Rectification using a Gyrator Circuit

Posted on Sep 4, 2012

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.

Rectification using a Gyrator Circuit
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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

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