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I started reading "Build Your Own Transistor Radios" by Ronald Quan and I don't get how the envelope detector works in his TRF radio.

enter image description here

I know that a "classic" envelope detector looks like this:

enter image description here

But in the first example the D3, C7 and R7 looks like they have been "reversed" and I don't understand the way current flows and why it works. Why didn't the author choose the "classic" representation where D3 faces the C7 and R7?

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    \$\begingroup\$ Ok, they are "reversed", but if you think that in fact the power supply is the "ground", it is ok. \$\endgroup\$
    – Antonio51
    Commented Dec 2, 2023 at 16:44
  • \$\begingroup\$ @Antonio51 If you put it like that it starting to make sense but can you please elaborate why do you consider the power supply to be "ground"? \$\endgroup\$ Commented Dec 2, 2023 at 16:52
  • \$\begingroup\$ These circuits are functional in DC or AC. In AC, we “short-circuit” the DC sources to study them. \$\endgroup\$
    – Antonio51
    Commented Dec 2, 2023 at 16:56

2 Answers 2

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It doesn't matter which polarity half of the modulated waveform you choose to rectify in order to detect the audio. It works just fine for both but, in the particular case of your circuit, the author chose to rectify the portion of the carrier that was more negative at the collector of Q. It makes no difference at all in reality: -

enter image description here

Image from here and, as you can see, the upper envelope and the lower envelope are mirror images of each other but, they will sound exactly the same when processed into audio.

If I'd have shown the detector circuit using two diodes like this would you have been confused: -

enter image description here

Maybe this circuit won't confuse you: -

enter image description here

They can all be made to work as envelope detectors.

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  • \$\begingroup\$ Thank you for the answer, I get that the negative peak is symetric to the positive peak, what I don't understand is how the current is flowing during rectification. The way I see it, during the positive peak of the signal (when the voltage accross R5 || L3 is at maximum) current flows through the collector and charges C7, right? But how does it flow during the negative peak (when voltage across R5 || L3 is at minimum)? \$\endgroup\$ Commented Dec 2, 2023 at 16:56
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    \$\begingroup\$ It flows back through the 100 k resistor. \$\endgroup\$
    – Andy aka
    Commented Dec 2, 2023 at 17:07
  • \$\begingroup\$ Aha that was the clincher \$\endgroup\$
    – Andy aka
    Commented Dec 2, 2023 at 17:34
  • \$\begingroup\$ Antonio51's comment on the question also shed some light, my confusion was due to not understanding fully the voltage changes across the (R5 || L3), in my misunderstanding I thought that the voltage peaks were at the junction of (R5 || L3) with the DC voltage source, and the junction of the D3 with the collector was at the same potential, but it's the other way around. \$\endgroup\$ Commented Dec 2, 2023 at 17:46
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It's actually not that complex. When you take an average value of an AC signal of zero mean by low-pass filtering, the result is near zero.

By getting rid of the negative (or positive -- it doesn't make a difference which) part of the signal, the low-pass filtering results are much closer to the envelope.

The diode in your circuit is a half wave rectifier. The LPF result will be even closer to the envelope by using a full wave rectifier (i.e., inverting the negative part of the signal instead of just zeroing it out), but often, half wave gives you a "good enough" answer.

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