How does this metal detector circuit work with an AM receiver?

Question

I do not understand how this circuit works. From Fun Electronics Projects for the Experimenter by Newton C. Braga.

It is described as using the beat-frequency method to detect an object. The frequency is said to change when metals come within a short distance, which when picked up by an AM receiver changes the pitch of the audible sound.

This gives me 2 questions:

1. Does the frequency actually change? If yes, how does this work with an AM radio? I thought amplitude-modulated radio worked through changes in amplitude, not frequency.
2. How does the circuit produce oscillations? Which part of the circuit converts DC into AC?

Answer 1

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. V1 is 1 V, 100 kHz. V2 is 0.5 V, 102 kHz.

Figure 2. Note that V1 and V2 start and end in phase but are out of phase in the middle of the trace.

Does the frequency actually change?

Yes. The nearby metal changes the inductance of L1 and that changes the oscillator frequency.

I thought amplitude-modulated radio worked through changes in amplitude, not frequency.

It does, but this trick is interfering with the carrier wave to create some modulation on it. In Figure 1 V1 represents a radio station broadcasting on 100 kHz while V2 represents the metal detector "broadcasting" on 102 kHz. The AM radio's tuner isn't so selective that it could be tuned to ignore a very close frequency to that selected on the tuner dial. The result is that the receiver responds to the sum of the signals.

Figure 2 shows us that two close carrier signals will "beat" - periodically in phase and drifting to out-of-phase and back in-phase again. The result is an AM signal! You should be able to hear these once the difference between the two carrier frequencies is in the audible range. So, for a 100 kHz reference I would expect to be able to hear beats if the detector oscillates somewhere between 100,050 Hz to 105,000 Hz giving an audible beat tone of 50 Hz to 5 kHz.

Answer 2

How does the circuit produce oscillations? Which part of the circuit converts DC into AC?

The entire circuit is an oscillator. It is generally a tuned collector feedback oscillator, and specifically a variant called the Hartley oscillator.

The coil fulfills two functions: it is the L element in a tuned LC circuit, and it is also a radiator (antenna). So this circuit is an oscillator and a weak transmitter at the same time.

Does the frequency actually change?

Yes. Conductive material near coils that are a part of tuned circuits changes the tuning of the circuit. This is used as one of many techniques of electrical position measurement.

If yes, how does this work with an AM radio?

An AM radio is just a tuned receiver.

The beat frequency is obtained between the oscillator output and some other reference frequency source. You can turn on a signal generator with a piece of wire on the output acting as an antenna, and tune it to beat with the "metal detector"'s oscillator. Or you can use an existing AM station as the source of the reference frequency, although you'll hear both the beat frequency and the AM station's baseband signal at the same time. So using a signal generator or another local oscillator may make life easier.