2
\$\begingroup\$

I’m fairly new to electronics / electrical engineering, and I’ve been reading about amplitude modulation because I find it really intriguing.

I found the following AM modulator circuit in this answer by Andy aka to a previous question and tried it out (Carrier is V2, Message Signal is V1). It works great in practice, but I don’t really understand how it works or why it’s capable of modulating the signal.

Circuit

LTSpice Simulation Results

In particular, I’m confused about:

  • How this circuit actually modulates the carrier with the input (message) signal
  • Which part of the circuit causes the carrier amplitude to vary
  • Why the peak-to-peak of the modulated signal is 1.25V instead of the messages 2.5V

I have a rough theoretical understanding that AM means changing the amplitude of a high-frequency carrier according to a lower-frequency signal, but I can’t quite connect that idea to what’s happening inside this circuit.

A simple, step-by-step explanation of the signal flow would be really helpful. Time-domain intuition would be especially appreciated, since I’m still learning how to think about these circuits.

Thanks a lot, I’m enjoying learning this stuff and would love to understand it more deeply instead of just knowing that it “works.”

\$\endgroup\$
1
  • \$\begingroup\$ Your example modulator is actually a double-sideband somewhat suppresed-carrier modulator. To make it into an AM modulator, add a D.C. offset to V1...something like PULSE(1.25 1.25 0 50u 50u 0 100u) \$\endgroup\$ Commented 14 hours ago

2 Answers 2

Reset to default
1
\$\begingroup\$

How this circuit actually modulates the carrier with the input (message) signal

The diode is used as a blocker so, if the modulating signal is different in amplitude than the carrier signal then it limits the carrier amplitude. In limiting the carrier amplitude it imposes the shape of the modulating signal onto the carrier envelope (on the left of the 10 kΩ resistor). The trick is to keep the modulating signal peak less than the carrier peak signal so that there is always carrier envelope shaping occurring.

Here's what the signal looks like on the input to the buffer for reference: -

enter image description here

This is amplitude modulation but, it also contains the modulating signal as well. That modulating signal is removed by the band-pass filter leaving the modulated carrier at the output.

Which part of the circuit causes the carrier amplitude to vary

The diode is the crucial part of the circuit.

Why the peak-to-peak of the modulated signal is 1.25V instead of the messages 2.5V

I think you mean "modulating" signal rather than "modulated" signal. The modulated signal is the modulated carrier so I think you will agree.

Anyway, the circuit is so very simple so you have to get amplitude levels in the right area for this to work without excessive distortion. For instance, in my original circuit I used DC offsets to tweak the modulation process: -

enter image description here

It wasn't a definitely thought-out design; I just wanted to show how really clean AM can be obtained with a low component count and little circuit sophistication.

You might get more insight by using a buffer (as per my original circuit) and looking at the waveforms on the input to said buffer.

\$\endgroup\$
3
  • \$\begingroup\$ Thanks a lot for this thorough explanation. For the last part, I did mean the modulated carrier (Vout), it goes from -0.75V to 0.75V, while the “message” (V1) goes from -1.25V to 1.25V. \$\endgroup\$ Commented 15 hours ago
  • 1
    \$\begingroup\$ @GabrielStan the two resistors also form a potential divider so thinks reduce a bit more. I added the waveforms at the output of the diode for ease of understanding. \$\endgroup\$ Commented 15 hours ago
  • \$\begingroup\$ Thanks a lot, I think I understand it now. \$\endgroup\$ Commented 15 hours ago
1
\$\begingroup\$

Another approach to Andy's fine answer:

Andy says, "The diode is used as a blocker".

A modulator can be made by thinking of the diode as a switch: it is in one of two states: open or else closed.
You can use a switch in a few different ways to either pass the baseband signal on to a load, or force the baseband signal to zero. It switches on/off at a rate determined by the carrier frequency, which is much higher than the baseband frequency.
Note that the baseband signal generator (V1, V2, below) has a peak amplitude of one volt added to a DC offset of one volt. So it swings between +2V on its top peak down to 0V on its bottom peak. This represents 100% modulation. With no AC peak amplitude at V1,V2, baseband would be steadily at +1V due to the D.C. offset:

schematic

simulate this circuit – Schematic created using CircuitLab time domain of "out_1" When you send the buffer's output on to an LC tuned circuit resonant at the carrier frequency of 25 kHz, the 1V offset is rejected, and the bottom half becomes inverse of top half.
If you were to send the output "out_1" to an antenna that can radiate a 25 kHz signal, the D.C. offset of 1V would not be radiated, but the 25 kHz carrier wave (along with its modulation) would be radiated.
However, harmonics of the 25 kHz signal would also be radiated, because it is squarely chopped. Using a 25 kHz resonator before passing to the antenna should reject harmonics, leaving the 25 kHz component (with its modulation) alone.

An electronic switch made from a FET transistor:

schematic

simulate this circuit

\$\endgroup\$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.