0
\$\begingroup\$

I have an assignment on AM modulation.

I implemented a circuit and the output wasn't exactly correct, the output is not symmetrical about the x-axis.

I'll post screenshots which might explain the situation better. enter image description here

\$\endgroup\$
  • 2
    \$\begingroup\$ so, sadly your screenshot doesn't explain the problem at all. You need to describe the problem, then say what you think is wrong, show what you've tried to narrow down the problem or solve it, and most importantly, ask an exact question! \$\endgroup\$ – Marcus Müller Aug 14 '17 at 22:23
1
\$\begingroup\$

The emitter of Q1 is connected to ground which means if the collector has a negative voltage due to the V3 input voltage, the transistor is in reverse active mode.

You don't want that to happen as the reverse active mode has vastly different parameters than the forward active mode. Most manufacturers don't even specify parameters for this mode as they can't control them properly during manufacturing. Creating poor symetric transistors is simply not their goal.

You have to put the low frequency source V3 between the emitter and ground and connect the collector to V+, so the emitter is at a lower voltage than base and collector for any polarity of the low frequency source V3.

\$\endgroup\$
0
\$\begingroup\$

What you're seeing is not actually a problem. It is just a question of some of the modulating signal being added to the AM output signal. A better high-pass or band-pass filter at the carrier frequency would clean this right up.

Try doing a frequency-domain analysis of the output signal to see what I'm talking about.

\$\endgroup\$
  • \$\begingroup\$ Can you please explain the difference between applying the message signal to the emitter or the collector, and what is the purpose of each component in this circuit. \$\endgroup\$ – awstw Aug 15 '17 at 18:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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