According to Wikipedia, amplitude modulation is achieved by multiplying the carrier signal with message signal.

Where is this multiplication performed in this AM transmitter circuit?

enter image description here

It seems to me (noob) the transformer only steps up the voltage of the message but does not multiply it with the carrier signal (1MHz) generated by the crystal?

Here is a video of the circuit in action.

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    \$\begingroup\$ Traditionally, in some sort of mixer, either explicit or implicit. Your circuit appears to be modulating the power supply voltage. \$\endgroup\$ – Chris Stratton Jun 28 '18 at 19:42
  • \$\begingroup\$ This is the complete circuit and works as is. \$\endgroup\$ – Jet Blue Jun 28 '18 at 19:45
  • \$\begingroup\$ It's dependent on a large unidentified black box which does all the work. We can guess what that is (likely with some accuracy), but without specifics of the internals that would only be a guess. Also, if it's what it likely is, it's not legal to operate, due to harmonics. \$\endgroup\$ – Chris Stratton Jun 28 '18 at 19:46
  • \$\begingroup\$ @ChrisStratton These are all the components and connections used...In this video you can see it assembled and working...no black box... \$\endgroup\$ – Jet Blue Jun 28 '18 at 19:52
  • \$\begingroup\$ @ChrisStratton No traditionally AM is not done by a mixer. This incorrect statement has been made several times in ESE in answer to this question. \$\endgroup\$ – Henry Crun Jun 28 '18 at 19:56

What you have drawn shows how it is hooked up and does not show a proper circuit.

With some guessing I see how it can work though.

The supply voltage of the crystal oscillator is varied (modulated) and that gives very weak amplitude modulation because those changes in the oscillator's supply voltage will result in changes in the amplitude of the oscillator's output signal.

This is not a proper AM transmitter, it's a hack. For example, a proper AM transmitter would transmit a sinewave containing only one frequency, for example 1 MHz. This circuit outputs a square wave which means it outputs a whole bunch of frequencies, all the uneven multiples of 1 MHz. So it will transmit 1 MHz, 3 MHz, 5 MHz .... etc.

It's a "poor man's" circuit and behaves as such. Don't expect miracles from it and don't connect a long (efficient) antenna to it as you will disturb AM reception in your area.

  • \$\begingroup\$ The modulation arrangement is a proper , and traditional, AM transmitter. But you are quite right that its anti-social to connect an antenna. And modulating the oscillator is just wrong. \$\endgroup\$ – Henry Crun Jun 28 '18 at 20:43
  • \$\begingroup\$ If you look at the amateur radio frequencies you see that they are even harmonics of each other, so that the hard to avoid 2nd harmonic, falls into another ham band, not on a commercial user \$\endgroup\$ – Henry Crun Jun 28 '18 at 20:45
  • \$\begingroup\$ A perfect square wave only has odd harmonics 1,3,5,7 etc. In practice the oscillator module will have some 2nd, but it should be well less than the 3rd \$\endgroup\$ – Henry Crun Jun 28 '18 at 20:46
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    \$\begingroup\$ You're right about the uneven harmonics, but my point is that this isn't a proper AM modulator circuit so I would never call it "proper and traditional". \$\endgroup\$ – Bimpelrekkie Jun 28 '18 at 20:50
  • \$\begingroup\$ I really meant the modulation arrangement. You are quite right this is a hack. But that is the great beauty of AM - so fabulously easy to make. My first transmitter was this: 1 transistor oscillator on CB band with heising modulation. Gave about 10km range - until the transistor metal turned grey from the heat after half an hours or so. \$\endgroup\$ – Henry Crun Jun 28 '18 at 20:54

AM is not made by a multiplier or mixer. This is often repeated here, and completely incorrect. It probably comes from mathematical explanations given in text books, and taught at uni, but not really understood by those repeating it.

AM is done in the first instance, by changing the power supply voltage to the transmitter output stage, so that the AMPLITUDE is changed up and down.

Your arrangement is called Heising Modulation. It is the easiest arrangement. DC passes through transformer to the transmitter. With no modulation there is 9V on the transmitter.

Now lets say we have +/-5V peak AC modulation and a 1:1 transformer. The (AC) modulation will be superimposed onto the DC battery supply.

At +ve peaks, this adds to the 9V, so we get +12V on the transmitter, and the RF amplitude goes up.

On negative peaks, we get 4V on the transmitter, and the RF amplitude drops.

The level of the AC is set to get enough modulation. At some voltage >0, the RF output drops to 0. At this point we get overmodulation.

You would not of course modulate the power to your oscillator. But this arrangement works excellently feeding the power to VDD of 74HC04 cmos invertor as an output stage. You would feed 4V through the transformer, and modulation would swing VDD from 2-6V

Now you can make AM using an analog multiplier or mixer and carrier injection (i.e. multiplication), but you don't. If you did you would need to amplify it with a linear amplifier. These are expensive, relatively difficult, and have low efficiency (<25%), and worse, all the modulation power is also dissipated in the RF amplifiers.

A Class C amplifier that was traditionally used has 60% efficiency, and importantly, the modulation power is being dissipated in a (cheap) audio amplifier stage, so the RF devices required can be 4x smaller.

Modern transmitters use class D,E and can get even better overall efficiency.

Frequency sidebands (mixing products) are a side-effect of amplitude modulating a transmitter output stage.

In receivers and transmitters this effect can and is be used to make frequency convertors (mixers) at low power levels.

Noise is a side-effect of a two stroke motorcycle. It does not make the motorcycle a "noise generator" nor the RF amplifier a "mixer".

In fact in an AM signal, while it is interesting that we get sidebands (mixing products), they are not actually a functional part of the whole AM system. In the receiver all the frequency products are lumped together, and the amplitude measured with a diode. An AM system can be made with no regard to any mixing products and the consequences (bandwidth) whatsoever. (and in the early days before decent selectivity, it was).

  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Dave Tweed Jun 28 '18 at 21:21
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    \$\begingroup\$ The mechanism described here as different than mixer, is in fact only a type of mixer. The actual fact, readily seen in the frequency domain, is information is contained only in the sidebands and put there by mixing, and it is only because of this that AM works at all. One doesn't need to understand that fact to build a simple radio from instructions, but it is unarguably a fact. Attempts to claim otherwise only proclaim the speaker's lack of understanding of the subject. \$\endgroup\$ – Chris Stratton Jun 28 '18 at 21:30

This circuit adds audio transformed with some step up ratio to make the audio more sensitive at low levels .

The FOX 5V 1MHz XO was a clipped sine out with a level somewhat proportional to supply voltage from perhaps <3V , underspec to about 7V (overspec) .

I would expect it so sound crappy but it transmits on any long wire for a short range.

So in effect the XO supply voltage performs the AM.

Don’t use 9V unless you feel lucky? Or bought spares.

Nowadays 3.3V XO’s are more common from Fox. Look around. For any brand.

  • square waves means you get harmonics. But the transformer needs to be tiny or better ferrite to be less lossy. Otherwise you can skip the transformer and AC couple across a 100~1k Ohm in series from 5V or 4 alkaline to the XO from the headphone jack
  • \$\begingroup\$ the transformer is carrying audio, and would be an iron core audio output transformer of the kind formerly used in radios as a speaker transformer. \$\endgroup\$ – Henry Crun Jun 29 '18 at 1:18
  • \$\begingroup\$ the old clipped sines were just a bipolar transistor so quite happy with 9V I expect. For small modulation index, it will be linear, and the audio quality would be good - probably excellent in fact, because with 11:1 turns ratio, and (say)2 V in, you would only have 2% modulation - its got to be linear. Although, I suspect the turns ratio is actually the other way around 8ohm to the phone, 1000 to the osc, and as you say - distortion. \$\endgroup\$ – Henry Crun Jun 29 '18 at 1:59

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