I have built an Amplitude Modulator using Proteus. Here is the schematic and output. The output does not seem to be correct however. Output


Can anyone point out what is the problem?

  • 1
    \$\begingroup\$ There's some amplitude modulation in there ... at a guess about 10% depth. If you were to high pass filter the output to eliminate the modulating waveform (or bandpass centred on carrier freq) you'd see it more clearly. Alternatively, subtract a percentage of the modulating signal from the outputi \$\endgroup\$ – Brian Drummond Jul 11 '16 at 13:21
  • \$\begingroup\$ I changed C3 to 0.1uF to form a high pass filter with a cut-off f=1.6kHz. Still makes no difference. \$\endgroup\$ – hacker804 Jul 11 '16 at 15:58
  • \$\begingroup\$ You'd need a better HPF than that. \$\endgroup\$ – Brian Drummond Jul 11 '16 at 17:24
  • \$\begingroup\$ The problem with the high pass filtering is that you have to pretty much decimate the 100 Hz signal to start seeing the wood for the trees. Try lowering the modulation to 10 Hz and upping the carrier to 100 kHz then have a 100 kHz high passs filter. \$\endgroup\$ – Andy aka Jul 11 '16 at 18:18
  • \$\begingroup\$ The other thing is that you may be modulating on the "early effect" and this produces less modulation than in the base saturation region. Try altering the bias points on the base this way or that way to see if you get an improvement in mod depth. You want the average/dc collector voltage to be no more than a couple of volts above the emitter to operate in the saturation region but it can be tricky to get optimized. \$\endgroup\$ – Andy aka Jul 11 '16 at 18:20

A single transistor used as a multiplier (modulator) is basically imperfect. Yes it will amplitude modulate the carrier but it will also inject the output with the modulation signal and this is what you have. As Brian Drummond says in his comment, there is amplitude modulation occurring AND regular additive mixing.

Look at the envelope valleys; the upper is about 2.2 big squares and the lower is about 3 squares: -

enter image description here

If these were equal there would be no modulation. Usually, to get rid of the base-band artifacts two emitter fed transistors are used with a differential output taken from both collectors. That differential output will contain identical base-band artifacts and therefore they will cancel out. However, to do this it's better to drive the bases differentially. This will work reasonably well: -

enter image description here

You can replace the transformer with collector resistors and look at the diff voltage across both collecotrs.

Or do what Brian says and high pass filter the output to remove the base-band signal.


You appear to have swapped the carrier and modulation inputs.

  • The carrier is the primary signal and should be applied to C1.
  • The modulation signal is applied to R4 and shifts the operating point of the transistor to modulate the amplification of the carrier signal.
  • \$\begingroup\$ In the whole scheme of analogue multiplication swapping those inputs shouldn't be a problem especially when the carrier is only 10 kHz \$\endgroup\$ – Andy aka Jul 11 '16 at 14:44
  • \$\begingroup\$ @hacker: Andy is better on this stuff than me. My first crystal radio (over 40 years ago) didn't work and I stayed away from radio pretty much after that. ;^) Can you post a screen grab of the result if you swap the inputs? \$\endgroup\$ – Transistor Jul 11 '16 at 15:18

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