5
\$\begingroup\$

In preparation for the amateur radio exam, I have been following the preparation course material by DJ4UF. In the section about modulation it presents this diagram of a simplified AM modulator.

enter image description here

The low frequency signal (NF) and the high frequency carrier wave (HF) are added, the diode cuts off one half-wave and the resonant circuit "recreates" the previously cut off half-wave. The result is a amplitude-modulated signal. So far so good.

What is the purpose of the 47k resistors? The explanatory text mentions that they are necessary to "add the currents in the diode". What would change if we remove them and just directly connect them to the diode?

\$\endgroup\$
9
\$\begingroup\$

The concept here is that you are adding the signal currents before feeding the sum to the diode; the resistors are there to convert the voltage sources "NF" and "HF" to current sources.

You'd get exactly the same effect by adding the signal voltages directly — simply connect the "NF" and "HF" boxes in series, without any resistors. The only downside to this is that "NF" and "HF" can't share a common ground, and that's often a desirable feature of a practical system. But some AM transmitters isolate the NF signal with a transformer, which solves that problem.


Note that the circuit as given is not at all practical — you would not want to feed a parallel-tuned circuit, which has a high impedance at resonance, from a current source. Instead, you would use a series-tuned circuit that keeps the diode cathode close to ground potential.

|improve this answer|||||
\$\endgroup\$
  • 1
    \$\begingroup\$ Two voltage sources fighting one another when connected in parallel could cause intermodulation distortion within one or both generators. One of those distortion products is exactly what you want out of the modulator. But other distortion products are not desired, and "Siebung" won't reject them. The series-connection of sources of Dave's answer is one solution and is far less lossy than those 47k series resistors. \$\endgroup\$ – glen_geek Oct 19 '18 at 14:13
  • 1
    \$\begingroup\$ @glen_geek Aha! So in a way these resistors also prevent that the signal of one frequency source interferes with the other source before being added, if I understood that correctly. \$\endgroup\$ – ahemmetter Oct 19 '18 at 14:30
  • 1
    \$\begingroup\$ @ahemmetter Yes. And those resistors also prevent the diode's non-linear behaviour from causing problems of intermodulation back in those generators. Consider this a conceptual modulator - it isn't very practical because its power efficiency is poor. \$\endgroup\$ – glen_geek Oct 19 '18 at 14:57
  • \$\begingroup\$ "Consider this a conceptual modulator" -- it wouldn't be totally unreasonable if it were being implemented at the low level suggested by the 47k-ohm resistors. There's probably better uses of circuit board space than something that would implement this particular block diagram, though. \$\endgroup\$ – TimWescott Oct 19 '18 at 15:46
  • \$\begingroup\$ @Dave-Tweed I strongly disagree with your comment about parallel vs serie tuned circuit. For this circuit to work, you need the high impedance at resonance and since you have this diode, it is mostly for DC that you need to draw the cathode to ground potential, which wouldn't work with a serie tuned circuit. \$\endgroup\$ – Camion Oct 21 '18 at 19:07
4
\$\begingroup\$

The problem is essentially that both HF and LF sources have a (relatively) low output our internal impedance (because if it wasn't the case and both had the same impedance, you might consider the 47K resistance to be inside both circuits).

So would happen what happen if you put two different voltage generator in parallel. the resulting loop would be one generator = LF-HF with no resistance. You would get a short circuit...

In the case those would have higher but different internal impedance, the result would then be that the output of the generator with the lowest internal impedance would take precedence.

With the two equal resistances bridge, you make sure the output is the average between both generators. Then the diode cuts a part of the HF according to the LF level, and the parallel tuned circuits eliminate everything which in not around the HF frequency. i.e. the continuous component, and the harmonics resulting from the diode cut.

|improve this answer|||||
\$\endgroup\$

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.