Question about electrostatic pickups in old electro-acoustic pianos

Oct 02, 2004 2 Replies

Hi, Some old electro-acoustic pianos have metal reeds which vibrate. One end only of each reed is secured to a metal bar, with the other end of the reed able to vibrate freely. At the free end of each reed, there is a plate of metal (a pickup), which is very close to the end of the reed. All these pickups are connected to together. The whole assembly of the bar, reeds, and pickups, forms a giant capacitor - it is charged to hundreds of volts (DC), and a preamplifier amplifies the voltage which is developed across the capacitor when the reeds are set into vibration with a plucking mechanism.



Now, some say that the reeds need to be *magnetised*. This is something I don't understand. Do you think that magnetising the reeds (along the length of the reed), would cause any appreciable change in signal amplitude? Assume that the magnetic attraction between the end of the reed and the pickup is neglible, for the sakes of this discussion.



Thanks, Greg.


Note that the source impedance of the DC supply across the electrodes is high.

Greg.

No- this is an electrostatic device- no magnetisation needed. There are also (or have been) electrostatic speakers. Generally electrostatic devices for electromechanical conversion suffer from low energy density compeared to electromagnetic devices. - that is- it must be big to produce appreciable power. In this case only a few microwatts (if that) would be involved.

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