Electret microphone ?

P

phil

Guest
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 ą3dB (0dB = 1V/ľbar at 1kHz, Vcc = 4ˇ5V, Rl = 1k
Ohms)
Impedance: 1k Ohms maximum
S/N ratio: >40dB
Sound pressure level: 120dB maximum
Power supply: 1ˇ5V to 10Vdc
Recommended voltage: 4ˇ5V (optimum performance)
Current drain: 0ˇ4mA


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
 
On Thu, 29 Apr 2004 12:46:19 +0100, "phil" <aaa@aaaaaaaa.com> wrote:

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 ą3dB (0dB = 1V/ľbar at 1kHz, Vcc = 4ˇ5V, Rl = 1k
Ohms)
Impedance: 1k Ohms maximum
S/N ratio: >40dB
Sound pressure level: 120dB maximum
Power supply: 1ˇ5V to 10Vdc
Recommended voltage: 4ˇ5V (optimum performance)
Current drain: 0ˇ4mA


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.
Click to expand...
SOME FORMULAS FOR WORKING WITH SOUND:

1 Pascal (Pa) = 1 Newton/m˛

= 10 dyne/cm˛

= 10 ľbar

= 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 ľPa (or ľNewton/m˛)

= 0.0002 ľbar (or dyne/cm˛)

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.


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 ?

I'm not sure what you are asking here. The PC can provide
Click to expand...
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
dqatechATdaqartaDOTcom

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com
 
"Bob Masta" <NoSpam@daqarta.com> wrote in message
news:40910472.5362119@news.itd.umich.edu...
On Thu, 29 Apr 2004 12:46:19 +0100, "phil" <aaa@aaaaaaaa.com> wrote:

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 ą3dB (0dB = 1V/ľbar at 1kHz, Vcc = 4ˇ5V, Rl = 1k
Ohms)
Impedance: 1k Ohms maximum
S/N ratio: >40dB
Sound pressure level: 120dB maximum
Power supply: 1ˇ5V to 10Vdc
Recommended voltage: 4ˇ5V (optimum performance)
Current drain: 0ˇ4mA


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.


SOME FORMULAS FOR WORKING WITH SOUND:

1 Pascal (Pa) = 1 Newton/m˛

= 10 dyne/cm˛

= 10 ľbar

= 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 ľPa (or ľNewton/m˛)

= 0.0002 ľbar (or dyne/cm˛)

OK, so your mic sensitivity is -60 dB re 1V/ubar. That means
Click to expand...

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


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.


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 ?

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.
Click to expand...
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 ?


Hope this helps!
Click to expand...
Yes, very much so, thanks!!
Bob Masta
dqatechATdaqartaDOTcom

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Click to expand...
 
Hi Phil,


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 ?
Click to expand...
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.


--
Regards,
Soeren

* If it puzzles you dear... Reverse engineer *
 
Soeren <Look@iNO-SPAMt.dk.invalid> wrote in message news:<Xns94DB23F40A254o8oLOOKatHOMEo8o@212.242.40.196>...
Hi Phil,


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 ?

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


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

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


--
Regards,
Soeren

* If it puzzles you dear... Reverse engineer *
 
You must log in or register to reply here.

Welcome to EDABoard.com

Sponsor

Back
Top