Marik
Well-known member
[quote author="crazydoc"][quote author="xvlk"]
... and the ribbon itself {current path perpendicular to magnetic field}
voltage=B*l*velocity = B*l*diff(x,t)
where B is induction in the gap {you can measure it
by Hall probe from old videorecorder}
l is efffective length of the ribbon
diff(x,t) is ribbon velocity
This is one part of equations of the electrodynamic conversion.
[/quote]
So the faster the ribbon moves, the greater its output, (voltage or current, assuming a constant impedance and a constant displacement of the ribbon.) So intuitively (to me) this means that for electromagnetic portion of this system, output is proportional to frequency.
[/quote]
Crazydoc,
Your intuition is right!
The "v" (velocity of ribbon) identifies as (P1-P2)/zA, where P1-P2 is pressure difference, z-acoustic impedance of the ribbon, and A is ribbon area. If z is a mass reactance, then v=(P1*d/c*M*A)*cos0, where M=mass of the ribbon, and c=velocity of sound.
Because the ribbon workes as a mass controlled system, there is a natural fall with frequency increase.
... and the ribbon itself {current path perpendicular to magnetic field}
voltage=B*l*velocity = B*l*diff(x,t)
where B is induction in the gap {you can measure it
by Hall probe from old videorecorder}
l is efffective length of the ribbon
diff(x,t) is ribbon velocity
This is one part of equations of the electrodynamic conversion.
[/quote]
So the faster the ribbon moves, the greater its output, (voltage or current, assuming a constant impedance and a constant displacement of the ribbon.) So intuitively (to me) this means that for electromagnetic portion of this system, output is proportional to frequency.
[/quote]
Crazydoc,
Your intuition is right!
The "v" (velocity of ribbon) identifies as (P1-P2)/zA, where P1-P2 is pressure difference, z-acoustic impedance of the ribbon, and A is ribbon area. If z is a mass reactance, then v=(P1*d/c*M*A)*cos0, where M=mass of the ribbon, and c=velocity of sound.
So something must be damping this frequency dependent increased output, which I assume is a decreased displacement (excursion) of the ribbon as frequency increases, because of either the mass of the ribbon or the mass of air it has to displace.
Because the ribbon workes as a mass controlled system, there is a natural fall with frequency increase.