[ProAudio] Microphones question

Bill Whitlock engineer_bill at verizon.net
Sat Jun 12 16:49:27 PDT 2021

Power is irrelevant here!  Mic preamps and virtually every electronic amplifier I'm aware of is a voltage amplifier.  Power transfer had meaning when signal sources and loads were passive ... in the earliest days of telephone systems - there were no amplifiers, so power transfer was an issue.  We certainly don't add an 8 Ω resistor in series with the output of a so-called audio "power" amplifier so it can have its maximum power transfer!  The "power" in audio power amplifier is a long-standing misnomer ... technically, it should be called a voltage amplifier with high output power capability.  The reason IEC and many other standards specify low device output impedances and high input impedances (for example, <2.2 kΩ and >22 kΩ respectively for consumer gear) is to transfer 90% or more of the signal voltage.  This is one of the most commonly heard misunderstandings when I teach my seminars on signal interfaces and grounding!  Right next to "balanced lines require equal and opposite signal swings" ... which is completely indefensible BS!
Bill WhitlockAES Life FellowVentura, CA

-----Original Message-----
From: Dan Lavry via ProAudio <proaudio at bach.pgm.com>
To: Bill Whitlock <engineer_bill at verizon.net>; proaudio at bach.pgm.com; chris at chriscaudle.org <chris at chriscaudle.org>; 6807.chris at pop.powweb.com <6807.chris at pop.powweb.com>
Sent: Sat, Jun 12, 2021 4:25 pm
Subject: Re: [ProAudio] Microphones question

Scott is talking about maximum power transfer. A voltage source V with series source resistance RS is connected to load RL. If RL is open, there is no load current. If the load is a short, there is no load voltage. In either case no power on the load, it takes both current and voltage.So somewhere between zero and open there is a resistance for peak power transfer. It happen to be the peak of a parabola curve at RS=RL.I do agree that peak power transfer is not the only consideration, and different tradeoff is in order.
Dan Lavry

Sent from Samsung Galaxy smartphone.

-------- Original message --------From: Bill Whitlock via ProAudio <proaudio at bach.pgm.com> Date: 6/12/21 3:12 PM (GMT-08:00) To: proaudio at bach.pgm.com, chris at chriscaudle.org, 6807.chris at pop.powweb.com Subject: Re: [ProAudio] Microphones question 
Where does that come from?  The lowest noise at the output of an amplifier results from the source impedance being equal to or less than the "optimum" for a particular amplifier - where e-noise and I-noise are equal contributors to total noise. This is determined by the first-stage device and its operating point.  The amplifier's input impedance is irrelevant.  For example, a transistor with a collector current of 1 mA may have equal e and i noise with a 500 Ω source impedance, but the input impedance of the stage could be 100 kΩ!  In most amplifiers, an intentional resistor is added only to lower input impedance for mic loading purposes 
See Motchenbacher & Fitchen, "Low Noise Electronic System Design" page 62 and others.
Bill Whitlock

-----Original Message (in part)-----
From: Scott Dorsey via ProAudio <proaudio at bach.pgm.com>
To: proaudio at bach.pgm.com; chris at chriscaudle.org; 6807.chris at pop.powweb.com
Sent: Sat, Jun 12, 2021 2:42 pm
Subject: Re: [ProAudio] Microphones question

In a perfect world the lowest noise would be when the input impedance of the
preamp matches the output impedance of the microphone, ....

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