I have been reading up on semiconductors and all of the references that I have found say that the first practical application of the semiconductor diode was in crystal radios, and that semiconductor-based rectifiers quickly gave way to tube-based amplifiers.

So I am trying to understand why the rectifier is necessary at all. An excellent explanation of how a crystal radio works (and why it is now hard to get the components to build them) can be found here. For those who don't want to click, here is the circuit diagram:

Crystal Radio Circuit

So the coil and capacitor form a resonating circuit. Frequencies below a threshold go through the coil to ground, and those above a threshold go through the capacitor to ground, but those at the resonating frequency are stuck and have to go through the diode to the headphones. Every description of this circuit I have read say that the diode somehow demodulates the signal, and I just don't understand how it can do that. There is, say, a 88Khz carrier frequency which is AM modulated with a 300Hz-3KHz signal of the human voice. How does the diode, by chopping off the parts of the signal under the zero, do that?

  • 2
    \$\begingroup\$ The rectifier, er, rectifies -- turns high frequency AC into something closer to DC. \$\endgroup\$
    – Hot Licks
    Dec 6, 2017 at 18:39
  • 1
    \$\begingroup\$ Note that the diode alone doesn't do it. We also need a lowpass filter to remove the 88KHz signal from the output. In the above diagram, the large inductance of the headphones filters, leaving only audio. More common is to add a capacitor across the output. \$\endgroup\$
    – wbeaty
    Dec 7, 2017 at 10:19
  • \$\begingroup\$ @wbeaty can you please explain to me how the headphones act as a lowpass filter removig the carrier signal ...thank you....my name is Julius \$\endgroup\$
    – Sedumjoy
    Jun 4, 2019 at 23:46

4 Answers 4


The diode demodulates the AM radio signal. To demodulate (recover the audio signal) from an AM radio signal all that is needed is to retrieve the amplitude of the signal:

enter image description here

Source: this article

That's what the diode does.

It blocks the negative part of the wave but lets the positive part pass. This together with the capacitor recovers the audio signal.

Your example does not contain a resistor and a capacitor, they are present though. The headphones can only work on audio signals so it basically performs the same function (a low pass filter) without needing those components.


It's called an envelope detector. The diode prevents the base frequency from going negative. The original signal had an average value of 0. If you fed this through a low-pass filter (aka a capacitor), the output signal would be 0. With the diode in place, the signal can never go negative and now if you average out your signal using a low-pass filter, you get a slowly varying signal (relative to the base frequency) that no longer has an average of 0. This signal is now useful for the speaker.

enter image description here https://en.wikipedia.org/wiki/Envelope_detector

  • 1
    \$\begingroup\$ Wow. That's really cool! \$\endgroup\$
    – vy32
    Dec 6, 2017 at 15:36

Here's a physical description that might help intuitively -

Hum a 1kHz tone into a microphone, and broadcast it on a 100kHz AM carrier.

At your receiver, ideally you would like the earpiece diaphragm to alternately displace outwards and then displace inwards every millisecond, and for decent sound quality maybe you'll settle for having it alternately displace outwards and then rebound to equilibrium every millisecond.

Without the diode, your earpiece diaphragm will attempt to vibrate at 100kHz strongly for half a millisecond, and then more weakly or not at all for the next half millisecond. Even if the earpiece responds slightly at that frequency, your ear will not and you will hear nothing.

With the diode, for half a millisecond your earpiece diaphragm will be nudged outwards every 10 microseconds (5 microseconds at a time). Even without any extra filtering capacitors and thus with all those 5 microsecond gaps in the current, 500 straight microseconds of having the diaphragm continually nudged in the same direction at such close intervals should accomplish some displacement. That is, the mechanical characteristics of your earpiece will probably accomplish some of the actual demodulation when operating on a rectified signal. When operating on an unrectified signal however, those same mechanical characteristics will demodulate it to something close to silence.


Without the diode the average current in the headphones (1) would be 0, so there would be nothing to hear.
The diode acts as a non linear component (2) that create a no-null current in the headphones.
It happens that this current is proportional to the amplitude of the wave received by the antenna. This correspond exactly (3) the audio signal.

(1) average over say 0.1ms (what a hear can perceive)
(2) more precisely : non linear and not odd (that is to say, even, or with a certain "even effect")
(3) in amplitude modulation (AM)

  • 1
    \$\begingroup\$ The coil and tuning capacitor regenerate the original carrier. The regenerated carrier + the modulated carrier get mixed by the non-linear action of the diode. The mixing process -->a series of products - n.Fc +/- m.Fm; n, m are integers, Fc is carrier frequency and Fm modulating frequency. One result will be Fm and much of the small n and m sums and differences will remove most of the carrier, Fc. A small capacitor, say 100 pf, across your headphones will suitably attenuate Fc so it has no effect on the Fm signal reaching the headphones. Exactly the same process occurs in any mixing circuit. \$\endgroup\$
    – Brian
    Dec 13, 2017 at 2:06
  • 1
    \$\begingroup\$ @Brian What you say true, but one don't need reasoning in the frequency domain to understand what's happening. I think that, because people invest a lot of time do understand the frequency domain, they trend to forget the simple temporal domain \$\endgroup\$
    – andre314
    Dec 13, 2017 at 11:49
  • 1
    \$\begingroup\$ By the way, my answer was oriented to show that a lot of kind of nonlinearities are able to detect the AM. There's no need to detect exactly the envelope of the carrier. Of course there's the problem of gain, eventually necessity of amplification etc.., but it's not the subject here (nobody speaks of that here) \$\endgroup\$
    – andre314
    Dec 13, 2017 at 11:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.