On Fri, 18 Jun 2004 10:03:13 GMT, Chris Hornbeck
wrote:

My choice of models here is most likely wrong, so I'm
getting some sleep. Please correct me and I'll check in
later.

In the clear light of day, *of course* the small-signal
model and constant charge apply. Thanks to all for
correcting me.

Chris Hornbeck

Jay Kadis wrote:

In article ,
Logan Shaw wrote:

Physics seems to tell me that pressure will be proportional to force,
and force is what accelerates the diaphgram. Therefore, shouldn't
the pressure and the diaphgram's instantaneous acceleration be
proportional? If so, then the diaphragm's velocity is the
integral of the pressure over time.

You seem to be ignoring the restoring force, since the diaphragm is restrained
and not free to move in response to the pressure exerted.

Hmm, I am starting to learn that all this stuff is way more complicated
than I thought it was. Thought I understood it, but apparently only in
a very shallow sense.

How big is the restoring force compared to the force due to pressure?
I'm assuming significantly smaller, so that ignoring it can still give
you a workable first-order approximation. Or is that not true?

- Logan

In article ,
Logan Shaw wrote:

Jay Kadis wrote:

In article ,
Logan Shaw wrote:

Physics seems to tell me that pressure will be proportional to force,
and force is what accelerates the diaphgram. Therefore, shouldn't
the pressure and the diaphgram's instantaneous acceleration be
proportional? If so, then the diaphragm's velocity is the
integral of the pressure over time.

You seem to be ignoring the restoring force, since the diaphragm is
restrained
and not free to move in response to the pressure exerted.

Hmm, I am starting to learn that all this stuff is way more complicated
than I thought it was. Thought I understood it, but apparently only in
a very shallow sense.

How big is the restoring force compared to the force due to pressure?
I'm assuming significantly smaller, so that ignoring it can still give
you a workable first-order approximation. Or is that not true?

- Logan


It is the sum of all forces resisting the movement of the diaphragm: it would
depend on the tension applied to the diaphragm in the case of a condensor
element and the stiffness of the suspension for a dynamic element. In the case
of the condensor element, displacement is very small whereas with a dynamic
element, there is more significant movement. This would imply that the
restoring force is large for a condensor element and smaller for a dynamic
transducer.

-Jay
--
x------- Jay Kadis ------- x---- Jay's Attic Studio ------x
x Lecturer, Audio Engineer x Dexter Records x
x CCRMA, Stanford University x http://www.offbeats.com/ x
x-------- http://ccrma-www.stanford.edu/~jay/ ----------x

On Fri, 18 Jun 2004 16:49:08 GMT, Logan Shaw
wrote:

How big is the restoring force compared to the force due to pressure?
I'm assuming significantly smaller, so that ignoring it can still give
you a workable first-order approximation. Or is that not true?

There are two useful choices for the mass/ compliance resonance,
above and below the passband. The usual choice for open (pressure-
differential sensitive) microphones is below passband and the usual
choice for closed (pressure sensitive) microphones is above
passband.

Chris Hornbeck

Chris Hornbeck wrote:

On Fri, 18 Jun 2004 10:37:43 +0100, Don Pearce
wrote:


And what is capacitance's relation to displacement?


Capacitance = E0 A / d Farads

Where E0 = 8.86 e-12

A = diaphragm area (square metres)
d = diaphragm / backplate separation (metres)

So capacitance varies linearly with the displacement.


I've misspoken, and it's so late here that I probably will again,
but generally, the assumption of constant charge is a small-
signal one.

In the generalized case of a pressure sensitive diaphragm,
with constant voltage rather than constant charge, capacitance
is a square function of displacement.

My choice of models here is most likely wrong, so I'm
getting some sleep. Please correct me and I'll check in
later.


You'll still be right. The constant charge is achieved by
huge series resistance such that the time constant is long
compared to a cycle at the lowest frequency of normal
interest. Thus Q=C*V, a constant, and C=E0*A/d so that
V=d*Q/(E0*A).

Where this breaks down is that under larger displacements
the charge moves around on the surface of a conductive
coated diaphragm, migrating rapidly to where it gets closest
and this will introduce non-linearities. A high resistance
coating would eliminate this but they don't use that.

The RF mic eliminates dependancy on stored charge so is
lower distortion and reaches lower frequencies.


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

Chris Hornbeck wrote:

On Fri, 18 Jun 2004 10:03:13 GMT, Chris Hornbeck
wrote:


My choice of models here is most likely wrong, so I'm
getting some sleep. Please correct me and I'll check in
later.


In the clear light of day, *of course* the small-signal
model and constant charge apply. Thanks to all for
correcting me.


Didn't help me. I read you backwards which cancelled your
error. :-)


Bob
--

"Things should be described as simply as possible, but no
simpler."

A. Einstein

"Bob Cain"


Where this breaks down is that under larger displacements
the charge moves around on the surface of a conductive
coated diaphragm, migrating rapidly to where it gets closest
and this will introduce non-linearities.


** As you saw earlier in the thread, the KM130 can handle up to 150 dB SPL
with only 0.5 % THD - and even this amount is mostly from the electronics.

It is easy to say that some problem exists with a purely verbal
ument - then offer no proof.

That is just what all the hi-fi snake oil merchants and their addicts do.




............. Phil

In article writes:

"Mike Rivers"
Phil Allison:

All right. You win. It's your newsgroup.

I'll be back after you finish your work here.

--
I'm really Mike Rivers )
However, until the spam goes away or Hell freezes over,
lots of IP addresses are blocked from this system. If
you e-mail me and it bounces, use your secret decoder ring
and reach me he double-m-eleven-double-zero at yahoo

(Mike Rivers) wrote:


writes:

"Mike Rivers"
Phil Allison:

All right. You win. It's your newsgroup.
I'll be back after you finish your work here.

Nobody here gives a flying **** about Phil Allison, Mike.

Harvey Gerst
Indian Trail Recording Studio
http://www.ITRstudio.com/

Logan Shaw wrote:

You know all this, so I'm only getting all snippy to emphasize
that it's "Pressure drop. De pressure's gonna drop. Pressure
drop...."

The wha? Huh?

See soundtrack: _The Harder They Come_

"de pressure gonna drop on you..."

--
ha

x-no archive: yes

Harvey Gerst wrote:

(Mike Rivers) wrote:

Nobody here gives a flying **** about Phil Allison, Mike.

If his mother had been smart enough to work a flush toilet, he wouldn't
be soiling up the newsgroup.

--
ha

On 19 Jun 2004 08:17:45 -0400, (Mike Rivers)
wrote:

All right. You win. It's your newsgroup.

I'll be back after you finish your work here.

The real shame is that Phil has a useful and interesting point to
make, and the loudspeaker analogy would help in describing the
relationship between RMS pressure, velocity and displacement.

I for one wish he would make it. Phil, you can probably say it
better than I can. Step up, guy.

Chris Hornbeck

On Fri, 18 Jun 2004 17:09:56 GMT, Chris Hornbeck
wrote:

There are two useful choices for the mass/ compliance resonance,
above and below the passband. The usual choice for open (pressure-
differential sensitive) microphones is below passband and the usual
choice for closed (pressure sensitive) microphones is above
passband.

This is so poorly worded that I can't let it drop. Imagine a
loudspeaker woofer in a sealed box and driven by a constant
RMS voltage at all frequencies.

If we measure its sound pressure output, we see it rise at
12 dB/ octave from very low frequencies up to its resonant
frequency, where pressure response levels off to flat.

If we measure its diaphragm excursion, we see that it's flat
from low frequencies up to resonance, then it falls at 12 dB/
octave.

There are two components to the diaphragm excursion. The
diaphragm's velocity increases at 6 dB/ octave up to resonance,
then decreases at 6 dB/ octave above resonance. And
wavelength, which falls continuously at 6 dB/ octave.

Microphones are also four terminal devices designed to
translate voltage and pressure. All the same rules apply.
To make non-equalized microphones with flat pressure input vs.
voltage output some strategies emerge:

If our electrical generating mechanism has a flat relation of
excursion to voltage, we simply operate below resonance.

If our generator is velocity sensitive, then our output will
rise at 6 dB/ octave below resonance and fall at 6 dB/ octave
above resonance. Above resonance this is useful with open
diaphragms ("pressure-gradient microphones") whose diaphragm
excursion increases at 6 dB/ octave in their major working
frequency range.

Now I promise the shut the heck up.

Chris Hornbeck