Good day Guys,

I am working on an AM Modulator and Demodulator circuit using the diode method. My modulation signal is a 10kHz Square Wave and I have a carrier frequency of 850 kHz. My modulation works just fine, according to my general knowledge, however, my demodulator is currently outputting a very weird looking wave, it resembles a square wave but is not close enough to know it was one, at first sight, my question is how can I improve the demodulation?

I have attached a screen shot of my circuit and the various outputs(Input Signal - Blue, Modulated signal - Red, Demodulated Signal - Green) below,

Note 1: I am using LtSpice.

Note 2: This circuit is just for me to learn how the modulation and demodulation works, and thus it has implemented ideal diodes and none practical component values.

Simulation Output Simulated Circuit

  • \$\begingroup\$ Look at the modulated waves envelope, it already is a low pass filtered version of the square wave. \$\endgroup\$ – PlasmaHH Feb 23 '17 at 12:19
  • \$\begingroup\$ @PlasmaHH Is there a way to improve this evelope ? I have added the 1 Ohm resistor in parallel to my LC bridge to reduce the transient response of my Modulation signal to try and improve the envelope. Is there more ways to improve this ? \$\endgroup\$ – 0x436f72647265 Feb 23 '17 at 12:22
  • \$\begingroup\$ it seems this is a good opportunity to learn how to use parameter stepping in ltspice to try out and compare the outcome of different values for components. look around at various nodes how your signal looks like there. I am not quite sure what you try to achieve with the filter. \$\endgroup\$ – PlasmaHH Feb 23 '17 at 12:24
  • \$\begingroup\$ The problem with a square wave is harmonic content (massive side bands), by transmitting just the fundamental you can reconstruct the square wave at the receiver using a Schmitt trigger. \$\endgroup\$ – JIm Dearden Feb 23 '17 at 12:55
  • \$\begingroup\$ Your screenshot quality makes it almost impossible to read the values which might make quite a difference, like for C1 is it 5u or 6u? Try making it 5uF and 5nH in that filter, then in the final filter make it 500kΩ 15p, a terrible amplitude but rather squarish it should be. \$\endgroup\$ – PlasmaHH Feb 23 '17 at 13:15

R8 (1 Mohm) and C4 (15 pF) have a time constant that is too long for decent 10 kHz demodulation. Try lowering R3 to 1 kohm. You will get more carrier ripple coming through but the square wave shape will be much improved.

Regards your modulator, it is a very practical use of a simple diode to achieve AM but your inductor is too small and your capacitor is too big for a practical design to use as a bandpass filter at 800 kHz. Try raising the inductor value by a thousand and lowering C by a thousand, then recompute R4 to be about a kohm. That output filter and its Q is fundamental to optimizing the modulation shape in this type of diode modulator.

  • \$\begingroup\$ Is there a rule of thumb for how long your time constant should be ? \$\endgroup\$ – 0x436f72647265 Feb 23 '17 at 12:58
  • \$\begingroup\$ Well, yes but it usually relates to sinewaves and slew rate. With a square wave, the resistor has to discharge the capacitor when the modulated carrier falls. Your current time constant is 15 us and the 10 kHz square wave has a half period of 50 us so you are significantly eating into the fall time. I'd go for one tenth of the half period as a rule of thumb but it's really down to what is important to you. \$\endgroup\$ – Andy aka Feb 23 '17 at 13:01
  • \$\begingroup\$ Thank you man, I appreciate the help. I have reduced my time constant to 3 us, and one can start to see the Square wave shape a lot nicer. Thanks for the help! \$\endgroup\$ – 0x436f72647265 Feb 23 '17 at 13:12

The signal modulating the carrier is a \$f_0=10\ kHz\$ square wave. Note that it has a quite high bandwidth: theoretically infinite, practically it's \$BW\approx10\ f_0\$.

A 100 kHz bandwidth on a 850 kHz carrier is way too much for the low order filters that a simple modulation-demodulation system like this uses. You have a ratio of \$\frac{BW}{f_c}\approx12\%\$

Either reduce the square wave frequency or increase the carrier frequency. Try with a ratio \$\frac{BW}{f_c}=1\%\$ and look what happens. Remember to readjust the filters according to the changes you make.

Also, try using a sine wave or a triangular wave as modulating signals. Their bandwidth is lower than that of the square wave.

  • \$\begingroup\$ Unfortunately I can't change the specification as this is an assignment for Uni, but I'll remember your advice in future! \$\endgroup\$ – 0x436f72647265 Feb 23 '17 at 13:08

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.