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For a typical mic a singer uses in a concert, how much current is generated when the singer sings? The current transmitted from the diaphragm into the wire.

What does the following specification tell us about the value I am looking for, for example?

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    \$\begingroup\$ Most microphones don't work like that. There's an amplifier inside the microphone as the actual current in the microphone's diaphragm is quite small, that small current is across a high impedance and that results in a voltage that is amplified. \$\endgroup\$ – Bimpelrekkie Aug 23 at 10:20
  • \$\begingroup\$ @Bimpelrekkie: Sounds like an answer. \$\endgroup\$ – JRE Aug 23 at 10:52
  • \$\begingroup\$ Thanks! I'd reckon there's an amplifier in the microphone. Coming to the point about that small current across a high impedance, is work done in amplifying the voltage? Could you pleas explain the mechanism? Pardon me, I have forgotten a lot of my physics. \$\endgroup\$ – Benjamin T Aug 23 at 14:09
  • \$\begingroup\$ by the way, if you want 32-bit noise floors in 40KHz music, you need 2,500 watts out of the transducer. For 24-bit noise floors in 50KHz music, you only need about 40 milliWatts from the transducer. \$\endgroup\$ – analogsystemsrf Aug 24 at 4:38
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All mics work by converting energy in a tiny vibrating diaphragm to electricity. There are two common ways that this is done. In a dynamic mic the most common type works like a loudspeaker in reverse (in fact you can use a LS as a poor quality mic, it works). Here the mic inpedance itself is typically around 200R, and it is designed to work into an impedance of about 2k. The actual sensitivity varies a lot, but under "normal" conditions we might see a voltage of around 5mV across the mic. This means that it is generating a current in the order of 2.5uA.

The other common type of mic is a capacitor (condenser) mic. There are various types but in these the diaphragm (which is part of a charged capacitor) is connected to a FET preamp in the mic itself. Not sure what the current would be for these (I expect it varies quite a lot with diaphragm size and so on) but anyway you never see that current at the XLR connector on the mic body. What you see there is the preamp output, which is designed to be of similar magnitude to a dynamic mic, so that it can use the same preamp (which is now perhaps better described as a secondary preamp).

The mic in your question is a condenser mic by the way. It quotes 33mV/Pa. According to this chart, 2Pa is the sound of a jackhammer (they don't say from how close, but loud!) - so 1Pa is still going loud - louder than most singers, except maybe that guy in ACDC. So I would think that our 5mV figure is in the right ballpark here, or perhaps slightly on the low side. (As there is really no "standard vocalist" you are not going to get a precise answer anyway.)

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    \$\begingroup\$ "there is really no standard vocalist" -> honestly, I wouldn't be surprised if there's an ISO standard for that. They have a standard cup of tea for goodness' sake! \$\endgroup\$ – Hearth Aug 23 at 12:23
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    \$\begingroup\$ Thanks for the very detailed reply. I need some time to synthesize it. \$\endgroup\$ – Benjamin T Aug 23 at 14:11
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The current transmitted from the diaphragm into the wire.

No, it isn't. This kind of microphone has an internal amplifier in addition to the diaphragm.

This amplifier operates from the DC "phantom power" provided by the mixing console (48 VDC @ 2 mA), and converts some of that to an AC current/voltage on the wire.

In this case, the rated load impedance is 1000Ω and the output voltage is given as 33 mV/Pa, so the output current would be 33 mV / 1000Ω = 33 µA/Pa.

Is that what you're looking for?

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  • \$\begingroup\$ Thanks, Dave! Yes, pretty much that's what I'm looking at. Oh, I was too fast; I didn't mean current transmitted from the diaphragm to the wire; I meant energy conversion. And oh, so this also answers my question on your reply above. Work is indeed done in the amplification. But in the absence of the amplifier, how much current can a typical microphone produce when it's diaphragm is receiving sound waves from a typical vocalist? \$\endgroup\$ – Benjamin T Aug 23 at 14:15
  • \$\begingroup\$ For a condenser element, basically zero, it is generally loaded by a fet gate and at least 1 Giga ohm of resistor! Voltage level? Maybe 10mV or so, but into the 1 Gig (sometimes 10 Gig) || 10pF (Fet gate), that is not a lot. Of course the output of the built in preamp is another story... The dynamic elements do a bit better generally having lower outputs, but at a MUCH lower impedance, so more current (and power) is available. \$\endgroup\$ – Dan Mills Aug 24 at 10:47

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