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I demodulated a signal using this simple envelope detector. When I checked the output, the amplitude was decreased - it was not the same as the original signal that was modulated (from the forward voltage of the diode. Please correct me if my understanding is wrong.) Do the resistor and capacitor also contribute to this attenuation of amplitude? If so, is there a way where I can reduce the attenuation by choosing the right pairs of R and C?

enter image description here

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  • \$\begingroup\$ The demodulated amplitude look pretty much the same in your image of before and after. Try drawing the waveforms you saw \$\endgroup\$
    – Andy aka
    Dec 30, 2022 at 16:06
  • \$\begingroup\$ In general, with signal processing equipment like this, the last thing you care about is how much a stage amplifies or attenuates the signal. You care about how much the signal is distorted and how much it is corrupted by noise -- but unless you're building a crystal radio, it can come out of your demodulator weak as can be and as long as it can be amplified up to a useful level you shouldn't care. \$\endgroup\$
    – TimWescott
    Dec 30, 2022 at 19:12

2 Answers 2

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The output will be decreased, the diode impedance and resistor form a voltage divider so you're not going to get the same voltage out as you put in.

The capacitor is to remove the RF component of the signal. The diode allows the positive half of the signal through which charges the capacitor, the resistor allows the charge to drain off. the RC time constant will affect how well the detector works, so if you change the resistor to reduce the signal attenuation you'll probably need to change the capacitor value as well. You would increase the resistance and decrease the capacitance to keep the RC time constant the same.

The best option would be to follow the detector with an amplifier to bring the signal back up to where you want it. This could be a single transistor stage or something like an opamp.

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Do the resistor and capacitor also contribute to this attenuation of amplitude?

"Attenuation" is the fact of the diode (non-ideal).
It can be better with a Germanium diode, but it is always there.

If so, is there a way where I can reduce the attenuation by choosing the right pairs of R and C?

The choice of R & C is dictated by the highest frequency of the signal modulator.
If not well chosen, you get what is called "diagonal clipping" as shown in picture.

enter image description here

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