# Electret microphone ?

Discussion in 'Electronic Basics' started by phil, Apr 29, 2004.

1. ### philGuest

Hello all,

This is actually a continuation of my previous post. The circuit in question
is http://www.uoguelph.ca/~antoon/circ/fmt1.htm and the question is about
electret microphones. I have pulled the specifications of an electret from
maplin, as follows:

Response: Omni-directional
Frequency response: 50Hz to 13kHz
Sensitivity: -60dB ±3dB (0dB = 1V/µbar at 1kHz, Vcc = 4·5V, Rl = 1k
Ohms)
Impedance: 1k Ohms maximum
S/N ratio: >40dB
Sound pressure level: 120dB maximum
Power supply: 1·5V to 10Vdc
Recommended voltage: 4·5V (optimum performance)
Current drain: 0·4mA

My question is what sort of ac voltage output range can I expect from the
microphone and what does it mean 0db = 1V/ubar and do we (and if so , how)
use it to find our output from the electret.

Also, is it possile to remove the electret and the 1st stage amplifier, and
connect the output from e.g. my computer sound card directly to C2
and control the gain from the pc volume control ?

Tia

2. ### Bob MastaGuest

SOME FORMULAS FOR WORKING WITH SOUND:

1 Pascal (Pa) = 1 Newton/m²

= 10 dyne/cm²

= 10 µbar

= 94 dB SPL

Sound Power Level (dB SPL):

SPL = 20 * log(P / P0)

where P is the measured pressure and P0 is
a reference pressure in the same system of
units:

P0 = 20 µPa (or µNewton/m²)

= 0.0002 µbar (or dyne/cm²)

OK, so your mic sensitivity is -60 dB re 1V/ubar. That means
that if you get an output that is 60 dB below 1 volt (ie 1 mV)
then the sound pressure is 1 ubar. Since 10 ubar = 94 dB SPL, then
1 ubar = 74 dB SPL.

So if you feed it 74 dB SPL, expect about 1 mV.
For 94 dB SPL, expect about 10 mV.

I'm not sure what you are asking here. The PC can provide
a controlled amount of amplification of the mic signal. Why
do you want to take the mic apart? If you try to take it
apart, for whatever reason, you will probably destroy its
usefullness as a mic. This is a mechanical system that
is sensitive to tiny vibrations, so it's not terribly robust
inside.

Hope this helps!

Bob Masta

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com

3. ### philGuest

When you say -60 dB re 1V/ubar, what does "re" stand for

What I meant was not to actually dismantle the microphone, simply to
remove all components up to C2, and feed the signal into C2 from
my PC sound card. This is just to get a prototype running a bit faster,
and I can control the gain of the audio from the PC volume control,
which sounds rather convenient ?

Yes, very much so, thanks!!

4. ### Ian BellGuest

phil wrote:

snip
relative to or with respect to

Ian

5. ### SoerenGuest

Hi Phil,

You would need a potential divider like this:

PC-soundcard O---[33k]---+
_|_
| |
| |<-----O to left side of C2
|_| 1k
PC |
Gnd O--------------------+-------O to trnsmitter gnd.

6. ### Glenn GundlachGuest

It appears to me to be a bit 'reverse engineered' right now. Aren't
the ins and outs swapped?
GG

7. ### SoerenGuest

Hi Glenn,

Nope.
The input from the soundcard should be as light a load as possible
(within reason), while the output must have a low impedance for driving
the next stage (the x-mitter).