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I do not completely understand how this circuit works:

enter image description here

First, I do not understand how the LC filter selects the frequency. I know that it's impedance is 'infinity' at the resonant frequency, but I don't understand how it rejects the other frequencies: no matter the impedance of the LC filter for a given frequency, the voltage is applied at the 'X' node and hence every frequency is applied at the input of the diode.

I do not understand the concept of 'shunting a signal to ground.'

Second, I do not understand how the R1/C2 network works.

It is said that its purpose is to 'smooth' the input signal in order to get only its envelope at the amplifier input. How does it work? Because the output signal of the diode (the AM signal of the selected frequency) is directly applied at the input of the amplifier, so I don't see how the R1/C2 network can smooth anything.

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First, I do not understand how the LC filter selects the frequency. I know that it's impedance is 'infinity' at the resonant frequency, but I don't understand how it rejects the other frequencies: no matter the impedance of the LC filter for a given frequency, the voltage is applied at the 'X' node and hence every frequency is applied at the input of the diode? I do not understand the concept of 'shunting a signal to ground'.

You forget that your antenna doesn't have zero impedance (it's not a perfect voltage source, it's more of a pretty imperfect current source): all the current that flows through the LC filter can't flow through the diode – it's simple as that.

Second, I do not understand how the R1/C2 network works? It is said that its purpose is to 'smooth' the input signal in order to get only its envelope at the amplifier input.

Ah, ok, whoever said that mixed a few things up.

Again, C2 shunts high frequencies to ground (makes sense, right? for high frequencies, the capacitor becomes nearly a short), and for that to reduce the voltage that the input of the amplifier sees, there needs to be a source impedance (a series resistance). That's why a RC lowpass filter looks like

input voltage >----| R | --+------> output voltage
                           |
                         C =
                           |
 GND ----------------------+------

Probably whoever claimed R1 was involved in the smoothing thought it was the R in that RC filter network. It's not, because it's not in series, but in parallel to the C. The role of the R in that filter is taken by the voltage drop over the diode.

The R in the circuit above probably just serves to bias the voltage at the output of the diode around ground.

How does it works? Because the output signal of the diode (the AM signal of the selected frequency) is directly applied at the input of the amplifier, so I don't see how the R1/C2 network can smooth anything.

As said, current through the diode leads to voltage drop over the diode. High frequency voltages over a capacitor lead to high current through the capacitor. So, this attenuates/suppresses high frequency voltages.

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  • \$\begingroup\$ Thank you for your explanation! However I have some questions: 1) I'm not sure to understand the sentence: 'The role of the R in that filter is taken by the voltage drop over the diode'. I would have understood if the V/I relation was linear, but the relation is exponential and we approximate V ~ 0.7V. Does it mean we replace the diode by it's dynamic resistance? 2) How does the R in the circuit bias the voltage at the output of the diode, and why is it required to do so? Thank you very much. \$\endgroup\$ – Wheatley Jun 20 at 14:41
  • \$\begingroup\$ 1) yeah, for the purpose of the RC filter, that's OK, the diode-C filter doesn't really need to be a "well-defined" and mathematically easy-to-calculate filter. 2) assume you've got zero signal at the antenna for a while. No matter what the charge over C2 was before, after that it's zero. You need that because no current flows into the amplifier or "backwards" through the diode. \$\endgroup\$ – Marcus Müller Jun 20 at 15:18
  • \$\begingroup\$ If you treat the antenna as a current source (like you suggest in your 1st paragraph), the function of the R in the filter might make more sense. \$\endgroup\$ – The Photon Jun 20 at 15:27
  • \$\begingroup\$ @ThePhoton absolutely true, but I'm almost certain R1 was meant to avoid simply only continuously charging C1. Also, I'd like to consider the output of the diode as a "high-impedance voltage source", because otherwise it gets unnecessarily complex to mentally model what the rectification does; but that's mostly a personal thing, I guess. \$\endgroup\$ – Marcus Müller Jun 20 at 15:33
  • \$\begingroup\$ Or, just regard the diode, capacitor and resistor is a half wave rectifier with smoothing; the R adds a touch of ripple at the carrier frequency but, largely, the carrier becomes the envelope of the carrier just like in a regular AC half wave rectifier circuit. \$\endgroup\$ – Andy aka Jun 20 at 15:35
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This circuit captures a frequency which is banded with the LC filter. Then the diode work as a detector. The signal drops across r1 and the c1 filters high frequency signals to the ground and let the low frequency signal go inside the amplifier. (I have a confusion though what is the c1 doing there) This while system is a diode detector AM receiver(Demodulator).

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  • \$\begingroup\$ Maybe add in "envelope" in there somewhere. \$\endgroup\$ – DKNguyen Jun 20 at 22:33
  • \$\begingroup\$ I would like to add that too. But I don't have full concept about envelope too. This could maje things more complex. \$\endgroup\$ – Sohan Arafat Jun 20 at 22:36
  • \$\begingroup\$ I was just thinking that mentioning the diode works to detect the envelope of the signal would be enough. But if you want more detailed, the diode rejects the negative polarity of the signal so that C1 can be charged only by the positive polarity. This results in C1 smoothing out waveform which gets rid of the carrier and leaves only the amplitude (aka the positive envelope). R1 gives the cap a chance to discharge so the envelope can go low where required or else the capacitor could only get higher and higher. and never go low. \$\endgroup\$ – DKNguyen Jun 20 at 22:40

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