"Jim" wrote in message
...
i need a variable for a receiving loop antenna. does anyone know how to
determine the area of plates for a given value of cap? for example i
need 365 pf. two half circle plates would be handy, separated by a piece
of plastic or card stock. what surface area for each plate would
approximate that value? does two stator plates with one rotor between
them increase the value if compared to one plate each?
=========================
In the May 2005 issue of 'Electron' ( mag of VERON ,the IARU affiliated
society in the Netherlands) there is an article , by PE1LKT ,on a homebrew
'sliding 'capacitor made of pieces of double sided printed circuit board.
5 plates are fixed with 4 plates sliding in between.
The fixed and sliding plates fit inside grooves ,made in 2 pieces of timber
or polystyrene (alternatively inside glued strips of such material). The
sliding plates are tapered at one end and straight at the other end where
they are also bolted together with double sided PCB material spacers or
bolts and washers. At that end is also a soldered nut which suits a threaded
bar which is supported by a simple 'hose type bearing' with the end coupled
via a piece of rubber or plastic hose to a fixed in-line electric screw
driver.
The electric motor is operated remotely via a three position switch for '
right hand rotation' - 'off ' -'left hand rotation'
The fixed plates are obviously also galvanically connected at the far end
away from the moving plates .

This variable capacitor is used for a transmitting magloop and hence is
subjected to high RF voltages ,hence the plates are appropriately spaced .
Spacing could be minimised for a RX loop

The above variable cap has a capacitance of approx 400 pF ,requiring fixed
plates each with an area of 210 x 170 mm ( 357 cmsq equals approx 55 inch
sq). The sliding plates are slightly shorter.

For a RX loop requiring less plate spacing the plate area can be reduced.

Frank GM0CSZ / KN6WH