal, because the diode only conduct current in one direction (theoretically speaking).
Now, in that second part of circuit should be added another resistor, otherwise the circuit would not be completed and no current would circulate through it. So if you add a resistor to the cathode of that diode and to ground, there will appear a current quickly increasing from zero to max, only in one direction and it will do that many times per second, based on the tuned frequency of the first stage of the circuit.
Sending this rectified signal to an acoustic transducer will then emit some sound pressure representing the transmitted modulation of the RF energy transmitted.

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    \$\begingroup\$ interesting unit there, MHz per second! \$\endgroup\$ – Neil_UK Apr 8 '17 at 4:21
  • \$\begingroup\$ Good idea to split it up like that. \$\endgroup\$ – tomnexus Apr 8 '17 at 8:17
  • \$\begingroup\$ @Neil_UK Sounds similar to meters per second per second. \$\endgroup\$ – a CVn Apr 8 '17 at 11:57
  • \$\begingroup\$ Mhz per second! - Yeap, Brain fog, 1:00 AM, should have re-read my post :) \$\endgroup\$ – Fred Cailloux Apr 8 '17 at 20:05

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