I’ve always been under the impression that speakers require an AC signal. The reason being that positive and negative current are required for the electromagnet to properly vibrate the diaphragm.

Recently I’ve been doing some reading about AM modulation and envelope detectors. Most of the resources I’ve found about this show that the output of the envelope detector is a sinusoidal DC signal, but they frequently show that signal being output directly into an audio output or speaker. Will that work?

Here’s a link: https://www.electronics-notes.com/articles/radio/modulation/am-diode-detector-demodulator.php

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    \$\begingroup\$ A DC sine wave is basically an AC wave with a non- zero reference point and a frequency > 0. \$\endgroup\$
    – Passerby
    Nov 18, 2018 at 6:59
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    \$\begingroup\$ 'DC sine wave' needs to be defined. Do you mean DC + sine wave? Also, it depends whether a 0 Hz signal qualifies as a sinewave. \$\endgroup\$
    – Chu
    Nov 18, 2018 at 9:05
  • \$\begingroup\$ Is this what you are asking: electronics.stackexchange.com/questions/469627/… ? \$\endgroup\$
    – ocrdu
    Oct 16, 2020 at 21:38

2 Answers 2


it's a voice coil, so you theoretically can, but really, DON'T, if you do you may end up burning it, and it won't produce any sound except the click you will hear when connecting it. The click happens because you have a step signal when you connect the thing actually moving the speaker, however when the current stabilizes the speaker won't move at all producing no sound, the current however will keep flowing, and since the current is DC and there is no motion, you won't have inductive reactance nor counter electromotive force meaning that the only thing limiting the current is the resistance of the coil, which is far lower than the impedance of the speaker, so the thing will heat up until the insulation starts to burn, at that point you have ruined your device


That answer does not make sense. If you have a typical AC audio signal, and you cut off the lower half of the waveform, say with a diode, you have a DC signal of no greater power. Plus, current is at zero for half of the time, therefore it would enjoy the same OR BETTER cooling time as the full AC.

The only real difference then is that the natural tension of the spider and surround are solely responsible for returning the speaker's cone to its original position. The lower half of the AC sinewave not being present to power the cone back.

This might work at lower frequencies, albeit not very efficiently, but certainly the physical structure could not respond in time for a high frequency signal. Plus, if the physical structure had enough strength to pull the cone back during the last half of a cycle, then it indoubtedly would have too much resistance to motion to operate efficiently during the first half, requiring MUCH more power to be applied for the same output.

There is one other issue: only a half-sinewave signal (AC with a diode) could be used with a speaker, not an actual DC signal. A "sinusoidal DC" signal IS an AC signal, just with the reference point moved. (A stepped DC could also be used because once agsin it is the essentially a rough sine wave or AC). Pure unmodified DC is a "square wave", which means current goes from zero to full power and back to zero in an instant, for each cycle. This creates enormous stresses compared to the gentle up and down motion of a sine wave (AC). DC literally will tear the cone apart at any good amount of power applied.

DC also does not sound "musical", so even if you could create a speaker cone that wouldn't be torn apart by reversing direction, it would not sound good.

So it appears that the issue is a more practical one of the physical, not an electrical one. The first answer above is false.

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    \$\begingroup\$ Welcome :-) Please read the tour and help center to see the rules and site philosophy. FYI, phrases like "The first answer above [...]" and "That answer [...]" don't work here, since the order of answers changes depending on things like activity (e.g. edits & new answers), votes and user settings. To refer to another answer, copy the URL from the "share" link under that answer and use that as a reference. \$\endgroup\$
    – SamGibson
    Oct 16, 2020 at 19:56
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    \$\begingroup\$ I also think you may have interpreted diegogmx's answer incorrectly, I think they assume that the signal is literally DC (eg. speaker connected to a battery and absolutely nothing else) \$\endgroup\$
    – Oskar Skog
    Oct 16, 2020 at 20:46
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    \$\begingroup\$ There is probably some confusion here about the definition of "DC" (constant vs. possibly varying but always positive) and "AC" (varying but possibly all positive vs. varying between positive and negative). \$\endgroup\$
    – ocrdu
    Oct 16, 2020 at 21:45
  • \$\begingroup\$ To clarify: I was referring to diego's comments. Further breaking down what they said; unless the current of the DC signal surpasses the current of the AC signal, the ONLY difference in coil heating would be the time during each quarter of the AC cycle that the signal is not at full amplitude. However, all but the cheapest speakers are designed with a LOT of room for overcurrent situations, so unless you are running the thing hot enough that the AC current ALSO burns the coil, the DC signal should be particularly damaging. It is the extreme stress at the start of the dc pulse that harms it. \$\endgroup\$ Oct 18, 2020 at 0:08
  • \$\begingroup\$ BTW, this is why we older audio techs laugh at the "digital speakers" marketing claims. Even "digital ready" means nothing practical in a speaker, which is inherently analog and AC. \$\endgroup\$ Oct 18, 2020 at 0:15

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