The goal of this question. Demodulate the AM signal.

(Please assume a diode demodulating AM radio of the most imaginable simplicity.)

I am assuming an alternating current sine wave is coming into a wire from the antenna? In order for an envelope to coexist with the sine wave the amplitude is varying I assume?

If the antenna is the correct length then resonance of some fundamental frequency of the EM wave allows the signal to increase in strength in the antenna before it enters the diode? An assumption on my part.

The signal is then converted to DC but since the strength of the signal determines the fluctuations in the "envelope" the voice information is in "duplicate" so the bottom half is cut out of the rectifier , no? From this point I am lost.

The sketch shows a capacitor in parallel. I know what a capacitor does in principle. Is it attempting to maintain the voltage as the current changes since capacitors don't like to have their voltage changed and resist it, no? Please complete my picture to demodulate the AM signal.

  • 3
    \$\begingroup\$ Consider breaking your question into paragraphs, each containing a single thought, so that people will bother to read it. They'll be much more likely to answer if they don't have to struggle to read the question. \$\endgroup\$
    – The Photon
    Mar 20, 2017 at 16:38
  • 1
    \$\begingroup\$ See en.wikipedia.org/wiki/Envelope_detector \$\endgroup\$
    – Trevor_G
    Mar 20, 2017 at 16:46
  • 1
    \$\begingroup\$ quora.com/What-is-mean-diode-detector \$\endgroup\$ Mar 20, 2017 at 16:47
  • 1
    \$\begingroup\$ @JImDearden, don't don't use a mean diode in your circuit, it'll make all the other diodes feel bad. \$\endgroup\$
    – The Photon
    Mar 20, 2017 at 16:50
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    \$\begingroup\$ Just FYI: a capacitor is not needed. I used to work with a guy who grew up in West Berlin during the cold war. The americans operated a really high power AM broadcast transmitter to send news (and propaganda) to the people in East Germany. The guy I worked with told of "building" a crystal radio. Connect one terminal of a normal speaker to ground. Hold a diode in one hand by one leg, and touch the diode to the free terminal of the speaker - enjoy listening to Voice of America. \$\endgroup\$
    – JRE
    Mar 20, 2017 at 17:24

3 Answers 3


We need to start at the beginning.


At the source the music signal (represented by a sine wave) is combined (used to modulate) a high frequency carrier signal at the transmitter.

  • The music signal contains low frequencies (as humans hear from about 20hz to 20Khz)
  • The carrier signal is much higher frequency (say 100 KHz).

note: the carrier signal is a fixed frequency as each radio station operates at a different frequency.


This modulated signal is now transmitted a long way over the air.


Somewhere far away, say 20 km, we now need a way of recovering that original music signal from the 'received signal' we receive at the antenna.

To do this we are going to use a circuit called an envelope detector.

The envelope detector is first going to convert the AC signal into a DC signal.


Imagine the received signal (the modulated signal) is a 'messy' sine wave that varies 10V peak to peak (it goes from -5V to +5V).

We are going to ignore the negative half of this 'messy' sine wave.

A diode only allows current to flow one way, and hence has the effect of producing a constant voltage

  • the -5V to 5V signal has an average voltage of 0V
  • the 0V to 5V signal has an average voltage of 2.5V


Now imagine that this 'messy' sine wave is not a 'true sine wave' but is actually made up by a more complicated waveform that contains the original message signal (at low frequencies) and the 'messy' bits at high frequencies.

The envelope detector is now going to remove the high frequencies from the low frequencies. This is done with a lowpass filter (resistor and capacitor). A lowpass filter lets low frequencies pass but stops high frequencies from passing.

Now we have recovered the original signal!


Sometimes a picture can really help understanding: Image 1

I have tried to give you the bigger picture and have skipped some details here.

  • \$\begingroup\$ Ah ...the trick on the modulating end is the lower fundamental frequency will multiply the higher frequency to produce a new period. The new period is the mysterious "envelope" that everyone has been describing and eluding me. One other mystery is that when the signals are mixed why does it turn out to be a multiplication, could it not just as well be an addition? The demodulation appears more straight forward to me than the modulation. or am i missing the boat? \$\endgroup\$
    – Sedumjoy
    Mar 21, 2017 at 20:42
  • \$\begingroup\$ Several ways to implement AM modulation. (1) use a powerful audio amplifier to provide the plate-voltage to the RF-amplifier; (2) use a multiplier (usually performed on silicon, with closely located transistors for almost-identical behavior) to implement Vo = [1+sin(modulation)]*sin(carrier); (3) generate counter-rotating phasors at the sideband frequencies, and linearly sum with the carrier; \$\endgroup\$ Mar 22, 2017 at 16:47
  • \$\begingroup\$ It can't be an addition - because if you add a high frequency and a low frequency then you have a signal that contains high and low frequency. The idea of a carrier signal is that the high frequency "carries" the low frequency. Hence in order to carry the signal (to provide an envelope for the information it contains) we use one frequency to modulate the other (and modulation involves multiplication). \$\endgroup\$ Mar 22, 2017 at 23:37
  • \$\begingroup\$ If you generate phase-locked frequencies/sidebands, then a linear summation is the way to combine. \$\endgroup\$ Mar 23, 2017 at 3:06

The capacitor is not required for the demodulator to work. However it does improve its operation is two ways:-

  1. By charging up to the peak signal amplitude and holding most of that charge between peaks, it increases the audio output voltage.

enter image description here

enter image description here

Without any capacitance after the diode, the average output voltage of a half wave rectifier is Vpeak*0.318. With a suitably sized capacitor the audio output voltage increases closer to the peak value. The result is a stronger audio signal, potentially as much as 10*log(3.182) = 10dB louder.

In practice audio circuits have some capacitance of their own, which will affect the output to varying degrees depending on the rf frequency and complex impedance of the load. For example a magnetic headphone (commonly used in 'crystal' sets) has high inductance which resonates with stray capacitance at some unspecified frequency. An audio amplifier may be connected through a signal cable whose capacitance depends on length, and the amp's input capacitance may vary depending on the volume control setting. Having a parallel capacitor in the detector circuit reduces the effect of varying load capacitance and inductance, making the audio signal amplitude more stable and predictable.

However if the capacitance is too high then it will fill in the gaps between peaks enough to distort the troughs in the lower part of the audio waveform and reduce the amplitude at higher frequencies. To faithfully reproduce the original modulation there needs to be a balance between filter capacitance and load impedance.

  1. The capacitor filters out the rf frequency.

You can't hear rf frequencies but your amplifier might. Without low pass filtering the rf component of the signal is higher than the audio component. If the amplifier doesn't filter this out internally then its power consumption will be increased and it could produce distortion and spurious frequencies (which could be audible). Effective power output will also be reduced because the amp will overload at a lower audio signal level.

Even if the amp can handle this rf leakage without distortion, you now have high power rf signals in the audio circuit, which could feed back into sensitive rf stages to cause distortion or even oscillation. Since only a small amount of capacitance couples rf signals quite well, the amount of feedback could change depending on the position of speaker or headphone wires or the operator's body.


This circuit is a peak detector, with decay rate fast enough to allow the audio/music to appear on output.


simulate this circuit – Schematic created using CircuitLab

Here are some useful waveforms, showing a charge-fast, discharge-slow behavior enter image description here

  • \$\begingroup\$ This doesn't do much, if anything, to help explain the capacitor to the OP. \$\endgroup\$
    – Asmyldof
    Mar 20, 2017 at 18:01

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