Rectification using a Gyrator Circuit

Notes:
To avoid excess ripple output on a power supply feeding a heavy load, usually a large value capacitor is
chosen following the rectifier. In this circuit, C1 is only a 470uF capacitor.  The gyrator principle uses the
effect that the value of input capacitance at the base of a transitor is effectively multiplied by the current
gain of the transistor.  Here C2 which is 100u appears at the ouput ( Vreg ) to be 100 x current gain of
the 2N3055 power transistor. If you assume a dc current gain of 50, then the smoothing across the
supply, would be as though you had chosen a 5000uF capacitor. The graph below shows the output
voltage and current through the load :-

The load draws nearly 400mA.  With the output directly from the rectifier there is about 5v pk-pk ripple
in the output.   Using the output at the emitter of the transistor things are much better. The circuit will
take a few hundred milliseconds for the output voltage to stabilize and reach maximum value.  The
advantages are that a smaller, less costly reservoir capacitor can be used with this circuit to give a
high quality smoothed supply.

Source:www.zen22142.zen.co.uk