0
\$\begingroup\$

I'm designing AM modulation and demodulation circuits. The AM modulation works fine but when I connected it to the demodulator (I'm using envelope detection) the modulating signal changed to the original message.

Amplitude modulation circuit design:

Am modulation circuit design

the modulation output
the first one from the top is the carrier. the second one is the message. the third one is the modulation output

The modulation output the first one from the top is the carrier. The second one is the message. The third one is the modulation output.

When the envelope circuit is connected, this happens:

the third one from the top is the modulation output. The fourth one is the Demodulation output.

The third one from the top is the modulation output. The fourth one is the demodulation output.

AM modulation and demodulation circuit design:

AM Modulation and Demodulation Circuit design.

\$\endgroup\$
2
  • \$\begingroup\$ A capacitor before the diode should be used. If not, DC supply will "bypass" the diode. \$\endgroup\$
    – Antonio51
    Commented Jun 18, 2022 at 17:21
  • 2
    \$\begingroup\$ 500 ohm load is much, much too low. Increase it to 100k ohm at least. Reduce C2 by a similar factor. You will also have superimposed DC because the collector will be at VCC. The 1N4001 is a terrible diode to use for demodulation. \$\endgroup\$ Commented Jun 18, 2022 at 18:14

1 Answer 1

1
\$\begingroup\$

Simulating an "Additive" mixer is a bit difficult with BF at low frequency and LO (Local Oscillator) at high frequency.

Important thing: choose L & C for resonance @ frequency of the local oscillator (1 MHz).
Choose also resistance in parallel not too high.

enter image description here

And here is harmonic content of output.

enter image description here

Here is an example (another type of demodulator, quadratic).

Note the use of an emitter follower between the "mixer" and the "demodulator".
So, the "demodulator" does not interfere with the "mixer".

Note also the use of an L-C filter, example.

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.